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16

The Dependence of Amplitudes on Position

2

The wave function

Now that you have some idea about how things are going to look, we want to go back to the beginning and study the problem of describing the motion of an electron along a line without having to consider states connected with atoms on a lattice. We want to go back to the beginning and see what ideas we have to use if we want to describe the motion of a free particle in space. Since we are interested in the behavior of a particle, along a continuum, we will be dealing with an infinite number of possible states and, as you will see, the ideas we have developed for dealing with a finite number of states will need some technical modifications. We begin by letting the state vector $\ket{x}$ stand for a state in which a particle is located precisely at the coordinate $x$. For every value $x$ along the line—for instance $1.73$, or $9.67$, or $10.00$—there is the corresponding state. We will take these states $\ket{x}$ as our base states and, if we include all the points on the line, we will have a complete set for motion in one dimension. Now suppose we have a different kind of a state, say $\ket{\psi}$, in which an electron is distributed in some way along the line. One way of describing this state is to give all the amplitudes that the electron will be also found in each of the base states $\ket{x}$. We must give an infinite set of amplitudes, one for each value of $x$. We will write these amplitudes as $\braket{x}{\psi}$. Each of these amplitudes is a complex number and since there is one such complex number for each value of $x$, the amplitude $\braket{x}{\psi}$ is indeed just a function of $x$. We will also write it as $C(x)$, \begin{equation} \label{Eq:III:16:14} C(x)\equiv\braket{x}{\psi}. \end{equation} We have already considered such amplitudes which vary in a continuous way with the coordinates when we talked about the variations of amplitude with time in Chapter 7. We showed there, for example, that a particle with a definite momentum should be expected to have a particular variation of its amplitude in space. If a particle has a definite momentum $p$ and a corresponding definite energy $E$, the amplitude to be found at any position $x$ would look like \begin{equation} \label{Eq:III:16:15} \braket{x}{\psi}=C(x)\propto e^{+ipx/\hbar}. \end{equation} This equation expresses an important general principle of quantum mechanics which connects the base states corresponding to different positions in space to another system of base states—all the states of definite momentum. The definite momentum states are often more convenient than the states in $x$ for certain kinds of problems. Either set of base states is, of course, equally acceptable for a description of a quantum mechanical situation. We will come back later to the matter of the connection between them. For the moment we want to stick to our discussion of a description in terms of the states $\ket{x}$. Before proceeding, we want to make one small change in notation which we hope will not be too confusing. The function $C(x)$, defined in Eq. (16.14), will of course have a form which depends on the particular state $\ket{\psi}$ under consideration. We should indicate that in some way. We could, for example, specify which function $C(x)$ we are talking about by a subscript say, $C_\psi(x)$. Although this would be a perfectly satisfactory notation, it is a little bit cumbersome and is not the one you will find in most books. Most people simply omit the letter $C$ and use the symbol $\psi$ to define the function \begin{equation} \label{Eq:III:16:16} \psi(x)\equiv C_\psi(x)=\braket{x}{\psi}. \end{equation} Since this is the notation used by everybody else in the world, you might as well get used to it so that you will not be frightened when you come across it somewhere else. Remember though, that we will now be using $\psi$ in two different ways. In Eq. (16.14), $\psi$ stands for a label we have given to a particular physical state of the electron. On the left-hand side of Eq. (16.16), on the other hand, the symbol $\psi$ is used to define a mathematical function of $x$ which is equal to the amplitude to be associated with each point $x$ along the line. We hope it will not be too confusing once you get accustomed to the idea. Incidentally, the function $\psi(x)$ is usually called “the wave function”—because it more often than not has the form of a complex wave in its variables. Since we have defined $\psi(x)$ to be the amplitude that an electron in the state $\psi$ will be found at the location $x$, we would like to interpret the absolute square of $\psi$ to be the probability of finding an electron at the position $x$. Unfortunately, the probability of finding a particle exactly at any particular point is zero. The electron will, in general, be smeared out in a certain region of the line, and since, in any small piece of the line, there are an infinite number of points, the probability that it will be at any one of them cannot be a finite number. We can only describe the probability of finding an electron in terms of a probability distribution2 which gives the relative probability of finding the electron at various approximate locations along the line. Let’s let $\prob(x,\Delta x)$ stand for the chance of finding the electron in a small interval $\Delta x$ located near $x$. If we go to a small enough scale in any physical situation, the probability will be varying smoothly from place to place, and the probability of finding the electron in any small finite line segment $\Delta x$ will be proportional to $\Delta x$. We can modify our definitions to take this into account. We can think of the amplitude $\braket{x}{\psi}$ as representing a kind of “amplitude density” for all the base states $\ket{x}$ in a small region. Since the probability of finding an electron in a small interval $\Delta x$ at $x$ should be proportional to the interval $\Delta x$, we choose our definition of $\braket{x}{\psi}$ so that the following relation holds: \begin{equation*} \prob(x,\Delta x)=\abs{\braket{x}{\psi}}^2\,\Delta x. \end{equation*} The amplitude $\braket{x}{\psi}$ is therefore proportional to the amplitude that an electron in the state $\psi$ will be found in the base state $x$ and the constant of proportionality is chosen so that the absolute square of the amplitude $\braket{x}{\psi}$ gives the probability density of finding an electron in any small region. We can write, equivalently, \begin{equation} \label{Eq:III:16:17} \prob(x,\Delta x)=\abs{\psi(x)}^2\,\Delta x. \end{equation} We will now have to modify some of our earlier equations to make them compatible with this new definition of a probability amplitude. Suppose we have an electron in the state $\ket{\psi}$ and we want to know the amplitude for finding it in a different state $\ket{\phi}$ which may correspond to a different spread-out condition of the electron. When we were talking about a finite set of discrete states, we would have used Eq. (16.5). Before modifying our definition of the amplitudes we would have written \begin{equation} \label{Eq:III:16:18} \braket{\phi}{\psi}=\sum_{\text{all $x$}} \braket{\phi}{x}\braket{x}{\psi}. \end{equation} Now if both of these amplitudes are normalized in the same way as we have described above, then a sum of all the states in a small region of $x$ would be equivalent to multiplying by $\Delta x$, and the sum over all values of $x$ simply becomes an integral. With our modified definitions, the correct form becomes \begin{equation} \label{Eq:III:16:19} \braket{\phi}{\psi}=\int_{\text{all $x$}} \braket{\phi}{x}\braket{x}{\psi}\,dx. \end{equation} The amplitude $\braket{x}{\psi}$ is what we are now calling $\psi(x)$ and, in a similar way, we will choose to let the amplitude $\braket{x}{\phi}$ be represented by $\phi(x)$. Remembering that $\braket{\phi}{x}$ is the complex conjugate of $\braket{x}{\phi}$, we can write Eq. (16.19) as \begin{equation} \label{Eq:III:16:20} \braket{\phi}{\psi}=\int\phi\cconj(x)\psi(x)\,dx. \end{equation} With our new definitions everything follows with the same formulas as before if you always replace a summation sign by an integral over $x$. We should mention one qualification to what we have been saying. Any suitable set of base states must be complete if it is to be used for an adequate description of what is going on. For an electron in one dimension it is not really sufficient to specify only the base states $\ket{x}$, because for each of these states the electron may have a spin which is either up or down. One way of getting a complete set is to take two sets of states in $x$, one for up spin and the other for down spin. We will, however, not worry about such complications for the time being.

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16

The Dependence of Amplitudes on Position

3

States of definite momentum

Suppose we have an electron in a state $\ket{\psi}$ which is described by the probability amplitude $\braket{x}{\psi}=\psi(x)$. We know that this represents a state in which the electron is spread out along the line in a certain distribution so that the probability of finding the electron in a small interval $dx$ at the location $x$ is just \begin{equation*} \prob(x,dx)=\abs{\psi(x)}^2\,dx. \end{equation*} What can we say about the momentum of this electron? We might ask what is the probability that this electron has the momentum $p$? Let’s start out by calculating the amplitude that the state $\ket{\psi}$ is in another state $\ket{\mom p}$ which we define to be a state with the definite momentum $p$. We can find this amplitude by using our basic equation for the resolution of amplitudes, Eq. (16.19). In terms of the state $\ket{\mom p}$ \begin{equation} \label{Eq:III:16:21} \braket{\mom p}{\psi}=\int_{x=-\infty}^{+\infty} \braket{\mom p}{x}\braket{x}{\psi}\,dx. \end{equation} And the probability that the electron will be found with the momentum $p$ should be given in terms of the absolute square of this amplitude. We have again, however, a small problem about the normalizations. In general we can only ask about the probability of finding an electron with a momentum in a small range $dp$ at the momentum $p$. The probability that the momentum is exactly some value $p$ must be zero (unless the state $\ket{\psi}$ happens to be a state of definite momentum). Only if we ask for the probability of finding the momentum in a small range $dp$ at the momentum $p$ will we get a finite probability. There are several ways the normalizations can be adjusted. We will choose one of them which we think to be the most convenient, although that may not be apparent to you just now. We take our normalizations so that the probability is related to the amplitude by \begin{equation} \label{Eq:III:16:22} \prob(p,dp)=\abs{\braket{\mom p}{\psi}}^2\,\frac{dp}{2\pi\hbar}. \end{equation} With this definition the normalization of the amplitude $\braket{\mom p}{x}$ is determined. The amplitude $\braket{\mom p}{x}$ is, of course, just the complex conjugate of the amplitude $\braket{x}{\mom p}$, which is just the one we have written down in Eq. (16.15). With the normalization we have chosen, it turns out that the proper constant of proportionality in front of the exponential is just $1$. Namely, \begin{equation} \label{Eq:III:16:23} \braket{\mom p}{x}=\braket{x}{\mom p}\cconj=e^{-ipx/\hbar}. \end{equation} Equation (16.21) then becomes \begin{equation} \label{Eq:III:16:24} \braket{\mom p}{\psi}=\int_{-\infty}^{+\infty} e^{-ipx/\hbar}\braket{x}{\psi}\,dx. \end{equation} This equation together with Eq. (16.22) allows us to find the momentum distribution for any state $\ket{\psi}$. Let’s look at a particular example—for instance one in which an electron is localized in a certain region around $x=0$. Suppose we take a wave function which has the following form: \begin{equation} \label{Eq:III:16:25} \psi(x)=Ke^{-x^2/4\sigma^2}. \end{equation} The probability distribution in $x$ for this wave function is the absolute square, or \begin{equation} \label{Eq:III:16:26} \prob(x,dx)=P(x)\,dx=K^2e^{-x^2/2\sigma^2}\,dx. \end{equation} The probability density function $P(x)$ is the Gaussian curve shown in Fig. 16–1. Most of the probability is concentrated between $x=+\sigma$ and $x=-\sigma$. We say that the “half-width” of the curve is $\sigma$. (More precisely, $\sigma$ is equal to the root-mean-square of the coordinate $x$ for something spread out according to this distribution.) We would normally choose the constant $K$ so that the probability density $P(x)$ is not merely proportional to the probability per unit length in $x$ of finding the electron, but has a scale such that $P(x)\,\Delta x$ is equal to the probability of finding the electron in $\Delta x$ near $x$. The constant $K$ which does this can be found by requiring that $\int_{-\infty}^{+\infty}P(x)\,dx=1$, since there must be unit probability that the electron is found somewhere. Here, we get that $K=(2\pi\sigma^2)^{-1/4}$. [We have used the fact that $\int_{-\infty}^{+\infty}e^{-t^2}\,dt=\sqrt{\pi}$; see Vol. I, footnote 40-1.] Now let’s find the distribution in momentum. Let’s let $\phi(p)$ stand for the amplitude to find the electron with the momentum $p$, \begin{equation} \label{Eq:III:16:27} \phi(p)\equiv\braket{\mom p}{\psi}. \end{equation} Substituting Eq. (16.25) into Eq. (16.24) we get \begin{equation} \label{Eq:III:16:28} \phi(p)=\int_{-\infty}^{+\infty} e^{-ipx/\hbar}\cdot Ke^{-x^2/4\sigma^2}\,dx. \end{equation} The integral can also be rewritten as \begin{equation} \label{Eq:III:16:29} Ke^{-p^2\sigma^2/\hbar^2}\int_{-\infty}^{+\infty} e^{-(1/4\sigma^2)(x+2ip\sigma^2/\hbar)^2}\,dx. \end{equation} We can now make the substitution $u=x+2ip\sigma^2/\hbar$, and the integral is \begin{equation} \label{Eq:III:16:30} \int_{-\infty}^{+\infty}e^{-u^2/4\sigma^2}\,du=2\sigma\sqrt{\pi}. \end{equation} (The mathematicians would probably object to the way we got there, but the result is, nevertheless, correct.) \begin{equation} \label{Eq:III:16:31} \phi(p)=(8\pi\sigma^2)^{1/4}e^{-p^2\sigma^2/\hbar^2}. \end{equation} We have the interesting result that the amplitude function in $p$ has precisely the same mathematical form as the amplitude function in $x$; only the width of the Gaussian is different. We can write this as \begin{equation} \label{Eq:III:16:32} \phi(p)=(\eta^2/2\pi\hbar^2)^{-1/4}e^{-p^2/4\eta^2}, \end{equation} where the half-width $\eta$ of the $p$-distribution function is related to the half-width $\sigma$ of the $x$-distribution by \begin{equation} \label{Eq:III:16:33} \eta=\frac{\hbar}{2\sigma}. \end{equation} Our result says: if we make the width of the distribution in $x$ very small by making $\sigma$ small, $\eta$ becomes large and the distribution in $p$ is very much spread out. Or, conversely: if we have a narrow distribution in $p$, it must correspond to a spread-out distribution in $x$. We can, if we like, consider $\eta$ and $\sigma$ to be some measure of the uncertainty in the localization of the momentum and of the position of the electron in the state we are studying. If we call them $\Delta p$ and $\Delta x$ respectively Eq. (16.33) becomes \begin{equation} \label{Eq:III:16:34} \Delta p\,\Delta x=\frac{\hbar}{2}. \end{equation} Interestingly enough, it is possible to prove that for any other form of a distribution in $x$ or in $p$, the product $\Delta p\,\Delta x$ cannot be smaller than the one we have found here. The Gaussian distribution gives the smallest possible value for the product of the root-mean-square widths. In general, we can say \begin{equation} \label{Eq:III:16:35} \Delta p\,\Delta x\geq\frac{\hbar}{2}. \end{equation} This is a quantitative statement of the Heisenberg uncertainty principle, which we have discussed qualitatively many times before. We have usually made the approximate statement that the minimum value of the product $\Delta p\,\Delta x$ is of the same order as $\hbar$.

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The Dependence of Amplitudes on Position

4

Normalization of the states in $\boldsymbol{x}$

We return now to the discussion of the modifications of our basic equations which are required when we are dealing with a continuum of base states. When we have a finite number of discrete states, a fundamental condition which must be satisfied by the set of base states is \begin{equation} \label{Eq:III:16:36} \braket{i}{j}=\delta_{ij}. \end{equation} If a particle is in one base state, the amplitude to be in another base state is $0$. By choosing a suitable normalization, we have defined the amplitude $\braket{i}{i}$ to be $1$. These two conditions are described by Eq. (16.36). We want now to see how this relation must be modified when we use the base states $\ket{x}$ of a particle on a line. If the particle is known to be in one of the base states $\ket{x}$, what is the amplitude that it will be in another base state $\ket{x'}$? If $x$ and $x'$ are two different locations along the line, then the amplitude $\braket{x}{x'}$ is certainly $0$, so that is consistent with Eq. (16.36). But if $x$ and $x'$ are equal, the amplitude $\braket{x}{x'}$ will not be $1$, because of the same old normalization problem. To see how we have to patch things up, we go back to Eq. (16.19), and apply this equation to the special case in which the state $\ket{\phi}$ is just the base state $\ket{x'}$. We would have then \begin{equation} \label{Eq:III:16:37} \braket{x'}{\psi}=\int\braket{x'}{x}\psi(x)\,dx. \end{equation} Now the amplitude $\braket{x}{\psi}$ is just what we have been calling the function $\psi(x)$. Similarly the amplitude $\braket{x'}{\psi}$, since it refers to the same state $\ket{\psi}$, is the same function of the variable $x'$, namely $\psi(x')$. We can, therefore, rewrite Eq. (16.37) as \begin{equation} \label{Eq:III:16:38} \psi(x')=\int\braket{x'}{x}\psi(x)\,dx. \end{equation} This equation must be true for any state $\ket{\psi}$ and, therefore, for any arbitrary function $\psi(x)$. This requirement should completely determine the nature of the amplitude $\braket{x}{x'}$—which is, of course, just a function that depends on $x$ and $x'$. Our problem now is to find a function $f(x,x')$, which when multiplied into $\psi(x)$, and integrated over all $x$ gives just the quantity $\psi(x')$. It turns out that there is no mathematical function which will do this! At least nothing like what we ordinarily mean by a “function.” Suppose we pick $x'$ to be the special number $0$ and define the amplitude $\braket{0}{x}$ to be some function of $x$, let’s say $f(x)$. Then Eq. (16.38) would read as follows: \begin{equation} \label{Eq:III:16:39} \psi(0)=\int f(x)\psi(x)\,dx. \end{equation} What kind of function $f(x)$ could possibly satisfy this equation? Since the integral must not depend on what values $\psi(x)$ takes for values of $x$ other than $0$, $f(x)$ must clearly be $0$ for all values of $x$ except $0$. But if $f(x)$ is $0$ everywhere, the integral will be $0$, too, and Eq. (16.39) will not be satisfied. So we have an impossible situation: we wish a function to be $0$ everywhere but at a point, and still to give a finite integral. Since we can’t find a function that does this, the easiest way out is just to say that the function $f(x)$ is defined by Eq. (16.39). Namely, $f(x)$ is that function which makes (16.39) correct. The function which does this was first invented by Dirac and carries his name. We write it $\delta(x)$. All we are saying is that the function $\delta(x)$ has the strange property that if it is substituted for $f(x)$ in the Eq. (16.39), the integral picks out the value that $\psi(x)$ takes on when $x$ is equal $0$; and, since the integral must be independent of $\psi(x)$ for all values of $x$ other than $0$, the function $\delta(x)$ must be $0$ everywhere except at $x=0$. Summarizing, we write \begin{equation} \label{Eq:III:16:40} \braket{0}{x}=\delta(x), \end{equation} where $\delta(x)$ is defined by \begin{equation} \label{Eq:III:16:41} \psi(0)=\int\delta(x)\psi(x)\,dx. \end{equation} Notice what happens if we use the special function “$1$” for the function $\psi$ in Eq. (16.41). Then we have the result \begin{equation} \label{Eq:III:16:42} 1=\int\delta(x)\,dx. \end{equation} That is, the function $\delta(x)$ has the property that it is $0$ everywhere except at $x=0$ but has a finite integral equal to unity. We must imagine that the function $\delta(x)$ has such a fantastic infinity at one point that the total area comes out equal to one. One way of imagining what the Dirac $\delta$-function is like is to think of a sequence of rectangles—or any other peaked function you care to—which gets narrower and narrower and higher and higher, always keeping a unit area, as sketched in Fig. 16–2. The integral of this function from $-\infty$ to $+\infty$ is always $1$. If you multiply it by any function $\psi(x)$ and integrate the product, you get something which is approximately the value of the function at $x=0$, the approximation getting better and better as you use the narrower and narrower rectangles. You can if you wish, imagine the $\delta$-function in terms of this kind of limiting process. The only important thing, however, is that the $\delta$-function is defined so that Eq. (16.41) is true for every possible function $\psi(x)$. That uniquely defines the $\delta$-function. Its properties are then as we have described. If we change the argument of the $\delta$-function from $x$ to $x-x'$, the corresponding relations are \begin{gather} \delta(x-x')=0,\quad x'\neq x,\notag\\[2ex] \label{Eq:III:16:43} \int\delta(x-x')\psi(x)\,dx=\psi(x'). \end{gather} If we use $\delta(x-x')$ for the amplitude $\braket{x}{x'}$ in Eq. (16.38), that equation is satisfied. Our result then is that for our base states in $x$, the condition corresponding to (16.36) is \begin{equation} \label{Eq:III:16:44} \braket{x'}{x}=\delta(x-x'). \end{equation} We have now completed the necessary modifications of our basic equations which are necessary for dealing with the continuum of base states corresponding to the points along a line. The extension to three dimensions is fairly obvious; first we replace the coordinate $x$ by the vector $\FLPr$. Then integrals over $x$ become replaced by integrals over $x$, $y$, and $z$. In other words, they become volume integrals. Finally, the one-dimensional $\delta$-function must be replaced by just the product of three $\delta$-functions, one in $x$, one in $y$, and the other in $z$, $\delta(x-x')\,\delta(y-y')\,\delta(z-z')$. Putting everything together we get the following set of equations for the amplitudes for a particle in three dimensions: \begin{gather} \label{Eq:III:16:45} \braket{\phi}{\psi}=\int \braket{\phi}{\FLPr}\braket{\FLPr}{\psi}\, dV,\\[1.5ex] \label{Eq:III:16:46} \begin{aligned} \braket{\FLPr}{\psi}&=\psi(\FLPr),\\[1.5ex] \braket{\FLPr}{\phi}&=\phi(\FLPr), \end{aligned}\\[1.5ex] \label{Eq:III:16:47} \braket{\phi}{\psi}=\int \phi\cconj(\FLPr)\psi(\FLPr)\,dV,\\[1.5ex] \label{Eq:III:16:48} \braket{\FLPr'}{\FLPr}=\delta(x-x')\,\delta(y-y')\,\delta(z-z'). \end{gather} What happens when there is more than one particle? We will tell you about how to handle two particles and you will easily see what you must do if you want to deal with a larger number. Suppose there are two particles, which we can call particle No. $1$ and particle No. $2$. What shall we use for the base states? One perfectly good set can be described by saying that particle $1$ is at $x_1$ and particle $2$ is at $x_2$, which we can write as $\ket{x_1,x_2}$. Notice that describing the position of only one particle does not define a base state. Each base state must define the condition of the entire system. You must not think that each particle moves independently as a wave in three dimensions. Any physical state $\ket{\psi}$ can be defined by giving all of the amplitudes $\braket{x_1,x_2}{\psi}$ to find the two particles at $x_1$ and $x_2$. This generalized amplitude is therefore a function of the two sets of coordinates $x_1$ and $x_2$. You see that such a function is not a wave in the sense of an oscillation that moves along in three dimensions. Neither is it generally simply a product of two individual waves, one for each particle. It is, in general, some kind of a wave in the six dimensions defined by $x_1$ and $x_2$. If there are two particles in nature which are interacting, there is no way of describing what happens to one of the particles by trying to write down a wave function for it alone. The famous paradoxes that we considered in earlier chapters—where the measurements made on one particle were claimed to be able to tell what was going to happen to another particle, or were able to destroy an interference—have caused people all sorts of trouble because they have tried to think of the wave function of one particle alone, rather than the correct wave function in the coordinates of both particles. The complete description can be given correctly only in terms of functions of the coordinates of both particles.

3

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The Dependence of Amplitudes on Position

5

The Schrödinger equation

So far we have just been worrying about how we can describe states which may involve an electron being anywhere at all in space. Now we have to worry about putting into our description the physics of what can happen in various circumstances. As before, we have to worry about how states can change with time. If we have a state $\ket{\psi}$ which goes over into another state $\ket{\psi'}$ sometime later, we can describe the situation for all times by making the wave function—which is just the amplitude $\braket{\FLPr}{\psi}$—a function of time as well as a function of the coordinate. A particle in a given situation can then be described by giving a time-varying wave function $\psi(\FLPr,t)=\psi(x,y,z,t)$. This time-varying wave function describes the evolution of successive states that occur as time develops. This so-called “coordinate representation”—which gives the projections of the state $\ket{\psi}$ into the base states $\ket{\FLPr}$ may not always be the most convenient one to use—but we will consider it first. In Chapter 8 we described how states varied in time in terms of the Hamiltonian $H_{ij}$. We saw that the time variation of the various amplitudes was given in terms of the matrix equation \begin{equation} \label{Eq:III:16:49} i\hbar\,\ddt{C_i}{t}=\sum_jH_{ij}C_j. \end{equation} This equation says that the time variation of each amplitude $C_i$ is proportional to all of the other amplitudes $C_j$, with the coefficients $H_{ij}$. How would we expect Eq. (16.49) to look when we are using the continuum of base states $\ket{x}$? Let’s first remember that Eq. (16.49) can also be written as \begin{equation*} i\hbar\,\ddt{}{t}\,\braket{i}{\psi}= \sum_j\bracket{i}{\Hop}{j}\braket{j}{\psi}. \end{equation*} Now it is clear what we should do. For the $x$-representation we would expect \begin{equation} \label{Eq:III:16:50} i\hbar\,\ddp{}{t}\,\braket{x}{\psi}= \int\bracket{x}{\Hop}{x'}\braket{x'}{\psi}\,dx'. \end{equation} The sum over the base states $\ket{j}$, gets replaced by an integral over $x'$. Since $\bracket{x}{\Hop}{x'}$ should be some function of $x$ and $x'$, we can write it as $H(x,x')$—which corresponds to $H_{ij}$ in Eq. (16.49). Then Eq. (16.50) is the same as \begin{gather} \label{Eq:III:16:51} i\hbar\,\ddp{}{t}\,\psi(x)=\int \kern -.6ex H(x,x')\psi(x')\,dx' \end{gather} with \begin{gather*} H(x,x')\equiv\bracket{x}{\Hop}{x'}. \end{gather*} According to Eq. (16.51), the rate of change of $\psi$ at $x$ would depend on the value of $\psi$ at all other points $x'$; the factor $H(x,x')$ is the amplitude per unit time that the electron will jump from $x'$ to $x$. It turns out in nature, however, that this amplitude is zero except for points $x'$ very close to $x$. This means—as we saw in the example of the chain of atoms at the beginning of the chapter, Eq. (16.12)—that the right-hand side of Eq. (16.51) can be expressed completely in terms of $\psi$ and the derivatives of $\psi$ with respect to $x$, all evaluated at the position $x$. For a particle moving freely in space with no forces, no disturbances, the correct law of physics is \begin{equation*} \int\kern -.6ex H(x,x')\psi(x')\,dx'=-\frac{\hbar^2}{2m}\, \frac{\partial^2}{\partial x^2}\,\psi(x). \end{equation*} Where did we get that from? Nowhere. It’s not possible to derive it from anything you know. It came out of the mind of Schrödinger, invented in his struggle to find an understanding of the experimental observations of the real world. You can perhaps get some clue of why it should be that way by thinking of our derivation of Eq. (16.12) which came from looking at the propagation of an electron in a crystal. Of course, free particles are not very exciting. What happens if we put forces on the particle? Well, if the force of a particle can be described in terms of a scalar potential $V(x)$—which means we are thinking of electric forces but not magnetic forces—and if we stick to low energies so that we can ignore complexities which come from relativistic motions, then the Hamiltonian which fits the real world gives \begin{equation} \label{Eq:III:16:52} \int\kern -.6ex H(x,x')\psi(x')\,dx'=-\frac{\hbar^2}{2m}\, \frac{\partial^2}{\partial x^2}\,\psi(x) +V(x)\psi(x). \end{equation} \begin{equation} \begin{aligned} \int \kern -.6ex H&(x,x')\psi(x')\,dx'=\\ &-\frac{\hbar^2}{2m}\,\frac{\partial^2}{\partial x^2}\,\psi(x) +V(x)\psi(x).\notag \end{aligned} \label{Eq:III:16:52} \end{equation} Again, you can get some clue as to the origin of this equation if you go back to the motion of an electron in a crystal, and see how the equations would have to be modified if the energy of the electron varied slowly from one atomic site to the other—as it might do if there were an electric field across the crystal. Then the term $E_0$ in Eq. (16.7) would vary slowly with position and would correspond to the new term we have added in (16.52). [You may be wondering why we went straight from Eq. (16.51) to Eq. (16.52) instead of just giving you the correct function for the amplitude $H(x,x')=\bracket{x}{\Hop}{x'}$. We did that because $H(x,x')$ can only be written in terms of strange algebraic functions, although the whole integral on the right-hand side of Eq. (16.51) comes out in terms of things you are used to. If you are really curious, $H(x,x')$ can be written in the following way: \begin{equation*} H(x,x')=-\frac{\hbar^2}{2m}\,\delta''(x-x') +V(x)\,\delta(x-x'), \end{equation*} where $\delta''$ means the second derivative of the delta function. This rather strange function can be replaced by a somewhat more convenient algebraic differential operator, which is completely equivalent: \begin{equation*} H(x,x')=\biggl\{ -\frac{\hbar^2}{2m}\,\frac{\partial^2}{\partial x^2}+V(x)\biggr\} \delta(x-x'). \end{equation*} We will not be using these forms, but will work directly with the form in Eq. (16.52).] If we now use the expression we have in (16.52) for the integral in (16.50) we get the following differential equation for $\psi(x)=\braket{x}{\psi}$: \begin{equation} \label{Eq:III:16:53} i\hbar\,\ddp{\psi}{t}=-\frac{\hbar^2}{2m}\, \frac{\partial^2}{\partial x^2}\,\psi(x)+V(x)\psi(x). \end{equation} It is fairly obvious what we should use instead of Eq. (16.53) if we are interested in motion in three dimensions. The only changes are that $\partial^2/\partial x^2$ gets replaced by \begin{equation*} \nabla^2=\frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2}, \end{equation*} and $V(x)$ gets replaced by $V(x,y,z)$. The amplitude $\psi(x,y,z)$ for an electron moving in a potential $V(x,y,z)$ obeys the differential equation \begin{equation} \label{Eq:III:16:54} i\hbar\,\ddp{\psi}{t}=-\frac{\hbar^2}{2m}\, \nabla^2\psi+V\psi. \end{equation} It is called the Schrödinger equation, and was the first quantum-mechanical equation ever known. It was written down by Schrödinger before any of the other quantum equations we have described in this book were discovered. Although we have approached the subject along a completely different route, the great historical moment marking the birth of the quantum mechanical description of matter occurred when Schrödinger first wrote down his equation in 1926. For many years the internal atomic structure of matter had been a great mystery. No one had been able to understand what held matter together, why there was chemical binding, and especially how it could be that atoms could be stable. Although Bohr had been able to give a description of the internal motion of an electron in a hydrogen atom which seemed to explain the observed spectrum of light emitted by this atom, the reason that electrons moved in this way remained a mystery. Schrödinger’s discovery of the proper equations of motion for electrons on an atomic scale provided a theory from which atomic phenomena could be calculated quantitatively, accurately, and in detail. In principle, Schrödinger’s equation is capable of explaining all atomic phenomena except those involving magnetism and relativity. It explains the energy levels of an atom, and all the facts of chemical binding. This is, however, true only in principle—the mathematics soon becomes too complicated to solve exactly any but the simplest problems. Only the hydrogen and helium atoms have been calculated to a high accuracy. However, with various approximations, some fairly sloppy, many of the facts of more complicated atoms and of the chemical binding of molecules can be understood. We have shown you some of these approximations in earlier chapters. The Schrödinger equation as we have written it does not take into account any magnetic effects. It is possible to take such effects into account in an approximate way by adding some more terms to the equation. However, as we have seen in Volume II, magnetism is essentially a relativistic effect, and so a correct description of the motion of an electron in an arbitrary electromagnetic field can only be discussed in a proper relativistic equation. The correct relativistic equation for the motion of an electron was discovered by Dirac a year after Schrödinger brought forth his equation, and takes on quite a different form. We will not be able to discuss it at all here. Before we go on to look at some of the consequences of the Schrödinger equation, we would like to show you what it looks like for a system with a large number of particles. We will not be making any use of the equation, but just want to show it to you to emphasize that the wave function $\psi$ is not simply an ordinary wave in space, but is a function of many variables. If there are many particles, the equation becomes \begin{equation} \label{Eq:III:16:55} i\hbar\,\ddp{\psi(\FLPr_1,\FLPr_2,\FLPr_3,\dotsc)}{t}= \sum_i-\frac{\hbar^2}{2m_i}\biggl\{\! \frac{\partial^2\psi}{\partial x_i^2} +\frac{\partial^2\psi}{\partial y_i^2} +\frac{\partial^2\psi}{\partial z_i^2}\!\biggr\} +V(\FLPr_1,\FLPr_2,\dotsc)\psi. \end{equation} \begin{gather} \label{Eq:III:16:55} i\hbar\,\ddp{\psi(\FLPr_1,\FLPr_2,\FLPr_3,\dotsc)}{t}=\\[1.2ex] \sum_i-\frac{\hbar^2}{2m_i}\biggl\{\! \frac{\partial^2\psi}{\partial x_i^2} +\frac{\partial^2\psi}{\partial y_i^2} +\frac{\partial^2\psi}{\partial z_i^2}\!\biggr\} +V(\FLPr_1,\FLPr_2,\dotsc)\psi.\notag \end{gather} The potential function $V$ is what corresponds classically to the total potential energy of all the particles. If there are no external forces acting on the particles, the function $V$ is simply the electrostatic energy of interaction of all the particles. That is, if the $i$th particle carries the charge $Z_iq_e$, then the function $V$ is simply3 \begin{equation} \label{Eq:III:16:56} V(\FLPr_1,\FLPr_2,\FLPr_3,\dotsc)= \sum_{\substack{\text{all}\\\text{pairs}}} \frac{Z_iZ_j}{r_{ij}}\,e^2. \end{equation}

3

16

The Dependence of Amplitudes on Position

6

Quantized energy levels

In a later chapter we will look in detail at a solution of Schrödinger’s equation for a particular example. We would like now, however, to show you how one of the most remarkable consequence of Schrödinger’s equation comes about—namely, the surprising fact that a differential equation involving only continuous functions of continuous variables in space can give rise to quantum effects such as the discrete energy levels in an atom. The essential fact to understand is how it can be that an electron which is confined to a certain region of space by some kind of a potential “well” must necessarily have only one or another of a certain well-defined set of discrete energies. Suppose we think of an electron in a one-dimensional situation in which its potential energy varies with $x$ in a way described by the graph in Fig. 16–3. We will assume that this potential is static—it doesn’t vary with time. As we have done so many times before, we would like to look for solutions corresponding to states of definite energy, which means, of definite frequency. Let’s try a solution of the form \begin{equation} \label{Eq:III:16:57} \psi=a(x)e^{-iEt/\hbar}. \end{equation} If we substitute this function into the Schrödinger equation, we find that the function $a(x)$ must satisfy the following differential equation: \begin{equation} \label{Eq:III:16:58} \frac{d^2a(x)}{dx^2}=\frac{2m}{\hbar^2} [V(x)-E]a(x). \end{equation} This equation says that at each $x$ the second derivative of $a(x)$ with respect to $x$ is proportional to $a(x)$, the coefficient of proportionality being given by the quantity $(2m/\hbar^2)(V-E)$. The second derivative of $a(x)$ is the rate of change of its slope. If the potential $V$ is greater than the energy $E$ of the particle, the rate of change of the slope of $a(x)$ will have the same sign as $a(x)$. That means that the curve of $a(x)$ will be concave away from the $x$-axis. That is, it will have, more or less, the character of the positive or negative exponential function, $e^{\pm x}$. This means that in the region to the left of $x_1$, in Fig. 16–3, where $V$ is greater than the assumed energy $E$, the function $a(x)$ would have to look like one or another of the curves shown in part (a) of Fig. 16–4. If, on the other hand, the potential function $V$ is less than the energy $E$, the second derivative of $a(x)$ with respect to $x$ has the opposite sign from $a(x)$ itself, and the curve of $a(x)$ will always be concave toward the $x$-axis like one of the pieces shown in part (b) of Fig. 16–4. The solution in such a region has, piece-by-piece, roughly the form of a sinusoidal curve. Now let’s see if we can construct graphically a solution for the function $a(x)$ which corresponds to a particle of energy $E_a$ in the potential $V$ shown in Fig. 16–3. Since we are trying to describe a situation in which a particle is bound inside the potential well, we want to look for solutions in which the wave amplitude takes on very small values when $x$ is way outside the potential well. We can easily imagine a curve like the one shown in Fig. 16–5 which tends toward zero for large negative values of $x$, and grows smoothly as it approaches $x_1$. Since $V$ is equal to $E_a$ at $x_1$, the curvature of the function becomes zero at this point. Between $x_1$ and $x_2$, the quantity $V-E_a$ is always a negative number, so the function $a(x)$ is always concave toward the axis, and the curvature is larger the larger the difference between $E_a$ and $V$. If we continue the curve into the region between $x_1$ and $x_2$, it should go more or less as shown in Fig. 16–5. Now let’s continue this curve into the region to the right of $x_2$. There it curves away from the axis and takes off toward large positive values, as drawn in Fig. 16–6. For the energy $E_a$ we have chosen, the solution for $a(x)$ gets larger and larger with increasing $x$. In fact, its curvature is also increasing (if the potential continues to stay flat). The amplitude rapidly grows to immense proportions. What does this mean? It simply means that the particle is not “bound” in the potential well. It is infinitely more likely to be found outside of the well, than inside. For the solution we have manufactured, the electron is more likely to be found at $x=+\infty$ than anywhere else. We have failed to find a solution for a bound particle. Let’s try another energy, say one a little bit higher than $E_a$—say the energy $E_b$ in Fig. 16–7. If we start with the same conditions on the left, we get the solution drawn in the lower half of Fig. 16–7. It looked at first as though it were going to be better, but it ends up just as bad as the solution for $E_a$—except that now $a(x)$ is getting more and more negative as we go toward large values of $x$. Maybe that’s the clue. Since changing the energy a little bit from $E_a$ to $E_b$ causes the curve to flip from one side of the axis to the other, perhaps there is some energy lying between $E_a$ and $E_b$ for which the curve will approach zero for large values of $x$. There is, indeed, and we have sketched how the solution might look in Fig. 16–8. You should appreciate that the solution we have drawn in the figure is a very special one. If we were to raise or lower the energy ever so slightly, the function would go over into curves like one or the other of the two broken-line curves shown in Fig. 16–8, and we would not have the proper conditions for a bound particle. We have obtained a result that if a particle is to be bound in a potential well, it can do so only if it has a very definite energy. Does that mean that there is only one energy for a particle bound in a potential well? No. Other energies are possible, but not energies too close to $E_c$. Notice that the wave function we have drawn in Fig. 16–8 crosses the axis four times in the region between $x_1$ and $x_2$. If we were to pick an energy quite a bit lower than $E_c$, we could have a solution which crosses the axis only three times, only two times, only once, or not at all. The possible solutions are sketched in Fig. 16–9. (There may also be other solutions corresponding to values of the energy higher than the ones shown.) Our conclusion is that if a particle is bound in a potential well, its energy can take on only the certain special values in a discrete energy spectrum. You see how a differential equation can describe the basic fact of quantum physics. We might remark one other thing. If the energy $E$ is above the top of the potential well, then there are no longer any discrete solutions, and any possible energy is permitted. Such solutions correspond to the scattering of free particles by a potential well. We have seen an example of such solutions when we considered the effects of impurity atoms in a crystal.

3

17

Symmetry and Conservation Laws

1

Symmetry

In classical physics there are a number of quantities which are conserved—such as momentum, energy, and angular momentum. Conservation theorems about corresponding quantities also exist in quantum mechanics. The most beautiful thing of quantum mechanics is that the conservation theorems can, in a sense, be derived from something else, whereas in classical mechanics they are practically the starting points of the laws. (There are ways in classical mechanics to do an analogous thing to what we will do in quantum mechanics, but it can be done only at a very advanced level.) In quantum mechanics, however, the conservation laws are very deeply related to the principle of superposition of amplitudes, and to the symmetry of physical systems under various changes. This is the subject of the present chapter. Although we will apply these ideas mostly to the conservation of angular momentum, the essential point is that the theorems about the conservation of all kinds of quantities are—in the quantum mechanics—related to the symmetries of the system. We begin, therefore, by studying the question of symmetries of systems. A very simple example is the hydrogen molecular ion—we could equally well take the ammonia molecule—in which there are two states. For the hydrogen molecular ion we took as our base states one in which the electron was located near proton number $1$, and another in which the electron was located near proton number $2$. The two states—which we called $\ketsl{\slOne}$ and $\ketsl{\slTwo}$—are shown again in Fig. 17–1(a). Now, so long as the two nuclei are both exactly the same, then there is a certain symmetry in this physical system. That is to say, if we were to reflect the system in the plane halfway between the two protons—by which we mean that everything on one side of the plane gets moved to the symmetric position on the other side—we would get the situations in Fig. 17–1(b). Since the protons are identical, the operation of reflection changes $\ketsl{\slOne}$ into $\ketsl{\slTwo}$ and $\ketsl{\slTwo}$ into $\ketsl{\slOne}$. We’ll call this reflection operation $\Pop$ and write \begin{equation} \label{Eq:III:17:1} \Pop\,\ketsl{\slOne}=\ketsl{\slTwo},\quad \Pop\,\ketsl{\slTwo}=\ketsl{\slOne}. \end{equation} So our $\Pop$ is an operator in the sense that it “does something” to a state to make a new state. The interesting thing is that $\Pop$ operating on any state produces some other state of the system. Now $\Pop$, like any of the other operators we have described, has matrix elements which can be defined by the usual obvious notation. Namely, \begin{equation*} P_{11}=\bracketsl{\slOne}{\Pop}{\slOne}\quad \text{and}\quad P_{12}=\bracketsl{\slOne}{\Pop}{\slTwo} \end{equation*} are the matrix elements we get if we multiply $\Pop\,\ketsl{\slOne}$ and $\Pop\,\ketsl{\slTwo}$ on the left by $\bra{\slOne}$. From Eq. (17.1) they are \begin{align} \bracketsl{\slOne}{\Pop}{\slOne}&=P_{11}= \braketsl{\slOne}{\slTwo}=0,\notag\\[.5ex] \label{Eq:III:17:2} \bracketsl{\slOne}{\Pop}{\slTwo}&=P_{12}= \braketsl{\slOne}{\slOne}=1. \end{align} In the same way we can get $P_{21}$ and $P_{22}$. The matrix of $\Pop$—with respect to the base system $\ketsl{\slOne}$ and $\ketsl{\slTwo}$—is \begin{equation} \label{Eq:III:17:3} P= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}. \end{equation} We see once again that the words operator and matrix in quantum mechanics are practically interchangeable. There are slight technical differences—like the difference between a “numeral” and a “number”—but the distinction is something pedantic that we don’t have to worry about. So whether $\Pop$ defines an operation, or is actually used to define a matrix of numbers, we will call it interchangeably an operator or a matrix. Now we would like to point out something. We will suppose that the physics of the whole hydrogen molecular ion system is symmetrical. It doesn’t have to be—it depends, for instance, on what else is near it. But if the system is symmetrical, the following idea should certainly be true. Suppose we start at $t=0$ with the system in the state $\ketsl{\slOne}$ and find after an interval of time $t$ that the system turns out to be in a more complicated situation—in some linear combination of the two base states. Remember that in Chapter 8 we used to represent “going for a period of time” by multiplying by the operator $\Uop$. That means that the system would after a while—say $15$ seconds to be definite—be in some other state. For example, it might be $\sqrt{2/3}$ parts of the state $\ketsl{\slOne}$ and $i\sqrt{1/3}$ parts of the state $\ketsl{\slTwo}$, and we would write \begin{equation} \label{Eq:III:17:4} \ket{\text{$\psi$ at $15$ sec}}= \Uop(15,0)\,\ketsl{\slOne}=\sqrt{2/3}\,\ketsl{\slOne} +i\sqrt{1/3}\,\ketsl{\slTwo}. \end{equation} \begin{align} \label{Eq:III:17:4} \ket{\text{$\psi$ at $15$ sec}}&=\Uop(15,0)\,\ketsl{\slOne}\\[.7ex] &=\sqrt{2/3}\,\ketsl{\slOne}+i\sqrt{1/3}\,\ketsl{\slTwo}.\notag \end{align} Now we ask what happens if we start the system in the symmetric state $\ketsl{\slTwo}$ and wait for $15$ seconds under the same conditions? It is clear that if the world is symmetric—as we are supposing—we should get the state symmetric to (17.4): \begin{equation} \label{Eq:III:17:5} \ket{\text{$\psi$ at $15$ sec}}= \Uop(15,0)\,\ketsl{\slTwo}=\sqrt{2/3}\,\ketsl{\slTwo} +i\sqrt{1/3}\,\ketsl{\slOne}. \end{equation} \begin{align} \label{Eq:III:17:5} \ket{\text{$\psi$ at $15$ sec}}&=\Uop(15,0)\,\ketsl{\slTwo}\\[.7ex] &=\sqrt{2/3}\,\ketsl{\slTwo}+i\sqrt{1/3}\,\ketsl{\slOne}.\notag \end{align} The same ideas are sketched diagrammatically in Fig. 17–2. So if the physics of a system is symmetrical with respect to some plane, and we work out the behavior of a particular state, we also know the behavior of the state we would get by reflecting the original state in the symmetry plane. We would like to say the same things a little bit more generally—which means a little more abstractly. Let $\Qop$ be any one of a number of operations that you could perform on a system without changing the physics. For instance, for $\Qop$ we might be thinking of $\Pop$, the operation of a reflection in the plane between the two atoms in the hydrogen molecule. Or, in a system with two electrons, we might be thinking of the operation of interchanging the two electrons. Another possibility would be, in a spherically symmetric system, the operation of a rotation of the whole system through a finite angle around some axis—which wouldn’t change the physics. Of course, we would normally want to give each special case some special notation for $\Qop$. Specifically, we will normally define the $\Rop_y(\theta)$ to be the operation “rotate the system about the $y$-axis by the angle $\theta$”. By $\Qop$ we mean just any one of the operators we have described or any other one—which leaves the basic physical situation unchanged. Let’s think of some more examples. If we have an atom with no external magnetic field or no external electric field, and if we were to turn the coordinates around any axis, it would be the same physical system. Again, the ammonia molecule is symmetrical with respect to a reflection in a plane parallel to that of the three hydrogens—so long as there is no electric field. When there is an electric field, when we make a reflection we would have to change the electric field also, and that changes the physical problem. But if we have no external field, the molecule is symmetrical. Now we consider a general situation. Suppose we start with the state $\ket{\psi_1}$ and after some time or other under given physical conditions it has become the state $\ket{\psi_2}$. We can write \begin{equation} \label{Eq:III:17:6} \ket{\psi_2}=\Uop\,\ket{\psi_1}. \end{equation} [You can be thinking of Eq. (17.4).] Now imagine we perform the operation $\Qop$ on the whole system. The state $\ket{\psi_1}$ will be transformed to a state $\ket{\psi_1'}$, which we can also write as $\Qop\,\ket{\psi_1}$. Also the state $\ket{\psi_2}$ is changed into $\ket{\psi_2'}=\Qop\,\ket{\psi_2}$. Now if the physics is symmetrical under $\Qop$ (don’t forget the if; it is not a general property of systems), then, waiting for the same time under the same conditions, we should have \begin{equation} \label{Eq:III:17:7} \ket{\psi_2'}=\Uop\,\ket{\psi_1'}. \end{equation} [Like Eq. (17.5).] But we can write $\Qop\,\ket{\psi_1}$ for $\ket{\psi_1'}$ and $\Qop\,\ket{\psi_2}$ for $\ket{\psi_2'}$ so (17.7) can also be written \begin{equation} \label{Eq:III:17:8} \Qop\,\ket{\psi_2}=\Uop\Qop\,\ket{\psi_1}. \end{equation} If we now replace $\ket{\psi_2}$ by $\Uop\,\ket{\psi_1}$—Eq. (17.6)—we get \begin{equation} \label{Eq:III:17:9} \Qop\Uop\,\ket{\psi_1}=\Uop\Qop\,\ket{\psi_1}. \end{equation} It’s not hard to understand what this means. Thinking of the hydrogen ion it says that: “making a reflection and waiting a while”—the expression on the right of Eq. (17.9)—is the same as “waiting a while and then making a reflection”—the expression on the left of (17.9). These should be the same so long as $U$ doesn’t change under the reflection. Since (17.9) is true for any starting state $\ket{\psi_1}$, it is really an equation about the operators: \begin{equation} \label{Eq:III:17:10} \Qop\Uop=\Uop\Qop. \end{equation} This is what we wanted to get—it is a mathematical statement of symmetry. When Eq. (17.10) is true, we say that the operators $\Uop$ and $\Qop$ commute. We can then define “symmetry” in the following way: A physical system is symmetric with respect to the operation $\Qop$ when $\Qop$ commutes with $\Uop$, the operation of the passage of time. [In terms of matrices, the product of two operators is equivalent to the matrix product, so Eq. (17.10) also holds for the matrices $Q$ and $U$ for a system which is symmetric under the transformation $Q$.] Incidentally, since for infinitesimal times $\epsilon$ we have $\Uop=1-i\Hop\epsilon/\hbar$—where $\Hop$ is the usual Hamiltonian (see Chapter 8)—you can see that if (17.10) is true, it is also true that \begin{equation} \label{Eq:III:17:11} \Qop\Hop=\Hop\Qop. \end{equation} So (17.11) is the mathematical statement of the condition for the symmetry of a physical situation under the operator $\Qop$. It defines a symmetry.

3

17

Symmetry and Conservation Laws

2

Symmetry and conservation

Before applying the result we have just found, we would like to discuss the idea of symmetry a little more. Suppose that we have a very special situation: after we operate on a state with $\Qop$, we get the same state. This is a very special case, but let’s suppose it happens to be true for a state $\ket{\psi_0}$ that $\ket{\psi'}=\Qop\,\ket{\psi_0}$ is physically the same state as $\ket{\psi_0}$. That means that $\ket{\psi'}$ is equal to $\ket{\psi_0}$ except for some phase factor.1 How can that happen? For instance, suppose that we have an $\text{H}_2^+$ ion in the state which we once called $\ketsl{\slI}$.2 For this state there is equal amplitude to be in the base states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$. The probabilities are shown as a bar graph in Fig. 17–3(a). If we operate on $\ketsl{\slI}$ with the reflection operator $\Pop$, it flips the state over changing $\ketsl{\slOne}$ to $\ketsl{\slTwo}$ and $\ketsl{\slTwo}$ to $\ketsl{\slOne}$—we get the probabilities shown in Fig. 17–3(b). But that’s just the state $\ketsl{\slI}$ all over again. If we start with state $\ketsl{\slII}$ the probabilities before and after reflection look just the same. However, there is a difference if we look at the amplitudes. For the state $\ketsl{\slI}$ the amplitudes are the same after the reflection, but for the state $\ketsl{\slII}$ the amplitudes have the opposite sign. In other words, \begin{equation} \begin{aligned} \Pop\,\ketsl{\slI}&=\Pop\biggl\{ \frac{\ketsl{\slOne}+\ketsl{\slTwo}}{\sqrt{2}}\biggr\}= \frac{\ketsl{\slTwo}+\ketsl{\slOne}}{\sqrt{2}}=\ketsl{\slI},\\[2ex] \Pop\,\ketsl{\slII}&=\Pop\biggl\{ \frac{\ketsl{\slOne}-\ketsl{\slTwo}}{\sqrt{2}}\biggr\}= \frac{\ketsl{\slTwo}-\ketsl{\slOne}}{\sqrt{2}}=-\,\ketsl{\slII}. \end{aligned} \label{Eq:III:17:12} \end{equation} \begin{equation} \begin{alignedat}{2} \Pop\,\ketsl{\slI}\,&=\Pop\biggl\{&\frac{\ketsl{\slOne}+\ketsl{\slTwo}}{\sqrt{2}}&\biggr\}\\[1ex] &=&\frac{\ketsl{\slTwo}+\ketsl{\slOne}}{\sqrt{2}}&=\ketsl{\slI},\\[2ex] \Pop\,\ketsl{\slII}\,&=\Pop\biggl\{&\frac{\ketsl{\slOne}-\ketsl{\slTwo}}{\sqrt{2}}&\biggr\}\\[1ex] &=&\frac{\ketsl{\slTwo}-\ketsl{\slOne}}{\sqrt{2}}&=-\,\ketsl{\slII}. \end{alignedat} \label{Eq:III:17:12} \end{equation} If we write $\Pop\,\ket{\psi_0}=e^{i\delta}\,\ket{\psi_0}$, we have that $e^{i\delta}=1$ for the state $\ketsl{\slI}$ and $e^{i\delta}=-1$ for the state $\ketsl{\slII}$. Let’s look at another example. Suppose we have a RHC polarized photon propagating in the $z$-direction. If we do the operation of a rotation around the $z$-axis, we know that this just multiplies the amplitude by $e^{i\phi}$ when $\phi$ is the angle of the rotation. So for the rotation operation in this case, $\delta$ is just equal to the angle of rotation. Now it is clear that if it happens to be true that an operator $\Qop$ just changes the phase of a state at some time, say $t=0$, it is true forever. In other words, if the state $\ket{\psi_1}$ goes over into the state $\ket{\psi_2}$ after a time $t$, or \begin{equation} \label{Eq:III:17:13} \Uop(t,0)\,\ket{\psi_1}=\ket{\psi_2} \end{equation} and if the symmetry of the situation makes it so that \begin{equation} \label{Eq:III:17:14} \Qop\,\ket{\psi_1}=e^{i\delta}\,\ket{\psi_1}, \end{equation} then it is also true that \begin{equation} \label{Eq:III:17:15} \Qop\,\ket{\psi_2}=e^{i\delta}\,\ket{\psi_2}. \end{equation} This is clear, since \begin{equation*} \Qop\,\ket{\psi_2}=\Qop\Uop\,\ket{\psi_1}=\Uop\Qop\,\ket{\psi_1},\notag \end{equation*} and if $\Qop\,\ket{\psi_1}=e^{i\delta}\,\ket{\psi_1}$, then \begin{equation*} \Qop\,\ket{\psi_2}=\Uop e^{i\delta}\,\ket{\psi_1}= e^{i\delta}\Uop\,\ket{\psi_1}=e^{i\delta}\,\ket{\psi_2}. \end{equation*} [The sequence of equalities follows from (17.13) and (17.10) for a symmetrical system, from (17.14), and from the fact that a number like $e^{i\delta}$ commutes with an operator.] So with certain symmetries something which is true initially is true for all times. But isn’t that just a conservation law? Yes! It says that if you look at the original state and by making a little computation on the side discover that an operation which is a symmetry operation of the system produces only a multiplication by a certain phase, then you know that the same property will be true of the final state—the same operation multiplies the final state by the same phase factor. This is always true even though we may not know anything else about the inner mechanism of the universe which changes a system from the initial to the final state. Even if we do not care to look at the details of the machinery by which the system gets from one state to another, we can still say that if a thing is in a state with a certain symmetry character originally, and if the Hamiltonian for this thing is symmetrical under that symmetry operation, then the state will have the same symmetry character for all times. That’s the basis of all the conservation laws of quantum mechanics. Let’s look at a special example. Let’s go back to the $\Pop$ operator. We would like first to modify a little our definition of $\Pop$. We want to take for $\Pop$ not just a mirror reflection, because that requires defining the plane in which we put the mirror. There is a special kind of a reflection that doesn’t require the specification of a plane. Suppose we redefine the operation $\Pop$ this way: First you reflect in a mirror in the $z$-plane so that $z$ goes to $-z$, $x$ stays $x$, and $y$ stays $y$; then you turn the system $180^\circ$ about the $z$-axis so that $x$ is made to go to $-x$ and $y$ to $-y$. The whole thing is called an inversion. Every point is projected through the origin to the diametrically opposite position. All the coordinates of everything are reversed. We will still use the symbol $\Pop$ for this operation. It is shown in Fig. 17–4. It is a little more convenient than a simple reflection because it doesn’t require that you specify which coordinate plane you used for the reflection—you need specify only the point which is at the center of symmetry. Now let’s suppose that we have a state $\ket{\psi_0}$ which under the inversion operation goes into $e^{i\delta}\,\ket{\psi_0}$—that is, \begin{equation} \label{Eq:III:17:16} \ket{\psi_0'}=\Pop\,\ket{\psi_0}=e^{i\delta}\,\ket{\psi_0}. \end{equation} Then suppose that we invert again. After two inversions we are right back where we started from—nothing is changed at all. We must have that \begin{equation*} \Pop\,\ket{\psi_0'}=\Pop\!\cdot\!\Pop\,\ket{\psi_0}=\ket{\psi_0}. \end{equation*} But \begin{equation*} \Pop\!\cdot\!\Pop\,\ket{\psi_0}=\Pop e^{i\delta}\,\ket{\psi_0} =e^{i\delta}\Pop\,\ket{\psi_0}=(e^{i\delta})^2\,\ket{\psi_0}. \end{equation*} It follows that \begin{equation*} (e^{i\delta})^2=1. \end{equation*} So if the inversion operator is a symmetry operation of a state, there are only two possibilities for $e^{i\delta}$: \begin{equation*} e^{i\delta}=\pm1, \end{equation*} which means that \begin{equation} \label{Eq:III:17:17} \Pop\,\ket{\psi_0}=\ket{\psi_0}\quad \text{or}\quad \Pop\,\ket{\psi_0}=-\,\ket{\psi_0}. \end{equation} Classically, if a state is symmetric under an inversion, the operation gives back the same state. In quantum mechanics, however, there are the two possibilities: we get the same state or minus the same state. When we get the same state, $\Pop\,\ket{\psi_0}=\ket{\psi_0}$, we say that the state $\ket{\psi_0}$ has even parity. When the sign is reversed so that $\Pop\,\ket{\psi_0}=-\,\ket{\psi_0}$, we say that the state has odd parity. (The inversion operator $\Pop$ is also known as the parity operator.) The state $\ketsl{\slI}$ of the $\text{H}_2^+$ ion has even parity; and the state $\ketsl{\slII}$ has odd parity—see Eq. (17.12). There are, of course, states which are not symmetric under the operation $\Pop$; these are states with no definite parity. For instance, in the $\text{H}_2^+$ system the state $\ketsl{\slI}$ has even parity, the state $\ketsl{\slII}$ has odd parity, and the state $\ketsl{\slOne}$ has no definite parity. When we speak of an operation like inversion being performed “on a physical system” we can think about it in two ways. We can think of physically moving whatever is at $\FLPr$ to the inverse point at $-\FLPr$, or we can think of looking at the same system from a new frame of reference $x',y',z'$ related to the old by $x'=-x$, $y'=-y$, and $z'=-z$. Similarly, when we think of rotations, we can think of rotating bodily a physical system, or of rotating the coordinate frame with respect to which we measure the system, keeping the “system” fixed in space. Generally, the two points of view are essentially equivalent. For rotation they are equivalent except that rotating a system by the angle $\theta$ is like rotating the reference frame by the negative of $\theta$. In these lectures we have usually considered what happens when a projection is made into a new set of axes. What you get that way is the same as what you get if you leave the axes fixed and rotate the system backwards by the same amount. When you do that, the signs of the angles are reversed.3 Many of the laws of physics—but not all—are unchanged by a reflection or an inversion of the coordinates. They are symmetric with respect to an inversion. The laws of electrodynamics, for instance, are unchanged if we change $x$ to $-x$, $y$ to $-y$, and $z$ to $-z$ in all the equations. The same is true for the laws of gravity, and for the strong interactions of nuclear physics. Only the weak interactions—responsible for $\beta$-decay—do not have this symmetry. (We discussed this in some detail in Chapter 52, Vol. I.) We will for now leave out any consideration of the $\beta$-decays. Then in any physical system where $\beta$-decays are not expected to produce any appreciable effect—an example would be the emission of light by an atom—the Hamiltonian $\Hop$ and the operator $\Pop$ will commute. Under these circumstances we have the following proposition. If a state originally has even parity, and if you look at the physical situation at some later time, it will again have even parity. For instance, suppose an atom about to emit a photon is in a state known to have even parity. You look at the whole thing—including the photon—after the emission; it will again have even parity (likewise if you start with odd parity). This principle is called the conservation of parity. You can see why the words “conservation of parity” and “reflection symmetry” are closely intertwined in the quantum mechanics. Although until a few years ago it was thought that nature always conserved parity, it is now known that this is not true. It has been discovered to be false because the $\beta$-decay reaction does not have the inversion symmetry which is found in the other laws of physics. Now we can prove an interesting theorem (which is true so long as we can disregard weak interactions): Any state of definite energy which is not degenerate must have a definite parity. It must have either even parity or odd parity. (Remember that we have sometimes seen systems in which several states have the same energy—we say that such states are degenerate. Our theorem will not apply to them.) For a state $\ket{\psi_0}$ of definite energy, we know that \begin{equation} \label{Eq:III:17:18} \Hop\,\ket{\psi_0}=E\,\ket{\psi_0}, \end{equation} where $E$ is just a number—the energy of the state. If we have any operator $\Qop$ which is a symmetry operator of the system we can prove that \begin{equation} \label{Eq:III:17:19} \Qop\,\ket{\psi_0}=e^{i\delta}\,\ket{\psi_0} \end{equation} so long as $\ket{\psi_0}$ is a unique state of definite energy. Consider the new state $\ket{\psi_0'}$ that you get from operating with $\Qop$. If the physics is symmetric, then $\ket{\psi_0'}$ must have the same energy as $\ket{\psi_0}$. But we have taken a situation in which there is only one state of that energy, namely $\ket{\psi_0}$, so $\ket{\psi_0'}$ must be the same state—it can only differ by a phase. That’s the physical argument. The same thing comes out of our mathematics. Our definition of symmetry is Eq. (17.10) or Eq. (17.11) (good for any state $\psi$), \begin{equation} \label{Eq:III:17:20} \Hop\Qop\,\ket{\psi}=\Qop\Hop\,\ket{\psi}. \end{equation} But we are considering only a state $\ket{\psi_0}$ which is a definite energy state, so that $\Hop\,\ket{\psi_0}=E\,\ket{\psi_0}$. Since $E$ is just a number that floats through $\Qop$ if we want, we have \begin{equation*} \Qop\Hop\,\ket{\psi_0}=\Qop E\,\ket{\psi_0}=E\Qop\,\ket{\psi_0}. \end{equation*} So \begin{equation} \label{Eq:III:17:21} \Hop\{\Qop\,\ket{\psi_0}\}=E\{\Qop\,\ket{\psi_0}\}. \end{equation} So $\ket{\psi_0'}=\Qop\,\ket{\psi_0}$ is also a definite energy state of $\Hop$—and with the same $E$. But by our hypothesis, there is only one such state; it must be that $\ket{\psi_0'}=e^{i\delta}\,\ket{\psi_0}$. What we have just proved is true for any operator $\Qop$ that is a symmetry operator of the physical system. Therefore, in a situation in which we consider only electrical forces and strong interactions—and no $\beta$-decay—so that inversion symmetry is an allowed approximation, we have that $\Pop\,\ket{\psi}=e^{i\delta}\,\ket{\psi}$. But we have also seen that $e^{i\delta}$ must be either $+1$ or $-1$. So any state of a definite energy (which is not degenerate) has got either an even parity or an odd parity.

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Symmetry and Conservation Laws

3

The conservation laws

We turn now to another interesting example of an operation: a rotation. We consider the special case of an operator that rotates an atomic system by angle $\phi$ around the $z$-axis. We will call this operator4 $\Rop_z(\phi)$. We are going to suppose that we have a physical situation where we have no influences lined up along the $x$- and $y$-axes. Any electric field or magnetic field is taken to be parallel to the $z$-axis5 so that there will be no change in the external conditions if we rotate the whole physical system about the $z$-axis. For example, if we have an atom in empty space and we turn the atom around the $z$-axis by an angle $\phi$, we have the same physical system. Now then, there are special states which have the property that such an operation produces a new state which is the original state multiplied by some phase factor. Let us make a quick side remark to show you that when this is true the phase change must always be proportional to the angle $\phi$. Suppose that you would rotate twice by the angle $\phi$. That’s the same thing as rotating by the angle $2\phi$. If a rotation by $\phi$ has the effect of multiplying the state $\ket{\psi_0}$ by a phase $e^{i\delta}$ so that \begin{equation*} \Rop_z(\phi)\,\ket{\psi_0}=e^{i\delta}\,\ket{\psi_0}, \end{equation*} two such rotations in succession would multiply the state by the factor $(e^{i\delta})^2=e^{i2\delta}$, since \begin{equation*} \Rop_z(\phi)\Rop_z(\phi)\,\ket{\psi_0}= \Rop_z(\phi)e^{i\delta}\,\ket{\psi_0}= e^{i\delta}\Rop_z(\phi)\,\ket{\psi_0}= e^{i\delta}e^{i\delta}\,\ket{\psi_0}. \end{equation*} \begin{alignat*}{2} \Rop_z(\phi)\Rop_z(\phi)\,\ket{\psi_0}&=\;&\Rop_z(\phi)e^{i\delta}\,\ket{\psi_0}&\\[1ex] &=\;&e^{i\delta}\Rop_z(\phi)\,\ket{\psi_0}&\\[1ex] &=&e^{i\delta}e^{i\delta}\,\ket{\psi_0}&. \end{alignat*} The phase change $\delta$ must be proportional to $\phi$.6 We are considering then those special states $\ket{\psi_0}$ for which \begin{equation} \label{Eq:III:17:22} \Rop_z(\phi)\,\ket{\psi_0}=e^{im\phi}\,\ket{\psi_0}, \end{equation} where $m$ is some real number. We also know the remarkable fact that if the system is symmetrical for a rotation around $z$ and if the original state happens to have the property that (17.22) is true, then it will also have the same property later on. So this number $m$ is a very important one. If we know its value initially, we know its value at the end of the game. It is a number which is conserved—$m$ is a constant of the motion. The reason that we pull out $m$ is because it hasn’t anything to do with any special angle $\phi$, and also because it corresponds to something in classical mechanics. In quantum mechanics we choose to call $m\hbar$—for such states as $\ket{\psi_0}$—the angular momentum about the $z$-axis. If we do that we find that in the limit of large systems the same quantity is equal to the $z$-component of the angular momentum of classical mechanics. So if we have a state for which a rotation about the $z$-axis just produces a phase factor $e^{im\phi}$, then we have a state of definite angular momentum about that axis—and the angular momentum is conserved. It is $m\hbar$ now and forever. Of course, you can rotate about any axis, and you get the conservation of angular momentum for the various axes. You see that the conservation of angular momentum is related to the fact that when you turn a system you get the same state with only a new phase factor. We would like to show you how general this idea is. We will apply it to two other conservation laws which have exact correspondence in the physical ideas to the conservation of angular momentum. In classical physics we also have conservation of momentum and conservation of energy, and it is interesting to see that both of these are related in the same way to some physical symmetry. Suppose that we have a physical system—an atom, some complicated nucleus, or a molecule, or something—and it doesn’t make any difference if we take the whole system and move it over to a different place. So we have a Hamiltonian which has the property that it depends only on the internal coordinates in some sense, and does not depend on the absolute position in space. Under those circumstances there is a special symmetry operation we can perform which is a translation in space. Let’s define $\Dop_x(a)$ as the operation of a displacement by the distance $a$ along the $x$-axis. Then for any state we can make this operation and get a new state. But again there can be very special states which have the property that when you displace them by $a$ along the $x$-axis you get the same state except for a phase factor. It’s also possible to prove, just as we did above, that when this happens, the phase must be proportional to $a$. So we can write for these special states $\ket{\psi_0}$ \begin{equation} \label{Eq:III:17:23} \Dop_x(a)\,\ket{\psi_0}=e^{ika}\,\ket{\psi_0}. \end{equation} The coefficient $k$, when multiplied by $\hbar$, is called the $x$-component of the momentum. And the reason it is called that is that this number is numerically equal to the classical momentum $p_x$ when we have a large system. The general statement is this: If the Hamiltonian is unchanged when the system is displaced, and if the state starts with a definite momentum in the $x$-direction, then the momentum in the $x$-direction will remain the same as time goes on. The total momentum of a system before and after collisions—or after explosions or what not—will be the same. There is another operation that is quite analogous to the displacement in space: a delay in time. Suppose that we have a physical situation where there is nothing external that depends on time, and we start something off at a certain moment in a given state and let it roll. Now if we were to start the same thing off again (in another experiment) two seconds later—or/say, delayed by a time $\tau$—and if nothing in the external conditions depends on the absolute time, the development would be the same and the final state would be the same as the other final state, except that it will get there later by the time $\tau$. Under those circumstances we can also find special states which have the property that the development in time has the special characteristic that the delayed state is just the old, multiplied by a phase factor. Once more it is clear that for these special states the phase change must be proportional to $\tau$. We can write \begin{equation} \label{Eq:III:17:24} \Dop_t(\tau)\,\ket{\psi_0}=e^{-i\omega\tau}\,\ket{\psi_0}. \end{equation} It is conventional to use the negative sign in defining $\omega$; with this convention $\omega\hbar$ is the energy of the system, and it is conserved. So a system of definite energy is one which when displaced $\tau$ in time reproduces itself multiplied by $e^{-i\omega\tau}$. (That’s what we have said before when we defined a quantum state of definite energy, so we’re consistent with ourselves.) It means that if a system is in a state of definite energy, and if the Hamiltonian doesn’t depend on $t$, then no matter what goes on, the system will have the same energy at all later times. You see, therefore, the relation between the conservation laws and the symmetry of the world. Symmetry with respect to displacements in time implies the conservation of energy; symmetry with respect to position in $x$, $y$, or $z$ implies the conservation of that component of momentum. Symmetry with respect to rotations around the $x$-, $y$-, and $z$-axes implies the conservation of the $x$-, $y$-, and $z$-components of angular momentum. Symmetry with respect to reflection implies the conservation of parity. Symmetry with respect to the interchange of two electrons implies the conservation of something we don’t have a name for, and so on. Some of these principles have classical analogs and others do not. There are more conservation laws in quantum mechanics than are useful in classical mechanics—or, at least, than are usually made use of. In order that you will be able to read other books on quantum mechanics, we must make a small technical aside—to describe the notation that people use. The operation of a displacement with respect to time is, of course, just the operation $\Uop$ that we talked about before: \begin{equation} \label{Eq:III:17:25} \Dop_t(\tau)=\Uop(t+\tau,t). \end{equation} Most people like to discuss everything in terms of infinitesimal displacements in time, or in terms of infinitesimal displacements in space, or in terms of rotations through infinitesimal angles. Since any finite displacement or angle can be accumulated by a succession of infinitesimal displacements or angles, it is often easier to analyze first the infinitesimal case. The operator of an infinitesimal displacement $\Delta t$ in time is—as we have defined it in Chapter 8— \begin{equation} \label{Eq:III:17:26} \Dop_t(\Delta t)=1-\frac{i}{\hbar}\,\Delta t\Hop. \end{equation} Then $\Hop$ is analogous to the classical quantity we call energy, because if $\Hop\,\ket{\psi}$ happens to be a constant times $\ket{\psi}$ namely, $\Hop\,\ket{\psi}=E\,\ket{\psi}$ then that constant is the energy of the system. The same thing is done for the other operations. If we make a small displacement in $x$, say by the amount $\Delta x$, a state $\ket{\psi}$ will, in general, go over into some other state $\ket{\psi'}$. We can write \begin{equation} \label{Eq:III:17:27} \ket{\psi'}=\Dop_x(\Delta x)\,\ket{\psi}= \Bigl(1+\frac{i}{\hbar}\,\pop_x\Delta x\Bigr)\ket{\psi}, \end{equation} since as $\Delta x$ goes to zero, the $\ket{\psi'}$ should become just $\ket{\psi}$ or $\Dop_x(0)=1$, and for small $\Delta x$ the change of $\Dop_x(\Delta x)$ from $1$ should be proportional to $\Delta x$. Defined this way, the operator $\pop_x$ is called the momentum operator—for the $x$-component, of course. For identical reasons, people usually write for small rotations \begin{equation} \label{Eq:III:17:28} \Rop_z(\Delta\phi)\,\ket{\psi}= \Bigl(1+\frac{i}{\hbar}\,\Jop_z\Delta\phi\Bigr)\ket{\psi} \end{equation} and call $\Jop_z$ the operator of the $z$-component of angular momentum. For those special states for which $\Rop_z(\phi)\,\ket{\psi_0}=e^{im\phi}\,\ket{\psi_0}$, we can for any small angle—say $\Delta\phi$—expand the right-hand side to first order in $\Delta\phi$ and get \begin{equation*} \Rop_z(\Delta\phi)\,\ket{\psi_0}= e^{im\Delta\phi}\,\ket{\psi_0}= (1+im\Delta\phi)\,\ket{\psi_0}. \end{equation*} Comparing this with the definition of $\Jop_z$ in Eq. (17.28), we get that \begin{equation} \label{Eq:III:17:29} \Jop_z\,\ket{\psi_0}=m\hbar\,\ket{\psi_0}. \end{equation} In other words, if you operate with $\Jop_z$ on a state with a definite angular momentum about the $z$-axis, you get $m\hbar$ times the same state, where $m\hbar$ is the amount of $z$-component of angular momentum. It is quite analogous to operating on a definite energy state with $\Hop$ to get $E\,\ket{\psi}$. We would now like to make some applications of the ideas of the conservation of angular momentum—to show you how they work. The point is that they are really very simple. You knew before that angular momentum is conserved. The only thing you really have to remember from this chapter is that if a state $\ket{\psi_0}$ has the property that upon a rotation through an angle $\phi$ about the $z$-axis, it becomes $e^{im\phi}\,\ket{\psi_0}$; it has a $z$-component of angular momentum equal to $m\hbar$. That’s all we will need to do a number of interesting things.

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Symmetry and Conservation Laws

4

Polarized light

First of all we would like to check on one idea. In Section 11-4 we showed that when RHC polarized light is viewed in a frame rotated by the angle $\phi$ about the $z$-axis7 it gets multiplied by $e^{i\phi}$. Does that mean then that the photons of light that are right circularly polarized carry an angular momentum of one unit8 along the $z$-axis? Indeed it does. It also means that if we have a beam of light containing a large number of photons all circularly polarized the same way—as we would have in a classical beam—it will carry angular momentum. If the total energy carried by the beam in a certain time is $W$, then there are $N=W/\hbar\omega$ photons. Each one carries the angular momentum $\hbar$, so there is a total angular momentum of \begin{equation} \label{Eq:III:17:30} J_z=N\hbar=\frac{W}{\omega}. \end{equation} Can we prove classically that light which is right circularly polarized carries an energy and angular momentum in proportion to $W/\omega$? That should be a classical proposition if everything is right. Here we have a case where we can go from the quantum thing to the classical thing. We should see if the classical physics checks. It will give us an idea whether we have a right to call $m$ the angular momentum. Remember what right circularly polarized light is, classically. It’s described by an electric field with an oscillating $x$-component and an oscillating $y$-component $90^\circ$ out of phase so that the resultant electric vector $\Efieldvec$ goes in a circle—as drawn in Fig. 17–5(a). Now suppose that such light shines on a wall which is going to absorb it—or at least some of it—and consider an atom in the wall according to the classical physics. We have often described the motion of the electron in the atom as a harmonic oscillator which can be driven into oscillation by an external electric field. We’ll suppose that the atom is isotropic, so that it can oscillate equally well in the $x$- or $y$-directions. Then in the circularly polarized light, the $x$-displacement and the $y$-displacement are the same, but one is $90^\circ$ behind the other. The net result is that the electron moves in a circle, as shown in Fig. 17–5(b). The electron is displaced at some displacement $\FLPr$ from its equilibrium position at the origin and goes around with some phase lag with respect to the vector $\Efieldvec$. The relation between $\Efieldvec$ and $\FLPr$ might be as shown in Fig. 17–5(b). As time goes on, the electric field rotates and the displacement rotates with the same frequency, so their relative orientation stays the same. Now let’s look at the work being done on this electron. The rate that energy is being put into this electron is $v$, its velocity, times the component of $q\Efieldvec$ parallel to the velocity: \begin{equation} \label{Eq:III:17:31} \ddt{W}{t}=q\Efield_tv. \end{equation} But look, there is angular momentum being poured into this electron, because there is always a torque about the origin. The torque is $q\Efield_tr$, which must be equal to the rate of change of angular momentum $dJ_z/dt$: \begin{equation} \label{Eq:III:17:32} \ddt{J_z}{t}=q\Efield_tr. \end{equation} Remembering that $v=\omega r$, we have that \begin{equation*} \ddt{J_z}{W}=\frac{1}{\omega}. \end{equation*} Therefore, if we integrate the total angular momentum which is absorbed, it is proportional to the total energy—the constant of proportionality being $1/\omega$, which agrees with Eq. (17.30). Light does carry angular momentum—$1$ unit (times $\hbar$) if it is right circularly polarized along the $z$-axis, and $-1$ unit along the $z$-axis if it is left circularly polarized. Now let’s ask the following question: If light is linearly polarized in the $x$-direction, what is its angular momentum? Light polarized in the $x$-direction can be represented as the superposition of RHC and LHC polarized light. Therefore, there is a certain amplitude that the angular momentum is $+\hbar$ and another amplitude that the angular momentum is $-\hbar$, so it doesn’t have a definite angular momentum. It has an amplitude to appear with $+\hbar$ and an equal amplitude to appear with $-\hbar$. The interference of these two amplitudes produces the linear polarization, but it has equal probabilities to appear with plus or minus one unit of angular momentum. Macroscopic measurements made on a beam of linearly polarized light will show that it carries zero angular momentum, because in a large number of photons there are nearly equal numbers of RHC and LHC photons contributing opposite amounts of angular momentum—the average angular momentum is zero. And in the classical theory you don’t find the angular momentum unless there is some circular polarization. We have said that any spin-one particle can have three values of $J_z$, namely $+1,0,-1$ (the three states we saw in the Stern-Gerlach experiment). But light is screwy; it has only two states. It does not have the zero case. This strange lack is related to the fact that light cannot stand still. For a particle of spin $j$ which is standing still, there must be the $2j+1$ possible states with values of $j_z$ going in steps of $1$ from $-j$ to $+j$. But it turns out that for something of spin $j$ with zero mass only the states with the components $+j$ and $-j$ along the direction of motion exist. For example, light does not have three states, but only two—although a photon is still an object of spin one. How is this consistent with our earlier proofs—based on what happens under rotations in space—that for spin-one particles three states are necessary? For a particle at rest, rotations can be made about any axis without changing the momentum state. Particles with zero rest mass (like photons and neutrinos) cannot be at rest; only rotations about the axis along the direction of motion do not change the momentum state. Arguments about rotations around one axis only are insufficient to prove that three states are required, given that one of them varies as $e^{i\phi}$ under rotations by the angle $\phi$.9 One further side remark. For a zero rest mass particle, in general, only one of the two spin states with respect to the line of motion ($+j$, $-j$) is really necessary. For neutrinos—which are spin one-half particles—only the states with the component of angular momentum opposite to the direction of motion $(-\hbar/2)$ exist in nature [and only along the motion $(+\hbar/2)$ for antineutrinos]. When a system has inversion symmetry (so that parity is conserved, as it is for light) both components ($+j$, and $-j$) are required.

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Symmetry and Conservation Laws

5

The disintegration of the $\boldsymbol{\Lambda}^{\boldsymbol{0}}$

Now we want to give an example of how we use the theorem of conservation of angular momentum in a specifically quantum physical problem. We look at break-up of the lambda particle ($\Lambda^0$), which disintegrates into a proton and a $\pi^-$ meson by a “weak” interaction: \begin{equation*} \Lambda^0\to\text{p}+\pi^-. \end{equation*} Assume we know that the pion has spin zero, that the proton has spin one-half, and that the $\Lambda^0$ has spin one-half. We would like to solve the following problem: Suppose that a $\Lambda^0$ were to be produced in a way that caused it to be completely polarized—by which we mean that its spin is, say “up,” with respect to some suitably chosen $z$-axis—see Fig. 17–6(a). The question is, with what probability will it disintegrate so that the proton goes off at an angle $\theta$ with respect to the $z$-axis—as in Fig. 17–6(b)? In other words, what is the angular distribution of the disintegrations? We will look at the disintegration in the coordinate system in which the $\Lambda^0$ is at rest—we will measure the angles in this rest frame; then they can always be transformed to another frame if we want. We begin by looking at the special circumstance in which the proton is emitted into a small solid angle $\Delta\Omega$ along the $z$-axis (Fig. 17–7). Before the disintegration we have a $\Lambda^0$ with its spin “up,” as in part (a) of the figure. After a short time—for reasons unknown to this day, except that they are connected with the weak decays—the $\Lambda^0$ explodes into a proton and a pion. Suppose the proton goes up along the $+z$-axis. Then, from the conservation of momentum, the pion must go down. Since the proton is a spin one-half particle, its spin must be either “up” or “down”—there are, in principle, the two possibilities shown in parts (b) and (c) of the figure. The conservation of angular momentum, however, requires that the proton have spin “up.” This is most easily seen from the following argument. A particle moving along the $z$-axis cannot contribute any angular momentum about this axis by virtue of its motion; therefore, only the spins can contribute to $J_z$. The spin angular momentum about the $z$-axis is $+\hbar/2$ before the disintegration, so it must also be $+\hbar/2$ afterward. We can say that since the pion has no spin, the proton spin must be “up.” If you are worried that arguments of this kind may not be valid in quantum mechanics, we can take a moment to show you that they are. The initial state (before the disintegration), which we can call $\ket{\Lambda^0,\text{spin \(+z\)}}$ has the property that if it is rotated about the $z$-axis by the angle $\phi$, the state vector gets multiplied by the phase factor $e^{i\phi/2}$. (In the rotated system the state vector is $e^{i\phi/2}\,\ket{\Lambda^0,\text{spin \(+z\)}}$.) That’s what we mean by spin “up” for a spin one-half particle. Since nature’s behavior doesn’t depend on our choice of axes, the final state (the proton plus pion) must have the same property. We could write the final state as, say, \begin{equation*} \ket{\text{proton going $+z$, spin $+z$; pion going $-z$}}. \end{equation*} But we really do not need to specify the pion motion, since in the frame we have chosen the pion always moves opposite the proton; we can simplify our description of the final state to \begin{equation*} \ket{\text{proton going $+z$, spin $+z$}}. \end{equation*} Now what happens to this state vector if we rotate the coordinates about the $z$-axis by the angle $\phi$? Since the proton and pion are moving along the $z$-axis, their motion isn’t changed by the rotation. (That’s why we picked this special case; we couldn’t make the argument otherwise.) Also, nothing happens to the pion, because it is spin zero. The proton, however, has spin one-half. If its spin is “up” it will contribute a phase change of $e^{i\phi/2}$ in response to the rotation. (If its spin were “down” the phase change due to the proton would be $e^{-i\phi/2}$.) But the phase change with rotation before and after the excitement must be the same if angular momentum is to be conserved. (And it will be, since there are no outside influences in the Hamiltonian.) So the only possibility is that the proton spin will be “up.” If the proton goes up, its spin must also be “up.” We conclude, then, that the conservation of angular momentum permits the process shown in part (b) of Fig. 17–7, but does not permit the process shown in part (c). Since we know that the disintegration occurs, there is some amplitude for process (b)—proton going up with spin “up.” We’ll let $a$ stand for the amplitude that the disintegration occurs in this way in any infinitesimal interval of time.10 Now let’s see what would happen if the $\Lambda^0$ spin were initially “down.” Again we ask about the decays in which the proton goes up along the $z$-axis, as shown in Fig. 17–8. You will appreciate that in this case the proton must have spin “down” if angular momentum is conserved. Let’s say that the amplitude for such a disintegration is $b$. We can’t say anything more about the two amplitudes $a$ and $b$. They depend on the inner machinery of $\Lambda^0$, and the weak decays, and nobody yet knows how to calculate them. We’ll have to get them from experiment. But with just these two amplitudes we can find out all we want to know about the angular distribution of the disintegration. We only have to be careful always to define completely the states we are talking about. We want to know the probability that the proton will go off at the angle $\theta$ with respect to the $z$-axis (into a small solid angle $\Delta\Omega$) as drawn in Fig. 17–6. Let’s put a new $z$-axis in this direction and call it the $z'$-axis. We know how to analyze what happens along this axis. With respect to this new axis, the $\Lambda^0$ no longer has its spin “up,” but has a certain amplitude to have its spin “up” and another amplitude to have its spin “down.” We have already worked these out in Chapter 6, and again in Chapter 10, Eq. (10.30). The amplitude to be spin “up” is $\cos\theta/2$, and the amplitude to be spin “down” is11 $-\sin\theta/2$. When the $\Lambda^0$ spin is “up” along the $z'$-axis it will emit a proton in the $+z'$-direction with the amplitude $a$. So the amplitude to find an “up”-spinning proton coming out along the $z'$-direction is \begin{equation} \label{Eq:III:17:33} a\cos\frac{\theta}{2}. \end{equation} Similarly, the amplitude to find a “down”-spinning proton coming along the positive $z'$-axis is \begin{equation} \label{Eq:III:17:34} -b\sin\frac{\theta}{2}. \end{equation} The two processes that these amplitudes refer to are shown in Fig. 17–9. Let’s now ask the following easy question. If the $\Lambda^0$ has spin up along the $z$-axis, what is the probability that the decay proton will go off at the angle $\theta$? The two spin states (“up” or “down” along $z'$) are distinguishable even though we are not going to look at them. So to get the probability we square the amplitudes and add. The probability $f(\theta)$ of finding a proton in a small solid angle $\Delta\Omega$ at $\theta$ is \begin{equation} \label{Eq:III:17:35} f(\theta)=\abs{a}^2\cos^2\frac{\theta}{2}+ \abs{b}^2\sin^2\frac{\theta}{2}. \end{equation} Remembering that $\sin^2\theta/2=\tfrac{1}{2}(1-\cos\theta)$ and that $\cos^2\theta/2=\tfrac{1}{2}(1+\cos\theta)$, we can write $f(\theta)$ as \begin{equation} \label{Eq:III:17:36} f(\theta)=\!\biggl(\frac{\abs{a}^2\!+\abs{b}^2}{2}\biggr)\!+\! \biggl(\frac{\abs{a}^2\!-\abs{b}^2}{2}\biggr)\!\cos\theta. \end{equation} The angular distribution has the form \begin{equation} \label{Eq:III:17:37} f(\theta)=\beta(1+\alpha\cos\theta). \end{equation} The probability has one part that is independent of $\theta$ and one part that varies linearly with $\cos\theta$. From measuring the angular distribution we can get $\alpha$ and $\beta$, and therefore, $\abs{a}$ and $\abs{b}$. Now there are many other questions we can answer. Are we interested only in protons with spin “up” along the old $z$-axis? Each of the terms in (17.33) and (17.34) will give an amplitude to find a proton with spin “up” and with spin “down” with respect to the $z'$-axis ($+z'$ and $-z'$). Spin “up” with respect to the old axis $\ket{+z}$ can be expressed in terms of the base states $\ket{+z'}$ and $\ket{-z'}$. We can then combine the two amplitudes (17.33) and (17.34) with the proper coefficients ($\cos\theta/2$ and $-\sin\theta/2$) to get the total amplitude \begin{equation*} \biggl(a\cos^2\frac{\theta}{2}+b\sin^2\frac{\theta}{2}\biggr). \end{equation*} Its square is the probability that the proton comes out at the angle $\theta$ with its spin the same as the $\Lambda^0$ (“up” along the $z$-axis). If parity were conserved, we could say one more thing. The disintegration of Fig. 17–8 is just the reflection—in the $xy$-plane of the disintegration—of Fig. 17–7.12 If parity were conserved, $b$ would have to be equal to $a$ or to $-a$. Then the coefficient $\alpha$ of (17.37) would be zero, and the disintegration would be equally likely to occur in all directions. The experimental results show, however, that there is an asymmetry in the disintegration. The measured angular distribution does go as $\cos\theta$ as we predict—and not as $\cos^2\theta$ or any other power. In fact, since the angular distribution has this form, we can deduce from these measurements that the spin of the $\Lambda^0$ is $1/2$. Also, we see that parity is not conserved. In fact, the coefficient $\alpha$ is found experimentally to be $-0.62\pm0.05$, so $b$ is about twice as large as $a$. The lack of symmetry under a reflection is quite clear. You see how much we can get from the conservation of angular momentum. We will give some more examples in the next chapter. Parenthetical note. By the amplitude $a$ in this section we mean the amplitude that the state $\ket{\text{proton going \(+z\), spin \(+z\)}}$ is generated in an infinitesimal time $dt$ from the state $\ket{\text{\(\Lambda\), spin \(+z\)}}$, or, in other words, that \begin{equation} \label{Eq:III:17:38} \bracket{\text{proton going $+z$, spin $+z$}} {H} {\text{$\Lambda$, spin $+z$}} =i\hbar a, \end{equation} where $H$ is the Hamiltonian of the world—or, at least, of whatever is responsible for the $\Lambda$-decay. The conservation of angular momentum means that the Hamiltonian must have the property that \begin{equation} \label{Eq:III:17:39} \bracket{\text{proton going $+z$, spin $-z$}} {H} {\text{$\Lambda$, spin $+z$}} =0. \end{equation} By the amplitude $b$ we mean that \begin{equation} \label{Eq:III:17:40} \bracket{\text{proton going $+z$, spin $-z$}} {H} {\text{$\Lambda$, spin $-z$}} =i\hbar b. \end{equation} Conservation of angular momentum implies that \begin{equation} \label{Eq:III:17:41} \bracket{\text{proton going $+z$, spin $+z$}} {H} {\text{$\Lambda$, spin $-z$}} =0. \end{equation} If the amplitudes written in (17.33) and (17.34) are not clear, we can express them more mathematically as follows. By (17.33) we intend the amplitude that the $\Lambda$ with spin along $+z$ will disintegrate into a proton moving along the $+z'$-direction with its spin also in the $+z'$-direction, namely the amplitude \begin{equation} \label{Eq:III:17:42} \bracket{\text{proton going $+z'$, spin $+z'$}} {H} {\text{$\Lambda$, spin $+z$}}. \end{equation} By the general theorems of quantum mechanics, this amplitude can be written as \begin{equation} \label{Eq:III:17:43} \sum_i \bracket{\text{proton going $+z'$, spin $+z'$}} {H}{\Lambda,i} \braket{\Lambda,i} {\text{$\Lambda$, spin $+z$}}, \end{equation} where the sum is to be taken over the base states $\ket{\Lambda,i}$ of the $\Lambda$-particle at rest. Since the $\Lambda$-particle is spin one-half, there are two such base states which can be in any reference base we wish. If we use for base states spin “up” and spin “down” with respect to $z'$ ($+z'$, $-z'$), the amplitude of (17.43) is equal to the sum \begin{align} &\bracket{\text{proton going $+z'$, spin $+z'$}} {H}{\Lambda,+z'} \braket{\Lambda,+z'}{\Lambda,+z}\notag\\[.5ex] \label{Eq:III:17:44} +\,&\bracket{\text{proton going $+z'$, spin $+z'$}} {H}{\Lambda,-z'} \braket{\Lambda,-z'}{\Lambda,+z}. \end{align} The first factor of the first term is $a$, and the first factor of the second term is zero—from the definition of (17.38), and from (17.41), which in turn follows from angular momentum conservation. The remaining factor $\braket{\Lambda,+z'}{\Lambda,+z}$ of the first term is just the amplitude that a spin one-half particle which has spin “up” along one axis will also have spin “up” along an axis tilted at the angle $\theta$, which is $\cos\theta/2$—see Table 6–2. So (17.44) is just $a\cos\theta/2$, as we wrote in (17.33). The amplitude of (17.34) follows from the same kind of arguments for a spin “down” $\Lambda$-particle.

3

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Symmetry and Conservation Laws

6

Summary of the rotation matrices

We would like now to bring together in one place the various things we have learned about the rotations for particles of spin one-half and spin one—so they will be convenient for future reference. On the next page you will find tables of the two rotation matrices $R_z(\phi)$ and $R_y(\theta)$ for spin one-half particles, for spin-one particles, and for photons (spin-one particles with zero rest mass). For each spin we will give the terms of the matrix $\bracket{j}{R}{i}$ for rotations about the $z$-axis or the $y$-axis. They are, of course, exactly equivalent to the amplitudes like $\braket{+T}{\OS}$ we have used in earlier chapters. We mean by $R_z(\phi)$ that the state is projected into a new coordinate system which is rotated through the angle $\phi$ about the $z$-axis—using always the right-hand rule to define the positive sense of the rotation. By $R_y(\theta)$ we mean that the reference axes are rotated by the angle $\theta$ about the $y$-axis. Knowing these two rotations, you can, of course, work out any arbitrary rotation. As usual, we write the matrix elements so that the state on the left is a base state of the new (rotated) frame and the state on the right is a base state of the old (unrotated) frame. You can interpret the entries in the tables in many ways. For instance, the entry $e^{-i\phi/2}$ in Table 17–1 means that the matrix element $\bracket{-}{R}{-}=e^{-i\phi/2}$. It also means that $\Rop\,\ket{-}=e^{-i\phi/2}\,\ket{-}$, or that $\bra{-}\,\Rop=\bra{-}\,e^{-i\phi/2}$. It’s all the same thing.

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18

Angular Momentum

1

Electric dipole radiation

In the last chapter we developed the idea of the conservation of angular momentum in quantum mechanics, and showed how it might be used to predict the angular distribution of the proton from the disintegration of the $\Lambda$-particle. We want now to give you a number of other, similar, illustrations of the consequences of momentum conservation in atomic systems. Our first example is the radiation of light from an atom. The conservation of angular momentum (among other things) will determine the polarization and angular distribution of the emitted photons. Suppose we have an atom which is in an excited state of definite angular momentum—say with a spin of one—and it makes a transition to a state of angular momentum zero at a lower energy, emitting a photon. The problem is to figure out the angular distribution and polarization of the photons. (This problem is almost exactly the same as the $\Lambda^0$ disintegration, except that we have spin-one instead of spin one-half particles.) Since the upper state of the atom is spin one, there are three possibilities for its $z$-component of angular momentum. The value of $m$ could be $+1$, or $0$, or $-1$. We will take $m=+1$ for our example. Once you see how it goes, you can work out the other cases. We suppose that the atom is sitting with its angular momentum along the $+z$-axis—as in Fig. 18–1(a)—and ask with what amplitude it will emit right circularly polarized light upward along the $z$-axis, so that the atom ends up with zero angular momentum—as shown in part (b) of the figure. Well, we don’t know the answer to that. But we do know that right circularly polarized light has one unit of angular momentum about its direction of propagation. So after the photon is emitted, the situation would have to be as shown in Fig. 18–1(b)—the atom is left with zero angular momentum about the $z$-axis, since we have assumed an atom whose lower state is spin zero. We will let $a$ stand for the amplitude for such an event. More precisely, we let $a$ be the amplitude to emit a photon into a certain small solid angle $\Delta\Omega$, centered on the $z$-axis, during a time $dt$. Notice that the amplitude to emit a LHC photon in the same direction is zero. The net angular momentum about the $z$-axis would be $-1$ for such a photon and zero for the atom for a total of $-1$, which would not conserve angular momentum. Similarly, if the spin of the atom is initially “down” ($-1$ along the $z$-axis), it can emit only a LHC polarized photon in the direction of the $+z$-axis, as shown in Fig. 18–2. We will let $b$ stand for the amplitude for this event—meaning again the amplitude that the photon goes into a certain solid angle $\Delta\Omega$. On the other hand, if the atom is in the $m=0$ state, it cannot emit a photon in the $+z$-direction at all, because a photon can have only the angular momentum $+1$ or $-1$ along its direction of motion. Next, we can show that $b$ is related to $a$. Suppose we perform an inversion of the situation in Fig. 18–1, which means that we should imagine what the system would look like if we were to move each part of the system to an equivalent point on the opposite side of the origin. This does not mean that we should reflect the angular momentum vectors, because they are artificial. We should, rather, invert the actual character of the motion that would correspond to such an angular momentum. In Fig. 18–3(a) and (b) we show what the process of Fig. 18–1 looks like before and after an inversion with respect to the center of the atom. Notice that the sense of rotation of the atom is unchanged.1 In the inverted system of Fig. 18–3(b) we have an atom with $m=+1$ emitting a LHC photon downward. If we now rotate the system of Fig. 18–3(b) by $180^\circ$ about the $x$- or $y$-axis, it becomes identical to Fig. 18–2. The combination of the inversion and rotation turns the second process into the first. Using Table 17–2, we see that a rotation of $180^\circ$ about the $y$-axis just throws an $m=-1$ state into an $m=+1$ state, so the amplitude $b$ must be equal to the amplitude $a$ except for a possible sign change due to the inversion. The sign change in the inversion will depend on the parities of the initial and final state of the atom. In atomic processes, parity is conserved, so the parity of the whole system must be the same before and after the photon emission. What happens will depend on whether the parities of the initial and final states of the atom are even or odd—the angular distribution of the radiation will be different for different cases. We will take the common case of odd parity for the initial state and even parity for the final state; it will give what is called “electric dipole radiation.” (If the initial and final states have the same parity we say there is “magnetic dipole radiation,” which has the character of the radiation from an oscillating current in a loop.) If the parity of the initial state is odd, its amplitude reverses its sign in the inversion which takes the system from (a) to (b) of Fig. 18–3. The final state of the atom has even parity, so its amplitude doesn’t change sign. If the reaction is going to conserve parity, the amplitude $b$ must be equal to $a$ in magnitude but of the opposite sign. We conclude that if the amplitude is $a$ that an $m=+1$ state will emit a photon upward, then for the assumed parities of the initial and final states the amplitude that an $m=-1$ state will emit a LHC photon upward is $-a$.2 We have all we need to know to find the amplitude for a photon to be emitted at any angle $\theta$ with respect to the $z$-axis. Suppose we have an atom originally polarized with $m=+1$. We can resolve this state into $+1$, $0$, and $-1$ states with respect to a new $z'$-axis in the direction of the photon emission. The amplitudes for these three states are just the ones given in the lower half of Table 17–2. The amplitude that a RHC photon is emitted in the direction $\theta$ is then $a$ times the amplitude to have $m=+1$ in that direction, namely, \begin{equation} \label{Eq:III:18:1} a\bracket{+}{R_y(\theta)}{+}=\frac{a}{2}\,(1+\cos\theta). \end{equation} The amplitude that a LHC photon is emitted in the same direction is $-a$ times the amplitude to have $m=-1$ in the new direction. Using Table 17–2, it is \begin{equation} \label{Eq:III:18:2} -a\bracket{-}{R_y(\theta)}{+}=-\frac{a}{2}\,(1-\cos\theta). \end{equation} If you are interested in other polarizations you can find out the amplitude for them from the superposition of these two amplitudes. To get the intensity of any component as a function of angle, you must, of course, take the absolute square of the amplitudes.

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Angular Momentum

2

Light scattering

Let’s use these results to solve a somewhat more complicated problem—but also one which is somewhat more real. We suppose that the same atoms are sitting in their ground state ($j=0$), and scatter an incoming beam of light. Let’s say that the light is going initially in the $+z$-direction, so that we have photons coming up to the atom from the $-z$-direction, as shown in Fig. 18–4(a). We can consider the scattering of light as a two-step process: The photon is absorbed, and then is re-emitted. If we start with a RHC photon as in Fig. 18–4(a), and angular momentum is conserved, the atom will be in an $m=+1$ state after the absorption—as shown in Fig. 18–4(b). We call the amplitude for this process $c$. The atom can then emit a RHC photon in the direction $\theta$—as in Fig. 18–4(c). The total amplitude that a RHC photon is scattered in the direction $\theta$ is just $c$ times (18.1). Let’s call this scattering amplitude $\bracket{R'}{S}{R}$; we have \begin{equation} \label{Eq:III:18:3} \bracket{R'}{S}{R}=\frac{ac}{2}\,(1+\cos\theta). \end{equation} There is also an amplitude that a RHC photon will be absorbed and that a LHC photon will be emitted. The product of the two amplitudes is the amplitude $\bracket{L'}{S}{R}$ that a RHC photon is scattered as a LHC photon. Using (18.2), we have \begin{equation} \label{Eq:III:18:4} \bracket{L'}{S}{R}=-\frac{ac}{2}\,(1-\cos\theta). \end{equation} Now let’s ask about what happens if a LHC photon comes in. When it is absorbed, the atom will go into an $m=-1$ state. By the same kind of arguments we used in the preceding section, we can show that this amplitude must be $-c$. The amplitude that an atom in the $m=-1$ state will emit a RHC photon at the angle $\theta$ is $a$ times the amplitude $\bracket{+}{R_y(\theta)}{-}$, which is $\tfrac{1}{2}(1-\cos\theta)$. So we have \begin{equation} \label{Eq:III:18:5} \bracket{R'}{S}{L}=-\frac{ac}{2}\,(1-\cos\theta). \end{equation} Finally, the amplitude for a LHC photon to be scattered as a LHC photon is \begin{equation} \label{Eq:III:18:6} \bracket{L'}{S}{L}=\frac{ac}{2}\,(1+\cos\theta). \end{equation} (There are two minus signs which cancel.) If we make a measurement of the scattered intensity for any given combination of circular polarizations it will be proportional to the square of one of our four amplitudes. For instance, with an incoming beam of RHC light the intensity of the RHC light in the scattered radiation will vary as $(1+\cos\theta)^2$. That’s all very well, but suppose we start out with linearly polarized light. What then? If we have $x$-polarized light, it can be represented as a superposition of RHC and LHC light. We write (see Section 11-4) \begin{equation} \label{Eq:III:18:7} \ket{x}=\frac{1}{\sqrt{2}}\,(\ket{R}+\ket{L}). \end{equation} Or, if we have $y$-polarized light, we would have \begin{equation} \label{Eq:III:18:8} \ket{y}=-\frac{i}{\sqrt{2}}\,(\ket{R}-\ket{L}). \end{equation} Now what do you want to know? Do you want the amplitude that an $x$-polarized photon will scatter into a RHC photon at the angle $\theta$? You can get it by the usual rule for combining amplitudes. First, multiply (18.7) by $\bra{R'}\,S$ to get \begin{equation} \label{Eq:III:18:9} \bracket{R'}{S}{x}=\!\frac{1}{\sqrt{2}}\, (\bracket{R'}{S}{R}\!+\!\bracket{R'}{S}{L}), \end{equation} and then use (18.3) and (18.5) for the two amplitudes. You get \begin{equation} \label{Eq:III:18:10} \bracket{R'}{S}{x}=\frac{ac}{\sqrt{2}}\cos\theta. \end{equation} If you wanted the amplitude that an $x$-photon would scatter into a LHC photon, you would get \begin{equation} \label{Eq:III:18:11} \bracket{L'}{S}{x}=\frac{ac}{\sqrt{2}}\cos\theta. \end{equation} Finally, suppose you wanted to know the amplitude that an $x$-polarized photon will scatter while keeping its $x$-polarization. What you want is $\bracket{x'}{S}{x}$. This can be written as \begin{equation} \label{Eq:III:18:12} \bracket{x'}{S}{x}=\braket{x'}{R'}\bracket{R'}{S}{x}+ \braket{x'}{L'}\bracket{L'}{S}{x}. \end{equation} \begin{align} \bracket{x'}{S}{x}=\,&\braket{x'}{R'}\bracket{R'}{S}{x}\notag\\[.5ex] \label{Eq:III:18:12} +\,&\braket{x'}{L'}\bracket{L'}{S}{x}. \end{align} If you then use the relations \begin{align} \label{Eq:III:18:13} \ket{R'}&=\frac{1}{\sqrt{2}}\,(\ket{x'}+i\,\ket{y'}),\\[1ex] \label{Eq:III:18:14} \ket{L'}&=\frac{1}{\sqrt{2}}\,(\ket{x'}-i\,\ket{y'}) \end{align} it follows that \begin{align} \label{Eq:III:18:15} \braket{x'}{R'}&=\frac{1}{\sqrt{2}},\\[1ex] \label{Eq:III:18:16} \braket{x'}{L'}&=\frac{1}{\sqrt{2}}. \end{align} So you get that \begin{equation} \label{Eq:III:18:17} \bracket{x'}{S}{x}=ac\cos\theta. \end{equation} The answer is that a beam of $x$-polarized light will be scattered at the direction $\theta$ (in the $xz$-plane) with an intensity proportional to $\cos^2\theta$. If you ask about $y$-polarized light, you find that \begin{equation} \label{Eq:III:18:18} \bracket{y'}{S}{x}=0. \end{equation} So the scattered light is completely polarized in the $x$-direction. Now we notice something interesting. The results (18.17) and (18.18) correspond exactly to the classical theory of light scattering we gave in Vol. I, Section 32-5, where we imagined that the electron was bound to the atom by a linear restoring force—so that it acted like a classical oscillator. Perhaps you are thinking: “It’s so much easier in the classical theory; if it gives the right answer why bother with the quantum theory?” For one thing, we have considered so far only the special—though common—case of an atom with a $j=1$ excited state and a $j=0$ ground state. If the excited state had spin two, you would get a different result. Also, there is no reason why the model of an electron attached to a spring and driven by an oscillating electric field should work for a single photon. But we have found that it does in fact work, and that the polarization and intensities come out right. So in a certain sense we are bringing the whole course around to the real truth. Whereas we have, in Vol. I, done the theory of the index of refraction, and of light scattering, by the classical theory, we have now shown that the quantum theory gives the same result for the most common case. In effect we have now done the polarization of sky light, for instance, by quantum mechanical arguments, which is the only truly legitimate way. It should be, of course, that all the classical theories which work are supported ultimately by legitimate quantum arguments. Naturally, those things which we have spent a great deal of time in explaining to you were selected from just those parts of classical physics which still maintain validity in quantum mechanics. You’ll notice that we did not discuss in great detail any model of the atom which has electrons going around in orbits. That’s because such a model doesn’t give results which agree with the quantum mechanics. But the electron on a spring—which is not, in a sense, at all the way an atom “looks”—does work, and so we used that model for the theory of the index of refraction.

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18

Angular Momentum

3

The annihilation of positronium

We would like next to take an example which is very pretty. It is quite interesting and, although somewhat complicated, we hope not too much so. Our example is the system called positronium, which is an “atom” made up of an electron and a positron—a bound state of an e$^+$ and an e$^-$. It is like a hydrogen atom, except that a positron replaces the proton. This object has—like the hydrogen atom—many states. Also like the hydrogen, the ground state is split into a “hyperfine structure” by the interaction of the magnetic moments. The spins of the electron and positron are each one-half, and they can be either parallel or antiparallel to any given axis. (In the ground state there is no other angular momentum due to orbital motion.) So there are four states: three are the substates of a spin-one system, all with the same energy; and one is a state of spin zero with a different energy. The energy splitting is, however, much larger than the $1420$ megacycles of hydrogen because the positron magnetic moment is so much stronger—$1000$ times stronger—than the proton moment. The most important difference, however, is that positronium cannot last forever. The positron is the antiparticle of the electron; they can annihilate each other. The two particles disappear completely—converting their rest energy into radiation, which appears as $\gamma$-rays (photons). In the disintegration, two particles with a finite rest mass go into two or more objects which have zero rest mass.3 We begin by analyzing the disintegration of the spin-zero state of the positronium. It disintegrates into two $\gamma$-rays with a lifetime of about $10^{-10}$ second. Initially, we have a positron and an electron close together and with spins antiparallel, making the positronium system. After the disintegration there are two photons going out with equal and opposite momenta (Fig. 18–5). The momenta must be equal and opposite, because the total momentum after the disintegration must be zero, as it was before, if we are taking the case of annihilation at rest. If the positronium is not at rest, we can ride with it, solve the problem, and then transform everything back to the lab system. (See, we can do anything now; we have all the tools.) First, we note that the angular distribution is not very interesting. Since the initial state has spin zero, it has no special axis—it is symmetric under all rotations. The final state must then also be symmetric under all rotations. That means that all angles for the disintegration are equally likely—the amplitude is the same for a photon to go in any direction. Of course, once we find one of the photons in some direction the other must be opposite. The only remaining question, which we now want to look at, is about the polarization of the photons. Let’s call the directions of motion of the two photons the plus and minus $z$-axes. We can use any representations we want for the polarization states of the photons; we will choose for our description right and left circular polarization—always with respect to the directions of motion.4 Right away, we can see that if the photon going upward is RHC, then angular momentum will be conserved if the downward going photon is also RHC. Each will carry $+1$ unit of angular momentum with respect to its momentum direction, which means plus and minus one unit about the $z$-axis. The total will be zero, and the angular momentum after the disintegration will be the same as before. See Fig. 18–6. The same arguments show that if the upward going photon is RHC, the downward cannot be LHC. Then the final state would have two units of angular momentum. This is not permitted if the initial state has spin zero. Note that such a final state is also not possible for the other positronium ground state of spin one, because it can have a maximum of one unit of angular momentum in any direction. Now we want to show that two-photon annihilation is not possible at all from the spin-one state. You might think that if we took the $j=1$, $m=0$ state—which has zero angular momentum about the $z$-axis—it should be like the spin-zero state, and could disintegrate into two RHC photons. Certainly, the disintegration sketched in Fig. 18–7(a) conserves angular momentum about the $z$-axis. But now look what happens if we rotate this system around the $y$-axis by $180^\circ$; we get the picture shown in Fig. 18–7(b). It is exactly the same as in part (a) of the figure. All we have done is interchange the two photons. Now photons are Bose particles; if we interchange them, the amplitude has the same sign, so the amplitude for the disintegration in part (b) must be the same as in part (a). But we have assumed that the initial object is spin one. And when we rotate a spin-one object in a state with $m=0$ by $180^\circ$ about the $y$-axis, its amplitudes change sign (see Table 17–2 for $\theta=\pi$). So the amplitudes for (a) and (b) in Fig. 18–7 should have opposite signs; the spin-one state cannot disintegrate into two photons. When positronium is formed you would expect it to end up in the spin-zero state $1/4$ of the time and in the spin-one state (with $m=-1$, $0$, or $+1$) $3/4$ of the time. So $1/4$ of the time you would get two-photon annihilations. The other $3/4$ of the time there can be no two-photon annihilations. There is still an annihilation, but it has to go with three photons. It is harder for it to do that and the lifetime is $1000$ times longer—about $10^{-7}$ second. This is what is observed experimentally. We will not go into any more of the details of the spin-one annihilation. So far we have that if we only worry about angular momentum, the spin-zero state of the positronium can go into two RHC photons. There is also another possibility: it can go into two LHC photons as shown in Fig. 18–8. The next question is, what is the relation between the amplitudes for these two possible decay modes? We can find out from the conservation of parity. To do that, however, we need to know the parity of the positronium. Now theoretical physicists have shown in a way that is not easy to explain that the parity of the electron and the positron—its antiparticle—must be opposite, so that the spin-zero ground state of positronium must be odd. We will just assume that it is odd, and since we will get agreement with experiment, we can take that as sufficient proof. Let’s see then what happens if we make an inversion of the process in Fig. 18–6. When we do that, the two photons reverse directions and polarizations. The inverted picture looks just like Fig. 18–8. Assuming that the parity of the positronium is odd, the amplitudes for the two processes in Figs. 18–6 and 18–8 must have the opposite sign. Let’s let $\ket{R_1R_2}$ stand for the final state of Fig. 18–6 in which both photons are RHC, and let $\ket{L_1L_2}$ stand for the final state of Fig. 18–8, in which both photons are LHC. The true final state—let’s call it $\ket{F}$—must be \begin{equation} \label{Eq:III:18:19} \ket{F}=\ket{R_1R_2}-\ket{L_1L_2}. \end{equation} Then an inversion changes the $R$’s into $L$’s and gives the state \begin{equation} \label{Eq:III:18:20} P\,\ket{F}=\ket{L_1L_2}-\ket{R_1R_2}=-\,\ket{F}, \end{equation} which is the negative of (18.19). So the final state $\ket{F}$ has negative parity, which is the same as the initial spin-zero state of the positronium. This is the only final state that conserves both angular momentum and parity. There is some amplitude that the disintegration into this state will occur, which we don’t need to worry about now, however, since we are only interested in questions about the polarization. What does the final state of (18.19) mean physically? One thing it means is the following: If we observe the two photons in two detectors which can be set to count separately the RHC or LHC photons, we will always see two RHC photons together, or two LHC photons together. That is, if you stand on one side of the positronium and someone else stands on the opposite side, you can measure the polarization and tell the other guy what polarization he will get. You have a $50$-$50$ chance of catching a RHC photon or a LHC photon; whichever one you get, you can predict that he will get the same. Since there is a $50$-$50$ chance for RHC or LHC polarization, it sounds as though it might be like linear polarization. Let’s ask what happens if we observe the photon in counters that accept only linearly polarized light. For $\gamma$-rays it is not as easy to measure the polarization as it is for light; there is no polarizer which works well for such short wavelengths. But let’s imagine that there is, to make the discussion easier. Suppose that you have a counter that only accepts light with $x$-polarization, and that there is a guy on the other side that also looks for linear polarized light with, say, $y$-polarization. What is the chance you will pick up the two photons from an annihilation? What we need to ask is the amplitude that $\ket{F}$ will be in the state $\ket{x_1y_2}$. In other words, we want the amplitude \begin{equation*} \braket{x_1y_2}{F}, \end{equation*} which is, of course, just \begin{equation} \label{Eq:III:18:21} \braket{x_1y_2}{R_1R_2}-\braket{x_1y_2}{L_1L_2}. \end{equation} Now although we are working with two-particle amplitudes for the two photons, we can handle them just as we did the single particle amplitudes, since each particle acts independently of the other. That means that the amplitude $\braket{x_1y_2}{R_1R_2}$ is just the product of the two independent amplitudes $\braket{x_1}{R_1}$ and $\braket{y_2}{R_2}$. Using Table 17–3, these two amplitudes are $1/\sqrt{2}$ and $i/\sqrt{2}$—so \begin{equation*} \braket{x_1y_2}{R_1R_2}=+\frac{i}{2}. \end{equation*} Similarly, we find that \begin{equation*} \braket{x_1y_2}{L_1L_2}=-\frac{i}{2}. \end{equation*} Subtracting these two amplitudes according to (18.21), we get that \begin{equation} \label{Eq:III:18:22} \braket{x_1y_2}{F}=+i. \end{equation} So there is a unit probability5 that if you get a photon in your $x$-polarized detector, the other guy will get a photon in his $y$-polarized detector. Now suppose that the other guy sets his counter for $x$-polarization the same as yours. He would never get a count when you got one. If you work it through, you will find that \begin{equation} \label{Eq:III:18:23} \braket{x_1x_2}{F}=0. \end{equation} It will, naturally, also work out that if you set your counter for $y$-polarization he will get coincident counts only if he is set for $x$-polarization. Now this all leads to an interesting situation. Suppose you were to set up something like a piece of calcite which separated the photons into $x$-polarized and $y$-polarized beams, and put a counter in each beam. Let’s call one the $x$-counter and the other the $y$-counter. If the guy on the other side does the same thing, you can always tell him which beam his photon is going to go into. Whenever you and he get simultaneous counts, you can see which of your detectors caught the photon and then tell him which of his counters had a photon. Let’s say that in a certain disintegration you find that a photon went into your $x$-counter; you can tell him that he must have had a count in his $y$-counter. Now many people who learn quantum mechanics in the usual (old-fashioned) way find this disturbing. They would like to think that once the photons are emitted it goes along as a wave with a definite character. They would think that since “any given photon” has some “amplitude” to be $x$-polarized or to be $y$-polarized, there should be some chance of picking it up in either the $x$- or $y$-counter and that this chance shouldn’t depend on what some other person finds out about a completely different photon. They argue that “someone else making a measurement shouldn’t be able to change the probability that I will find something.” Our quantum mechanics says, however, that by making a measurement on photon number one, you can predict precisely what the polarization of photon number two is going to be when it is detected. This point was never accepted by Einstein, and he worried about it a great deal—it became known as the “Einstein-Podolsky-Rosen paradox.” But when the situation is described as we have done it here, there doesn’t seem to be any paradox at all; it comes out quite naturally that what is measured in one place is correlated with what is measured somewhere else. The argument that the result is paradoxical runs something like this: If you have a counter which tells you whether your photon is RHC or LHC, you can predict exactly what kind of a photon (RHC or LHC) he will find. The photons he receives must, therefore, each be purely RHC or purely LHC, some of one kind and some of the other. Surely you cannot alter the physical nature of his photons by changing the kind of observation you make on your photons. No matter what measurements you make on yours, his must still be either RHC or LHC. Now suppose he changes his apparatus to split his photons into two linearly polarized beams with a piece of calcite so that all of his photons go either into an $x$-polarized beam or into a $y$-polarized beam. There is absolutely no way, according to quantum mechanics, to tell into which beam any particular RHC photon will go. There is a 50% probability it will go into the $x$-beam and a 50% probability it will go into the $y$-beam. And the same goes for a LHC photon. Since each photon is RHC or LHC—according to (2) and (3)—each one must have a 50-50 chance of going into the $x$-beam or the $y$-beam and there is no way to predict which way it will go. Yet the theory predicts that if you see your photon go through an $x$-polarizer you can predict with certainty that his photon will go into his $y$-polarized beam. This is in contradiction to (5) so there is a paradox. Nature apparently doesn’t see the “paradox,” however, because experiment shows that the prediction in (6) is, in fact, true. We have already discussed the key to this “paradox” in our very first lecture on quantum mechanical behavior in Chapter 37, Vol. I.6 In the argument above, steps (1), (2), (4), and (6) are all correct, but (3), and its consequence (5), are wrong; they are not a true description of nature. Argument (3) says that by your measurement (seeing a RHC or a LHC photon) you can determine which of two alternative events occurs for him (seeing a RHC or a LHC photon), and that even if you do not make your measurement you can still say that his event will occur either by one alternative or the other. But it was precisely the point of Chapter 37, Vol. I, to point out right at the beginning that this is not so in Nature. Her way requires a description in terms of interfering amplitudes, one amplitude for each alternative. A measurement of which alternative actually occurs destroys the interference, but if a measurement is not made you cannot still say that “one alternative or the other is still occurring.” If you could determine for each one of your photons whether it was RHC or LHC, and also whether it was $x$-polarized (all for the same photon) there would indeed be a paradox. But you cannot do that—it is an example of the uncertainty principle. Do you still think there is a “paradox”? Make sure that it is, in fact, a paradox about the behavior of Nature, by setting up an imaginary experiment for which the theory of quantum mechanics would predict inconsistent results via two different arguments. Otherwise the “paradox” is only a conflict between reality and your feeling of what reality “ought to be.” Do you think that it is not a “paradox,” but that it is still very peculiar? On that we can all agree. It is what makes physics fascinating.

3

18

Angular Momentum

4

Rotation matrix for any spin

By now you can see, we hope, how important the idea of the angular momentum is in understanding atomic processes. So far, we have considered only systems with spins—or “total angular momentum”—of zero, one-half, or one. There are, of course, atomic systems with higher angular momenta. For analyzing such systems we would need to have tables of rotation amplitudes like those in Section 17-6. That is, we would need the matrix of amplitudes for spin $\tfrac{3}{2}$, $2$, $\tfrac{5}{2}$, $3$, etc. Although we will not work out these tables in detail, we would like to show you how it is done, so that you can do it if you ever need to. As we have seen earlier, any system which has the spin or “total angular momentum” $j$ can exist in any one of $(2j+1)$ states for which the $z$-component of angular momentum can have any one of the discrete values in the sequence $j$, $j-1$, $j-2$, $\ldots\,$, $-(j-1)$, $-j$ (all in units of $\hbar$). Calling the $z$-component of angular momentum of any particular state $m\hbar$, we can define a particular angular momentum state by giving the numerical values of the two “angular momentum quantum numbers” $j$ and $m$. We can indicate such a state by the state vector $\ket{j,m}$. In the case of a spin one-half particle, the two states are then $\ket{\tfrac{1}{2},\tfrac{1}{2}}$ and $\ket{\tfrac{1}{2},-\tfrac{1}{2}}$; or for a spin-one system, the states would be written in this notation as $\ket{1,+1}$, $\ket{1,0}$, $\ket{1,-1}$. A spin-zero particle has, of course, only the one state $\ket{0,0}$. Now we want to know what happens when we project the general state $\ket{j,m}$ into a representation with respect to a rotated set of axes. First, we know that $j$ is a number which characterizes the system, so it doesn’t change. If we rotate the axes, all we do is get a mixture of the various $m$-values for the same $j$. In general, there will be some amplitude that in the rotated frame the system will be in the state $\ket{j,m'}$, where $m'$ gives the new $z$-component of angular momentum. So what we want are all the matrix elements $\bracket{j,m'}{R}{j,m}$ for various rotations. We already know what happens if we rotate by an angle $\phi$ about the $z$-axis. The new state is just the old one multiplied by $e^{im\phi}$—it still has the same $m$-value. We can write this by \begin{equation} \label{Eq:III:18:24} R_z(\phi)\,\ket{j,m}=e^{im\phi}\,\ket{j,m}. \end{equation} Or, if you prefer, \begin{equation} \label{Eq:III:18:25} \bracket{j,m'}{R_z(\phi)}{j,m}=\delta_{m,m'}e^{im\phi} \end{equation} (where $\delta_{m,m'}$ is $1$ if $m'=m$, or zero otherwise). For a rotation about any other axis there will be a mixing of the various $m$-states. We could, of course, try to work out the matrix elements for an arbitrary rotation described by the Euler angles $\beta$, $\alpha$, and $\gamma$. But it is easier to remember that the most general such rotation can be made up of the three rotations $R_z(\gamma)$, $R_y(\alpha)$, $R_z(\beta)$; so if we know the matrix elements for a rotation about the $y$-axis, we will have all we need. How can we find the rotation matrix for a rotation by the angle $\theta$ about the $y$-axis for a particle of spin $j$? We can’t tell you how to do it in a basic way (with what we have had). We did it for spin one-half by a complicated symmetry argument. We then did it for spin one by taking the special case of a spin-one system which was made up of two spin one-half particles. If you will go along with us and accept the fact that in the general case the answers depend only on the spin $j$, and are independent of how the inner guts of the object of spin $j$ are put together, we can extend the spin-one argument to an arbitrary spin. We can, for example, cook up an artificial system of spin $\tfrac{3}{2}$ out of three spin one-half objects. We can even avoid complications by imagining that they are all distinct particles—like a proton, an electron, and a muon. By transforming each spin one-half object, we can see what happens to the whole system—remembering that the three amplitudes are multiplied for the combined state. Let’s see how it goes in this case. Suppose we take the three spin one-half objects all with spins “up”; we can indicate this state by $\ket{+\,+\,+}$. If we look at this system in a frame rotated about the $z$-axis by the angle $\phi$, each plus stays a plus, but gets multiplied by $e^{i\phi/2}$. We have three such factors, so \begin{equation} \label{Eq:III:18:26} R_z(\phi)\,\ket{+\,+\,+}=e^{i(3\phi/2)}\,\ket{+\,+\,+}. \end{equation} Evidently the state $\ket{+\,+\,+}$ is just what we mean by the $m=+\tfrac{3}{2}$ state, or the state $\ket{\tfrac{3}{2},+\tfrac{3}{2}}$. If we now rotate this system about the $y$-axis, each of the spin one-half objects will have some amplitude to be plus or to be minus, so the system will now be a mixture of the eight possible combinations $\ket{+\,+\,+}$, $\ket{+\,+\,-}$, $\ket{+\,-\,+}$, $\ket{-\,+\,+}$, $\ket{+\,-\,-}$, $\ket{-\,+\,-}$, $\ket{-\,-\,+}$, or $\ket{-\,-\,-}$. It is clear, however, that these can be broken up into four sets, each set corresponding to a particular value of $m$. First, we have $\ket{+\,+\,+}$, for which $m=\tfrac{3}{2}$. Then there are the three states $\ket{+\,+\,-}$, $\ket{+\,-\,+}$, and $\ket{-\,+\,+}$—each with two plusses and one minus. Since each spin one-half object has the same chance of coming out minus under the rotation, the amounts of each of these three combinations should be equal. So let’s take the combination \begin{equation} \label{Eq:III:18:27} \frac{1}{\sqrt{3}}\,\{ \ket{+\,+\,-}+\ket{+\,-\,+}+\ket{-\,+\,+}\} \end{equation} with the factor $1/\sqrt{3}$ put in to normalize the state. If we rotate this state about the $z$-axis, we get a factor $e^{i\phi/2}$ for each plus, and $e^{-i\phi/2}$ for each minus. Each term in (18.27) is multiplied by $e^{i\phi/2}$, so there is the common factor $e^{i\phi/2}$. This state satisfies our idea of an $m=+\tfrac{1}{2}$ state; we can conclude that \begin{equation} \label{Eq:III:18:28} \frac{1}{\sqrt{3}}\,\{ \ket{+\,+\,-}+\ket{+\,-\,+}+\ket{-\,+\,+}\}= \ket{\tfrac{3}{2},+\tfrac{1}{2}}. \end{equation} \begin{gather} \frac{1}{\sqrt{3}}\,\{ \ket{+\,+\,-}+\ket{+\,-\,+}+\ket{-\,+\,+}\}\notag\\ \label{Eq:III:18:28} =\ket{\tfrac{3}{2},+\tfrac{1}{2}}. \end{gather} Similarly, we can write \begin{equation} \label{Eq:III:18:29} \frac{1}{\sqrt{3}}\,\{ \ket{+\,-\,-}+\ket{-\,+\,-}+\ket{-\,-\,+}\}= \ket{\tfrac{3}{2},-\tfrac{1}{2}}, \end{equation} \begin{gather} \frac{1}{\sqrt{3}}\,\{ \ket{+\,-\,-}+\ket{-\,+\,-}+\ket{-\,-\,+}\}\notag\\ \label{Eq:III:18:29} =\ket{\tfrac{3}{2},-\tfrac{1}{2}}, \end{gather} which corresponds to a state with $m=-\tfrac{1}{2}$. Notice that we take only the symmetric combinations—we do not take any combinations with minus signs. They would correspond to states of the same $m$ but a different $j$. (It’s just like the spin-one case, where we found that $(1/\sqrt{2})\{\ket{+\,-}+\ket{-\,+}\}$ was the state $\ket{1,0}$, but the state $(1/\sqrt{2})\{\ket{+\,-}-\ket{-\,+}\}$ was the state $\ket{0,0}$.) Finally, we would have that \begin{equation} \label{Eq:III:18:30} \ket{\tfrac{3}{2},-\tfrac{3}{2}}=\ket{-\,-\,-}. \end{equation} We summarize our four states in Table 18–1. Now all we have to do is take each state and rotate it about the $y$-axis and see how much of the other states it gives—using our known rotation matrix for the spin one-half particles. We can proceed in exactly the same way we did for the spin-one case in Section 12-6. (It just takes a little more algebra.) We will follow directly the ideas of Chapter 12, so we won’t repeat all the explanations in detail. The states in the system $S$ will be labelled $\ket{\tfrac{3}{2},+\tfrac{3}{2},S}=\ket{+\,+\,+}$, $\ket{\tfrac{3}{2},+\tfrac{1}{2},S}= (1/\sqrt{3})\{\ket{+\,+\,-}+\ket{+\,-\,+}+\ket{-\,+\,+}\}$, and so on. The $T$-system will be one rotated about the $y$-axis of $S$ by the angle $\theta$. States in $T$ will be labelled $\ket{\tfrac{3}{2},+\tfrac{3}{2},T}$, $\ket{\tfrac{3}{2},+\tfrac{1}{2},T}$, and so on. Of course, $\ket{\tfrac{3}{2},+\tfrac{3}{2},T}$ is the same as $\ket{+'\,+'\,+'}$, the primes referring always to the $T$-system. Similarly, $\ket{\tfrac{3}{2},+\tfrac{1}{2},T}$ will be equal to $(1/\sqrt{3})\{\ket{+'\,+'\,-'}+\ket{+'\,-'\,+'}+\ket{-'\,+'\,+'}\}$, and so on. Each $\ket{+'}$ state in the $T$-frame comes from both the $\ket{+}$ and $\ket{-}$ states in $S$ via the matrix elements of Table 12–4. When we have three spin one-half particles, Eq. (12.47) gets replaced by \begin{align} \ket{+\,+\,+}=&\;a^3\ket{+'\,+'\,+'}+a^2b\,\{\ket{+'\,+'\,-'}+ \ket{+'\,-'\,+'}+\ket{-'\,+'\,+'}\}\notag\\[1ex] &+\,ab^2\{\ket{+'\,-'\,-'}+\ket{-'\,+'\,-'}+ \ket{-'\,-'\,+'}\}\notag\\[1ex] \label{Eq:III:18:31} &+\,b^3\ket{-'\,-'\,-'}. \end{align} \begin{align} &\ket{+\,+\,+}=\notag\\[1ex] &\phantom{+a^2b} a^3\ket{+'\,+'\,+'}\notag\\[.5ex] &+a^2b\{\ket{+'\,+'\,-'}\!+\!\ket{+'\,-'\,+'}+\ket{-'\,+'\,+'}\}\notag\\[.5ex] &+ab^2\!\{\ket{+'\,-'\,-'}\!+\!\ket{-'\,+'\,-'}+\ket{-'\,-'\,+'}\}\notag\\[.5ex] \label{Eq:III:18:31} &+\phantom{a\!\{}b^3\ket{-'\,-'\,-'}. \end{align} Using the transformation of Table 12–4, we get instead of (12.48) the equation \begin{align} \ket{\tfrac{3}{2},+\tfrac{3}{2},S}&= a^3\,\ket{\tfrac{3}{2},+\tfrac{3}{2},T}+ \sqrt{3}\,a^2b\,\ket{\tfrac{3}{2},+\tfrac{1}{2},T}\notag\\[1ex] \label{Eq:III:18:32} &\quad+\sqrt{3}\,ab^2\,\ket{\tfrac{3}{2},-\tfrac{1}{2},T} +b^3\,\ket{\tfrac{3}{2},-\tfrac{3}{2},T}. \end{align} \begin{align} \ket{\tfrac{3}{2},+\tfrac{3}{2},S}&=a^3\,\ket{\tfrac{3}{2},+\tfrac{3}{2},T}\notag\\[.5ex] &+\sqrt{3}\,a^2b\,\ket{\tfrac{3}{2},+\tfrac{1}{2},T}\notag\\[.5ex] &+\sqrt{3}\,ab^2\,\ket{\tfrac{3}{2},-\tfrac{1}{2},T}\notag\\[.5ex] \label{Eq:III:18:32} &+b^3\,\ket{\tfrac{3}{2},-\tfrac{3}{2},T}. \end{align} This already gives us several of our matrix elements $\braket{jT}{iS}$. To get the expression for $\ket{\tfrac{3}{2},+\tfrac{1}{2},S}$ we begin with the transformation of a state with two “$+$” and one “$-$” pieces. For instance, \begin{align} \ket{+\,+\,-}&=a^2c\,\ket{+'\,+'\,+'}+a^2d\,\ket{+'\,+'\,-'}+ abc\,\ket{+'\,-'\,+'}\notag\\[1ex] &\quad+bac\,\ket{-'\,+'\,+'}+abd\,\ket{+'\,-'\,-'}+ bad\,\ket{-'\,+'\,-'}\notag\\[1ex] \label{Eq:III:18:33} &\quad+b^2c\,\ket{-'\,-'\,+'}+b^2d\,\ket{-'\,-'\,-'}. \end{align} \begin{align} \ket{+\,+\,-}&=a^2c\,\ket{+'\,+'\,+'}\notag\\[.5ex] &+a^2d\,\ket{+'\,+'\,-'}\notag\\[.5ex] &+abc\,\ket{+'\,-'\,+'}\notag\\[.5ex] &+bac\,\ket{-'\,+'\,+'}+abd\,\ket{+'\,-'\,-'}\notag\\[.5ex] &+bad\,\ket{-'\,+'\,-'}\notag\\[.5ex] \label{Eq:III:18:33} &+b^2c\,\ket{-'\,-'\,+'}+b^2d\,\ket{-'\,-'\,-'}. \end{align} Adding two similar expressions for $\ket{+\,-\,+}$ and $\ket{-\,+\,+}$ and dividing by $\sqrt{3}$, we find \begin{align} \ket{\tfrac{3}{2},+\tfrac{1}{2},S}&= \sqrt{3}\,a^2c\,\ket{\tfrac{3}{2},+\tfrac{3}{2},T}\notag\\[.5ex] &+(a^2d+2abc)\,\ket{\tfrac{3}{2},+\tfrac{1}{2},T}\notag\\[.5ex] &+(2bad+b^2c)\,\ket{\tfrac{3}{2},-\tfrac{1}{2},T}\notag\\[.5ex] \label{Eq:III:18:34} &+\sqrt{3}\,b^2d\,\ket{\tfrac{3}{2},-\tfrac{3}{2},T}. \end{align} Continuing the process we find all the elements $\braket{jT}{iS}$ of the transformation matrix as given in Table 18–2. The first column comes from Eq. (18.32); the second from (18.34). The last two columns were worked out in the same way. (The coefficients $a$, $b$, $c$, and $d$ are given in Table 12–4.) Now suppose the $T$-frame were rotated with respect to $S$ by the angle $\theta$ about their $y$-axes. Then $a$, $b$, $c$, and $d$ have the values [see (12.54)] $a=$ $d=$ $\cos\theta/2$, and $c=$ $-b=$ $\sin\theta/2$. Using these values in Table 18–2 we get the forms which correspond to the second part of Table 17–2, but now for a spin $\tfrac{3}{2}$ system. The arguments we have just gone through are readily generalized to a system of any spin $j$. The states $\ket{j,m}$ can be put together from $2j$ particles, each of spin one-half. (There are $j+m$ of them in the $\ket{+}$ state and $j-m$ in the $\ket{-}$ state.) Sums are taken over all the possible ways this can be done, and the state is normalized by multiplying by a suitable constant. Those of you who are mathematically inclined may be able to show that the following result comes out7: \begin{align} &\bracket{j,m'}{R_y(\theta)}{j,m}= [(j+m)!(j-m)!(j+m')!(j-m')!]^{1/2}\notag\\[2ex] \label{Eq:III:18:35} &\kern{4.5em}\times\sum_k \frac{(-1)^{k+m-m'}(\cos\theta/2)^{2j+m'-m-2k}(\sin\theta/2)^{m-m'+2k}} {(m-m'+k)!(j+m'-k)!(j-m-k)!k!}, \end{align} \begin{gather} \label{Eq:III:18:35} \bracket{j,m'}{R_y(\theta)}{j,m}=\\[1.5ex] [(j+m)!(j-m)!(j+m')!(j-m')!]^{1/2}\;\times\notag\\[1.25ex] \sum_k\!\! \frac{(-1)^{k+m-m'}\!(\cos\theta/2)^{2j+m'\!-m-2k}(\sin\theta/2)^{m-m'\!+2k}} {(m-m'+k)!(j+m'-k)!(j-m-k)!k!},\notag \end{gather} where $k$ is to go over all values which give terms $\geq0$ in all the factorials. This is quite a messy formula, but with it you can check Table 17–2 for $j=1$ and prepare tables of your own for larger $j$. Several special matrix elements are of extra importance and have been given special names. For example the matrix elements for $m=m'=0$ and integral $j$ are known as the Legendre polynomials and are called $P_j(\cos\theta)$: \begin{equation} \label{Eq:III:18:36} \bracket{j,0}{R_y(\theta)}{j,0}=P_j(\cos\theta). \end{equation} The first few of these polynomials are: \begin{align} \label{Eq:III:18:37} P_0(\cos\theta)&=1,\\[1ex] \label{Eq:III:18:38} P_1(\cos\theta)&=\cos\theta,\\[1ex] \label{Eq:III:18:39} P_2(\cos\theta)&=\tfrac{1}{2}(3\cos^2\theta-1),\\[1ex] \label{Eq:III:18:40} P_3(\cos\theta)&=\tfrac{1}{2}(5\cos^3\theta-3\cos\theta). \end{align}

3

18

Angular Momentum

5

Measuring a nuclear spin

We would like to show you one example of the application of the coefficients we have just described. It has to do with a recent, interesting experiment which you will now be able to understand. Some physicists wanted to find out the spin of a certain excited state of the Ne$^{20}$ nucleus. To do this, they bombarded a carbon target with a beam of accelerated carbon ions, and produced the desired excited state of Ne$^{20}$—called Ne$^{20*}$—in the reaction \begin{equation*} \text{C}^{12}+\text{C}^{12}\to \text{Ne}^{20*}+\alpha_1, \end{equation*} where $\alpha_1$ is the $\alpha$-particle, or He$^4$. Several of the excited states of Ne$^{20}$ produced this way are unstable and disintegrate in the reaction \begin{equation*} \text{Ne}^{20*}\to\text{O}^{16}+\alpha_2. \end{equation*} So experimentally there are two $\alpha$-particles which come out of the reaction. We call them $\alpha_1$ and $\alpha_2$; since they come off with different energies, they can be distinguished from each other. Also, by picking a particular energy for $\alpha_1$ we can pick out any particular excited state of the Ne$^{20}$. The experiment was set up as shown in Fig. 18–9. A beam of $16$-MeV carbon ions was directed onto a thin foil of carbon. The first $\alpha$-particle was counted in a silicon diffused junction detector marked $\alpha_1$—set to accept $\alpha$-particles of the proper energy moving in the forward direction (with respect to the incident C$^{12}$ beam). The second $\alpha$-particle was picked up in the counter $\alpha_2$ at the angle $\theta$ with respect to $\alpha_1$. The counting rate of coincidence signals from $\alpha_1$ and $\alpha_2$ were measured as a function of the angle $\theta$. The idea of the experiment is the following. First, you need to know that the spins of C$^{12}$, O$^{16}$, and the $\alpha$-particle are all zero. If we call the direction of motion of the initial C$^{12}$ the $+z$-direction, then we know that the Ne$^{20*}$ must have zero angular momentum about the $z$-axis. None of the other particles has any spin; the C$^{12}$ arrives along the $z$-axis and the $\alpha_1$ leaves along the $z$-axis so they can’t have any angular momentum about it. So whatever the spin $j$ of the Ne$^{20*}$ is, we know that it is in the state $\ket{j,0}$. Now what will happen when the Ne$^{20*}$ disintegrates into an O$^{16}$ and the second $\alpha$-particle? Well, the $\alpha$-particle is picked up in the counter $\alpha_2$ and to conserve momentum the O$^{16}$ must go off in the opposite direction.8 About the new axis through $\alpha_2$, there can be no component of angular momentum. The final state has zero angular momentum about the new axis, so the Ne$^{20*}$ can disintegrate this way only if it has some amplitude to have $m'$ equal to zero, where $m'$ is the quantum number of the component of angular momentum about the new axis. In fact, the probability of observing $\alpha_2$ at the angle $\theta$ is just the square of the amplitude (or matrix element) \begin{equation} \label{Eq:III:18:41} \bracket{j,0}{R_y(\theta)}{j,0}. \end{equation} To find the spin of the Ne$^{20*}$ state in question, the intensity of the second $\alpha$-particle was plotted as a function of angle and compared with the theoretical curves for various values of $j$. As we said in the last section, the amplitudes $\bracket{j,0}{R_y(\theta)}{j,0}$ are just the functions $P_j(\cos\theta)$. So the possible angular distributions are curves of $[P_j(\cos\theta)]^2$. The experimental results are shown in Fig. 18–10 for two of the excited states. You can see that the angular distribution for the $5.80$-MeV state fits very well the curve for $[P_1(\cos\theta)]^2$, and so it must be a spin-one state. The data for the $5.63$-MeV state, on the other hand, are quite different; they fit the curve $[P_3(\cos\theta)]^2$. The state has a spin of $3$. From this experiment we have been able to find out the angular momentum of two of the excited states of Ne$^{20*}$. This information can then be used for trying to understand what the configuration of protons and neutrons is inside this nucleus—one more piece of information about the mysterious nuclear forces.

3

18

Angular Momentum

6

Composition of angular momentum

When we studied the hyperfine structure of the hydrogen atom in Chapter 12 we had to work out the internal states of a system composed of two particles—the electron and the proton—each with a spin of one-half. We found that the four possible spin states of such a system could be put together into two groups—a group with one energy that looked to the external world like a spin-one particle, and one remaining state that behaved like a particle of zero spin. That is, putting together two spin one-half particles we can form a system whose “total spin” is one, or zero. In this section we want to discuss in more general terms the spin states of a system which is made up of two particles of arbitrary spin. It is another important problem about angular momentum in quantum mechanical systems. Let’s first rewrite the results of Chapter 12 for the hydrogen atom in a form that will be easier to extend to the more general case. We began with two particles which we will now call particle $a$ (the electron) and particle $b$ (the proton). Particle $a$ had the spin $j_a$ ($=\tfrac{1}{2}$), and its $z$-component of angular momentum $m_a$ could have one of several values (actually $2$, namely $m_a=+\tfrac{1}{2}$ or $m_a=-\tfrac{1}{2}$). Similarly, the spin state of particle $b$ is described by its spin $j_b$ and its $z$-component of angular momentum $m_b$. Various combinations of the spin states of the two particles could be formed. For instance, we could have particle $a$ with $m_a=\tfrac{1}{2}$ and particle $b$ with $m_b=-\tfrac{1}{2}$, to make a state $\ket{a,+\tfrac{1}{2};b,-\tfrac{1}{2}}$. In general, the combined states formed a system whose “system spin,” or “total spin,” or “total angular momentum” $J$ could be $1$, or $0$. And the system could have a $z$-component of angular momentum $M$, which was $+1$, $0$, or $-1$ when $J=1$, or $0$ when $J=0$. In this new language we can rewrite the formulas in (12.41) and (12.42) as shown in Table 18–3. In the table the left-hand column describes the compound state in terms of its total angular momentum $J$ and the $z$-component $M$. The right-hand column shows how these states are made up in terms of the $m$-values of the two particles $a$ and $b$. We want now to generalize this result to states made up of two objects $a$ and $b$ of arbitrary spins $j_a$ and $j_b$. We start by considering an example for which $j_a=\tfrac{1}{2}$ and $j_b=1$, namely, the deuterium atom in which particle $a$ is an electron (e) and particle $b$ is the nucleus—a deuteron (d). We have then that $j_a=j_{\text{e}}=\tfrac{1}{2}$. The deuteron is formed of one proton and one neutron in a state whose total spin is one, so $j_b=j_{\text{d}}=1$. We want to discuss the hyperfine states of deuterium—just the way we did for hydrogen. Since the deuteron has three possible states $m_b=m_{\text{d}}=+1$, $0$, $-1$, and the electron has two, $m_a=m_{\text{e}}=+\tfrac{1}{2}$, $-\tfrac{1}{2}$, there are six possible states as follows (using the notation $\ket{\text{e},m_{\text{e}};\text{d},m_{\text{d}}}$): \begin{equation} \begin{aligned} &\ket{\text{e},+\tfrac{1}{2};\text{d},+1},\\[.5ex] &\ket{\text{e},+\tfrac{1}{2};\text{d},0}; \ket{\text{e},-\tfrac{1}{2};\text{d},+1},\\[.5ex] &\ket{\text{e},+\tfrac{1}{2};\text{d},-1}; \ket{\text{e},-\tfrac{1}{2};\text{d},0},\\[.5ex] &\ket{\text{e},-\tfrac{1}{2};\text{d},-1}. \end{aligned} \label{Eq:III:18:42} \end{equation} You will notice that we have grouped the states according to the values of the sum of $m_{\text{e}}$ and $m_{\text{d}}$—arranged in descending order. Now we ask: What happens to these states if we project into a different coordinate system? If the new system is just rotated about the $z$-axis by the angle $\phi$, then the state $\ket{\text{e},m_{\text{e}};\text{d},m_{\text{d}}}$ gets multiplied by \begin{equation} \label{Eq:III:18:43} e^{im_{\text{e}}\phi}e^{im_{\text{d}}\phi}= e^{i(m_{\text{e}}+m_{\text{d}})\phi}. \end{equation} (The state may be thought of as the product $\ket{\text{e},m_{\text{e}}}\ket{\text{d},m_{\text{d}}}$, and each state vector contributes independently its own exponential factor.) The factor (18.43) is of the form $e^{iM\phi}$, so the state $\ket{\text{e},m_{\text{e}};\text{d},m_{\text{d}}}$ has a $z$-component of angular momentum equal to \begin{equation} \label{Eq:III:18:44} M=m_{\text{e}}+m_{\text{d}}. \end{equation} The $z$-component of the total angular momentum is the sum of the $z$-components of angular momentum of the parts. In the list of (18.42), therefore, the state in the top line has $M=+\tfrac{3}{2}$, the two in the second line have $M=+\tfrac{1}{2}$, the next two have $M=-\tfrac{1}{2}$, and the last state has $M=-\tfrac{3}{2}$. We see immediately one possibility for the spin $J$ of the combined state (the total angular momentum) must be $\tfrac{3}{2}$, and this will require four states with $M=+\tfrac{3}{2}$, $+\tfrac{1}{2}$, $-\tfrac{1}{2}$, and $-\tfrac{3}{2}$. There is only one candidate for $M=+\tfrac{3}{2}$, so we know already that \begin{equation} \label{Eq:III:18:45} \ket{J=\tfrac{3}{2},M=+\tfrac{3}{2}}= \ket{\text{e},+\tfrac{1}{2};\text{d},+1}. \end{equation} But what is the state $\ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}$? We have two candidates in the second line of (18.42), and, in fact, any linear combination of them would also have $M=\tfrac{1}{2}$. So, in general, we must expect to find that \begin{equation} \label{Eq:III:18:46} \ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}= \alpha\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}+ \beta\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1}, \end{equation} \begin{equation} \begin{gathered} \ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}=\\[.75ex] \alpha\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}+ \beta\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1}, \end{gathered} \label{Eq:III:18:46} \end{equation} where $\alpha$ and $\beta$ are two numbers. They are called the Clebsch-Gordan coefficients. Our next problem is to find out what they are. We can find out easily if we just remember that the deuteron is made up of a neutron and a proton, and write the deuteron states out more explicitly using the rules of Table 18–3. If we do that, the states listed in (18.42) then look as shown in Table 18–4. We want to form the four states of $J=\tfrac{3}{2}$, using the states in the table. But we already know the answer, because in Table 18–1 we have states of spin $\tfrac{3}{2}$ formed from three spin one-half particles. The first state in Table 18–1 has $\ket{J=\tfrac{3}{2},M=+\tfrac{3}{2}}$ and it is $\ket{+\,+\,+}$, which—in our present notation—is the same as $\ket{\text{e},+\tfrac{1}{2};\text{n},+\tfrac{1}{2};\text{p},+\tfrac{1}{ 2}}$, or the first state in Table 18–4. But this state is also the same as the first in the list of (18.42), confirming our statement in (18.45). The second line of Table 18–1 says—changing to our present notation—that \begin{align} \label{Eq:III:18:47} \ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}=&\frac{1}{\sqrt{3}}\,\{ \ket{\text{e},+\tfrac{1}{2};\text{n},+\tfrac{1}{2};\text{p},-\tfrac{1}{2}}\\[1ex] &+\;\ket{\text{e},+\tfrac{1}{2};\text{n},-\tfrac{1}{2};\text{p},+\tfrac{1}{2}} +\ket{\text{e},-\tfrac{1}{2};\text{n},+\tfrac{1}{2};\text{p},+\tfrac{1}{2}}\}.\notag \end{align} \begin{align} &\ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}=\notag\\[.5ex] &\quad\frac{1}{\sqrt{3}}\,\{ \ket{\text{e},+\tfrac{1}{2};\text{n},+\tfrac{1}{2};\text{p},-\tfrac{1}{2}}\notag\\[-.5ex] &\qquad+\,\ket{\text{e},+\tfrac{1}{2};\text{n},-\tfrac{1}{2};\text{p},+\tfrac{1}{2}}\notag\\[.75ex] \label{Eq:III:18:47} &\qquad+\,\ket{\text{e},-\tfrac{1}{2};\text{n},+\tfrac{1}{2};\text{p},+\tfrac{1}{2}}\}. \end{align} The right side can evidently be put together from the two entries in the second line of Table 18–4 by taking $\sqrt{2/3}$ of the first term with $\sqrt{1/3}$ of the second. That is, Eq. (18.47) is equivalent to \begin{equation} \label{Eq:III:18:48} \ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}= \sqrt{2/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}+ \sqrt{1/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1}. \end{equation} \begin{align} \ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}&= \sqrt{2/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}\notag\\[.5ex] \label{Eq:III:18:48} &+\sqrt{1/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1}. \end{align} We have found our two Clebsch-Gordan coefficients $\alpha$ and $\beta$ in Eq. (18.46): \begin{equation} \label{Eq:III:18:49} \alpha=\sqrt{2/3},\quad \beta=\sqrt{1/3}. \end{equation} Following the same procedure we can find that \begin{equation} \label{Eq:III:18:50} \ket{J=\tfrac{3}{2},M=-\tfrac{1}{2}}= \sqrt{1/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},-1}+ \sqrt{2/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},0}. \end{equation} \begin{align} \ket{J=\tfrac{3}{2},M=-\tfrac{1}{2}}&= \sqrt{1/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},-1}\notag\\[.5ex] \label{Eq:III:18:50} &+\sqrt{2/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},0}. \end{align} And, also, of course, \begin{equation} \label{Eq:III:18:51} \ket{J=\tfrac{3}{2},M=-\tfrac{3}{2}}= \ket{\text{e},-\tfrac{1}{2};\text{d},-1}. \end{equation} These are the rules for the composition of spin $1$ and spin $\tfrac{1}{2}$ to make a total $J=\tfrac{3}{2}$. We summarize (18.45), (18.48), (18.50), and (18.51) in Table 18–5. We have, however, only four states here while the system we are considering has six possible states. Of the two states in the second line of (18.42) we have used only one linear combination to form $\ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}$. There is another linear combination orthogonal to the one we have taken which also has $M=+\tfrac{1}{2}$, namely \begin{equation} \label{Eq:III:18:52} \sqrt{1/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}- \sqrt{2/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1}. \end{equation} Similarly, the two states in the third line of (18.42) can be combined to give two orthogonal states, each with $M=-\tfrac{1}{2}$. The one orthogonal to (18.52) is \begin{equation} \label{Eq:III:18:53} \sqrt{2/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},-1}- \sqrt{1/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},0}. \end{equation} These are the two remaining states. They have $M=m_{\text{e}}+m_{\text{d}}=\pm\tfrac{1}{2}$; and must be the two states corresponding to $J=\tfrac{1}{2}$. So we have \begin{equation} \begin{aligned} \ket{J=\tfrac{1}{2},M=+\tfrac{1}{2}}&= \sqrt{1/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}- \sqrt{2/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1},\\[2ex] \ket{J=\tfrac{1}{2},M=-\tfrac{1}{2}}&= \sqrt{2/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},-1}- \sqrt{1/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},0}. \end{aligned} \label{Eq:III:18:54} \end{equation} \begin{equation} \begin{aligned} \ket{J=\tfrac{1}{2},M=+\tfrac{1}{2}}&= \sqrt{1/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}\\[.5ex] &-\sqrt{2/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1},\\[2.25ex] \ket{J=\tfrac{1}{2},M=-\tfrac{1}{2}}&= \sqrt{2/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},-1}\\[.5ex] &-\sqrt{1/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},0}. \end{aligned} \label{Eq:III:18:54} \end{equation} We can verify that these two states do indeed behave like the states of a spin one-half object by writing out the deuterium parts in terms of the neutron and proton states—using Table 18–4. The first state in (18.52) is \begin{align} \sqrt{1/6}\,\{ \ket{\text{e},+\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}+ \ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}&\}\notag\\\notag\\ \label{Eq:III:18:55} {}-\sqrt{2/3}\, \ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}&, \end{align} \begin{align} \sqrt{1/6}\,\{ &\ket{\text{e},+\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\,+\notag\\ &\ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\notag\\[1ex] \label{Eq:III:18:55} -\,\sqrt{2/3}\, &\ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}, \end{align} which can also be written \begin{align} &\sqrt{1/3}\,\bigl[\sqrt{1/2}\,\{ \ket{\text{e},+\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}- \ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\notag\\\notag\\ \label{Eq:III:18:56} &{}+\sqrt{1/2}\{ \ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}- \ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\bigr]. \end{align} \begin{align} \sqrt{1/3}\biggl[\sqrt{1/2}\,\{ &\ket{\text{e},+\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\,-\notag\\ &\ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\notag\\[1ex] +\,\sqrt{1/2}\{ &\ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\,-\notag\\[-.5ex] &\ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\biggr]. \label{Eq:III:18:56} \end{align} Now look at the terms in the first curly brackets, and think of the e and p taken together. Together they form a spin-zero state (see the bottom line of Table 18–3), and contribute no angular momentum. Only the neutron is left, so the whole of the first curly bracket of (18.56) behaves under rotations like a neutron, namely as a state with $J=\tfrac{1}{2}$, $M=+\tfrac{1}{2}$. Following the same reasoning, we see that in the second curly bracket of (18.56) the electron and neutron team up to produce zero angular momentum, and only the proton contribution—with $m_{\text{p}}=\tfrac{1}{2}$—is left. The terms behave like an object with $J=\tfrac{1}{2}$, $M=+\tfrac{1}{2}$. So the whole expression of (18.56) transforms like $\ket{J=\tfrac{1}{2},M=+\tfrac{1}{2}}$ as it should. The $M=-\tfrac{1}{2}$ state which corresponds to (18.53) can be written down (by changing the proper $+\tfrac{1}{2}$'s to $-\tfrac{1}{2}$'s) to get \begin{align} &\sqrt{1/3}\,\bigl[\sqrt{1/2}\,\{ \ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}- \ket{\text{e},-\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\notag\\\notag\\ \label{Eq:III:18:57} &{}+\sqrt{1/2}\{ \ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}- \ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\}\bigr]. \end{align} \begin{align} \sqrt{1/3}\biggl[\sqrt{1/2}\,\{ &\ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\,-\notag\\ &\ket{\text{e},-\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\notag\\[1ex] \label{Eq:III:18:57} +\sqrt{1/2}\{ &\ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\,-\notag\\[-.5ex] &\ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\}\biggr]. \end{align} You can easily check that this is equal to the second line of (18.54), as it should be if the two terms of that pair are to be the two states of a spin one-half system. So our results are confirmed. A deuteron and an electron can exist in six spin states, four of which act like the states of a spin $\tfrac{3}{2}$ object (Table 18–5) and two of which act like an object of spin one-half (18.54). The results of Table 18–5 and of Eq. (18.54) were obtained by making use of the fact that the deuteron is made up of a neutron and a proton. The truth of the equations does not depend on that special circumstance. For any spin-one object put together with any spin one-half object the composition laws (and the coefficients) are the same. The set of equations in Table 18–5 means that if the coordinates are rotated about, say, the $y$-axis—so that the states of the spin one-half particle and of the spin-one particle change according to Table 17–1 and Table 17–2—the linear combinations on the right-hand side will change in the proper way for a spin $\tfrac{3}{2}$ object. Under the same rotation the states of (18.54) will change as the states of a spin one-half object. The results depend only on the rotation properties (that is, the spin states) of the two original particles but not in any way on the origins of their angular momenta. We have only made use of this fact to work out the formulas by choosing a special case in which one of the component parts is itself made up of two spin one-half particles in a symmetric state. We have put all our results together in Table 18–6, changing the notation “e” and “d” to “$a$” and “$b$” to emphasize the generality of the conclusions. Suppose we have the general problem of finding the states which can be formed when two objects of arbitrary spins are combined. Say one has $j_a$ (so its $z$-component $m_a$ runs over the $2j_a+1$ values from $-j_a$ to $+j_a$) and the other has $j_b$ (with $z$-component $m_b$ running over the values from $-j_b$ to $+j_b$). The combined states are $\ket{a,m_a;b,m_b}$, and there are $(2j_a+1)(2j_b+1)$ different ones. Now what states of total spin $J$ can be found? The total $z$-component of angular momentum $M$ is equal to $m_a+m_b$, and the states can all be listed according to $M$ [as in (18.42)]. The largest $M$ is unique; it corresponds to $m_a=j_a$ and $m_b=j_b$, and is, therefore, just $j_a+j_b$. That means that the largest total spin $J$ is also equal to the sum $j_a+j_b$: \begin{equation*} J=(M)_{\text{max}}=j_a+j_b. \end{equation*} For the first $M$ value smaller than $(M)_{\text{max}}$, there are two states (either $m_a$ or $m_b$ is one unit less than its maximum). They must contribute one state to the set that goes with $J=j_a+j_b$, and the one left over will belong to a new set with $J=j_a+j_b-1$. The next $M$-value—the third from the top of the list—can be formed in three ways. (From $m_a=j_a-2$, $m_b=j_b$; from $m_a=j_a-1$, $m_b=j_b-1$; and from $m_a=j_a$, $m_b=j_b-2$.) Two of these belong to groups already started above; the third tells us that states of $J=j_a+j_b-2$ must also be included. This argument continues until we reach a stage where in our list we can no longer go one more step down in one of the $m$’s to make new states. Let $j_b$ be the smaller of $j_a$ and $j_b$ (if they are equal take either one); then only $2j_b$ values of $J$ are required—going in integer steps from $j_a+j_b$ down to $j_a-j_b$. That is, when two objects of spin $j_a$ and $j_b$ are combined, the system can have a total angular momentum $J$ equal to any one of the values \begin{equation} J=\left\{ \begin{aligned} &j_a+j_b\\ &j_a+j_b-1\\ &j_a+j_b-2\\ &\;\vdots\\ |&j_a-j_b|. \end{aligned} \right. \label{Eq:III:18:58} \end{equation} (By writing $\abs{j_a-j_b}$ instead of $j_a-j_b$ we can avoid the extra admonition that $j_a\geq j_b$.) For each of these $J$ values there are the $2J+1$ states of different $M$-values—with $M$ going from $+J$ to $-J$. Each of these is formed from linear combinations of the original states $\ket{a,m_a;b,m_b}$ with appropriate factors—the Clebsch-Gordan coefficients for each particular term. We can consider that these coefficients give the “amount” of the state $\ket{j_a,m_a;j_b,m_b}$ which appears in the state $\ket{J,M}$. So each of the Clebsch-Gordan coefficients has, if you wish, six indices identifying its position in the formulas like those of Tables 18–3 and 18–6. That is, calling these coefficients $C(J,M;j_a,m_a;j_b,m_b)$, we could express the equality of the second line of Table 18–6 by writing \begin{align*} C(\tfrac{3}{2},+\tfrac{1}{2};\tfrac{1}{2},+\tfrac{1}{2};1,0)= \sqrt{2/3},\\[.5ex] C(\tfrac{3}{2},+\tfrac{1}{2};\tfrac{1}{2},-\tfrac{1}{2};1,+1)= \sqrt{1/3}. \end{align*} We will not calculate here the coefficients for any other special cases.9 You can, however, find tables in many books. You might wish to try another special case for yourself. The next one to do would be the composition of two spin-one particles. We give just the final result in Table 18–7. These laws of the composition of angular momenta are very important in particle physics—where they have innumerable applications. Unfortunately, we have no time to look at more examples here.

3

18

Angular Momentum

7

Added Note 1: Derivation of the rotation matrix

For those who would like to see the details, we work out here the general rotation matrix for a system with spin (total angular momentum) $j$. It is really not very important to work out the general case; once you have the idea, you can find the general results in tables in many books. On the other hand, after coming this far you might like to see that you can indeed understand even the very complicated formulas of quantum mechanics, such as Eq. (18.35), that come into the description of angular momentum. We extend the arguments of Section 18-4 to a system with spin $j$, which we consider to be made up of $2j$ spin one-half objects. The state with $m=j$ would be $\ket{+\,+\,+\dotsb+}$ (with $2j$ plus signs). For $m=j-1$, there will be $2j$ terms like $\ket{+\,+\dotsb+\,+\,-}$, $\ket{+\,+\dotsb+\,-\,+}$, and so on. Let’s consider the general case in which there are $r$ plusses and $s$ minuses—with $r+s=2j$. Under a rotation about the $z$-axis each of the $r$ plusses will contribute $e^{+i\phi/2}$. The result is a phase change of $(r/2-s/2)\phi$. You see that \begin{equation} \label{Eq:III:18:59} m=\frac{r-s}{2}. \end{equation} Just as for $j=\tfrac{3}{2}$, each state of definite $m$ must be the linear combination with plus signs of all the states with the same $r$ and $s$—that is, states corresponding to every possible arrangement which has $r$ plusses and $s$ minuses. We assume that you can figure out that there are $(r+s)!/r!s!$ such arrangements. To normalize each state, we should divide the sum by the square root of this number. We can write \begin{align} \biggl[\frac{(r+s)!}{r!s!}\biggr]^{-1/2}\kern{-3.5ex} \{\ket{\underbrace{+\,+\,+\dotsb+\,+}_r\, \underbrace{-\,-\,-\dotsb-\,-}_s}&\notag\\[1ex] +\,(\text{all rearrangements of order})\}&\notag\\[1.5ex] \label{Eq:III:18:60} =\ket{j,m}& \end{align} with \begin{equation} \label{Eq:III:18:61} j=\frac{r+s}{2},\quad m=\frac{r-s}{2}. \end{equation} It will help our work if we now go to still another notation. Once we have defined the states by Eq. (18.60), the two numbers $r$ and $s$ define a state just as well as $j$ and $m$. It will help us keep track of things if we write \begin{equation} \label{Eq:III:18:62} \ket{j,m}=\ket{\tover{r}{s}}, \end{equation} where, using the equalities of (18.61) \begin{equation*} r=j+m,\quad s=j-m. \end{equation*} Next, we would like to write Eq. (18.60) with a new special notation as \begin{equation} \label{Eq:III:18:63} \ket{j,m}=\ket{\tover{r}{s}}= \biggl[\frac{(r+s)!}{r!s!}\biggr]^{+1/2}\kern{-3.5ex} \{\ket{+}^r\ket{-}^s\}_{\text{perm}}. \end{equation} Note that we have changed the exponent of the factor in front to plus $\tfrac{1}{2}$. We do that because there are just $N=(r+s)!/r!s!$ terms inside the curly brackets. Comparing (18.63) with (18.60) it is clear that \begin{equation*} \{\ket{+}^r\ket{-}^s\}_{\text{perm}} \end{equation*} is just a shorthand way of writing \begin{equation*} \frac{\{\ket{+\,+\dotsb-\,-}+\text{all rearrangements}\}}{N}, \end{equation*} where $N$ is the number of different terms in the bracket. The reason that this notation is convenient is that each time we make a rotation, all of the plus signs contribute the same factor, so we get this factor to the $r$th power. Similarly, all together the $s$ minus terms contribute a factor to the $s$th power no matter what the sequence of the terms is. Now suppose we rotate our system by the angle $\theta$ about the $y$-axis. What we want is $R_y(\theta)\,\ket{\tover{r}{s}}$. When $R_y(\theta)$ operates on each $\ket{+}$ it gives \begin{equation} \label{Eq:III:18:64} R_y(\theta)\,\ket{+}=\ket{+}C+\ket{-}S, \end{equation} where $C=\cos\theta/2$ and $S=-\sin\theta/2$. When $R_y(\theta)$ operates on each $\ket{-}$ it gives \begin{equation*} R_y(\theta)\,\ket{-}=\ket{-}C-\ket{+}S. \end{equation*} So what we want is \begin{equation} \begin{aligned} &R_y(\theta)\ket{\tover{r}{s}}\\[.5ex] &=\biggl[\frac{(r\!+\!s)!}{r!s!}\biggr]^{1/2}\kern{-1ex} R_y(\theta)\,\{\ket{+}^r\ket{-}^s\}_{\text{perm}}\\[1ex] &=\biggl[\frac{(r\!+\!s)!}{r!s!}\biggr]^{1/2}\kern{-1ex} \{(R_y(\theta)\,\ket{+})^r(R_y(\theta)\,\ket{-})^s\}_{\text{perm}}\\[1ex] &=\biggl[\frac{(r\!+\!s)!}{r!s!}\biggr]^{1/2}\kern{-1ex} \{(\ket{+}C\!+\!\ket{-}S)^r(\ket{-}C\!-\!\ket{+}S)^s\}_{\text{perm}}. \end{aligned} \label{Eq:III:18:65} \end{equation} Now each binomial has to be expanded out to its appropriate power and the two expressions multiplied together. There will be terms with $\ket{+}$ to all powers from zero to $(r+s)$. Let’s look at all of the terms which have $\ket{+}$ to the $r'$ power. They will appear always multiplied with $\ket{-}$ to the $s'$ power, where $s'=2j-r'$. Suppose we collect all such terms. For each permutation they will have some numerical coefficient involving the factors of the binomial expansion as well as the factors $C$ and $S$. Suppose we call that factor $A_{r'}$. Then Eq. (18.65) will look like \begin{equation} \label{Eq:III:18:66} R_y(\theta)\,\ket{\tover{r}{s}}=\sum_{r'=0}^{r+s} \{A_{r'}\,\ket{+}^{r'}\ket{-}^{s'}\}_{\text{perm}}. \end{equation} Now let’s say that we divide $A_{r'}$ by the factor $[(r'+s')!/r'!s'!]^{1/2}$ and call the quotient $B_{r'}$. Equation (18.66) is then equivalent to \begin{equation} \label{Eq:III:18:67} R_y(\theta)\,\ket{\tover{r}{s}}\!=\!\sum_{r'=0}^{r+s} B_{r'}\!\biggl[\frac{(r'\!+\!s')!}{r'!s'!}\biggr]^{1/2}\kern{-2.5ex} \{\ket{+}^{r'}\ket{-}^{s'}\}_{\text{perm}}. \end{equation} (We could just say that this equation defines $B_{r'}$ by the requirement that (18.67) gives the same expression that appears in (18.65).) With this definition of $B_{r'}$ the remaining factors on the right-hand side of Eq. (18.67) are just the states $\ket{\tover{r'}{s'}}$. So we have that \begin{equation} \label{Eq:III:18:68} R_y(\theta)\,\ket{\tover{r}{s}}=\sum_{r'=0}^{r+s} B_{r'}\,\ket{\tover{r'}{s'}}, \end{equation} with $s'$ always equal to $r+s-r'$. This means, of course, that the coefficients $B_{r'}$ are just the matrix elements we want, namely \begin{equation} \label{Eq:III:18:69} \bracket{\tover{r'}{s'}}{R_y(\theta)}{\tover{r}{s}}=B_{r'}. \end{equation} Now we just have to push through the algebra to find the various $B_{r'}$. Comparing (18.65) with (18.67)—and remembering that $r'+s'=r+s$—we see that $B_{r'}$ is just the coefficient of $a^{r'}b^{s'}$ in the following expression: \begin{equation} \label{Eq:III:18:70} \biggl(\frac{r'!s'!}{r!s!}\biggr)^{1/2} (aC+bS)^r(bC-aS)^s. \end{equation} It is now only a dirty job to make the expansions by the binomial theorem, and collect the terms with the given power of $a$ and $b$. If you work it all out, you find that the coefficient of $a^{r'}b^{s'}$ in (18.70) is \begin{equation} \label{Eq:III:18:71} \biggl[\frac{r'!s'!}{r!s!}\biggr]^{1/2}\sum_k (-1)^kS^{r-r'+2k}C^{s+r'-2k}\cdot \frac{r!}{(r-r'+k)!(r'-k)!}\cdot \frac{s!}{(s-k)!k!}. \end{equation} \begin{align} \biggl[\frac{r'!s'!}{r!s!}\biggr]^{1/2}\kern{-2.5ex}&\lower .3ex {\sum_k (-1)^kS^{r-r'\!+2k}C^{s+r'\!-2k}}\notag\\[-2ex] \label{Eq:III:18:71} &\quad\;\times\frac{r!}{(r\!-\!r'\!+\!k)!(r'\!-\!k)!}\!\cdot\! \frac{s!}{(s\!-\!k)!k!}. \end{align} The sum is to be taken over all integers $k$ which give terms of zero or greater in the factorials. This expression is then the matrix element we wanted. Finally, we can return to our original notation in terms of $j$, $m$, and $m'$ using \begin{equation*} r=j+m,\quad r'=j+m',\quad s=j-m,\quad s'=j-m'. \end{equation*} \begin{alignat*}{2} &r&=&\,j+m,\quad &r'&=&\,j+m',\\[.5ex] &s&=&\,j-m,\quad &s'&=&\,j-m'. \end{alignat*} Making these substitutions, we get Eq. (18.35) in Section 18-4.

3

18

Angular Momentum

8

Added Note 2: Conservation of parity in photon emission

In Section 18-1 of this chapter we considered the emission of light by an atom that goes from an excited state of spin $1$ to a ground state of spin $0$. If the excited state has its spin up ($m=+1$), it can emit a RHC photon along the $+z$-axis or a LHC photon along the $-z$-axis. Let’s call these two states of the photon $\ket{R_{\text{up}}}$ and $\ket{L_{\text{dn}}}$. Neither of these states has a definite parity. Letting $\Pop$ be the parity operator, $\Pop\,\ket{R_{\text{up}}}=\ket{L_{\text{dn}}}$ and $\Pop\,\ket{L_{\text{dn}}}=\ket{R_{\text{up}}}$. What about our earlier proof that an atom in a state of definite energy must have a definite parity, and our statement that parity is conserved in atomic processes? Shouldn’t the final state in this problem (the state after the emission of a photon) have a definite parity? It does if we consider the complete final state which contains amplitudes for the emission of photons into all sorts of angles. In Section 18-1 we chose to consider only a part of the complete final state. If we wish we can look only at final states that do have a definite parity. For example, consider a final state $\ket{\psi_{F}}$ which has some amplitude $\alpha$ to be a RHC photon going along $+z$ and some amplitude $\beta$ to be a LHC photon going along $-z$. We can write \begin{equation} \label{Eq:III:18:72} \ket{\psi_{F}}=\alpha\,\ket{R_{\text{up}}}+ \beta\,\ket{L_{\text{dn}}}. \end{equation} The parity operation on this state gives \begin{equation} \label{Eq:III:18:73} \Pop\,\ket{\psi_{F}}=\alpha\,\ket{L_{\text{dn}}}+ \beta\,\ket{R_{\text{up}}}. \end{equation} This state will be $\pm\,\ket{\psi_{F}}$ if $\beta=\alpha$ or if $\beta=-\alpha$. So a final state of even parity is \begin{equation} \label{Eq:III:18:74} \ket{\psi_{F}^+}=\alpha\{\ket{R_{\text{up}}}+ \ket{L_{\text{dn}}}\}, \end{equation} and a state of odd parity is \begin{equation} \label{Eq:III:18:75} \ket{\psi_{F}^-}=\alpha\{\ket{R_{\text{up}}}- \ket{L_{\text{dn}}}\}. \end{equation} Next, we wish to consider the decay of an excited state of odd parity to a ground state of even parity. If parity is to be conserved, the final state of the photon must have odd parity. It must be the state in (18.75). If the amplitude to find $\ket{R_{\text{up}}}$ is $\alpha$, the amplitude to find $\ket{L_{\text{dn}}}$ is $-\alpha$. Now notice what happens when we perform a rotation of $180^\circ$ about the $y$-axis. The initial excited state of the atom becomes an $m=-1$ state (with no change in sign, according to Table 17–2). And the rotation of the final state gives \begin{equation} \label{Eq:III:18:76} R_y(180^\circ)\,\ket{\psi_{F}^-}=\alpha\{\ket{R_{\text{dn}}}- \ket{L_{\text{up}}}\}. \end{equation} Comparing this equation with (18.75), you see that for the assumed parity of the final state, the amplitude to get a LHC photon along $+z$ from the $m=-1$ initial state is the negative of the amplitude to get a RHC photon from the $m=+1$ initial state. This agrees with the result we found in Section 18-1.

3

19

The Hydrogen Atom and The Periodic Table

1

Schrödinger’s equation for the hydrogen atom

The most dramatic success in the history of the quantum mechanics was the understanding of the details of the spectra of some simple atoms and the understanding of the periodicities which are found in the table of chemical elements. In this chapter we will at last bring our quantum mechanics to the point of this important achievement, specifically to an understanding of the spectrum of the hydrogen atom. We will at the same time arrive at a qualitative explanation of the mysterious properties of the chemical elements. We will do this by studying in detail the behavior of the electron in a hydrogen atom—for the first time making a detailed calculation of a distribution-in-space according to the ideas we developed in Chapter 16. For a complete description of the hydrogen atom we should describe the motions of both the proton and the electron. It is possible to do this in quantum mechanics in a way that is analogous to the classical idea of describing the motion of each particle relative to the center of gravity, but we will not do so. We will just discuss an approximation in which we consider the proton to be very heavy, so we can think of it as fixed at the center of the atom. We will make another approximation by forgetting that the electron has a spin and should be described by relativistic laws of mechanics. Some small corrections to our treatment will be required since we will be using the nonrelativistic Schrödinger equation and will disregard magnetic effects. Small magnetic effects occur because from the electron’s point-of-view the proton is a circulating charge which produces a magnetic field. In this field the electron will have a different energy with its spin up than with it down. The energy of the atom will be shifted a little bit from what we will calculate. We will ignore this small energy shift. Also we will imagine that the electron is just like a gyroscope moving around in space always keeping the same direction of spin. Since we will be considering a free atom in space the total angular momentum will be conserved. In our approximation we will assume that the angular momentum of the electron spin stays constant, so all the rest of the angular momentum of the atom—what is usually called “orbital” angular momentum—will also be conserved. To an excellent approximation the electron moves in the hydrogen atom like a particle without spin—the angular momentum of the motion is a constant. With these approximations the amplitude to find the electron at different places in space can be represented by a function of position in space and time. We let $\psi(x,y,z,t)$ be the amplitude to find the electron somewhere at the time $t$. According to the quantum mechanics the rate of change of this amplitude with time is given by the Hamiltonian operator working on the same function. From Chapter 16, \begin{equation} \label{Eq:III:19:1} i\hbar\,\ddp{\psi}{t}=\Hcalop\psi, \end{equation} with \begin{equation} \label{Eq:III:19:2} \Hcalop=-\frac{\hbar^2}{2m}\,\nabla^2+V(\FLPr). \end{equation} Here, $m$ is the electron mass, and $V(\FLPr)$ is the potential energy of the electron in the electrostatic field of the proton. Taking $V=0$ at large distances from the proton we can write1 \begin{equation*} V=-\frac{e^2}{r}. \end{equation*} The wave function $\psi$ must then satisfy the equation \begin{equation} \label{Eq:III:19:3} i\hbar\,\ddp{\psi}{t}=-\frac{\hbar^2}{2m}\,\nabla^2\psi -\frac{e^2}{r}\,\psi. \end{equation} We want to look for definite energy states, so we try to find solutions which have the form \begin{equation} \label{Eq:III:19:4} \psi(\FLPr,t)=e^{-(i/\hbar)Et}\psi(\FLPr). \end{equation} The function $\psi(\FLPr)$ must then be a solution of \begin{equation} \label{Eq:III:19:5} -\frac{\hbar^2}{2m}\,\nabla^2\psi= \biggl(E+\frac{e^2}{r}\biggr)\psi, \end{equation} where $E$ is some constant—the energy of the atom. Since the potential energy term depends only on the radius, it turns out to be much more convenient to solve this equation in polar coordinates rather than rectangular coordinates. The Laplacian is defined in rectangular coordinates by \begin{equation*} \nabla^2=\frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}+ \frac{\partial^2}{\partial z^2}. \end{equation*} We want to use instead the coordinates $r$, $\theta$, $\phi$ shown in Fig. 19–1. These coordinates are related to $x$, $y$, $z$ by \begin{equation*} x=r\sin\theta\cos\phi;\quad y=r\sin\theta\sin\phi;\quad z=r\cos\theta. \end{equation*} It’s a rather tedious mess to work through the algebra, but you can eventually show that for any function $f(\FLPr)=f(r,\theta,\phi)$, \begin{equation} \label{Eq:III:19:6} \nabla^2f(r,\theta,\phi)=\frac{1}{r}\,\frac{\partial^2}{\partial r^2}\, (rf)+\frac{1}{r^2}\biggl\{ \frac{1}{\sin\theta}\,\ddp{}{\theta}\biggl( \sin\theta\,\ddp{f}{\theta}\biggr)+\frac{1}{\sin^2\theta}\, \frac{\partial^2f}{\partial\phi^2}\biggr\}. \end{equation} \begin{align} \label{Eq:III:19:6} &\nabla^2f(r,\theta,\phi)=\\[1ex] &\frac{1}{r}\,\frac{\partial^2}{\partial r^2}\,(rf)\;+\notag\\[1ex] &\frac{1}{r^2}\biggl\{ \frac{1}{\sin\theta}\,\ddp{}{\theta}\biggl( \sin\theta\,\ddp{f}{\theta}\biggr)+\frac{1}{\sin^2\theta}\, \frac{\partial^2f}{\partial\phi^2}\biggr\}.\notag \end{align} So in terms of the polar coordinates, the equation which is to be satisfied by $\psi(r,\theta,\phi)$ is \begin{equation} \label{Eq:III:19:7} \frac{1}{r}\,\frac{\partial^2}{\partial r^2}\, (r\psi)+\frac{1}{r^2}\biggl\{ \frac{1}{\sin\theta}\,\ddp{}{\theta}\biggl( \sin\theta\,\ddp{\psi}{\theta}\biggr)+\frac{1}{\sin^2\theta}\, \frac{\partial^2\psi}{\partial \phi^2}\biggr\}= -\frac{2m}{\hbar^2}\biggl(E+\frac{e^2}{r}\biggr)\psi. \end{equation} \begin{align} \label{Eq:III:19:7} &\frac{1}{r}\,\frac{\partial^2}{\partial r^2}\, (r\psi)\;+\\[1ex] &\frac{1}{r^2}\biggl\{ \frac{1}{\sin\theta}\,\ddp{}{\theta}\biggl( \sin\theta\,\ddp{\psi}{\theta}\biggr)+\frac{1}{\sin^2\theta}\, \frac{\partial^2\psi}{\partial \phi^2}\biggr\}\notag\\[1.5ex] &=-\frac{2m}{\hbar^2}\biggl(E+\frac{e^2}{r}\biggr)\psi.\notag \end{align}

3

19

The Hydrogen Atom and The Periodic Table

2

Spherically symmetric solutions

Let’s first try to find some very simple function that satisfies the horrible equation in (19.7). Although the wave function $\psi$ will, in general, depend on the angles $\theta$ and $\phi$ as well as on the radius $r$, we can see whether there might be a special situation in which $\psi$ does not depend on the angles. For a wave function that doesn’t depend on the angles, none of the amplitudes will change in any way if you rotate the coordinate system. That means that all of the components of the angular momentum are zero. Such a $\psi$ must correspond to a state whose total angular momentum is zero. (Actually, it is only the orbital angular momentum which is zero because we still have the spin of the electron, but we are ignoring that part.) A state with zero orbital angular momentum is called by a special name. It is called an “$s$-state”—you can remember “$s$ for spherically symmetric.”2 Now if $\psi$ is not going to depend on $\theta$ and $\phi$ then the entire Laplacian contains only the first term and Eq. (19.7) becomes much simpler: \begin{equation} \label{Eq:III:19:8} \frac{1}{r}\,\frac{d^2}{dr^2}\,(r\psi)= -\frac{2m}{\hbar^2}\biggl(E+\frac{e^2}{r}\biggr)\psi. \end{equation} Before you start to work on solving an equation like this, it’s a good idea to get rid of all excess constants like $e^2$, $m$, and $\hbar$, by making some scale changes. Then the algebra will be easier. If we make the following substitutions: \begin{equation} \label{Eq:III:19:9} r=\frac{\hbar^2}{me^2}\,\rho, \end{equation} and \begin{equation} \label{Eq:III:19:10} E=\frac{me^4}{2\hbar^2}\,\epsilon, \end{equation} then Eq. (19.8) becomes (after multiplying through by $\rho$) \begin{equation} \label{Eq:III:19:11} \frac{d^2(\rho\psi)}{d\rho^2}= -\biggl(\epsilon+\frac{2}{\rho}\biggr)\rho\psi. \end{equation} These scale changes mean that we are measuring the distance $r$ and energy $E$ as multiples of “natural” atomic units. That is, $\rho=r/r_B$, where $r_B=\hbar^2/me^2$, is called the “Bohr radius” and is about $0.528$ angstroms. Similarly, $\epsilon=E/E_R$, with $E_R=me^4/2\hbar^2$. This energy is called the “Rydberg” and is about $13.6$ electron volts. Since the product $\rho\psi$ appears on both sides, it is convenient to work with it rather than with $\psi$ itself. Letting \begin{equation} \label{Eq:III:19:12} \rho\psi=f, \end{equation} we have the more simple-looking equation \begin{equation} \label{Eq:III:19:13} \frac{d^2f}{d\rho^2}=-\biggl(\epsilon+\frac{2}{\rho}\biggr)f. \end{equation} Now we have to find some function $f$ which satisfies Eq. (19.13)—in other words, we just have to solve a differential equation. Unfortunately, there is no very useful, general method for solving any given differential equation. You just have to fiddle around. Our equation is not easy, but people have found that it can be solved by the following procedure. First, you replace $f$, which is some function of $\rho$, by a product of two functions \begin{equation} \label{Eq:III:19:14} f(\rho)=e^{-\alpha\rho}g(\rho). \end{equation} This just means that you are factoring $e^{-\alpha\rho}$ out of $f(\rho)$. You can certainly do that for any $f(\rho)$ at all. This just shifts our problem to finding the right function $g(\rho)$. Sticking (19.14) into (19.13), we get the following equation for $g$: \begin{equation} \label{Eq:III:19:15} \frac{d^2g}{d\rho^2}-2\alpha\,\ddt{g}{\rho}+\biggl( \frac{2}{\rho}+\epsilon+\alpha^2\biggr)g=0. \end{equation} Since we are free to choose $\alpha$, let’s make \begin{equation} \label{Eq:III:19:16} \alpha^2=-\epsilon, \end{equation} and get \begin{equation} \label{Eq:III:19:17} \frac{d^2g}{d\rho^2}-2\alpha\,\ddt{g}{\rho}+ \frac{2}{\rho}\,g=0. \end{equation} You may think we are no better off than we were at Eq. (19.13), but the happy thing about our new equation is that it can be solved easily in terms of a power series in $\rho$. (It is possible, in principle, to solve (19.13) that way too, but it is much harder.) We are saying that Eq. (19.17) can be satisfied by some $g(\rho)$ which can be written as a series, \begin{equation} \label{Eq:III:19:18} g(\rho)=\sum_{k=1}^\infty a_k\rho^k, \end{equation} in which the $a_k$ are constant coefficients. Now all we have to do is find a suitable infinite set of coefficients! Let’s check to see that such a solution will work. The first derivative of this $g(\rho)$ is \begin{equation*} \ddt{g}{\rho}=\sum_{k=1}^\infty a_kk\rho^{k-1}, \end{equation*} and the second derivative is \begin{equation*} \frac{d^2g}{d\rho^2}=\sum_{k=1}^\infty a_kk(k-1)\rho^{k-2}. \end{equation*} Using these expressions in (19.17) we have \begin{equation} \label{Eq:III:19:19} \sum_{k=1}^\infty \!k(k\!-\!1)a_k\rho^{k\!-\!2}\!-\!\! \sum_{k=1}^\infty \!2\alpha ka_k\rho^{k\!-\!1}\!+\!\! \sum_{k=1}^\infty \!2a_k\rho^{k\!-\!1}\!= 0. \end{equation} It’s not obvious that we have succeeded; but we forge onward. It will all look better if we replace the first sum by an equivalent. Since the first term of the sum is zero, we can replace each $k$ by $k+1$ without changing anything in the infinite series; with this change the first sum can equally well be written as \begin{equation*} \sum_{k=1}^\infty(k+1)ka_{k+1}\rho^{k-1}. \end{equation*} Now we can put all the sums together to get \begin{equation} \label{Eq:III:19:20} \sum_{k=1}^\infty[(k+1)ka_{k+1}-2\alpha ka_k+2a_k]\rho^{k-1}=0. \end{equation} This power series must vanish for all possible values of $\rho$. It can do that only if the coefficient of each power of $\rho$ is separately zero. We will have a solution for the hydrogen atom if we can find a set $a_k$ for which \begin{equation} \label{Eq:III:19:21} (k+1)ka_{k+1}-2(\alpha k-1)a_k=0 \end{equation} for all $k\geq1$. That is certainly easy to arrange. Pick any $a_1$ you like. Then generate all of the other coefficients from \begin{equation} \label{Eq:III:19:22} a_{k+1}=\frac{2(\alpha k-1)}{k(k+1)}\,a_k. \end{equation} With this you will get $a_2$, $a_3$, $a_4$, and so on, and each pair will certainly satisfy (19.21). We get a series for $g(\rho)$ which satisfies (19.17). With it we can make a $\psi$, that satisfies Schrödinger’s equation. Notice that the solutions depend on the assumed energy (through $\alpha$), but for each value of $\epsilon$, there is a corresponding series. We have a solution, but what does it represent physically? We can get an idea by seeing what happens far from the proton—for large values of $\rho$. Out there, the high-order terms of the series are the most important, so we should look at what happens for large $k$. When $k\gg1$, Eq. (19.22) is approximately the same as \begin{equation*} a_{k+1}=\frac{2\alpha}{k}\,a_k, \end{equation*} which means that \begin{equation} \label{Eq:III:19:23} a_{k+1}\approx\frac{(2\alpha)^k}{k!}. \end{equation} But these are just the coefficients of the series for $e^{+2\alpha\rho}$. The function of $g$ is a rapidly increasing exponential. Even coupled with $e^{-\alpha\rho}$ to produce $f(\rho)$—see Eq. (19.14)—it still gives a solution for $f(\rho)$ which goes like $e^{\alpha\rho}$ for large $\rho$. We have found a mathematical solution but not a physical one. It represents a situation in which the electron is least likely to be near the proton! It is always more likely to be found at a very large radius $\rho$. A wave function for a bound electron must go to zero for large $\rho$. We have to think whether there is some way to beat the game, and there is. Observe! If it just happened by luck that $\alpha$ were equal to $1/n$, where $n$ is any positive integer, then Eq. (19.22) would make $a_{n+1}=0$. All higher terms would also be zero. We wouldn’t have an infinite series but a finite polynomial. Any polynomial increases more slowly than $e^{\alpha\rho}$, so the term $e^{-\alpha\rho}$ will eventually beat it down, and the function $f$ will go to zero for large $\rho$. The only bound-state solutions are those for which $\alpha=1/n$, with $n=1$, $2$, $3$, $4$, and so on. Looking back to Eq. (19.16), we see that the bound-state solutions to the spherically symmetric wave equation can exist only when \begin{equation*} -\epsilon=1,\,\frac{1}{4},\,\frac{1}{9},\,\frac{1}{16},\dotsc, \frac{1}{n^2},\dotsc \end{equation*} The allowed energies are just these fractions times the Rydberg, $E_R=me^4/2\hbar^2$, or the energy of the $n$th energy level is \begin{equation} \label{Eq:III:19:24} E_n=-E_R\,\frac{1}{n^2}. \end{equation} There is, incidentally, nothing mysterious about negative numbers for the energy. The energies are negative because when we chose to write $V=-e^2/r$, we picked our zero point as the energy of an electron located far from the proton. When it is close to the proton, its energy is less, so somewhat below zero. The energy is lowest (most negative) for $n=1$, and increases toward zero with increasing $n$. Before the discovery of quantum mechanics, it was known from experimental studies of the spectrum of hydrogen that the energy levels could be described by Eq. (19.24), where $E_R$ was found from the observations to be about $13.6$ electron volts. Bohr then devised a model which gave the same equation and predicted that $E_R$ should be $me^4/2\hbar^2$. But it was the first great success of the Schrödinger theory that it could reproduce this result from a basic equation of motion for the electron. Now that we have solved our first atom, let’s look at the nature of the solution we got. Pulling all the pieces together, each solution looks like this: \begin{equation} \label{Eq:III:19:25} \psi_n=\frac{f_n(\rho)}{\rho}=\frac{e^{-\rho/n}}{\rho}\,g_n(\rho), \end{equation} where \begin{equation} \label{Eq:III:19:26} g_n(\rho)=\sum_{k=1}^na_k\rho^k \end{equation} and \begin{equation} \label{Eq:III:19:27} a_{k+1}=\frac{2(k/n-1)}{k(k+1)}\,a_k. \end{equation} So long as we are mainly interested in the relative probabilities of finding the electron at various places we can pick any number we wish for $a_1$. We may as well set $a_1=1$. (People often choose $a_1$ so that the wave function is “normalized,” that is, so that the integrated probability of finding the electron anywhere in the atom is equal to $1$. We have no need to do that just now.) For the lowest energy state, $n=1$, and \begin{equation} \label{Eq:III:19:28} \psi_1(\rho)=e^{-\rho}. \end{equation} For a hydrogen atom in its ground (lowest-energy) state, the amplitude to find the electron at any point drops off exponentially with the distance from the proton. It is most likely to be found right at the proton, and the characteristic spreading distance is about one unit in $\rho$, or about one Bohr radius, $r_B$. Putting $n=2$ gives the next higher level. The wave function for this state will have two terms. It is \begin{equation} \label{Eq:III:19:29} \psi_2(\rho)=\biggl(1-\frac{\rho}{2}\biggr)e^{-\rho/2}. \end{equation} The wave function for the next level is \begin{equation} \label{Eq:III:19:30} \psi_3(\rho)=\biggl(1-\frac{2\rho}{3}+\frac{2}{27}\,\rho^2\biggr) e^{-\rho/3}. \end{equation} The wave functions for these first three levels are plotted in Fig. 19–2. You can see the general trend. All of the wave functions approach zero rapidly for large $\rho$ after oscillating a few times. In fact, the number of “bumps” is just equal to $n$—or, if you prefer, the number of zero-crossings of $\psi_n$ is $n-1$.

3

19

The Hydrogen Atom and The Periodic Table

3

States with an angular dependence

In the states described by the $\psi_n(r)$ we have found that the probability amplitude for finding the electron is spherically symmetric—depending only on $r$, the distance for the proton. Such states have zero orbital angular momentum. We should now inquire about states which may have some angular dependences. We could, if we wished, just investigate the strictly mathematical problem of finding the functions of $r$, $\theta$, and $\phi$ which satisfy the differential equation (19.7)—putting in the additional physical conditions that the only acceptable functions are ones which go to zero for large $r$. You will find this done in many books. We are going to take a short cut by using the knowledge we already have about how amplitudes depend on angles in space. The hydrogen atom in any particular state is a particle with a certain “spin” $j$—the quantum number of the total angular momentum. Part of this spin comes from the electron’s intrinsic spin, and part from the electron’s motion. Since each of these two components acts independently (to an excellent approximation) we will again ignore the spin part and think only about the “orbital” angular momentum. This orbital motion behaves, however, just like a spin. For example, if the orbital quantum number is $l$, the $z$-component of angular momentum can be $l$, $l-1$, $l-2$, …, $-l$. (We are, as usual, measuring in units of $\hbar$.) Also, all the rotation matrices and other properties we have worked out still apply. (From now on we will really ignore the electron’s spin; when we speak of “angular momentum” we will mean only the orbital part.) Since the potential $V$ in which the electron moves depends only on $r$ and not on $\theta$ or $\phi$, the Hamiltonian is symmetric under all rotations. It follows that the angular momentum and all its components are conserved. (This is true for motion in any “central field”—one which depends only on $r$—so is not a special feature of the Coulomb $e^2/r$ potential.) Now let’s think of some possible state of the electron; its internal angular structure will be characterized by the quantum number $l$. Depending on the “orientation” of the total angular momentum with respect to the $z$-axis, the $z$-component of angular momentum will be $m$, which is one of the $2l+1$ possibilities between $+l$ and $-l$. Let’s say $m=1$. With what amplitude will the electron be found on the $z$-axis at some distance $r$? Zero. An electron on the $z$-axis cannot have any orbital angular momentum around that axis. Alright, suppose $m$ is zero, then there can be some nonzero amplitude to find the electron at each distance from the proton. We’ll call this amplitude $F_l(r)$. It is the amplitude to find the electron at the distance $r$ up along the $z$-axis, when the atom is in the state $\ket{l,0}$, by which we mean orbital spin $l$ and $z$-component $m=0$. If we know $F_l(r)$ everything is known. For any state $\ket{l,m}$, we know the amplitude $\psi_{l,m}(\FLPr)$ to find the electron anywhere in the atom. How? Watch. Suppose we have the atom in the state $\ket{l,m}$, what is the amplitude to find the electron at the angle $\theta,\phi$ and the distance $r$ from the origin? Put a new $z$-axis, say $z'$, at that angle (see Fig. 19–3), and ask what is the amplitude that the electron will be at the distance $r$ along the new axis $z'$? We know that it cannot be found along $z'$ unless its $z'$-component of angular momentum, say $m'$, is zero. When $m'$ is zero, however, the amplitude to find the electron along $z'$ is $F_l(r)$. Therefore, the result is the product of two factors. The first is the amplitude that an atom in the state $\ket{l,m}$ along the $z$-axis will be in the state $\ket{l,m'=0}$ with respect to the $z'$-axis. Multiply that amplitude by $F_l(r)$ and you have the amplitude $\psi_{l,m}(\FLPr)$ to find the electron at $(r,\theta,\phi)$ with respect to the original axes. Let’s write it out. We have worked out earlier the transformation matrices for rotations. To go from the frame $x,y,z$ to the frame $x',y',z'$ of Fig. 19–3, we can rotate first around the $z$-axis by the angle $\phi$, and then rotate about the new $y$-axis ($y'$) by the angle $\theta$. This combined rotation is the product \begin{equation*} R_y(\theta)R_z(\phi). \end{equation*} The amplitude to find the state $l,m'=0$ after the rotation is \begin{equation} \label{Eq:III:19:31} \bracket{l,0}{R_y(\theta)R_z(\phi)}{l,m}. \end{equation} Our result, then, is \begin{equation} \label{Eq:III:19:32} \psi_{l,m}(\FLPr)=\bracket{l,0}{R_y(\theta)R_z(\phi)}{l,m}F_l(r). \end{equation} The orbital motion can have only integral values of $l$. (If the electron can be found anywhere at $r\neq0$, there is some amplitude to have $m=0$ in that direction. And $m=0$ states exist only for integral spins.) The rotation matrices for $l=1$ are given in Table 17–2. For larger $l$ you can use the general formulas we worked out in Chapter 18. The matrices for $R_z(\phi)$ and $R_y(\theta)$ appear separately, but you know how to combine them. For the general case you would start with the state $\ket{l,m}$ and operate with $R_z(\phi)$ to get the new state $R_z(\phi)\,\ket{l,m}$ (which is just $e^{im\phi}\,\ket{l,m}$). Then you operate on this state with $R_y(\theta)$ to get the state $R_y(\theta)R_z(\phi)\,\ket{l,m}$. Multiplying by $\bra{l,0}$ gives the matrix element (19.31). The matrix elements of the rotation operation are functions of $\theta$ and $\phi$. The particular functions which appear in (19.31) also show up in many kinds of problems which involve waves in spherical geometries and so has been given a special name. Not everyone uses the same convention; but one of the most common ones is \begin{equation} \label{Eq:III:19:33} \bracket{l,0}{R_y(\theta)R_z(\phi)}{l,m}\equiv a\,Y_{l,m}(\theta,\phi). \end{equation} The functions $Y_{l,m}(\theta,\phi)$ are called the spherical harmonics, and $a$ is just a numerical factor which depends on the definition chosen for $Y_{l,m}$. For the usual definition, \begin{equation} \label{Eq:III:19:34} a=\sqrt{\frac{4\pi}{2l+1}}. \end{equation} With this notation, the hydrogen wave functions can be written \begin{equation} \label{Eq:III:19:35} \psi_{l,m}(\FLPr)=a\,Y_{l,m}(\theta,\phi)F_l(r). \end{equation} The angle functions $Y_{l,m}(\theta,\phi)$ are important not only in many quantum-mechanical problems, but also in many areas of classical physics in which the $\nabla^2$ operator appears, such as electromagnetism. As another example of their use in quantum mechanics, consider the disintegration of an excited state of Ne$^{20}$ (such as we discussed in the last chapter) which decays by emitting an $\alpha$-particle and going into O$^{16}$: \begin{equation*} \text{Ne}^{20*}\to\text{O}^{16}+\text{He}^4. \end{equation*} Suppose that the excited state has some spin $l$ (necessarily an integer) and that the $z$-component of angular momentum is $m$. We might now ask the following: given $l$ and $m$, what is the amplitude that we will find the $\alpha$-particle going off in a direction which makes the angle $\theta$ with respect to the $z$-axis and the angle $\phi$ with respect to the $xz$-plane—as shown in Fig. 19–4. To solve this problem we make, first, the following observation. A decay in which the $\alpha$-particle goes straight up along $z$ must come from a state with $m=0$. This is so because both O$^{16}$ and the $\alpha$-particle have spin zero, and because their motion cannot have any angular momentum about the $z$-axis. Let’s call this amplitude $a$ (per unit solid angle). Then, to find the amplitude for a decay at the arbitrary angle of Fig. 19–4, all we need to know is what amplitude the given initial state has zero angular momentum about the decay direction. The amplitude for the decay at $\theta$ and $\phi$ is then $a$ times the amplitude that a state $\ket{l,m}$ with respect to the $z$-axis will be in the state $\ket{l,0}$ with respect to $z'$—the decay direction. This latter amplitude is just what we have written in (19.31). The probability to see the $\alpha$-particle at $\theta,\phi$ is \begin{equation*} P(\theta,\phi)=a^2\abs{\bracket{l,0}{R_y(\theta)R_z(\phi)}{l,m}}^2. \end{equation*} As an example, consider an initial state with $l=1$ and various values of $m$. From Table 17–2 we know the necessary amplitudes. They are \begin{equation} \begin{aligned} \bracket{1,0}{R_y(\theta)R_z(\phi)}{1,+1}&= -\frac{1}{\sqrt{2}}\sin\theta e^{i\phi},\\[1ex] \bracket{1,0}{R_y(\theta)R_z(\phi)}{1,0}&=\cos\theta,\\[1ex] \bracket{1,0}{R_y(\theta)R_z(\phi)}{1,-1}&= \frac{1}{\sqrt{2}}\sin\theta e^{-i\phi}. \end{aligned} \label{Eq:III:19:36} \end{equation} These are the three possible angular distribution amplitudes—depending on the $m$-value of the initial nucleus. Amplitudes such as the ones in (19.36) appear so often and are sufficiently important that they are given several names. If the angular distribution amplitude is proportional to any one of the three functions or any linear combination of them, we say, “The system has an orbital angular momentum of one.” Or we may say, “The Ne$^{20*}$ emits a $p$-wave $\alpha$-particle.” Or we say, “The $\alpha$-particle is emitted in an $l=1$ state.” Because there are so many ways of saying the same thing it is useful to have a dictionary. If you are going to understand what other physicists are talking about, you will just have to memorize the language. In Table 19–1 we give a dictionary of orbital angular momentum. ($l=j=$ an integer) If the orbital angular momentum is zero, then there is no change when you rotate the coordinate system and there is no variation with angle—the “dependence” on angle is as a constant, say $1$. This is also called an “$s$-state”, and there is only one such state—as far as angular dependence is concerned. If the orbital angular momentum is $1$, then the amplitude of the angular variation may be any one of the three functions given—depending on the value of $m$—or it may be a linear combination. These are called “$p$-states,” and there are three of them. If the orbital angular momentum is $2$ then there are the five functions shown. Any linear combination is called an “$l=2$,” or a “$d$-wave” amplitude. Now you can immediately guess what the next letter is—what should come after $s$, $p$, $d$? Well, of course, $f$, $g$, $h$, and so on down the alphabet! The letters don’t mean anything. (They did once mean something—they meant “sharp” lines, “principal” lines, “diffuse” lines and “fundamental” lines of the optical spectra of atoms. But those were in the days when people did not know where the lines came from. After $f$ there were no special names, so we now just continue with $g$, $h$, and so on.) The angular functions in the table go by several names—and are sometimes defined with slightly different conventions about the numerical factors that appear out in front. Sometimes they are called “spherical harmonics,” and written as $Y_{l,m}(\theta,\phi)$. Sometimes they are written $P_l^m(\cos\theta)e^{im\phi}$, and if $m=0$, simply as $P_l(\cos\theta)$. The functions $P_l(\cos\theta)$ are called the “Legendre polynomials” in $\cos\theta$, and the functions $P_l^m(\cos\theta)$ are called the “associated Legendre functions.” You will find tables of these functions in many books. Notice, incidentally, that all the functions for a given $l$ have the property that they have the same parity—for odd $l$ they change sign under an inversion and for even $l$ they don’t change. So we can write that the parity of a state of orbital angular momentum $l$ is $(-1)^l$. As we have seen, these angular distributions may refer to a nuclear disintegration or some other process, or to the distribution of the amplitude to find an electron at some place in the hydrogen atom. For instance, if an electron is in a $p$-state ($l=1$) the amplitude to find it can depend on the angle in many possible ways—but all are linear combinations of the three functions for $l=1$ in Table 19–1. Let’s take the case $\cos\theta$. That’s interesting. That means that the amplitude is positive, say, in the upper part ($\theta < \pi/2$), is negative in the lower part ($\theta > \pi/2$), and is zero when $\theta$ is $90^\circ$. Squaring this amplitude we see that the probability of finding the electron varies with $\theta$ as shown in Fig. 19–5—and is independent of $\phi$. This angular distribution is responsible for the fact that in molecular binding the attraction of an electron in an $l=1$ state for another atom depends on direction—it is the origin of the directed valences of chemical attraction.

3

19

The Hydrogen Atom and The Periodic Table

4

The general solution for hydrogen

In Eq. (19.35) we have written the wave functions for the hydrogen atom as \begin{equation} \label{Eq:III:19:37} \psi_{l,m}(\FLPr)=a\,Y_{l,m}(\theta,\phi)F_l(r). \end{equation} These wave functions must be solutions of the differential equation (19.7). Let’s see what that means. Put (19.37) into (19.7); you get \begin{align} \frac{Y_{l,m}}{r}\,\frac{\partial^2}{\partial r^2}\, (rF_l)+\frac{F_l}{r^2\sin\theta}\,\ddp{}{\theta}&\biggl( \sin\theta\,\ddp{Y_{l,m}}{\theta}\biggr)+ \frac{F_l}{r^2\sin^2\theta}\,\frac{\partial^2Y_{l,m}} {\partial\phi^2}\notag\\[1ex] \label{Eq:III:19:38} &=\;-\frac{2m}{\hbar^2}\biggl(E+\frac{e^2}{r}\biggr) Y_{l,m}F_l. \end{align} \begin{align} \frac{Y_{l,m}}{r}\frac{\partial^2}{\partial r^2} (rF_l)\!&+\!\frac{F_l}{r^2\sin\theta}\ddp{}{\theta}\biggl(\! \sin\theta\ddp{Y_{l,m}}{\theta}\!\biggr)\notag\\[1ex] \label{Eq:III:19:38} &+\!\frac{F_l}{r^2\sin^2\theta}\frac{\partial^2Y_{l,m}} {\partial\phi^2}\\[1.5ex] &=-\frac{2m}{\hbar^2}\biggl(\!E+\!\frac{e^2}{r}\!\biggr) Y_{l,m}F_l.\notag \end{align} Now multiply through by $r^2/F_l$ and rearrange terms. The result is \begin{align} \frac{1}{\sin\theta}\,\ddp{}{\theta}&\biggl( \sin\theta\,\ddp{Y_{l,m}}{\theta}\biggr)+ \frac{1}{\sin^2\theta}\,\frac{\partial^2Y_{l,m}} {\partial\phi^2}\notag\\[1ex] \label{Eq:III:19:39} &=\;-\biggl[ \frac{r^2}{F_l}\biggl\{ \frac{1}{r}\,\frac{d^2}{dr^2}\,(rF_l)+\frac{2m}{\hbar^2} \biggl(E+\frac{e^2}{r}\biggr)F_l\biggr\}\biggr]Y_{l,m}. \end{align} \begin{equation} \begin{gathered} \frac{1}{\sin\theta}\ddp{}{\theta}\biggl(\! \sin\theta\ddp{Y_{l,m}}{\theta}\!\biggr)\!+\! \frac{1}{\sin^2\theta}\frac{\partial^2Y_{l,m}} {\partial\phi^2}\\[2ex] =-\biggl[ \frac{r^2}{F_l}\biggl\{\! \frac{1}{r}\frac{d^2}{dr^2}(rF_l)\!+\!\frac{2m}{\hbar^2} \biggl(\!E+\frac{e^2}{r}\!\biggr)F_l\!\biggr\}\biggr]Y_{l,m}. \end{gathered} \label{Eq:III:19:39} \end{equation} The left-hand side of this equation depends on $\theta$ and $\phi$, but not on $r$. No matter what value we choose for $r$, the left side doesn’t change. This must also be true for the right-hand side. Although the quantity in the square brackets has $r$’s all over the place, the whole quantity cannot depend on $r$, otherwise we wouldn’t have an equation good for all $r$. As you can see, the bracket also does not depend on $\theta$ or $\phi$. It must be some constant. Its value may well depend on the $l$-value of the state we are studying, since the function $F_l$ must be the one appropriate to that state; we’ll call the constant $K_l$. Equation (19.39) is therefore equivalent to two equations: \begin{gather} \label{Eq:III:19:40} \frac{1}{\sin\theta}\,\ddp{}{\theta}\biggl( \sin\theta\,\ddp{Y_{l,m}}{\theta}\biggr)+ \frac{1}{\sin^2\theta}\,\frac{\partial^2Y_{l,m}} {\partial\phi^2}=-K_lY_{l,m},\\ \notag\\[2pt] \label{Eq:III:19:41} \frac{1}{r}\,\frac{d^2}{dr^2}\,(rF_l)+ \frac{2m}{\hbar^2}\biggl(E+\frac{e^2}{r}\biggr)F_l= K_l\,\frac{F_l}{r^2}. \end{gather} \begin{align} \frac{1}{\sin\theta}\ddp{}{\theta}\biggl(\! \sin\theta\ddp{Y_{l,m}}{\theta}\!\biggr)\!+\! \frac{1}{\sin^2\theta}\frac{\partial^2Y_{l,m}} {\partial\phi^2}&\notag\\[1.5ex] \label{Eq:III:19:40} =-K_lY_{l,m},&\\[2.5ex] \frac{1}{r}\frac{d^2}{dr^2}(rF_l)+ \frac{2m}{\hbar^2}\biggl(\!E+\!\frac{e^2}{r}\!\biggr)F_l&\notag\\[.5ex] \label{Eq:III:19:41} =K_l\frac{F_l}{r^2}.& \end{align} Now look at what we’ve done. For any state described by $l$ and $m$, we know the functions $Y_{l,m}$; we can use Eq. (19.40) to determine the constant $K_l$. Putting $K_l$ into Eq. (19.41) we have a differential equation for the function $F_l(r)$. If we can solve that equation for $F_l(r)$, we have all of the pieces to put into (19.37) to give $\psi(\FLPr)$. What is $K_l$? First, notice that it must be the same for all $m$ (which go with a particular $l$), so we can pick any $m$ we want for $Y_{l,m}$ and plug it into (19.40) to solve for $K_l$. Perhaps the easiest one to use is $Y_{l,l}$. From Eq. (18.24), \begin{equation} \label{Eq:III:19:42} R_z(\phi)\,\ket{l,l}=e^{il\phi}\,\ket{l,l}. \end{equation} The matrix element for $R_y(\theta)$ is also quite simple: \begin{equation} \label{Eq:III:19:43} \bracket{l,0}{R_y(\theta)}{l,l}=b\,(\sin\theta)^l, \end{equation} where $b$ is some number.3 Combining the two, we obtain \begin{equation} \label{Eq:III:19:44} Y_{l,l}\propto e^{il\phi}\sin^l\theta. \end{equation} Putting this function into (19.40) gives \begin{equation} \label{Eq:III:19:45} K_l=l(l+1). \end{equation} Now that we have determined $K_l$, Eq. (19.41) tells us about the radial function $F_l(r)$. It is, of course, just the Schrödinger equation with the angular part replaced by its equivalent $K_lF_l/r^2$. Let’s rewrite (19.41) in the form we had in Eq. (19.8), as follows: \begin{equation} \label{Eq:III:19:46} \frac{1}{r}\frac{d^2}{dr^2}(rF_l)= -\frac{2m}{\hbar^2}\biggl\{\! E\!+\!\frac{e^2}{r}\!-\!\frac{l(l\!+\!1)\hbar^2}{2mr^2} \!\biggr\}F_l. \end{equation} A mysterious term has been added to the potential energy. Although we got this term by some mathematical shenanigan, it has a simple physical origin. We can give you an idea about where it comes from in terms of a semi-classical argument. Then perhaps you will not find it quite so mysterious. Think of a classical particle moving around some center of force. The total energy is conserved and is the sum of the potential and kinetic energies \begin{equation*} U = V(r) + \tfrac{1}{2}mv^2 = \text{constant}. \end{equation*} In general, $v$ can be resolved into a radial component $v_r$ and a tangential component $r\dot{\theta}$; then \begin{equation*} v^2=v_r^2+(r\dot{\theta})^2. \end{equation*} Now the angular momentum $mr^2\dot{\theta}$ is also conserved; say it is equal to $L$. We can then write \begin{equation*} mr^2\dot{\theta}=L,\quad \text{or}\quad r\dot{\theta}=\frac{L}{mr}, \end{equation*} and the energy is \begin{equation*} U = \tfrac{1}{2}mv_r^2+V(r)+\frac{L^2}{2mr^2}. \end{equation*} If there were no angular momentum we would have just the first two terms. Adding the angular momentum $L$ does to the energy just what adding a term $L^2/2mr^2$ to the potential energy would do. But this is almost exactly the extra term in (19.46). The only difference is that $l(l+1)\hbar^2$ appears for the angular momentum instead of $l^2\hbar^2$ as we might expect. But we have seen before (for example, Volume II, Section 34-7) that this is just the substitution that is usually required to make a quasi-classical argument agree with a correct quantum-mechanical calculation. We can, then, understand the new term as a “pseudo-potential” which gives the “centrifugal force” term that appears in the equations of radial motion for a rotating system. (See the discussion of “pseudo-forces” in Volume I, Section 12-5.) We are now ready to solve Eq. (19.46) for $F_l(r)$. It is very much like Eq. (19.8), so the same technique will work again. Everything goes as before until you get to Eq. (19.19) which will have the additional term \begin{equation} \label{Eq:III:19:47} -l(l+1)\sum_{k=1}^\infty a_k\rho^{k-2}. \end{equation} This term can also be written as \begin{equation} \label{Eq:III:19:48} -l(l+1)\biggl\{\frac{a_1}{\rho}+ \sum_{k=1}^\infty a_{k+1}\rho^{k-1}\biggr\}. \end{equation} (We have taken out the first term and then shifted the running index $k$ down by $1$.) Instead of Eq. (19.20) we have \begin{equation} \begin{aligned} \sum_{k=1}^\infty&[\{k(k+1)-l(l+1)\}a_{k+1}-2(\alpha k-1)a_k]\rho^{k-1}\\ &-\frac{l(l+1)a_1}{\rho}=0. \end{aligned} \label{Eq:III:19:49} \end{equation} There is only one term in $\rho^{-1}$, so it must be zero. The coefficient $a_1$ must be zero (unless $l=0$ and we have our previous solution). Each of the other terms is made zero by having the square bracket come out zero for every $k$. This condition replaces Eq. (19.22) by \begin{equation} \label{Eq:III:19:50} a_{k+1}=\frac{2(\alpha k-1)}{k(k+1)-l(l+1)}\,a_k. \end{equation} This is the only significant change from the spherically symmetric case. As before the series must terminate if we are to have solutions which can represent bound electrons. The series will end at $k=n$ if $\alpha n=1$. We get again the same condition on $\alpha$, that it must be equal to $1/n$, where $n$ is some positive integer. However, Eq. (19.50) also gives a new restriction. The index $k$ cannot be equal to $l$, the denominator becomes zero and $a_{l+1}$ is infinite. That is, since $a_1=0$, Eq. (19.50) implies that all successive $a_k$ are zero until we get to $a_{l+1}$, which can be nonzero. This means that $k$ must start at $l+1$ and end at $n$. Our final result is that for any $l$ there are many possible solutions which we can call $F_{n,l}$ where $n\geq l+1$. Each solution has the energy \begin{equation} \label{Eq:III:19:51} E_n=-\frac{me^4}{2\hbar^2}\biggl(\frac{1}{n^2}\biggr). \end{equation} The wave function for the state of this energy with the angular quantum numbers $l$ and $m$ is \begin{equation} \label{Eq:III:19:52} \psi_{n,l,m}=a\,Y_{l,m}(\theta,\phi)F_{n,l}(\rho), \end{equation} with \begin{equation} \label{Eq:III:19:53} \rho F_{n,l}(\rho)=e^{-\alpha\rho}\sum_{k=l+1}^na_k\rho^k. \end{equation} The coefficients $a_k$ are obtained from (19.50). We have, finally, a complete description of the states of a hydrogen atom.

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The Hydrogen Atom and The Periodic Table

5

The hydrogen wave functions

Let’s review what we have discovered. The states which satisfy Schrödinger’s equation for an electron in a Coulomb field are characterized by three quantum numbers $n$, $l$, $m$, all integers. The angular distribution of the electron amplitude can have only certain forms which we call $Y_{l,m}$. They are labeled by $l$, the quantum number of total angular momentum, and $m$, the “magnetic” quantum number, which can range from $-l$ to $+l$. For each angular configuration, various possible radial distributions $F_{n,l}(r)$ of the electron amplitude are possible; they are labeled by the principal quantum number $n$—which can range from $l+1$ to $\infty$. The energy of the state depends only on $n$, and increases with increasing $n$. The lowest energy, or ground, state is an $s$-state. It has $l=0$, $n=1$, and $m=0$. It is a “nondegenerate” state—there is only one with this energy, and its wave function is spherically symmetric. The amplitude to find the electron is a maximum at the center, and falls off monotonically with increasing distance from the center. We can visualize the electron amplitude as a blob as shown in Fig. 19–6(a). There are other $s$-states with higher energies, for $n=2$, $3$, $4$, … For each energy there is only one version ($m=0$), and they are all spherically symmetric. These states have amplitudes which alternate in sign one or more times with increasing $r$. There are $n-1$ spherical nodal surfaces—the places where $\psi$ goes through zero. The $2s$-state ($l=0$, $n=2$), for example, will look as sketched in Fig. 19–6(b). (The dark areas indicate regions where the amplitude is large, and the plus and minus signs indicate the relative phases of the amplitude.) The energy levels of the $s$-states are shown in the first column of Fig. 19–7. Then there are the $p$-states—with $l=1$. For each $n$, which must be $2$ or greater, there are three states of the same energy, one each for $m=+1$, $m=0$, and $m=-1$. The energy levels are as shown in Fig. 19–7. The angular dependences of these states are given in Table 19–1. For instance, for $m=0$, if the amplitude is positive for $\theta$ near zero, it will be negative for $\theta$ near $180^\circ$. There is a nodal plane coincident with the $xy$-plane. For $n>2$ there are also spherical nodes. The $n=2$, $m=0$ amplitude is sketched in Fig. 19–6(c), and the $n=3$, $m=0$ wave function is sketched in Fig. 19–6(d). You might think that since $m$ represents a kind of “orientation” in space, there should be similar distributions with the peaks of amplitude along the $x$-axis or along the $y$-axis. Are these perhaps the $m=+1$ and $m=-1$ states? No. But since we have three states with equal energies, any linear combinations of the three will also be stationary states of the same energy. It turns out that the “$x$”-state—which corresponds to the “$z$”-state, or $m=0$ state, of Fig. 19–6(c)—is a linear combination of the $m=+1$ and $m=-1$ states. The corresponding “$y$”-state is another combination. Specifically, we mean that \begin{align*} \text{“$z$”}&=\ket{1,0},\\[1ex] \text{“$x$”}&=-\frac{\ket{1,+1}-\ket{1,-1}}{\sqrt{2}},\\[1ex] \text{“$y$”}&=-\frac{\ket{1,+1}+\ket{1,-1}}{i\sqrt{2}}. \end{align*} These states all look the same when referred to their particular axes. The $d$-states ($l=2$) have five possible values of $m$ for each energy, the lowest energy has $n=3$. The levels go as shown in Fig. 19–7. The angular dependences get more complicated. For instance the $m=0$ states have two conical nodes, so the wave function reverses phase from $+$, to $-$, to $+$ as you go around from the north pole to the south pole. The rough form of the amplitude is sketched in (e) and (f) of Fig. 19–6 for the $m=0$ states with $n=3$ and $n=4$. Again, the larger $n$’s have spherical nodes. We will not try to describe any more of the possible states. You will find the hydrogen wave functions described in more detail in many books. Two good references are L. Pauling and E. B. Wilson, Introduction to Quantum Mechanics, McGraw-Hill (1935); and R. B. Leighton, Principles of Modern Physics, McGraw-Hill (1959). You will find in them graphs of some of the functions and pictorial representations of many states. We would like to mention one particular feature of the wave functions for higher $l$: for $l>0$ the amplitudes are zero at the center. That is not surprising, since it’s hard for an electron to have angular momentum when its radius arm is very small. For this reason, the higher the $l$, the more the amplitudes are “pushed away” from the center. If you look at the way the radial functions $F_{n,l}(r)$ vary for small $r$, you find from (19.53) that \begin{equation*} F_{n,l}(r)\approx r^l. \end{equation*} Such a dependence on $r$ means that for larger $l$’s you have to go farther from $r=0$ before you get an appreciable amplitude. This behavior is, incidentally, determined by the centrifugal force term in the radial equation, so the same thing will apply for any potential that varies slower than $1/r^2$ for small $r$—which most atomic potentials do.

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The Hydrogen Atom and The Periodic Table

6

The periodic table

We would like now to apply the theory of the hydrogen atom in an approximate way to get some understanding of the chemist’s periodic table of the elements. For an element with atomic number $Z$ there are $Z$ electrons held together by the electric attraction of the nucleus but with mutual repulsion of the electrons. To get an exact solution we would have to solve Schrödinger’s equation for $Z$ electrons in a Coulomb field. For helium the equation is \begin{equation*} -\frac{\hbar}{i}\ddp{\psi}{t}=-\frac{\hbar^2}{2m} (\nabla_1^2\psi\!+\!\nabla_2^2\psi)\!+\!\biggl(\! -\!\frac{2e^2}{r_1}\!-\!\frac{2e^2}{r_2}\!+\!\frac{e^2}{r_{12}}\!\biggr)\psi, \end{equation*} where $\nabla_1^2$ is a Laplacian which operates on $\FLPr_1$, the coordinate of one electron; $\nabla_2^2$ operates on $\FLPr_2$; and $r_{12}=\abs{\FLPr_1-\FLPr_2}$. (We are again neglecting the spin of the electrons.) To find the stationary states and energy levels we would have to find solutions of the form \begin{equation*} \psi=f(\FLPr_1,\FLPr_2)e^{-(i/\hbar)Et}. \end{equation*} The geometrical dependence is contained in $f$, which is a function of six variables—the simultaneous positions of the two electrons. No one has found an analytic solution, although solutions for the lowest energy states have been obtained by numerical methods. With $3$, $4$, or $5$ electrons it is hopeless to try to obtain exact solutions, and it is going too far to say that quantum mechanics has given a precise understanding of the periodic table. It is possible, however, even with a sloppy approximation—and some fixing—to understand, at least qualitatively, many chemical properties which show up in the periodic table. The chemical properties of atoms are determined primarily by their lowest energy states. We can use the following approximate theory to find these states and their energies. First, we neglect the electron spin, except that we adopt the exclusion principle and say that any particular electronic state can be occupied by only one electron. This means that any particular orbital configuration can have up to two electrons—one with spin up, the other with spin down. Next we disregard the details of the interactions between the electrons in our first approximation, and say that each electron moves in a central field which is the combined field of the nucleus and all the other electrons. For neon, which has $10$ electrons, we say that one electron sees an average potential due to the nucleus plus the other nine electrons. We imagine then that in the Schrödinger equation for each electron we put a $V(r)$ which is a $1/r$ field modified by a spherically symmetric charge density coming from the other electrons. In this model each electron acts like an independent particle. The angular dependence of its wave function will be just the same as the ones we had for the hydrogen atom. There will be $s$-states, $p$-states, and so on; and they will have the various possible $m$-values. Since $V(r)$ no longer goes as $1/r$, the radial part of the wave functions will be somewhat different, but it will be qualitatively the same, so we will have the same radial quantum numbers, $n$. The energies of the states will also be somewhat different. With these ideas, let’s see what we get. The ground state of hydrogen has $l=m=0$ and $n=1$; we say the electron configuration is $1s$. The energy is $-13.6$ eV. This means that it takes $13.6$ electron volts to pull the electron off the atom. We call this the “ionization energy”, $W_I$. A large ionization energy means that it is harder to pull the electron off and, in general, that the material is chemically less active. Now take helium. Both electrons can be in the same lowest state (one spin up and the other spin down). In this lowest state the electron moves in a potential which is for small $r$ like a Coulomb field for $z=2$ and for large $r$ like a Coulomb field for $z=1$. The result is a “hydrogen-like” $1s$ state with a somewhat lower energy. Both electrons occupy identical $1s$ states ($l=0$, $m=0$). The observed ionization energy (to remove one electron) is $24.6$ electron volts. Since the $1s$ “shell” is now filled—we allow only two electrons—there is practically no tendency for an electron to be attracted from another atom. Helium is chemically inert. The lithium nucleus has a charge of $3$. The electron states will again be hydrogen-like, and the three electrons will occupy the lowest three energy levels. Two will go into $1s$ states and the third will go into an $n=2$ state. But with $l=0$ or $l=1$? In hydrogen these states have the same energy, but in other atoms they don’t, for the following reason. Remember that a $2s$ state has some amplitude to be near the nucleus while the $2p$ state does not. That means that a $2s$ electron will feel some of the triple electric charge of the Li nucleus, but that a $2p$ electron will stay out where the field looks like the Coulomb field of a single charge. The extra attraction lowers the energy of the $2s$ state relative to the $2p$ state. The energy levels will be roughly as shown in Fig. 19–8—which you should compare with the corresponding diagram for hydrogen in Fig. 19–7. So the lithium atom will have two electrons in $1s$ states and one in a $2s$. Since the $2s$ electron has a higher energy than a $1s$ electron it is relatively easily removed. The ionization energy of lithium is only $5.4$ electron volts, and it is quite active chemically. So you can see the patterns which develop; we have given in Table 19–2 a list of the first $36$ elements, showing the states occupied by the electrons in the ground state of each atom. The Table gives the ionization energy for the most loosely bound electron, and the number of electrons occupying each “shell”—by which we mean states with the same $n$. Since the different $l$-states have different energies, each $l$-value corresponds to a sub-shell of $2(2l+1)$ possible states (of different $m$ and electron spin). These all have the same energy—except for some very small effects we are neglecting. Beryllium is like lithium except that it has two electrons in the $2s$ state as well as two in the filled $1s$ shell. Boron has $5$ electrons. The fifth must go into a $2p$ state. There are $2\times3=6$ different $2p$ states, so we can keep adding electrons until we get to a total of $8$. This takes us to neon. As we add these electrons we are also increasing $Z$, so the whole electron distribution gets pulled in closer and closer to the nucleus and the energy of the $2p$ states goes down. By the time we get to neon the ionization energy is up to $21.6$ electron volts. Neon does not easily give up an electron. Also there are no more low-energy slots to be filled, so it won’t try to grab an extra electron. Neon is chemically inert. Fluorine, on the other hand, does have an empty position where an electron can drop into a state of low energy, so it is quite active in chemical reactions. With sodium the eleventh electron must start a new shell—going into a $3s$ state. The energy level of this state is much higher; the ionization energy jumps down; and sodium is an active chemical. From sodium to argon the $s$ and $p$ states with $n=3$ are occupied in exactly the same sequence as for lithium to neon. Angular configurations of the electrons in the outer unfilled shell have the same sequence, and the progression of ionization energies is quite similar. You can see why the chemical properties repeat with increasing atomic number. Magnesium acts chemically much like beryllium, silicon like carbon, and chlorine like fluorine. Argon is inert like neon. You may have noticed that there is a slight peculiarity in the sequence of ionization energies between lithium and neon, and a similar one between sodium and argon. The last electron is bound to the oxygen atom somewhat less than we might expect. And sulfur is similar. Why should that be? We can understand it if we put in just a little bit of the effects of the interactions between individual electrons. Think of what happens when we put the first $2p$ electron onto the boron atom. It has six possibilities—three possible $p$-states, each with two spins. Imagine that the electron goes with spin up into the $m=0$ state, which we have also called the “$z$” state because it hugs the $z$-axis. Now what will happen in carbon? There are now two $2p$ electrons. If one of them goes into the “$z$” state, where will the second one go? It will have lower energy if it stays away from the first electron, which it can do by going into, say, the “$x$” state of the $2p$ shell. (This state is, remember, just a linear combination of the $m=+1$ and $m=-1$ states.) Next, when we go to nitrogen, the three $2p$ electrons will have the lowest energy of mutual repulsion if they go one each into the “$x$,” “$y$,” and “$z$” configurations. For oxygen, however, the jig is up. The fourth electron must go into one of the filled states—with opposite spin. It is strongly repelled by the electron already in that state, so its energy will not be as low as it might otherwise be, and it is more easily removed. That explains the break in the sequence of binding energies which appears between nitrogen and oxygen, and between phosphorus and sulfur. After argon, you would, at first, think that the new electrons would start to fill up the $3d$ states. But they don’t. As we described earlier—and illustrated in Fig. 19–8—the higher angular momentum states get pushed up in energy. By the time we get to the $3d$ states they are pushed to an energy a little bit above the energy of the $4s$ state. So in potassium the last electron goes into the $4s$ state. After this shell is filled (with two electrons) at calcium, the $3d$ states begin to be filled for scandium, titanium, and vanadium. The energies of the $3d$ and $4s$ states are so close together that small effects can shift the balance either way. By the time we get to put four electrons into the $3d$ states, their repulsion raises the energy of the $4s$ state just enough that its energy is slightly above the $3d$ energy, so one electron shifts over. For chromium we don’t get a $4$, $2$ combination as we would have expected, but instead a $5$, $1$ combination. The new electron added to get manganese fills up the $4s$ shell again, and the states of the $3d$ shell are then occupied one by one until we reach copper. Since the outermost shell of manganese, iron, cobalt, and nickel have the same configurations, however, they all tend to have similar chemical properties. (This effect is much more pronounced in the rare-earth elements which all have the same outer shell but a progressively filling inner shell which has much less influence on their chemical properties.) In copper an electron is robbed from the $4s$ shell, finally completing the $3d$ shell. The energy of the $10$, $1$ combination is, however, so close to the $9$, $2$ configuration for copper that just the presence of another atom nearby can shift the balance. For this reason the two last electrons of copper are nearly equivalent, and copper can have a valence of either $1$ or $2$. (It sometimes acts as though its electrons were in the $9$, $2$ combination.) Similar things happen at other places and account for the fact that other metals, such as iron, combine chemically with either of two valences. By zinc, both the $3d$ and $4s$ shells are filled once and for all. From gallium to krypton the sequence proceeds normally again, filling the $4p$ shell. The outer shells, the energies, and the chemical properties repeat the pattern of boron to neon and aluminum to argon. Krypton, like argon and neon, is known as “noble” gas. All three are chemically “inert.” This means only that, having filled shells of relatively low energy, there are few situations in which there is an energy advantage for them to join in a simple combination with other elements. Having a filled shell is not enough. Beryllium and magnesium have filled $s$-shells, but the energy of these shells is too high to lead to stability. Similarly, one would have expected another “noble” element at nickel, if the energy of the $3d$ shell had been lower (or the $4s$ higher). On the other hand, krypton is not completely inert; it will form a weakly-bound compound with chlorine. Since our sample has turned up most of the main features of the periodic table, we stop our examination at element number $36$—there are still seventy or so more! We would like to bring up only one more point—that we not only can understand the valences to some extent but also can say something about the directional properties of the chemical bonds. Take an atom like oxygen which has four $2p$ electrons. The first three go into “$x$,” “$y$,” and “$z$” states and the fourth will double one of these states, leaving two—say “$x$” and “$y$”—vacant. Consider then what happens in H$_2$O. Each of the two hydrogens are willing to share an electron with the oxygen, helping the oxygen to fill a shell. These electrons will tend to go into the “$x$” and “$y$” vacancies. So the water molecule should have the two hydrogen atoms making a right angle with respect to the center of the oxygen. The angle is actually $105^\circ$. We can even understand why the angle is larger than $90^\circ$. In sharing their electrons the hydrogens end up with a net positive charge. The electric repulsion “strains” the wave functions and pushes the angle out to $105^\circ$. The same situation occurs in H$_2$S. But because the sulfur atom is larger, the two hydrogen atoms are farther apart, there is less repulsion, and the angle is only pushed out to about $93^\circ$. Selenium is even larger, so in H$_2$Se the angle is very nearly $90^\circ$. We can use the same arguments to understand the geometry of ammonia, H$_3$N. Nitrogen has room for three more $2p$ electrons, one each for the “$x$,” “$y$,” and “$z$” type states. The three hydrogens should join on at right angles to each other. The angles come out a little larger than $90^\circ$—again from the electric repulsion—but at least we see why the molecule of H$_3$N is not flat. The angles in phosphene, H$_3$P, are close to $90^\circ$, and in H$_3$As are still closer. We assumed that NH$_3$ was not flat when we described it as a two-state system. And the nonflatness is what makes the ammonia maser possible. Now we see that also that shape can be understood from our quantum mechanics. The Schrödinger equation has been one of the great triumphs of physics. By providing the key to the underlying machinery of atomic structure it has given an explanation for atomic spectra, for chemistry, and for the nature of matter.

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20

Operators

1

Operations and operators

All the things we have done so far in quantum mechanics could be handled with ordinary algebra, although we did from time to time show you some special ways of writing quantum-mechanical quantities and equations. We would like now to talk some more about some interesting and useful mathematical ways of describing quantum-mechanical things. There are many ways of approaching the subject of quantum mechanics, and most books use a different approach from the one we have taken. As you go on to read other books you might not see right away the connections of what you will find in them to what we have been doing. Although we will also be able to get a few useful results, the main purpose of this chapter is to tell you about some of the different ways of writing the same physics. Knowing them you should be able to understand better what other people are saying. When people were first working out classical mechanics they always wrote all the equations in terms of $x$-, $y$-, and $z$-components. Then someone came along and pointed out that all of the writing could be made much simpler by inventing the vector notation. It’s true that when you come down to figuring something out you often have to convert the vectors back to their components. But it’s generally much easier to see what’s going on when you work with vectors and also easier to do many of the calculations. In quantum mechanics we were able to write many things in a simpler way by using the idea of the “state vector.” The state vector $\ket{\psi}$ has, of course, nothing to do with geometric vectors in three dimensions but is an abstract symbol that stands for a physical state, identified by the “label,” or “name,” $\psi$. The idea is useful because the laws of quantum mechanics can be written as algebraic equations in terms of these symbols. For instance, our fundamental law that any state can be made up from a linear combination of base states is written as \begin{equation} \label{Eq:III:20:1} \ket{\psi}=\sum_iC_i\,\ket{i}, \end{equation} where the $C_i$ are a set of ordinary (complex) numbers—the amplitudes $C_i=\braket{i}{\psi}$—while $\ket{1}$, $\ket{2}$, $\ket{3}$, and so on, stand for the base states in some base, or representation. If you take some physical state and do something to it—like rotating it, or like waiting for the time $\Delta t$—you get a different state. We say, “performing an operation on a state produces a new state.” We can express the same idea by an equation: \begin{equation} \label{Eq:III:20:2} \ket{\phi}=\Aop\,\ket{\psi}. \end{equation} An operation on a state produces another state. The operator $\Aop$ stands for some particular operation. When this operation is performed on any state, say $\ket{\psi}$, it produces some other state $\ket{\phi}$. What does Eq. (20.2) mean? We define it this way. If you multiply the equation by $\bra{i}$ and expand $\ket{\psi}$ according to Eq. (20.1), you get \begin{equation} \label{Eq:III:20:3} \braket{i}{\phi}=\sum_j\bracket{i}{\Aop}{j}\braket{j}{\psi}. \end{equation} (The states $\ket{j}$ are from the same set as $\ket{i}$.) This is now just an algebraic equation. The numbers $\braket{i}{\phi}$ give the amount of each base state you will find in $\ket{\phi}$, and it is given in terms of a linear superposition of the amplitudes $\braket{j}{\psi}$ that you find $\ket{\psi}$ in each base state. The numbers $\bracket{i}{\Aop}{j}$ are just the coefficients which tell how much of $\braket{j}{\psi}$ goes into each sum. The operator $\Aop$ is described numerically by the set of numbers, or “matrix,” \begin{equation} \label{Eq:III:20:4} A_{ij}\equiv\bracket{i}{\Aop}{j}. \end{equation} So Eq. (20.2) is a high-class way of writing Eq. (20.3). Actually it is a little more than that; something more is implied. In Eq. (20.2) we do not make any reference to a set of base states. Equation (20.3) is an image of Eq. (20.2) in terms of some set of base states. But, as you know, you may use any set you wish. And this idea is implied in Eq. (20.2). The operator way of writing avoids making any particular choice. Of course, when you want to get definite you have to choose some set. When you make your choice, you use Eq. (20.3). So the operator equation (20.2) is a more abstract way of writing the algebraic equation (20.3). It’s similar to the difference between writing \begin{equation*} \FLPc=\FLPa\times\FLPb \end{equation*} instead of \begin{aligned} c_x&=a_yb_z-a_zb_y,\\[.5ex] c_y&=a_zb_x-a_xb_z,\\[.5ex] c_z&=a_xb_y-a_yb_x. \end{aligned} The first way is much handier. When you want results, however, you will eventually have to give the components with respect to some set of axes. Similarly, if you want to be able to say what you really mean by $\Aop$, you will have to be ready to give the matrix $A_{ij}$ in terms of some set of base states. So long as you have in mind some set $\ket{i}$, Eq. (20.2) means just the same as Eq. (20.3). (You should remember also that once you know a matrix for one particular set of base states you can always calculate the corresponding matrix that goes with any other base. You can transform the matrix from one “representation” to another.) The operator equation in (20.2) also allows a new way of thinking. If we imagine some operator $\Aop$, we can use it with any state $\ket{\psi}$ to create a new state $\Aop\,\ket{\psi}$. Sometimes a “state” we get this way may be very peculiar—it may not represent any physical situation we are likely to encounter in nature. (For instance, we may get a state that is not normalized to represent one electron.) In other words, we may at times get “states” that are mathematically artificial. Such artificial “states” may still be useful, perhaps as the mid-point of some calculation. We have already shown you many examples of quantum-mechanical operators. We have had the rotation operator $\Rop_y(\theta)$ which takes a state $\ket{\psi}$ and produces a new state, which is the old state as seen in a rotated coordinate system. We have had the parity (or inversion) operator $\Pop$, which makes a new state by reversing all coordinates. We have had the operators $\sigmaop_x$, $\sigmaop_y$, and $\sigmaop_z$ for spin one-half particles. The operator $\Jop_z$ was defined in Chapter 17 in terms of the rotation operator for a small angle $\epsilon$. \begin{equation} \label{Eq:III:20:5} \Rop_z(\epsilon)=1+\frac{i}{\hbar}\,\epsilon\,\Jop_z. \end{equation} This just means, of course, that \begin{equation} \label{Eq:III:20:6} \Rop_z(\epsilon)\,\ket{\psi}=\ket{\psi}+ \frac{i}{\hbar}\,\epsilon\,\Jop_z\,\ket{\psi}. \end{equation} In this example, $\Jop_z\,\ket{\psi}$ is $\hbar/i\epsilon$ times the state you get if you rotate $\ket{\psi}$ by the small angle $\epsilon$ and then subtract the original state. It represents a “state” which is the difference of two states. One more example. We had an operator $\pop_x$—called the momentum operator ($x$-component) defined in an equation like (20.6). If $\Dop_x(L)$ is the operator which displaces a state along $x$ by the distance $L$, then $\pop_x$ is defined by \begin{equation} \label{Eq:III:20:7} \Dop_x(\delta)=1+\frac{i}{\hbar}\,\delta\pop_x, \end{equation} where $\delta$ is a small displacement. Displacing the state $\ket{\psi}$ along $x$ by a small distance $\delta$ gives a new state $\ket{\psi'}$. We are saying that this new state is the old state plus a small new piece \begin{equation*} \frac{i}{\hbar}\,\delta\pop_x\,\ket{\psi}. \end{equation*} The operators we are talking about work on a state vector like $\ket{\psi}$, which is an abstract description of a physical situation. They are quite different from algebraic operators which work on mathematical functions. For instance, $d/dx$ is an “operator” that works on $f(x)$ by changing it to a new function $f'(x)=df/dx$. Another example is the algebraic operator $\nabla^2$. You can see why the same word is used in both cases, but you should keep in mind that the two kinds of operators are different. A quantum-mechanical operator $\Aop$ does not work on an algebraic function, but on a state vector like $\ket{\psi}$. Both kinds of operators are used in quantum mechanics and often in similar kinds of equations, as you will see a little later. When you are first learning the subject it is well to keep the distinction always in mind. Later on, when you are more familiar with the subject, you will find that it is less important to keep any sharp distinction between the two kinds of operators. You will, indeed, find that most books generally use the same notation for both! We’ll go on now and look at some useful things you can do with operators. But first, one special remark. Suppose we have an operator $\Aop$ whose matrix in some base is $A_{ij}\equiv\bracket{i}{\Aop}{j}$. The amplitude that the state $\Aop\,\ket{\psi}$ is also in some other state $\ket{\phi}$ is $\bracket{\phi}{\Aop}{\psi}$. Is there some meaning to the complex conjugate of this amplitude? You should be able to show that \begin{equation} \label{Eq:III:20:8} \bracket{\phi}{\Aop}{\psi}\cconj=\bracket{\psi}{\Aop\adj}{\phi}, \end{equation} where $\Aop\adj$ (read “A dagger”) is an operator whose matrix elements are \begin{equation} \label{Eq:III:20:9} A_{ij}\adj=(A_{ji})\cconj. \end{equation} To get the $i,j$ element of $A\adj$ you go to the $j,i$ element of $A$ (the indexes are reversed) and take its complex conjugate. The amplitude that the state $\Aop\adj\,\ket{\phi}$ is in $\ket{\psi}$ is the complex conjugate of the amplitude that $\Aop\,\ket{\psi}$ is in $\ket{\phi}$. The operator $\Aop\adj$ is called the “Hermitian adjoint” of $\Aop$. Many important operators of quantum mechanics have the special property that when you take the Hermitian adjoint, you get the same operator back. If $\Bop$ is such an operator, then \begin{equation*} \Bop\adj=\Bop, \end{equation*} and it is called a “self-adjoint” or “Hermitian,” operator.

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2

Average energies

So far we have reminded you mainly of what you already know. Now we would like to discuss a new question. How would you find the average energy of a system—say, an atom? If an atom is in a particular state of definite energy and you measure the energy, you will find a certain energy $E$. If you keep repeating the measurement on each one of a whole series of atoms which are all selected to be in the same state, all the measurements will give $E$, and the “average” of your measurements will, of course, be just $E$. Now, however, what happens if you make the measurement on some state $\ket{\psi}$ which is not a stationary state? Since the system does not have a definite energy, one measurement would give one energy, the same measurement on another atom in the same state would give a different energy, and so on. What would you get for the average of a whole series of energy measurements? We can answer the question by projecting the state $\ket{\psi}$ onto the set of states of definite energy. To remind you that this is a special base set, we’ll call the states $\ket{\eta_i}$. Each of the states $\ket{\eta_i}$ has a definite energy $E_i$. In this representation, \begin{equation} \label{Eq:III:20:10} \ket{\psi}=\sum_iC_i\,\ket{\eta_i}. \end{equation} When you make an energy measurement and get some number $E_i$, you have found that the system was in the state $\eta_i$. But you may get a different number for each measurement. Sometimes you will get $E_1$, sometimes $E_2$, sometimes $E_3$, and so on. The probability that you observe the energy $E_1$ is just the probability of finding the system in the state $\ket{\eta_1}$, which is, of course, just the absolute square of the amplitude $C_1=\braket{\eta_1}{\psi}$. The probability of finding each of the possible energies $E_i$ is \begin{equation} \label{Eq:III:20:11} P_i=\abs{C_i}^2. \end{equation} How are these probabilities related to the mean value of a whole sequence of energy measurements? Let’s imagine that we get a series of measurements like this: $E_1$, $E_7$, $E_{11}$, $E_9$, $E_1$, $E_{10}$, $E_7$, $E_2$, $E_3$, $E_9$, $E_6$, $E_4$, and so on. We continue for, say, a thousand measurements. When we are finished we add all the energies and divide by one thousand. That’s what we mean by the average. There’s also a short-cut to adding all the numbers. You can count up how many times you get $E_1$, say that is $N_1$, and then count up the number of times you get $E_2$, call that $N_2$, and so on. The sum of all the energies is certainly just \begin{equation*} N_1E_1+N_2E_2+N_3E_3+\dotsb{}=\sum_iN_iE_i. \end{equation*} The average energy is this sum divided by the total number of measurements which is just the sum of all the $N_i$'s, which we can call $N$; \begin{equation} \label{Eq:III:20:12} E_{\text{av}}=\frac{\sum_iN_iE_i}{N}. \end{equation} We are almost there. What we mean by the probability of something happening is just the number of times we expect it to happen divided by the total number of tries. The ratio $N_i/N$ should—for large $N$—be very near to $P_i$, the probability of finding the state $\ket{\eta_i}$, although it will not be exactly $P_i$ because of the statistical fluctuations. Let’s write the predicted (or “expected”) average energy as $\av{E}$; then we can say that \begin{equation} \label{Eq:III:20:13} \av{E}=\sum_iP_iE_i. \end{equation} The same arguments apply for any measurement. The average value of a measured quantity $A$ should be equal to \begin{equation*} \av{A}=\sum_iP_iA_i, \end{equation*} where $A_i$ are the various possible values of the observed quantity, and $P_i$ is the probability of getting that value. Let’s go back to our quantum-mechanical state $\ket{\psi}$. Its average energy is \begin{equation} \label{Eq:III:20:14} \av{E}=\sum_i\abs{C_i}^2E_i=\sum_iC_i\cconj C_iE_i. \end{equation} Now watch this trickery! First, we write the sum as \begin{equation} \label{Eq:III:20:15} \sum_i\braket{\psi}{\eta_i}E_i\braket{\eta_i}{\psi}. \end{equation} Next we treat the left-hand $\bra{\psi}$ as a common “factor.” We can take this factor out of the sum, and write it as \begin{equation*} \bra{\psi}\,\biggl\{\sum_i\ket{\eta_i}E_i\braket{\eta_i}{\psi}\biggr\}. \end{equation*} This expression has the form \begin{equation*} \braket{\psi}{\phi}, \end{equation*} where $\ket{\phi}$ is some “cooked-up” state defined by \begin{equation} \label{Eq:III:20:16} \ket{\phi}=\sum_i\ket{\eta_i}E_i\braket{\eta_i}{\psi}. \end{equation} It is, in other words, the state you get if you take each base state $\ket{\eta_i}$ in the amount $E_i\braket{\eta_i}{\psi}$. Now remember what we mean by the states $\ket{\eta_i}$. They are supposed to be the stationary states—by which we mean that for each one, \begin{equation*} \Hop\,\ket{\eta_i}=E_i\,\ket{\eta_i}. \end{equation*} Since $E_i$ is just a number, the right-hand side is the same as $\ket{\eta_i}E_i$, and the sum in Eq. (20.16) is the same as \begin{equation*} \sum_i\Hop\,\ket{\eta_i}\braket{\eta_i}{\psi}. \end{equation*} Now $i$ appears only in the famous combination that contracts to unity, so \begin{equation*} \sum_i\Hop\,\ket{\eta_i}\braket{\eta_i}{\psi}= \Hop\sum_i\ket{\eta_i}\braket{\eta_i}{\psi}=\Hop\,\ket{\psi}. \end{equation*} Magic! Equation (20.16) is the same as \begin{equation} \label{Eq:III:20:17} \ket{\phi}=\Hop\,\ket{\psi}. \end{equation} The average energy of the state $\ket{\psi}$ can be written very prettily as \begin{equation} \label{Eq:III:20:18} \av{E}=\bracket{\psi}{\Hop}{\psi}. \end{equation} To get the average energy you operate on $\ket{\psi}$ with $\Hop$, and then multiply by $\bra{\psi}$. A simple result. Our new formula for the average energy is not only pretty. It is also useful, because now we don’t need to say anything about any particular set of base states. We don’t even have to know all of the possible energy levels. When we go to calculate, we’ll need to describe our state in terms of some set of base states, but if we know the Hamiltonian matrix $H_{ij}$ for that set we can get the average energy. Equation (20.18) says that for any set of base states $\ket{i}$, the average energy can be calculated from \begin{equation} \label{Eq:III:20:19} \av{E}=\sum_{ij}\braket{\psi}{i}\bracket{i}{\Hop}{j}\braket{j}{\psi}, \end{equation} where the amplitudes $\bracket{i}{\Hop}{j}$ are just the elements of the matrix $H_{ij}$. Let’s check this result for the special case that the states $\ket{i}$ are the definite energy states. For them, $\Hop\,\ket{j}=E_j\,\ket{j}$, so $\bracket{i}{\Hop}{j}=E_j\,\delta_{ij}$ and \begin{equation*} \av{E}=\sum_{ij}\braket{\psi}{i}E_j\delta_{ij}\braket{j}{\psi}= \sum_iE_i\braket{\psi}{i}\braket{i}{\psi}, \end{equation*} which is right. Equation (20.19) can, incidentally, be extended to other physical measurements which you can express as an operator. For instance, $\Lop_z$ is the operator of the $z$-component of the angular momentum $\FLPL$. The average of the $z$-component for the state $\ket{\psi}$ is \begin{equation*} \av{L_z}=\bracket{\psi}{\Lop_z}{\psi}. \end{equation*} One way to prove it is to think of some situation in which the energy is proportional to the angular momentum. Then all the arguments go through in the same way. In summary, if a physical observable $A$ is related to a suitable quantum-mechanical operator $\Aop$, the average value of $A$ for the state $\ket{\psi}$ is given by \begin{equation} \label{Eq:III:20:20} \av{A}=\bracket{\psi}{\Aop}{\psi}. \end{equation} By this we mean that \begin{equation} \label{Eq:III:20:21} \av{A}=\braket{\psi}{\phi}, \end{equation} with \begin{equation} \label{Eq:III:20:22} \ket{\phi}=\Aop\,\ket{\psi}. \end{equation}

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The average energy of an atom

Suppose we want the average energy of an atom in a state described by a wave function $\psi(\FLPr)$; How do we find it? Let’s first think of a one-dimensional situation with a state $\ket{\psi}$ defined by the amplitude $\braket{x}{\psi}=\psi(x)$. We are asking for the special case of Eq. (20.19) applied to the coordinate representation. Following our usual procedure, we replace the states $\ket{i}$ and $\ket{j}$ by $\ket{x}$ and $\ket{x'}$, and change the sums to integrals. We get \begin{equation} \label{Eq:III:20:23} \av{E}=\!\iint\braket{\psi}{x}\bracket{x}{\Hop}{x'} \braket{x'}{\psi}\,dx\,dx'\!. \end{equation} This integral can, if we wish, be written in the following way: \begin{equation} \label{Eq:III:20:24} \int\braket{\psi}{x}\braket{x}{\phi}\,dx, \end{equation} with \begin{equation} \label{Eq:III:20:25} \braket{x}{\phi}=\int\bracket{x}{\Hop}{x'}\braket{x'}{\psi}\,dx'. \end{equation} The integral over $x'$ in (20.25) is the same one we had in Chapter 16—see Eq. (16.50) and Eq. (16.52)—and is equal to \begin{equation*} -\frac{\hbar^2}{2m}\,\frac{d^2}{dx^2}\,\psi(x)+V(x)\psi(x). \end{equation*} We can therefore write \begin{equation} \label{Eq:III:20:26} \braket{x}{\phi}=\biggl\{\! -\frac{\hbar^2}{2m}\,\frac{d^2}{dx^2}+V(x)\biggr\}\psi(x). \end{equation} Remember that $\braket{\psi}{x}=$ $\braket{x}{\psi}\cconj=$ $\psi\cconj(x)$; using this equality, the average energy in Eq. (20.23) can be written as \begin{equation} \label{Eq:III:20:27} \av{E}=\!\int\!\psi\cconj(x)\biggl\{\! -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\!+\!V(x)\!\biggr\}\psi(x)\,dx. \end{equation} Given a wave function $\psi(x)$, you can get the average energy by doing this integral. You can begin to see how we can go back and forth from the state-vector ideas to the wave-function ideas. The quantity in the braces of Eq. (20.27) is an algebraic operator.1 We will write it as $\Hcalop$ \begin{equation*} \Hcalop=-\frac{\hbar^2}{2m}\,\frac{d^2}{dx^2}+V(x). \end{equation*} With this notation Eq. (20.23) becomes \begin{equation} \label{Eq:III:20:28} \av{E}=\int\psi\cconj(x)\Hcalop\psi(x)\,dx. \end{equation} The algebraic operator $\Hcalop$ defined here is, of course, not identical to the quantum-mechanical operator $\Hop$. The new operator works on a function of position $\psi(x)=\braket{x}{\psi}$ to give a new function of $x$, $\phi(x)=\braket{x}{\phi}$; while $\Hop$ operates on a state vector $\ket{\psi}$ to give another state vector $\ket{\phi}$, without implying the coordinate representation or any particular representation at all. Nor is $\Hcalop$ strictly the same as $\Hop$ even in the coordinate representation. If we choose to work in the coordinate representation, we would interpret $\Hop$ in terms of a matrix $\bracket{x}{\Hop}{x'}$ which depends somehow on the two “indices” $x$ and $x'$; that is, we expect—according to Eq. (20.25)—that $\braket{x}{\phi}$ is related to all the amplitudes $\braket{x}{\psi}$ by an integration. On the other hand, we find that $\Hcalop$ is a differential operator. We have already worked out in Section 16–5 the connection between $\bracket{x}{\Hop}{x'}$ and the algebraic operator $\Hcalop$. We should make one qualification on our results. We have been assuming that the amplitude $\psi(x)=\braket{x}{\psi}$ is normalized. By this we mean that the scale has been chosen so that \begin{equation*} \int\abs{\psi(x)}^2\,dx=1; \end{equation*} so the probability of finding the electron somewhere is unity. If you should choose to work with a $\psi(x)$ which is not normalized you should write \begin{equation} \label{Eq:III:20:29} \av{E}=\frac{\int\psi\cconj(x)\Hcalop\psi(x)\,dx} {\int\psi\cconj(x)\psi(x)\,dx}. \end{equation} It’s the same thing. Notice the similarity in form between Eq. (20.28) and Eq. (20.18). These two ways of writing the same result appear often when you work with the $x$-representation. You can go from the first form to the second with any $\Aop$ which is a local operator, where a local operator is one which in the integral \begin{equation*} \int\bracket{x}{\Aop}{x'}\braket{x'}{\psi}\,dx' \end{equation*} can be written as $\Acalop\psi(x)$, where $\Acalop$ is a differential algebraic operator. There are, however, operators for which this is not true. For them you must work with the basic equations in (20.21) and (20.22). You can easily extend the derivation to three dimensions. The result is that2 \begin{equation} \label{Eq:III:20:30} \av{E}=\int\psi\cconj(\FLPr)\Hcalop\psi(\FLPr)\,dV, \end{equation} with \begin{equation} \label{Eq:III:20:31} \Hcalop=-\frac{\hbar^2}{2m}\,\nabla^2+V(\FLPr), \end{equation} and with the understanding that \begin{equation} \label{Eq:III:20:32} \int\abs{\psi}^2\,dV=1. \end{equation} The same equations can be extended to systems with several electrons in a fairly obvious way, but we won’t bother to write down the results. With Eq. (20.30) we can calculate the average energy of an atomic state even without knowing its energy levels. All we need is the wave function. It’s an important law. We’ll tell you about one interesting application. Suppose you want to know the ground-state energy of some system—say the helium atom, but it’s too hard to solve Schrödinger’s equation for the wave function, because there are too many variables. Suppose, however, that you take a guess at the wave function—pick any function you like—and calculate the average energy. That is, you use Eq. (20.29)—generalized to three dimensions-to find what the average energy would be if the atom were really in the state described by this wave function. This energy will certainly be higher than the ground-state energy which is the lowest possible energy the atom can have.3 Now pick another function and calculate its average energy. If it is lower than your first choice you are getting closer to the true ground-state energy. If you keep on trying all sorts of artificial states you will be able to get lower and lower energies, which come closer and closer to the ground-state energy. If you are clever, you will try some functions which have a few adjustable parameters. When you calculate the energy it will be expressed in terms of these parameters. By varying the parameters to give the lowest possible energy, you are trying out a whole class of functions at once. Eventually you will find that it is harder and harder to get lower energies and you will begin to be convinced that you are fairly close to the lowest possible energy. The helium atom has been solved in just this way—not by solving a differential equation, but by making up a special function with a lot of adjustable parameters which are eventually chosen to give the lowest possible value for the average energy.

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The position operator

What is the average value of the position of an electron in an atom? For any particular state $\ket{\psi}$ what is the average value of the coordinate $x$? We’ll work in one dimension and let you extend the ideas to three dimensions or to systems with more than one particle: We have a state described by $\psi(x)$, and we keep measuring $x$ over and over again. What is the average? It is \begin{equation*} \int xP(x)\,dx, \end{equation*} where $P(x)\,dx$ is the probability of finding the electron in a little element $dx$ at $x$. Suppose the probability density $P(x)$ varies with $x$ as shown in Fig. 20–1. The electron is most likely to be found near the peak of the curve. The average value of $x$ is also somewhere near the peak. It is, in fact, just the center of gravity of the area under the curve. We have seen earlier that $P(x)$ is just $\abs{\psi(x)}^2=\psi\cconj(x)\psi(x)$, so we can write the average of $x$ as \begin{equation} \label{Eq:III:20:33} \av{x}=\int\psi\cconj(x)x\psi(x)\,dx. \end{equation} Our equation for $\av{x}$ has the same form as Eq. (20.28). For the average energy, the energy operator $\Hcalop$ appears between the two $\psi$’s, for the average position there is just $x$. (If you wish you can consider $x$ to be the algebraic operator “multiply by $x$.”) We can carry the parallelism still further, expressing the average position in a form which corresponds to Eq. (20.18). Suppose we just write \begin{equation} \label{Eq:III:20:34} \av{x}=\braket{\psi}{\alpha} \end{equation} with \begin{equation} \label{Eq:III:20:35} \ket{\alpha}=\xop\,\ket{\psi}, \end{equation} and then see if we can find the operator $\xop$ which generates the state $\ket{\alpha}$, which will make Eq. (20.34) agree with Eq. (20.33). That is, we must find a $\ket{\alpha}$, so that \begin{equation} \label{Eq:III:20:36} \braket{\psi}{\alpha}=\av{x}= \int\braket{\psi}{x}x\braket{x}{\psi}\,dx. \end{equation} First, let’s expand $\braket{\psi}{\alpha}$ in the $x$-representation. It is \begin{equation} \label{Eq:III:20:37} \braket{\psi}{\alpha}=\int\braket{\psi}{x}\braket{x}{\alpha}\,dx. \end{equation} Now compare the integrals in the last two equations. You see that in the $x$-representation \begin{equation} \label{Eq:III:20:38} \braket{x}{\alpha}=x\braket{x}{\psi}. \end{equation} Operating on $\ket{\psi}$ with $\xop$ to get $\ket{\alpha}$ is equivalent to multiplying $\psi(x)=\braket{x}{\psi}$ by $x$ to get $\alpha(x)=\braket{x}{\alpha}$. We have a definition of $\xop$ in the coordinate representation.4 [We have not bothered to try to get the $x$-representation of the matrix of the operator $\xop$. If you are ambitious you can try to show that \begin{equation} \label{Eq:III:20:39} \bracket{x}{\xop}{x'}=x\,\delta(x-x'). \end{equation} You can then work out the amusing result that \begin{equation} \label{Eq:III:20:40} \xop\,\ket{x}=x\,\ket{x}. \end{equation} The operator $\xop$ has the interesting property that when it works on the base states $\ket{x}$ it is equivalent to multiplying by $x$.] Do you want to know the average value of $x^2$? It is \begin{equation} \label{Eq:III:20:41} \av{x^2}=\int\psi\cconj(x)x^2\psi(x)\,dx. \end{equation} Or, if you prefer you can write \begin{equation} \av{x^2}=\braket{\psi}{\alpha'}\notag \end{equation} with \begin{equation} \label{Eq:III:20:42} \ket{\alpha'}=\xop^2\,\ket{\psi}. \end{equation} By $\xop^2$ we mean $\xop\xop$—the two operators are used one after the other. With the second form you can calculate $\av{x^2}$, using any representation (base-states) you wish. If you want the average of $x^n$, or of any polynomial in $x$, you can see how to get it.

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5

The momentum operator

Now we would like to calculate the mean momentum of an electron—again, we’ll stick to one dimension. Let $P(p)\,dp$ be the probability that a measurement will give a momentum between $p$ and $p+dp$. Then \begin{equation} \label{Eq:III:20:43} \av{p}=\int p\,P(p)\,dp. \end{equation} Now we let $\braket{p}{\psi}$ be the amplitude that the state $\ket{\psi}$ is in a definite momentum state $\ket{p}$. This is the same amplitude we called $\braket{\mom p}{\psi}$ in Section 16–3 and is a function of $p$ just as $\braket{x}{\psi}$ is a function of $x$. There we chose to normalize the amplitude so that \begin{equation} \label{Eq:III:20:44} P(p)=\frac{1}{2\pi\hbar}\,\abs{\braket{p}{\psi}}^2. \end{equation} We have, then, \begin{equation} \label{Eq:III:20:45} \av{p}=\int\braket{\psi}{p}p\braket{p}{\psi}\,\frac{dp}{2\pi\hbar}. \end{equation} The form is quite similar to what we had for $\av{x}$. If we want, we can play exactly the same game we did with $\av{x}$. First, we can write the integral above as \begin{equation} \label{Eq:III:20:46} \int\braket{\psi}{p}\braket{p}{\beta}\,\frac{dp}{2\pi\hbar}. \end{equation} You should now recognize this equation as just the expanded form of the amplitude $\braket{\psi}{\beta}$—expanded in terms of the base states of definite momentum. From Eq. (20.45) the state $\ket{\beta}$ is defined in the momentum representation by \begin{equation} \label{Eq:III:20:47} \braket{p}{\beta}=p\braket{p}{\psi} \end{equation} That is, we can now write \begin{equation} \label{Eq:III:20:48} \av{p}=\braket{\psi}{\beta} \end{equation} with \begin{equation} \label{Eq:III:20:49} \ket{\beta}=\pop\,\ket{\psi}, \end{equation} where the operator $\pop$ is defined in terms of the $p$-representation by Eq. (20.47). [Again, you can if you wish show that the matrix form of $\pop$ is \begin{equation} \label{Eq:III:20:50} \bracket{p}{\pop}{p'}=p\,\delta(p-p'), \end{equation} and that \begin{equation} \label{Eq:III:20:51} \pop\,\ket{p}=p\,\ket{p}. \end{equation} It works out the same as for $x$.] Now comes an interesting question. We can write $\av{p}$, as we have done in Eqs. (20.45) and (20.48), and we know the meaning of the operator $\pop$ in the momentum representation. But how should we interpret $\pop$ in the coordinate representation? That is what we will need to know if we have some wave function $\psi(x)$, and we want to compute its average momentum. Let’s make clear what we mean. If we start by saying that $\av{p}$ is given by Eq. (20.48), we can expand that equation in terms of the $p$-representation to get back to Eq. (20.46). If we are given the $p$-description of the state—namely the amplitude $\braket{p}{\psi}$, which is an algebraic function of the momentum $p$—we can get $\braket{p}{\beta}$ from Eq. (20.47) and proceed to evaluate the integral. The question now is: What do we do if we are given a description of the state in the $x$-representation, namely the wave function $\psi(x)=\braket{x}{\psi}$? Well, let’s start by expanding Eq. (20.48) in the $x$-representation. It is \begin{equation} \label{Eq:III:20:52} \av{p}=\int\braket{\psi}{x}\braket{x}{\beta}\,dx. \end{equation} Now, however, we need to know what the state $\ket{\beta}$ is in the $x$-representation. If we can find it, we can carry out the integral. So our problem is to find the function $\beta(x)=\braket{x}{\beta}$. We can find it in the following way. In Section 16–3 we saw how $\braket{p}{\beta}$ was related to $\braket{x}{\beta}$. According to Eq. (16.24), \begin{equation} \label{Eq:III:20:53} \braket{p}{\beta}=\int e^{-ipx/\hbar}\braket{x}{\beta}\,dx. \end{equation} If we know $\braket{p}{\beta}$ we can solve this equation for $\braket{x}{\beta}$. What we want, of course, is to express the result somehow in terms of $\psi(x)=\braket{x}{\psi}$, which we are assuming to be known. Suppose we start with Eq. (20.47) and again use Eq. (16.24) to write \begin{equation} \label{Eq:III:20:54} \braket{p}{\beta}=p\braket{p}{\psi}= p\int e^{-ipx/\hbar}\psi(x)\,dx. \end{equation} Since the integral is over $x$ we can put the $p$ inside the integral and write \begin{equation} \label{Eq:III:20:55} \braket{p}{\beta}=\int e^{-ipx/\hbar}p\psi(x)\,dx. \end{equation} Compare this with (20.53). You would say that $\braket{x}{\beta}$ is equal to $p\psi(x)$. No, No! The wave function $\braket{x}{\beta}=\beta(x)$ can depend only on $x$—not on $p$. That’s the whole problem. However, some ingenious fellow discovered that the integral in (20.55) could be integrated by parts. The derivative of $e^{-ipx/\hbar}$ with respect to $x$ is $(-i/\hbar)pe^{-ipx/\hbar}$, so the integral in (20.55) is equivalent to \begin{equation*} -\frac{\hbar}{i}\int\ddt{}{x}\,(e^{-ipx/\hbar})\psi(x)\,dx. \end{equation*} If we integrate by parts, it becomes \begin{equation*} -\frac{\hbar}{i}\bigl[e^{-ipx/\hbar}\psi(x)\bigr]_{-\infty}^{+\infty}+ \frac{\hbar}{i}\int e^{-ipx/\hbar}\ddt{\psi}{x}\,dx. \end{equation*} So long as we are considering bound states, so that $\psi(x)$ goes to zero at $x=\pm\infty$, the bracket is zero and we have \begin{equation} \label{Eq:III:20:56} \braket{p}{\beta}=\frac{\hbar}{i}\int e^{-ipx/\hbar}\ddt{\psi}{x}\,dx. \end{equation} Now compare this result with Eq. (20.53). You see that \begin{equation} \label{Eq:III:20:57} \braket{x}{\beta}=\frac{\hbar}{i}\,\ddt{}{x}\,\psi(x). \end{equation} We have the necessary piece to be able to complete Eq. (20.52). The answer is \begin{equation} \label{Eq:III:20:58} \av{p}=\!\int\!\psi\cconj(x)\,\frac{\hbar}{i}\,\ddt{}{x}\,\psi(x)\,dx. \end{equation} We have found how Eq. (20.48) looks in the coordinate representation. Now you should begin to see an interesting pattern developing. When we asked for the average energy of the state $\ket{\psi}$ we said it was \begin{equation*} \av{E}=\braket{\psi}{\phi}, \text{ with } \ket{\phi}=\Hop\,\ket{\psi}. \end{equation*} The same thing is written in the coordinate world as \begin{equation*} \av{E}=\!\int\!\psi\cconj(x)\phi(x)\,dx, \text{ with } \phi(x)=\Hcalop\psi(x). \end{equation*} Here $\Hcalop$ is an algebraic operator which works on a function of $x$. When we asked about the average value of $x$, we found that it could also be written \begin{equation*} \av{x}=\braket{\psi}{\alpha}, \text{ with } \ket{\alpha}=\xop\,\ket{\psi}. \end{equation*} In the coordinate world the corresponding equations are \begin{equation*} \av{x}=\!\int\!\psi\cconj(x)\alpha(x)\,dx, \text{ with } \alpha(x)=x\psi(x). \end{equation*} When we asked about the average value of $p$, we wrote \begin{equation*} \av{p}=\braket{\psi}{\beta}, \text{ with } \ket{\beta}=\pop\,\ket{\psi}. \end{equation*} In the coordinate world the equivalent equations were \begin{equation*} \av{p}=\!\int\!\psi\cconj(x)\beta(x)\,dx, \text{ with } \beta(x)=\frac{\hbar}{i}\,\ddt{}{x}\,\psi(x). \end{equation*} In each of our three examples we start with the state $\ket{\psi}$ and produce another (hypothetical) state by a quantum-mechanical operator. In the coordinate representation we generate the corresponding wave function by operating on the wave function $\psi(x)$ with an algebraic operator. There are the following one-to-one correspondences (for one-dimensional problems): \begin{equation} \label{Eq:III:20:59} \begin{aligned} \Hop&\to\Hcalop=-\frac{\hbar^2}{2m}\,\frac{d^2}{dx^2}+V(x),\\[1pt] \xop&\to x,\\[1pt] \pop_x&\to\Pcalop_x=\frac{\hbar}{i}\,\ddp{}{x}. \end{aligned} \end{equation} In this list, we have introduced the symbol $\Pcalop_x$ for the algebraic operator $(\hbar/i)\ddpl{}{x}$: \begin{equation} \label{Eq:III:20:60} \Pcalop_x=\frac{\hbar}{i}\,\ddp{}{x}, \end{equation} and we have inserted the $x$ subscript on $\Pcalop$ to remind you that we have been working only with the $x$-component of momentum. You can easily extend the results to three dimensions. For the other components of the momentum, \begin{align*} \pop_y&\to\Pcalop_y=\frac{\hbar}{i}\,\ddp{}{y},\\[1ex] \pop_z&\to\Pcalop_z=\frac{\hbar}{i}\,\ddp{}{z}. \end{align*} If you want, you can even think of an operator of the vector momentum and write \begin{equation*} \pvecop\to\Pcalvecop=\frac{\hbar}{i}\biggl( \FLPe_x\,\ddp{}{x}+\FLPe_y\,\ddp{}{y}+\FLPe_z\,\ddp{}{z}\biggr), \end{equation*} where $\FLPe_x$, $\FLPe_y$, and $\FLPe_z$ are the unit vectors in the three directions. It looks even more elegant if we write \begin{equation} \label{Eq:III:20:61} \pvecop\to\Pcalvecop=\frac{\hbar}{i}\,\FLPnabla. \end{equation} Our general result is that for at least some quantum-mechanical operators, there are corresponding algebraic operators in the coordinate representation. We summarize our results so far—extended to three dimensions—in Table 20–1. For each operator we have the two equivalent forms:5 \begin{equation} \label{Eq:III:20:62} \ket{\phi}=\Aop\,\ket{\psi} \end{equation} or \begin{equation} \label{Eq:III:20:63} \phi(\FLPr)=\Acalop\psi(\FLPr). \end{equation} We will now give a few illustrations of the use of these ideas. The first one is just to point out the relation between $\Pcalop$ and $\Hcalop$. If we use $\Pcalop_x$ twice, we get \begin{equation*} \Pcalop_x\Pcalop_x=-\hbar^2\,\frac{\partial^2}{\partial x^2}. \end{equation*} This means that we can write the equality \begin{equation*} \Hcalop=\frac{1}{2m}\,\{ \Pcalop_x\Pcalop_x+\Pcalop_y\Pcalop_y+\Pcalop_z\Pcalop_z\} +V(\FLPr). \end{equation*} Or, using the vector notation, \begin{equation} \label{Eq:III:20:64} \Hcalop=\frac{1}{2m}\,\Pcalvecop\cdot\Pcalvecop+V(\FLPr). \end{equation} (In an algebraic operator, any term without the operator symbol ($\op{\enspace}$) means just a straight multiplication.) This equation is nice because it’s easy to remember if you haven’t forgotten your classical physics. Everyone knows that the energy is (nonrelativistically) just the kinetic energy $p^2/2m$ plus the potential energy, and $\Hcalop$ is the operator of the total energy. This result has impressed people so much that they try to teach students all about classical physics before quantum mechanics. (We think differently!) But such parallels are often misleading. For one thing, when you have operators, the order of various factors is important; but that is not true for the factors in a classical equation. In Chapter 17 we defined an operator $\pop_x$ in terms of the displacement operator $\Dop_x$ by [see Eq. (17.27)] \begin{equation} \label{Eq:III:20:65} \ket{\psi'}=\Dop_x(\delta)\,\ket{\psi}= \biggl(1+\frac{i}{\hbar}\,\pop_x\delta\biggr)\ket{\psi}, \end{equation} where $\delta$ is a small displacement. We should show you that this is equivalent to our new definition. According to what we have just worked out, this equation should mean the same as \begin{equation*} \psi'(x)=\psi(x)+\ddp{\psi}{x}\,\delta. \end{equation*} But the right-hand side is just the Taylor expansion of $\psi(x+\delta)$, which is certainly what you get if you displace the state to the left by $\delta$ (or shift the coordinates to the right by the same amount). Our two definitions of $\pop$ agree! Let’s use this fact to show something else. Suppose we have a bunch of particles which we label $1$, $2$, $3$, …, in some complicated system. (To keep things simple we’ll stick to one dimension.) The wave function describing the state is a function of all the coordinates $x_1$, $x_2$, $x_3$, … We can write it as $\psi(x_1,x_2,x_3,\dotsc)$. Now displace the system (to the left) by $\delta$. The new wave function \begin{equation*} \psi'(x_1,x_2,x_3,\dotsc)=\psi(x_1+\delta,x_2+\delta,x_3+\delta,\dotsc) \end{equation*} can be written as \begin{align} \enspace\psi'(x_1&,x_2,x_3,\dotsc)=\,\psi(x_1,x_2,x_3,\dotsc)\notag\\[1ex] \label{Eq:III:20:66} &+\,\biggl\{\delta\,\ddp{\psi}{x_1}+\delta\,\ddp{\psi}{x_2}+ \delta\,\ddp{\psi}{x_3}+\dotsb\biggr\}.\enspace \end{align} According to Eq. (20.65) the operator of the momentum of the state $\ket{\psi}$ (let’s call it the total momentum) is equal to \begin{equation*} \Pcalop_{\text{total}}=\frac{\hbar}{i}\,\biggl\{ \ddp{}{x_1}+\ddp{}{x_2}+\ddp{}{x_3}+\dotsb\biggr\}. \end{equation*} But this is just the same as \begin{equation} \label{Eq:III:20:67} \Pcalop_{\text{total}}=\Pcalop_{x_1}+\Pcalop_{x_2}+\Pcalop_{x_3}+\dotsb. \end{equation} The operators of momentum obey the rule that the total momentum is the sum of the momenta of all the parts. Everything holds together nicely, and many of the things we have been saying are consistent with each other.

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20

Operators

6

Angular momentum

Let’s for fun look at another operation—the operation of orbital angular momentum. In Chapter 17 we defined an operator $\Jop_z$ in terms of $\Rop_z(\phi)$, the operator of a rotation by the angle $\phi$ about the $z$-axis. We consider here a system described simply by a single wave function $\psi(\FLPr)$, which is a function of coordinates only, and does not take into account the fact that the electron may have its spin either up or down. That is, we want for the moment to disregard intrinsic angular momentum and think about only the orbital part. To keep the distinction clear, we’ll call the orbital operator $\Lop_z$, and define it in terms of the operator of a rotation by an infinitesimal angle $\epsilon$ by \begin{equation*} \Rop_z(\epsilon)\,\ket{\psi}= \biggl(1+\frac{i}{\hbar}\,\epsilon\,\Lop_z\biggr)\ket{\psi}. \end{equation*} (Remember, this definition applies only to a state $\ket{\psi}$ which has no internal spin variables, but depends only on the coordinates $\FLPr=x,y,z$.) If we look at the state $\ket{\psi}$ in a new coordinate system, rotated about the $z$-axis by the small angle $\epsilon$, we see a new state \begin{equation*} \ket{\psi'}=\Rop_z(\epsilon)\,\ket{\psi}. \end{equation*} If we choose to describe the state $\ket{\psi}$ in the coordinate representation—that is, by its wave function $\psi(\FLPr)$, we would expect to be able to write \begin{equation} \label{Eq:III:20:68} \psi'(\FLPr)= \biggl(1+\frac{i}{\hbar}\,\epsilon\,\Lcalop_z\biggr)\psi(\FLPr). \end{equation} What is $\Lcalop_z$? Well, a point $P$ at $x$ and $y$ in the new coordinate system (really $x'$ and $y'$, but we will drop the primes) was formerly at $x-\epsilon y$ and $y+\epsilon x$, as you can see from Fig. 20–2. Since the amplitude for the electron to be at $P$ isn’t changed by the rotation of the coordinates we can write \begin{equation*} \psi'(x,y,z)=\psi(x-\epsilon y,y+\epsilon x,z)= \psi(x,y,z)-\epsilon y\,\ddp{\psi}{x}+ \epsilon x\,\ddp{\psi}{y} \end{equation*} \begin{align*} \psi'(x,y,z)&=\psi(x-\epsilon y,y+\epsilon x,z)\\[1ex] &=\psi(x,y,z)-\epsilon y\,\ddp{\psi}{x}+ \epsilon x\,\ddp{\psi}{y} \end{align*} (remembering that $\epsilon$ is a small angle). This means that \begin{equation} \label{Eq:III:20:69} \Lcalop_z=\frac{\hbar}{i}\biggl(x\,\ddp{}{y}-y\,\ddp{}{x}\biggr). \end{equation} That’s our answer. But notice. It is equivalent to \begin{equation} \label{Eq:III:20:70} \Lcalop_z=x\Pcalop_y-y\Pcalop_x. \end{equation} Returning to our quantum-mechanical operators, we can write \begin{equation} \label{Eq:III:20:71} \Lop_z=x\pop_y-y\pop_x. \end{equation} This formula is easy to remember because it looks like the familiar formula of classical mechanics; it is the $z$-component of \begin{equation} \label{Eq:III:20:72} \FLPL=\FLPr\times\FLPp. \end{equation} One of the fun parts of this operator business is that many classical equations get carried over into a quantum-mechanical form. Which ones don’t? There had better be some that don’t come out right, because if everything did, then there would be nothing different about quantum mechanics. There would be no new physics. Here is one equation which is different. In classical physics \begin{equation*} xp_x-p_xx=0. \end{equation*} What is it in quantum mechanics? \begin{equation*} \xop\pop_x-\pop_x\xop=? \end{equation*} Let’s work it out in the $x$-representation. So that we’ll know what we are doing we put in some wave function $\psi(x)$. We have \begin{equation*} x\Pcalop_x\psi(x)-\Pcalop_xx\psi(x), \end{equation*} or \begin{equation*} x\,\frac{\hbar}{i}\,\ddp{}{x}\,\psi(x)- \frac{\hbar}{i}\,\ddp{}{x}\,x\psi(x). \end{equation*} Remember now that the derivatives operate on everything to the right. We get \begin{equation} \label{Eq:III:20:73} x\,\frac{\hbar}{i}\,\ddp{\psi}{x}- \frac{\hbar}{i}\,\psi(x)- \frac{\hbar}{i}\,x\,\ddp{\psi}{x}=-\frac{\hbar}{i}\,\psi(x). \end{equation} The answer is not zero. The whole operation is equivalent simply to multiplication by $-\hbar/i$: \begin{equation} \label{Eq:III:20:74} \xop\pop_x-\pop_x\xop=-\frac{\hbar}{i}. \end{equation} If Planck’s constant were zero, the classical and quantum results would be the same, and there would be no quantum mechanics to learn! Incidentally, if any two operators $\Aop$ and $\Bop$, when taken together like this: \begin{equation*} \Aop\Bop-\Bop\Aop, \end{equation*} do not give zero, we say that “the operators do not commute.” And an equation such as (20.74) is called a “commutation rule.” You can see that the commutation rule for $p_x$ and $y$ is \begin{equation*} \pop_x\yop-\yop\pop_x=0. \end{equation*} There is another very important commutation rule that has to do with angular momenta. It is \begin{equation} \label{Eq:III:20:75} \Lop_x\Lop_y-\Lop_y\Lop_x=i\hbar\Lop_z. \end{equation} You can get some practice with $\xop$ and $\pop$ operators by proving it for yourself. It is interesting to notice that operators which do not commute can also occur in classical physics. We have already seen this when we have talked about rotation in space. If you rotate something, such as a book, by $90^\circ$ around $x$ and then $90^\circ$ around $y$, you get something different from rotating first by $90^\circ$ around $y$ and then by $90^\circ$ around $x$. It is, in fact, just this property of space that is responsible for Eq. (20.75).

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Operators

7

The change of averages with time

Now we want to show you something else. How do averages change with time? Suppose for the moment that we have an operator $\Aop$, which does not itself have time in it in any obvious way. We mean an operator like $\xop$ or $\pop$. (We exclude things like, say, the operator of some external potential that was being varied with time, such as $V(x,t)$.) Now suppose we calculate $\av{A}$, in some state $\ket{\psi}$, which is \begin{equation} \label{Eq:III:20:76} \av{A}=\bracket{\psi}{\Aop}{\psi}. \end{equation} How will $\av{A}$, depend on time? Why should it? One reason might be that the operator itself depended explicitly on time—for instance, if it had to do with a time-varying potential like $V(x,t)$. But even if the operator does not depend on $t$, say, for example, the operator $\Aop=\xop$, the corresponding average may depend on time. Certainly the average position of a particle could be moving. How does such a motion come out of Eq. (20.76) if $\Aop$ has no time dependence? Well, the state $\ket{\psi}$ might be changing with time. For nonstationary states we have often shown a time dependence explicitly by writing a state as $\ket{\psi(t)}$. We want to show that the rate of change of $\av{A}$, is given by a new operator we will call $\Adotop$. Remember that $\Aop$ is an operator, so that putting a dot over the $A$ does not here mean taking the time derivative, but is just a way of writing a new operator $\Adotop$ which is defined by \begin{equation} \label{Eq:III:20:77} \ddt{}{t}\,\av{A}=\bracket{\psi}{\Adotop}{\psi}. \end{equation} Our problem is to find the operator $\Adotop$. First, we know that the rate of change of a state is given by the Hamiltonian. Specifically, \begin{equation} \label{Eq:III:20:78} i\hbar\,\ddt{}{t}\,\ket{\psi(t)}=\Hop\,\ket{\psi(t)}. \end{equation} This is just the abstract way of writing our original definition of the Hamiltonian: \begin{equation} \label{Eq:III:20:79} i\hbar\,\ddt{C_i}{t}=\sum_jH_{ij}C_j. \end{equation} If we take the complex conjugate of Eq. (20.78), it is equivalent to \begin{equation} \label{Eq:III:20:80} -i\hbar\,\ddt{}{t}\,\bra{\psi(t)}=\bra{\psi(t)}\,\Hop. \end{equation} Next, see what happens if we take the derivatives with respect to $t$ of Eq. (20.76). Since each $\psi$ depends on $t$, we have \begin{equation} \label{Eq:III:20:81} \ddt{}{t}\av{A}=\!\biggl(\!\ddt{}{t}\bra{\psi}\!\biggr)\Aop\ket{\psi}\!+\! \bra{\psi}\Aop\biggl(\!\ddt{}{t}\ket{\psi}\!\biggr). \end{equation} Finally, using the two equations in (20.78) and (20.80) to replace the derivatives, we get \begin{equation*} \ddt{}{t}\,\av{A}=\frac{i}{\hbar}\,\{ \bracket{\psi}{\Hop\Aop}{\psi}- \bracket{\psi}{\Aop\Hop}{\psi}\}. \end{equation*} This equation is the same as \begin{equation*} \ddt{}{t}\,\av{A}=\frac{i}{\hbar}\, \bracket{\psi}{\Hop\Aop-\Aop\Hop}{\psi}. \end{equation*} Comparing this equation with Eq. (20.77), you see that \begin{equation} \label{Eq:III:20:82} \Adotop=\frac{i}{\hbar}\,(\Hop\Aop-\Aop\Hop). \end{equation} That is our interesting proposition, and it is true for any operator $\Aop$. Incidentally, if the operator $\Aop$ should itself be time dependent, we would have had \begin{equation} \label{Eq:III:20:83} \Adotop=\frac{i}{\hbar}\,(\Hop\Aop-\Aop\Hop) +\ddp{\Aop}{t}. \end{equation} Let us try out Eq. (20.82) on some example to see whether it really makes sense. For instance, what operator corresponds to $\xdotop$? We say it should be \begin{equation} \label{Eq:III:20:84} \xdotop=\frac{i}{\hbar}\,(\Hop\xop-\xop\Hop). \end{equation} What is this? One way to find out is to work it through in the coordinate representation using the algebraic operator for $\Hcalop$. In this representation the commutator is \begin{equation*} \Hcalop x-x\Hcalop=\biggl\{-\frac{\hbar^2}{2m}\,\frac{d^2}{dx^2}+ V(x)\biggr\}x-x\biggl\{-\frac{\hbar^2}{2m}\,\frac{d^2}{dx^2}+ V(x)\biggr\}. \end{equation*} \begin{align*} \Hcalop x&-x\Hcalop\,=\\[1.5ex] \!\biggl\{\!-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\!+\! V(x)\!\biggr\}x&-x\biggl\{\!-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\!+\! V(x)\!\biggr\}. \end{align*} If you operate with this on any wave function $\psi(x)$ and work out all of the derivatives where you can, you end up after a little work with \begin{equation} -\frac{\hbar^2}{m}\,\ddt{\psi}{x}.\notag \end{equation} But this is just the same as \begin{equation} -i\,\frac{\hbar}{m}\,\Pcalop_x\psi,\notag \end{equation} so we find that \begin{equation} \label{Eq:III:20:85} \Hop\xop-\xop\Hop=-i\,\frac{\hbar}{m}\,\pop_x \end{equation} or that \begin{equation} \label{Eq:III:20:86} \xdotop=\frac{\pop_x}{m}. \end{equation} A pretty result. It means that if the mean value of $x$ is changing with time the drift of the center of gravity is the same as the mean momentum divided by $m$. Exactly like classical mechanics. Another example. What is the rate of change of the average momentum of a state? Same game. Its operator is \begin{equation} \label{Eq:III:20:87} \pdotop=\frac{i}{\hbar}\,(\Hop\pop-\pop\Hop). \end{equation} Again you can work it out in the $x$ representation. Remember that $\pop$ becomes $d/dx$, and this means that you will be taking the derivative of the potential energy $V$ (in the $\Hcalop$)—but only in the second term. It turns out that it is the only term which does not cancel, and you find that \begin{equation} \Hcalop\Pcalop-\Pcalop\Hcalop=i\hbar\,\ddt{V}{x}\notag \end{equation} or that \begin{equation} \label{Eq:III:20:88} \pdotop=-\ddt{V}{x}. \end{equation} Again the classical result. The right-hand side is the force, so we have derived Newton’s law! But remember—these are the laws for the operators which give the average quantities. They do not describe what goes on in detail inside an atom. Quantum mechanics has the essential difference that $\pop\xop$ is not equal to $\xop\pop$. They differ by a little bit—by the small number $i\hbar$. But the whole wondrous complications of interference, waves, and all, result from the little fact that $\xop\pop-\pop\xop$ is not quite zero. The history of this idea is also interesting. Within a period of a few months in 1926, Heisenberg and Schrödinger independently found correct laws to describe atomic mechanics. Schrödinger invented his wave function $\psi(x)$ and found his equation. Heisenberg, on the other hand, found that nature could be described by classical equations, except that $xp-px$ should be equal to $i\hbar$, which he could make happen by defining them in terms of special kinds of matrices. In our language he was using the energy-representation, with its matrices. Both Heisenberg’s matrix algebra and Schrödinger’s differential equation explained the hydrogen atom. A few months later Schrödinger was able to show that the two theories were equivalent—as we have seen here. But the two different mathematical forms of quantum mechanics were discovered independently.

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21

The Schrödinger Equation in a Classical Context: A Seminar on Superconductivity

1

Schrödinger’s equation in a magnetic field

This lecture is only for entertainment. I would like to give the lecture in a somewhat different style—just to see how it works out. It’s not a part of the course—in the sense that it is not supposed to be a last minute effort to teach you something new. But, rather, I imagine that I’m giving a seminar or research report on the subject to a more advanced audience, to people who have already been educated in quantum mechanics. The main difference between a seminar and a regular lecture is that the seminar speaker does not carry out all the steps, or all the algebra. He says: “If you do such and such, this is what comes out,” instead of showing all of the details. So in this lecture I’ll describe the ideas all the way along but just give you the results of the computations. You should realize that you’re not supposed to understand everything immediately, but believe (more or less) that things would come out if you went through the steps. All that aside, this is a subject I want to talk about. It is recent and modern and would be a perfectly legitimate talk to give at a research seminar. My subject is the Schrödinger equation in a classical setting—the case of superconductivity. Ordinarily, the wave function which appears in the Schrödinger equation applies to only one or two particles. And the wave function itself is not something that has a classical meaning—unlike the electric field, or the vector potential, or things of that kind. The wave function for a single particle is a “field”—in the sense that it is a function of position—but it does not generally have a classical significance. Nevertheless, there are some situations in which a quantum mechanical wave function does have classical significance, and they are the ones I would like to take up. The peculiar quantum mechanical behavior of matter on a small scale doesn’t usually make itself felt on a large scale except in the standard way that it produces Newton’s laws—the laws of the so-called classical mechanics. But there are certain situations in which the peculiarities of quantum mechanics can come out in a special way on a large scale. At low temperatures, when the energy of a system has been reduced very, very low, instead of a large number of states being involved, only a very, very small number of states near the ground state are involved. Under those circumstances the quantum mechanical character of that ground state can appear on a macroscopic scale. It is the purpose of this lecture to show a connection between quantum mechanics and large-scale effects—not the usual discussion of the way that quantum mechanics reproduces Newtonian mechanics on the average, but a special situation in which quantum mechanics will produce its own characteristic effects on a large or “macroscopic” scale. I will begin by reminding you of some of the properties of the Schrödinger equation.1 I want to describe the behavior of a particle in a magnetic field using the Schrödinger equation, because the superconductive phenomena are involved with magnetic fields. An external magnetic field is described by a vector potential, and the problem is: what are the laws of quantum mechanics in a vector potential? The principle that describes the behavior of quantum mechanics in a vector potential is very simple. The amplitude that a particle goes from one place to another along a certain route when there’s a field present is the same as the amplitude that it would go along the same route when there’s no field, multiplied by the exponential of the line integral of the vector potential, times the electric charge divided by Planck’s constant2 (see Fig. 21–1): \begin{equation} \label{Eq:III:21:1} \braket{b}{a}_{\text{in $\FLPA$}}=\braket{b}{a}_{A=0}\cdot \exp\biggl[\frac{iq}{\hbar}\int_a^b\FLPA\cdot d\FLPs\biggr]. \end{equation} It is a basic statement of quantum mechanics. Now without the vector potential the Schrödinger equation of a charged particle (nonrelativistic, no spin) is \begin{equation} \label{Eq:III:21:2} -\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi= \frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\cdot \biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\psi+q\phi\psi, \end{equation} \begin{gather} \label{Eq:III:21:2} -\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi=\\[1ex] \frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\cdot \biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\psi+q\phi\psi,\notag \end{gather} where $\phi$ is the electric potential so that $q\phi$ is the potential energy.3 Equation (21.1) is equivalent to the statement that in a magnetic field the gradients in the Hamiltonian are replaced in each case by the gradient minus $q\FLPA$, so that Eq. (21.2) becomes \begin{equation} \label{Eq:III:21:3} -\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi= \frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\cdot \biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi+q\phi\psi. \end{equation} \begin{gather} \label{Eq:III:21:3} -\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi=\\[1ex] \frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\cdot \biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi+q\phi\psi.\notag \end{gather} This is the Schrödinger equation for a particle with charge $q$ moving in an electromagnetic field $\FLPA,\phi$ (nonrelativistic, no spin). To show that this is true I’d like to illustrate by a simple example in which instead of having a continuous situation we have a line of atoms along the $x$-axis with the spacing $b$ and we have an amplitude $iK\!/\hbar$ per unit time for an electron to jump from one atom to another when there is no field.4 Now according to Eq. (21.1) if there’s a vector potential in the $x$-direction $A_x(x,t)$, the amplitude to jump will be altered from what it was before by a factor $\exp[(iq/\hbar)\,A_xb]$, the exponent being $iq/\hbar$ times the vector potential integrated from one atom to the next. For simplicity we will write $(q/\hbar)A_x\equiv f(x)$, since $A_x$, will, in general, depend on $x$. If the amplitude to find the electron at the atom “$n$” located at $x$ is called $C(x)\equiv C_n$, then the rate of change of that amplitude is given by the following equation: \begin{align} -\frac{\hbar}{i}\ddp{}{t}C(x)=E_0C(x)&\!-\!Ke^{-ibf(x+b/2)}C(x\!+\!b)\notag\\ \label{Eq:III:21:4} &-\!Ke^{+ibf(x-b/2)}C(x\!-\!b). \end{align} There are three pieces. First, there’s some energy $E_0$ if the electron is located at $x$. As usual, that gives the term $E_0C(x)$. Next, there is the term $-KC(x+b)$, which is the amplitude for the electron to have jumped backwards one step from atom “$n+1$,” located at $x+b$. However, in doing so in a vector potential, the phase of the amplitude must be shifted according to the rule in Eq. (21.1). If $A_x$ is not changing appreciably in one atomic spacing, the integral can be written as just the value of $A_x$ at the midpoint, times the spacing $b$. So $(iq/\hbar)$ times the integral is just $ibf(x+b/2)$. Since the electron is jumping backwards, I showed this phase shift with a minus sign. That gives the second piece. In the same manner there’s a certain amplitude to have jumped from the other side, but this time we need the vector potential at a distance $(b/2)$ on the other side of $x$, times the distance $b$. That gives the third piece. The sum gives the equation for the amplitude to be at $x$ in a vector potential. Now we know that if the function $C(x)$ is smooth enough (we take the long wavelength limit), and if we let the atoms get closer together, Eq. (21.4) will approach the behavior of an electron in free space. So the next step is to expand the right-hand side of (21.4) in powers of $b$, assuming $b$ is very small. For example, if $b$ is zero the right-hand side is just $(E_0-2K)C(x)$, so in the zeroth approximation the energy is $E_0-2K$. Next comes the terms in $b$. But because the two exponentials have opposite signs, only even powers of $b$ remain. So if you make a Taylor expansion of $C(x)$, of $f(x)$, and of the exponentials, and then collect the terms in $b^2$, you get \begin{align} -\frac{\hbar}{i}\,\ddp{C(x)}{t}&=E_0C(x)-2KC(x)\notag\\ \label{Eq:III:21:5} &\quad-Kb^2\{C''(x)-2if(x)C'(x)-if'(x)C(x)-f^2(x)C(x)\}. \end{align} \begin{align} \label{Eq:III:21:5} -\frac{\hbar}{i}\,\ddp{C(x)}{t}&=E_0C(x)-2KC(x)\\[1ex] &-Kb^2\bigl\{C''(x)-2if(x)C'(x)\,-\notag\\[1ex] &\phantom{-Kb^2\{}\;if'(x)C(x)-f^2(x)C(x)\bigr\}.\notag \end{align} (The “primes” mean differentiation with respect to $x$.) Now this horrible combination of things looks quite complicated. But mathematically it’s exactly the same as \begin{equation} \label{Eq:III:21:6} -\frac{\hbar}{i}\,\ddp{C(x)}{t}=(E_0-2K)C(x)-Kb^2 \biggl[\ddp{}{x}-if(x)\biggr] \biggl[\ddp{}{x}-if(x)\biggr]C(x). \end{equation} \begin{align} \label{Eq:III:21:6} -\frac{\hbar}{i}\ddp{C(x)}{t}&=(E_0-2K)C(x)\\ &-Kb^2 \biggl[\ddp{}{x}-if(x)\biggr] \biggl[\ddp{}{x}-if(x)\biggr]C(x).\notag \end{align} The second bracket operating on $C(x)$ gives $C'(x)$ minus $if(x)C(x)$. The first bracket operating on these two terms gives the $C''$ term and terms in the first derivative of $f(x)$ and the first derivative of $C(x)$. Now remember that the solutions for zero magnetic field5 represent a particle with an effective mass $m_{\text{eff}}$ given by \begin{equation*} Kb^2=\frac{\hbar^2}{2m_{\text{eff}}}. \end{equation*} If you then set $E_0=2K$, and put back $f(x)=(q/\hbar)A_x$, you can easily check that Eq. (21.6) is the same as the first part of Eq. (21.3). (The origin of the potential energy term is well known, so I haven’t bothered to include it in this discussion.) The proposition of Eq. (21.1) that the vector potential changes all the amplitudes by the exponential factor is the same as the rule that the momentum operator, $(\hbar/i)\FLPnabla$ gets replaced by \begin{equation*} \frac{\hbar}{i}\,\FLPnabla-q\FLPA, \end{equation*} as you see in the Schrödinger equation of (21.3).

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The equation of continuity for probabilities

Now I turn to a second point. An important part of the Schrödinger equation for a single particle is the idea that the probability to find the particle at a position is given by the absolute square of the wave function. It is also characteristic of the quantum mechanics that probability is conserved in a local sense. When the probability of finding the electron somewhere decreases, while the probability of the electron being elsewhere increases (keeping the total probability unchanged), something must be going on in between. In other words, the electron has a continuity in the sense that if the probability decreases at one place and builds up at another place, there must be some kind of flow between. If you put a wall, for example, in the way, it will have an influence and the probabilities will not be the same. So the conservation of probability alone is not the complete statement of the conservation law, just as the conservation of energy alone is not as deep and important as the local conservation of energy.6 If energy is disappearing, there must be a flow of energy to correspond. In the same way, we would like to find a “current” of probability such that if there is any change in the probability density (the probability of being found in a unit volume), it can be considered as coming from an inflow or an outflow due to some current. This current would be a vector which could be interpreted this way—the $x$-component would be the net probability per second and per unit area that a particle passes in the $x$-direction across a plane parallel to the $yz$-plane. Passage toward $+x$ is considered a positive flow, and passage in the opposite direction, a negative flow. Is there such a current? Well, you know that the probability density $P(\FLPr,t)$ is given in terms of the wave function by \begin{equation} \label{Eq:III:21:7} P(\FLPr,t)=\psi\cconj(\FLPr,t)\psi(\FLPr,t). \end{equation} I am asking: Is there a current $\FLPJ$ such that \begin{equation} \label{Eq:III:21:8} \ddp{P}{t}=-\FLPdiv{\FLPJ}? \end{equation} If I take the time derivative of Eq. (21.7), I get two terms: \begin{equation} \label{Eq:III:21:9} \ddp{P}{t}=\psi\cconj\,\ddp{\psi}{t}+\psi\,\ddp{\psi\cconj}{t}. \end{equation} Now use the Schrödinger equation—Eq. (21.3)—for $\ddpl{\psi}{t}$; and take the complex conjugate of it to get $\ddpl{\psi\cconj}{t}$—each $i$ gets its sign reversed. You get \begin{equation} \begin{aligned} \ddp{P}{t}&=-\frac{i}{\hbar}\biggl[\psi\cconj\,\frac{1}{2m} \biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\cdot \biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi+ q\phi\psi\cconj\psi\\[.5ex] &\hphantom{{}={}}-\psi\,\frac{1}{2m} \biggl(-\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\cdot \biggl(-\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi\cconj -q\phi\psi\psi\cconj\biggr]. \end{aligned} \label{Eq:III:21:10} \end{equation} \begin{align} \label{Eq:III:21:10} &\ddp{P}{t}=\\ &-\frac{i}{\hbar}\biggl[\psi\cconj\!\frac{1}{2m} \biggl(\!\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\!\cdot\! \biggl(\!\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\psi\!+\! q\phi\psi\cconj\psi\notag\\[1ex] &\kern{2em}-\psi\frac{1}{2m} \biggl(\!-\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\!\cdot\! \biggl(\!-\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\psi\cconj\! \!-\!q\phi\psi\psi\cconj\biggr].\notag \end{align} The potential terms and a lot of other stuff cancel out. And it turns out that what is left can indeed be written as a perfect divergence. The whole equation is equivalent to \begin{equation} \label{Eq:III:21:11} \ddp{P}{t}=-\FLPdiv{\biggl\{ \frac{1}{2m}\,\psi\cconj \biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi+ \frac{1}{2m}\,\psi \biggl(-\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi\cconj \biggr\}}. \end{equation} \begin{align} \label{Eq:III:21:11} \ddp{P}{t}\!=-\FLPdiv{\biggl\{\! &\frac{1}{2m}\psi\cconj \biggl(\!\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\psi\,+\\[1ex] &\frac{1}{2m}\psi \biggl(\!-\frac{\hbar}{i}\FLPnabla\!-q\FLPA\!\biggr)\psi\cconj \!\biggr\}}.\notag \end{align} It is really not as complicated as it seems. It is a symmetrical combination of $\psi\cconj$ times a certain operation on $\psi$, plus $\psi$ times the complex conjugate operation on $\psi\cconj$. It is some quantity plus its own complex conjugate, so the whole thing is real—as it ought to be. The operation can be remembered this way: it is just the momentum operator $\Pcalvecop$ minus $q\FLPA$. I could write the current in Eq. (21.8) as \begin{equation} \label{Eq:III:21:12} \FLPJ\!=\!\frac{1}{2}\biggl\{\! \psi\cconj\biggl[\!\frac{\Pcalvecop\!-\!q\FLPA}{m}\!\biggr]\psi\!+\! \psi\biggl[\!\frac{\Pcalvecop\!-\!q\FLPA}{m}\!\biggr]\cconj\!\!\!\psi\cconj \!\biggr\}. \end{equation} There is then a current $\FLPJ$ which completes Eq. (21.8). Equation (21.11) shows that the probability is conserved locally. If a particle disappears from one region it cannot appear in another without something going on in between. Imagine that the first region is surrounded by a closed surface far enough out that there is zero probability to find the electron at the surface. The total probability to find the electron somewhere inside the surface is the volume integral of $P$. But according to Gauss’s theorem the volume integral of the divergence $\FLPJ$ is equal to the surface integral of its normal component. If $\psi$ is zero at the surface, Eq. (21.12) says that $\FLPJ$ is zero, so the total probability to find the particle inside can’t change. Only if some of the probability approaches the boundary can some of it leak out. We can say that it only gets out by moving through the surface—and that is local conservation.

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Two kinds of momentum

The equation for the current is rather interesting, and sometimes causes a certain amount of worry. You would think the current would be something like the density of particles times the velocity. The density should be something like $\psi\psi\cconj$, which is o.k. And each term in Eq. (21.12) looks like the typical form for the average-value of the operator \begin{equation} \label{Eq:III:21:13} \frac{\Pcalvecop-q\FLPA}{m}, \end{equation} so maybe we should think of it as the velocity of flow. It looks as though we have two suggestions for relations of velocity to momentum, because we would also think that momentum divided by mass, $\Pcalvecop/m$, should be a velocity. The two possibilities differ by the vector potential. It happens that these two possibilities were also discovered in classical physics, when it was found that momentum could be defined in two ways.7 One of them is called “kinematic momentum,” but for absolute clarity I will in this lecture call it the “$mv$-momentum.” This is the momentum obtained by multiplying mass by velocity. The other is a more mathematical, more abstract momentum, sometimes called the “dynamical momentum,” which I’ll call “$p$-momentum.” The two possibilities are \begin{equation} \label{Eq:III:21:14} \text{$mv$-momentum}=m\FLPv, \end{equation} \begin{equation} \label{Eq:III:21:15} \text{$p$-momentum}=m\FLPv + q\FLPA. \end{equation} It turns out that in quantum mechanics with magnetic fields it is the $p$-momentum which is connected to the gradient operator $\Pcalvecop$, so it follows that (21.13) is the operator of a velocity. I’d like to make a brief digression to show you what this is all about—why there must be something like Eq. (21.15) in the quantum mechanics. The wave function changes with time according to the Schrödinger equation in Eq. (21.3). If I would suddenly change the vector potential, the wave function wouldn’t change at the first instant; only its rate of change changes. Now think of what would happen in the following circumstance. Suppose I have a long solenoid, in which I can produce a flux of magnetic field ($\FLPB$-field), as shown in Fig. 21–2. And there is a charged particle sitting nearby. Suppose this flux nearly instantaneously builds up from zero to something. I start with zero vector potential and then I turn on a vector potential. That means that I produce suddenly a circumferential vector potential $\FLPA$. You’ll remember that the line integral of $\FLPA$ around a loop is the same as the flux of $\FLPB$ through the loop.8 Now what happens if I suddenly turn on a vector potential? According to the quantum mechanical equation the sudden change of $\FLPA$ does not make a sudden change of $\psi$; the wave function is still the same. So the gradient is also unchanged. But remember what happens electrically when I suddenly turn on a flux. During the short time that the flux is rising, there’s an electric field generated whose line integral is the rate of change of the flux with time: \begin{equation} \FLPE=-\ddp{\FLPA}{t}. \end{equation} That electric field is enormous if the flux is changing rapidly, and it gives a force on the particle. The force is the charge times the electric field, and so during the build up of the flux the particle obtains a total impulse (that is, a change in $m\FLPv$) equal to $-q\FLPA$. In other words, if you suddenly turn on a vector potential at a charge, this charge immediately picks up an $mv$-momentum equal to $-q\FLPA$. But there is something that isn’t changed immediately and that’s the difference between $m\FLPv$ and $-q\FLPA$. And so the sum $\FLPp=m\FLPv+q\FLPA$ is something which is not changed when you make a sudden change in the vector potential. This quantity $\FLPp$ is what we have called the $p$-momentum and is of importance in classical mechanics in the theory of dynamics, but it also has a direct significance in quantum mechanics. It depends on the character of the wave function, and it is the one to be identified with the operator \begin{equation*} \Pcalvecop=\frac{\hbar}{i}\,\FLPnabla. \end{equation*}

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The meaning of the wave function

When Schrödinger first discovered his equation he discovered the conservation law of Eq. (21.8) as a consequence of his equation. But he imagined incorrectly that $P$ was the electric charge density of the electron and that $\FLPJ$ was the electric current density, so he thought that the electrons interacted with the electromagnetic field through these charges and currents. When he solved his equations for the hydrogen atom and calculated $\psi$, he wasn’t calculating the probability of anything—there were no amplitudes at that time—the interpretation was completely different. The atomic nucleus was stationary but there were currents moving around; the charges $P$ and currents $\FLPJ$ would generate electromagnetic fields and the thing would radiate light. He soon found on doing a number of problems that it didn’t work out quite right. It was at this point that Born made an essential contribution to our ideas regarding quantum mechanics. It was Born who correctly (as far as we know) interpreted the $\psi$ of the Schrödinger equation in terms of a probability amplitude—that very difficult idea that the square of the amplitude is not the charge density but is only the probability per unit volume of finding an electron there, and that when you do find the electron some place the entire charge is there. That whole idea is due to Born. The wave function $\psi(\FLPr)$ for an electron in an atom does not, then, describe a smeared-out electron with a smooth charge density. The electron is either here, or there, or somewhere else, but wherever it is, it is a point charge. On the other hand, think of a situation in which there are an enormous number of particles in exactly the same state, a very large number of them with exactly the same wave function. Then what? One of them is here and one of them is there, and the probability of finding any one of them at a given place is proportional to $\psi\psi\cconj$. But since there are so many particles, if I look in any volume $dx\,dy\,dz$ I will generally find a number close to $\psi\psi\cconj\,dx\,dy\,dz$. So in a situation in which $\psi$ is the wave function for each of an enormous number of particles which are all in the same state, $\psi\psi\cconj$ can be interpreted as the density of particles. If, under these circumstances, each particle carries the same charge $q$, we can, in fact, go further and interpret $\psi\cconj\psi$ as the density of electricity. Normally, $\psi\psi\cconj$ is given the dimensions of a probability density, then $\psi$ should be multiplied by $q$ to give the dimensions of a charge density. For our present purposes we can put this constant factor into $\psi$, and take $\psi\psi\cconj$ itself as the electric charge density. With this understanding, $\FLPJ$ (the current of probability I have calculated) becomes directly the electric current density. So in the situation in which we can have very many particles in exactly the same state, there is possible a new physical interpretation of the wave functions. The charge density and the electric current can be calculated directly from the wave functions and the wave functions take on a physical meaning which extends into classical, macroscopic situations. Something similar can happen with neutral particles. When we have the wave function of a single photon, it is the amplitude to find a photon somewhere. Although we haven’t ever written it down there is an equation for the photon wave function analogous to the Schrödinger equation for the electron. The photon equation is just the same as Maxwell’s equations for the electromagnetic field, and the wave function is the same as the vector potential $\FLPA$. The wave function turns out to be just the vector potential. The quantum physics is the same thing as the classical physics because photons are noninteracting Bose particles and many of them can be in the same state—as you know, they like to be in the same state. The moment that you have billions in the same state (that is, in the same electromagnetic wave), you can measure the wave function, which is the vector potential, directly. Of course, it worked historically the other way. The first observations were on situations with many photons in the same state, and so we were able to discover the correct equation for a single photon by observing directly with our hands on a macroscopic level the nature of the wave function. Now the trouble with the electron is that you cannot put more than one in the same state. Therefore, it was long believed that the wave function of the Schrödinger equation would never have a macroscopic representation analogous to the macroscopic representation of the amplitude for photons. On the other hand, it is now realized that the phenomenon of superconductivity presents us with just this situation.

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The Schrödinger Equation in a Classical Context: A Seminar on Superconductivity

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Superconductivity

As you know, very many metals become superconducting below a certain temperature9—the temperature is different for different metals. When you reduce the temperature sufficiently the metals conduct electricity without any resistance. This phenomenon has been observed for a very large number of metals but not for all, and the theory of this phenomenon has caused a great deal of difficulty. It took a very long time to understand what was going on inside of superconductors, and I will only describe enough of it for our present purposes. It turns out that due to the interactions of the electrons with the vibrations of the atoms in the lattice, there is a small net effective attraction between the electrons. The result is that the electrons form together, if I may speak very qualitatively and crudely, bound pairs. Now you know that a single electron is a Fermi particle. But a bound pair would act as a Bose particle, because if I exchange both electrons in a pair I change the sign of the wave function twice, and that means that I don’t change anything. A pair is a Bose particle. The energy of pairing—that is, the net attraction—is very, very weak. Only a tiny temperature is needed to throw the electrons apart by thermal agitation, and convert them back to “normal” electrons. But when you make the temperature sufficiently low that they have to do their very best to get into the absolutely lowest state; then they do collect in pairs. I don’t wish you to imagine that the pairs are really held together very closely like a point particle. As a matter of fact, one of the great difficulties of understanding this phenomenon originally was that that is not the way things are. The two electrons which form the pair are really spread over a considerable distance; and the mean distance between pairs is relatively smaller than the size of a single pair. Several pairs are occupying the same space at the same time. Both the reason why electrons in a metal form pairs and an estimate of the energy given up in forming a pair have been a triumph of recent times. This fundamental point in the theory of superconductivity was first explained in the theory of Bardeen, Cooper, and Schrieffer,10 but that is not the subject of this seminar. We will accept, however, the idea that the electrons do, in some manner or other, work in pairs, that we can think of these pairs as behaving more or less like particles, and that we can therefore talk about the wave function for a “pair.” Now the Schrödinger equation for the pair will be more or less like Eq. (21.3). There will be one difference in that the charge $q$ will be twice the charge of an electron. Also, we don’t know the inertia—or effective mass—for the pair in the crystal lattice, so we don’t know what number to put in for $m$. Nor should we think that if we go to very high frequencies (or short wavelengths), this is exactly the right form, because the kinetic energy that corresponds to very rapidly varying wave functions may be so great as to break up the pairs. At finite temperatures there are always a few pairs which are broken up according to the usual Boltzmann theory. The probability that a pair is broken is proportional to $\exp(-E_{\text{pair}}/\kappa T)$. The electrons that are not bound in pairs are called “normal” electrons and will move around in the crystal in the ordinary way. I will, however, consider only the situation at essentially zero temperature—or, in any case, I will disregard the complications produced by those electrons which are not in pairs. Since electron pairs are bosons, when there are a lot of them in a given state there is an especially large amplitude for other pairs to go to the same state. So nearly all of the pairs will be locked down at the lowest energy in exactly the same state—it won’t be easy to get one of them into another state. There’s more amplitude to go into the same state than into an unoccupied state by the famous factor $\sqrt{n}$, where $n-1$ is the occupancy of the lowest state. So we would expect all the pairs to be moving in the same state. What then will our theory look like? I’ll call $\psi$ the wave function of a pair in the lowest energy state. However, since $\psi\psi\cconj$ is going to be proportional to the charge density $\rho$, I can just as well write $\psi$ as the square root of the charge density times some phase factor: \begin{equation} \label{Eq:III:21:17} \psi(\FLPr)=\rho^{1/2}(\FLPr)e^{i\theta(\FLPr)}, \end{equation} where $\rho$ and $\theta$ are real functions of $\FLPr$. (Any complex function can, of course, be written this way.) It’s clear what we mean when we talk about the charge density, but what is the physical meaning of the phase $\theta$ of the wave function? Well, let’s see what happens if we substitute $\psi(\FLPr)$ into Eq. (21.12), and express the current density in terms of these new variables $\rho$ and $\theta$. It’s just a change of variables and I won’t go through all the algebra, but it comes out \begin{equation} \label{Eq:III:21:18} \FLPJ=\frac{\hbar}{m}\biggl( \FLPgrad{\theta}-\frac{q}{\hbar}\,\FLPA\biggr)\rho. \end{equation} Since both the current density and the charge density have a direct physical meaning for the superconducting electron gas, both $\rho$ and $\theta$ are real things. The phase is just as observable as $\rho$; it is a piece of the current density $\FLPJ$. The absolute phase is not observable, but if the gradient of the phase is known everywhere, the phase is known except for a constant. You can define the phase at one point, and then the phase everywhere is determined. Incidentally, the equation for the current can be analyzed a little nicer, when you think that the current density $\FLPJ$ is in fact the charge density times the velocity of motion of the fluid of electrons, or $\rho\FLPv$. Equation (21.18) is then equivalent to \begin{equation} \label{Eq:III:21:19} m\FLPv=\hbar\,\FLPgrad{\theta}-q\FLPA. \end{equation} Notice that there are two pieces in the $mv$-momentum; one is a contribution from the vector potential, and the other, a contribution from the behavior of the wave function. In other words, the quantity $\hbar\,\FLPgrad{\theta}$ is just what we have called the $p$-momentum.

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6

The Meissner effect

Now we can describe some of the phenomena of superconductivity. First, there is no electrical resistance. There’s no resistance because all the electrons are collectively in the same state. In the ordinary flow of current you knock one electron or the other out of the regular flow, gradually deteriorating the general momentum. But here to get one electron away from what all the others are doing is very hard because of the tendency of all Bose particles to go in the same state. A current once started, just keeps on going forever. It’s also easy to understand that if you have a piece of metal in the superconducting state and turn on a magnetic field which isn’t too strong (we won’t go into the details of how strong), the magnetic field can’t penetrate the metal. If, as you build up the magnetic field, any of it were to build up inside the metal, there would be a rate of change of flux which would produce an electric field, and an electric field would immediately generate a current which, by Lenz’s law, would oppose the flux. Since all the electrons will move together, an infinitesimal electric field will generate enough current to oppose completely any applied magnetic field. So if you turn the field on after you’ve cooled a metal to the superconducting state, it will be excluded. Even more interesting is a related phenomenon discovered experimentally by Meissner.11 If you have a piece of the metal at a high temperature (so that it is a normal conductor) and establish a magnetic field through it, and then you lower the temperature below the critical temperature (where the metal becomes a superconductor), the field is expelled. In other words, it starts up its own current—and in just the right amount to push the field out. We can see the reason for that in the equations, and I’d like to explain how. Suppose that we take a piece of superconducting material which is in one lump. Then in a steady situation of any kind the divergence of the current must be zero because there’s no place for it to go. It is convenient to choose to make the divergence of $\FLPA$ equal to zero. (I should explain why choosing this convention doesn’t mean any loss of generality, but I don’t want to take the time.) Taking the divergence of Eq. (21.18), then gives that the Laplacian of $\theta$ is equal to zero. One moment. What about the variation of $\rho$? I forgot to mention an important point. There is a background of positive charge in this metal due to the atomic ions of the lattice. If the charge density $\rho$ is uniform there is no net charge and no electric field. If there would be any accumulation of electrons in one region the charge wouldn’t be neutralized and there would be a terrific repulsion pushing the electrons apart.12 So in ordinary circumstances the charge density of the electrons in the superconductor is almost perfectly uniform—I can take $\rho$ as a constant. Now the only way that $\nabla^2\theta$ can be zero everywhere inside the lump of metal is for $\theta$ to be a constant. And that means that there is no contribution to $\FLPJ$ from $p$-momentum. Equation (21.18) then says that the current is proportional to $\rho$ times $\FLPA$. So everywhere in a lump of superconducting material the current is necessarily proportional to the vector potential: \begin{equation} \label{Eq:III:21:20} \FLPJ=-\rho\,\frac{q}{m}\,\FLPA. \end{equation} Since $\rho$ and $q$ have the same (negative) sign, and since $\rho$ is a constant, I can set $-\rho q/m=-(\text{some positive constant})$; then \begin{equation} \label{Eq:III:21:21} \FLPJ=-(\text{some positive constant})\FLPA. \end{equation} This equation was originally proposed by London and London13 to explain the experimental observations of superconductivity—long before the quantum mechanical origin of the effect was understood. Now we can use Eq. (21.20) in the equations of electromagnetism to solve for the fields. The vector potential is related to the current density by \begin{equation} \label{Eq:III:21:22} \nabla^2\FLPA=-\frac{1}{\epsO c^2}\,\FLPJ. \end{equation} If I use Eq. (21.21) for $\FLPJ$, I have \begin{equation} \label{Eq:III:21:23} \nabla^2\FLPA=\lambda^2\FLPA, \end{equation} where $\lambda^2$ is just a new constant; \begin{equation} \label{Eq:III:21:24} \lambda^2=\rho\,\frac{q}{\epsO mc^2}. \end{equation} We can now try to solve this equation for $\FLPA$ and see what happens in detail. For example, in one dimension Eq. (21.23) has exponential solutions of the form $e^{-\lambda x}$ and $e^{+\lambda x}$. These solutions mean that the vector potential must decrease exponentially as you go from the surface into the material. (It can’t increase because there would be a blow up.) If the piece of metal is very large compared to $1/\lambda$, the field only penetrates to a thin layer at the surface—a layer about $1/\lambda$ in thickness. The entire remainder of the interior is free of field, as sketched in Fig. 21–3. This is the explanation of the Meissner effect. How big is the distance $1/\lambda$? Well, remember that $r_0$, the “electromagnetic radius” of the electron ($2.8\times10^{-13}$ cm), is given by \begin{equation*} mc^2=\frac{q_e^2}{4\pi\epsO r_0}. \end{equation*} Also, remember that $q$ in Eq. (21.24) is twice the charge of an electron, so \begin{equation*} \frac{q}{\epsO mc^2}=\frac{8\pi r_0}{q_e}. \end{equation*} Writing $\rho$ as $q_eN$, where $N$ is the number of electrons per cubic centimeter, we have \begin{equation} \label{Eq:III:21:25} \lambda^2=8\pi Nr_0. \end{equation} For a metal such as lead there are about $3\times10^{22}$ atoms per cm$^3$, so if each one contributed only one conduction electron, $1/\lambda$ would be about $2\times10^{-6}$ cm. That gives you the order of magnitude.

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The Schrödinger Equation in a Classical Context: A Seminar on Superconductivity

7

Flux quantization

The London equation (21.21) was proposed to account for the observed facts of superconductivity including the Meissner effect. In recent times, however, there have been some even more dramatic predictions. One prediction made by London was so peculiar that nobody paid much attention to it until recently. I will now discuss it. This time instead of taking a single lump, suppose we take a ring whose thickness is large compared to $1/\lambda$, and try to see what would happen if we started with a magnetic field through the ring, then cooled it to the superconducting state, and afterward removed the original source of $\FLPB$. The sequence of events is sketched in Fig. 21–4. In the normal state there will be a field in the body of the ring as sketched in part (a) of the figure. When the ring is made superconducting, the field is forced outside of the material (as we have just seen). There will then be some flux through the hole of the ring as sketched in part (b). If the external field is now removed, the lines of field going through the hole are “trapped” as shown in part (c). The flux $\Phi$ through the center can’t decrease because $\ddpl{\Phi}{t}$ must be equal to the line integral of $\FLPE$ around the ring, which is zero in a superconductor. As the external field is removed a super current starts flowing around the ring to keep the flux through the ring a constant. (It’s the old eddy-current idea, only with zero resistance.) These currents will, however, all flow near the surface (down to a depth $1/\lambda$), as can be shown by the same kind of analysis that I made for the solid block. These currents can keep the magnetic field out of the body of the ring, and produce the permanently trapped magnetic field as well. Now, however, there is an essential difference, and our equations predict a surprising effect. The argument I made above that $\theta$ must be a constant in a solid block does not apply for a ring, as you can see from the following arguments. Well inside the body of the ring the current density $\FLPJ$ is zero; so Eq. (21.18) gives \begin{equation} \label{Eq:III:21:26} \hbar\,\FLPgrad{\theta}=q\FLPA. \end{equation} Now consider what we get if we take the line integral of $\FLPA$ around a curve $\Gamma$, which goes around the ring near the center of its cross-section so that it never gets near the surface, as drawn in Fig. 21–5. From Eq. (21.26), \begin{equation} \label{Eq:III:21:27} \hbar\oint\FLPgrad{\theta}\cdot d\FLPs=q\oint\FLPA\cdot d\FLPs. \end{equation} Now you know that the line integral of $\FLPA$ around any loop is equal to the flux of $\FLPB$ through the loop \begin{equation*} \oint\FLPA\cdot d\FLPs=\Phi. \end{equation*} Equation (21.27) then becomes \begin{equation} \label{Eq:III:21:28} \oint\FLPgrad{\theta}\cdot d\FLPs=\frac{q}{\hbar}\,\Phi. \end{equation} The line integral of a gradient from one point to another (say from point $1$ to point $2$) is the difference of the values of the function at the two points. Namely, \begin{equation*} \int_1^2\FLPgrad{\theta}\cdot d\FLPs=\theta_2-\theta_1. \end{equation*} If we let the two end points $1$ and $2$ come together to make a closed loop you might at first think that $\theta_2$ would equal $\theta_1$, so that the integral in Eq. (21.28) would be zero. That would be true for a closed loop in a simply-connected piece of superconductor, but it is not necessarily true for a ring-shaped piece. The only physical requirement we can make is that there can be only one value of the wave function for each point. Whatever $\theta$ does as you go around the ring, when you get back to the starting point the $\theta$ you get must give the same value for the wave function \begin{equation*} \psi=\sqrt{\rho}e^{i\theta}. \end{equation*} This will happen if $\theta$ changes by $2\pi n$, where $n$ is any integer. So if we make one complete turn around the ring the left-hand side of Eq. (21.27) must be $\hbar\cdot2\pi n$. Using Eq. (21.28), I get that \begin{equation} \label{Eq:III:21:29} 2\pi n\hbar=q\Phi. \end{equation} The trapped flux must always be an integer times $2\pi\hbar/q$! If you would think of the ring as a classical object with an ideally perfect (that is, infinite) conductivity, you would think that whatever flux was initially found through it would just stay there—any amount of flux at all could be trapped. But the quantum-mechanical theory of superconductivity says that the flux can be zero, or $2\pi\hbar/q$, or $4\pi\hbar/q$, or $6\pi\hbar/q$, and so on, but no value in between. It must be a multiple of a basic quantum mechanical unit. London14 predicted that the flux trapped by a superconducting ring would be quantized and said that the possible values of the flux would be given by Eq. (21.29) with $q$ equal to the electronic charge. According to London the basic unit of flux should be $2\pi\hbar/q_e$, which is about $4\times10^{-7}$ $\text{gauss}\cdot\text{cm}^2$. To visualize such a flux, think of a tiny cylinder a tenth of a millimeter in diameter; the magnetic field inside it when it contains this amount of flux is about one percent of the earth’s magnetic field. It should be possible to observe such a flux by a sensitive magnetic measurement. In 1961 such a quantized flux was looked for and found by Deaver and Fairbank15 at Stanford University and at about the same time by Doll and Näbauer16 in Germany. In the experiment of Deaver and Fairbank, a tiny cylinder of superconductor was made by electroplating a thin layer of tin on a one-centimeter length of No. 56 ($1.3\times10^{-3}$ cm diameter) copper wire. The tin becomes superconducting below $3.8^\circ$K while the copper remains a normal metal. The wire was put in a small controlled magnetic field, and the temperature reduced until the tin became superconducting. Then the external source of field was removed. You would expect this to generate a current by Lenz’s law so that the flux inside would not change. The little cylinder should now have magnetic moment proportional to the flux inside. The magnetic moment was measured by jiggling the wire up and down (like the needle on a sewing machine, but at the rate of $100$ cycles per second) inside a pair of little coils at the ends of the tin cylinder. The induced voltage in the coils was then a measure of the magnetic moment. When the experiment was done by Deaver and Fairbank, they found that the flux was quantized, but that the basic unit was only one-half as large as London had predicted. Doll and Näbauer got the same result. At first this was quite mysterious,17 but we now understand why it should be so. According to the Bardeen, Cooper, and Schrieffer theory of superconductivity, the $q$ which appears in Eq. (21.29) is the charge of a pair of electrons and so is equal to $2q_e$. The basic flux unit is \begin{equation} \label{Eq:III:21:30} \Phi_0=\frac{\pi\hbar}{q_e}\approx2\times10^{-7}\text{ gauss}\cdot\text{cm}^2 \end{equation} or one-half the amount predicted by London. Everything now fits together, and the measurements show the existence of the predicted purely quantum-mechanical effect on a large scale.

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The Schrödinger Equation in a Classical Context: A Seminar on Superconductivity

8

The dynamics of superconductivity

The Meissner effect and the flux quantization are two confirmations of our general ideas. Just for the sake of completeness I would like to show you what the complete equations of a superconducting fluid would be from this point of view—it is rather interesting. Up to this point I have only put the expression for $\psi$ into equations for charge density and current. If I put it into the complete Schrödinger equation I get equations for $\rho$ and $\theta$. It should be interesting to see what develops, because here we have a “fluid” of electron pairs with a charge density $\rho$ and a mysterious $\theta$—we can try to see what kind of equations we get for such a “fluid”! So we substitute the wave function of Eq. (21.17) into the Schrödinger equation (21.3) and remember that $\rho$ and $\theta$ are real functions of $x$, $y$, $z$, and $t$. If we separate real and imaginary parts we obtain then two equations. To write them in a shorter form I will—following Eq. (21.19)—write \begin{equation} \label{Eq:III:21:31} \frac{\hbar}{m}\,\FLPgrad{\theta}-\frac{q}{m}\,\FLPA=\FLPv. \end{equation} One of the equations I get is then \begin{equation} \label{Eq:III:21:32} \ddp{\rho}{t}=-\FLPdiv{\rho\FLPv}. \end{equation} Since $\rho\FLPv$ is first $\FLPJ$, this is just the continuity equation once more. The other equation I obtain tells how $\theta$ varies; it is \begin{equation} \label{Eq:III:21:33} \hbar\,\ddp{\theta}{t}=-\frac{m}{2}\,v^2-q\phi+ \frac{\hbar^2}{2m}\biggl\{ \frac{1}{\sqrt{\rho}}\,\nabla^2(\sqrt{\rho})\biggr\}. \end{equation} Those who are thoroughly familiar with hydrodynamics (of which I’m sure few of you are) will recognize this as the equation of motion for an electrically charged fluid if we identify $\hbar\theta$ as the “velocity potential”—except that the last term, which should be the energy of compression of the fluid, has a rather strange dependence on the density $\rho$. In any case, the equation says that the rate of change of the quantity $\hbar\theta$ is given by a kinetic energy term, $-\tfrac{1}{2}mv^2$, plus a potential energy term, $-q\phi$, with an additional term, containing the factor $\hbar^2$, which we could call a “quantum mechanical energy.” We have seen that inside a superconductor $\rho$ is kept very uniform by the electrostatic forces, so this term can almost certainly be neglected in every practical application provided we have only one superconducting region. If we have a boundary between two superconductors (or other circumstances in which the value of $\rho$ may change rapidly) this term can become important. For those who are not so familiar with the equations of hydrodynamics, I can rewrite Eq. (21.33) in a form that makes the physics more apparent by using Eq. (21.31) to express $\theta$ in terms of $\FLPv$. Taking the gradient of the whole of Eq. (21.33) and expressing $\FLPgrad{\theta}$ in terms of $\FLPA$ and $\FLPv$ by using (21.31), I get \begin{equation} \label{Eq:III:21:34} \ddp{\FLPv}{t}=\frac{q}{m}\biggl(-\FLPgrad{\phi}-\ddp{\FLPA}{t}\biggr)- \FLPv\times(\FLPcurl{\FLPv})-(\FLPv\cdot\FLPnabla)\FLPv+ \FLPgrad{\frac{\hbar^2}{2m^2} \biggl(\frac{1}{\sqrt{\rho}}\,\nabla^2\sqrt{\rho}\biggr)}. \end{equation} \begin{align} \ddp{\FLPv}{t}&=\frac{q}{m}\biggl(-\FLPgrad{\phi}-\ddp{\FLPA}{t}\biggr) -\FLPv\times(\FLPcurl{\FLPv})\notag\\[1ex] \label{Eq:III:21:34} &-(\FLPv\cdot\FLPnabla)\FLPv + \FLPgrad{\frac{\hbar^2}{2m^2} \biggl(\frac{1}{\sqrt{\rho}}\,\nabla^2\sqrt{\rho}\biggr)}. \end{align} What does this equation mean? First, remember that \begin{equation} \label{Eq:III:21:35} -\FLPgrad{\phi}-\ddp{\FLPA}{t}=\FLPE. \end{equation} Next, notice that if I take the curl of Eq. (21.31), I get \begin{equation} \label{Eq:III:21:36} \FLPcurl{\FLPv}=-\frac{q}{m}\,\FLPcurl{\FLPA}, \end{equation} since the curl of a gradient is always zero. But $\FLPcurl{\FLPA}$ is the magnetic field $\FLPB$, so the first two terms can be written as \begin{equation*} \frac{q}{m}(\FLPE+\FLPv\times\FLPB). \end{equation*} Finally, you should understand that $\ddpl{\FLPv}{t}$ stands for the rate of change of the velocity of the fluid at a point. If you concentrate on a particular particle, its acceleration is the total derivative of $\FLPv$ (or, as it is sometimes called in fluid dynamics, the “comoving acceleration”), which is related to $\ddpl{\FLPv}{t}$ by18 \begin{equation} \label{Eq:III:21:37} \left.\ddt{\FLPv}{t}\right|_{\text{comoving}}\kern{-2ex}= \ddp{\FLPv}{t}+(\FLPv\cdot\FLPnabla)\FLPv. \end{equation} This extra term also appears as the third term on the right side of Eq. (21.34). Taking it to the left side, I can write Eq. (21.34) in the following way: \begin{equation} \label{Eq:III:21:38} \left.m\ddt{\FLPv}{t}\right|_{\text{comoving}}\kern{-3.5ex}= q(\FLPE\!+\!\FLPv\!\times\!\FLPB)\!+\!\FLPgrad{\frac{\hbar^2}{2m} \!\biggl(\!\frac{1}{\sqrt{\rho}}\nabla^2\!\!\sqrt{\rho}\!\biggr)}. \end{equation} We also have from Eq. (21.36) that \begin{equation} \label{Eq:III:21:39} \FLPcurl{\FLPv}=-\frac{q}{m}\,\FLPB. \end{equation} These two equations are the equations of motion of the superconducting electron fluid. The first equation is just Newton’s law for a charged fluid in an electromagnetic field. It says that the acceleration of each particle of the fluid whose charge is $q$ comes from the ordinary Lorentz force $q(\FLPE+\FLPv\times\FLPB)$ plus an additional force, which is the gradient of some mystical quantum mechanical potential—a force which is not very big except at the junction between two superconductors. The second equation says that the fluid is “ideal”—the curl of $\FLPv$ has zero divergence (the divergence of $\FLPB$ is always zero). That means that the velocity can be expressed in terms of velocity potential. Ordinarily one writes that $\FLPcurl{\FLPv}=\FLPzero$ for an ideal fluid, but for an ideal charged fluid in a magnetic field, this gets modified to Eq. (21.39). So, Schrödinger’s equation for the electron pairs in a superconductor gives us the equations of motion of an electrically charged ideal fluid. Superconductivity is the same as the problem of the hydrodynamics of a charged liquid. If you want to solve any problem about superconductors you take these equations for the fluid [or the equivalent pair, Eqs. (21.32) and (21.33)], and combine them with Maxwell’s equations to get the fields. (The charges and currents you use to get the fields must, of course, include the ones from the superconductor as well as from the external sources.) Incidentally, I believe that Eq. (21.38) is not quite correct, but ought to have an additional term involving the density. This new term does not depend on quantum mechanics, but comes from the ordinary energy associated with variations of density. Just as in an ordinary fluid there should be a potential energy density proportional to the square of the deviation of $\rho$ from $\rho_0$, the undisturbed density (which is, here, also equal to the charge density of the crystal lattice). Since there will be forces proportional to the gradient of this energy, there should be another term in Eq. (21.38) of the form: $(\text{const})\,\FLPgrad{(\rho-\rho_0)^2}$. This term did not appear from the analysis because it comes from the interactions between particles, which I neglected in using an independent-particle approximation. It is, however, just the force I referred to when I made the qualitative statement that electrostatic forces would tend to keep $\rho$ nearly constant inside a superconductor.

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The Schrödinger Equation in a Classical Context: A Seminar on Superconductivity

9

The Josephson junction

I would like to discuss next a very interesting situation that was noticed by Josephson19 while analyzing what might happen at a junction between two superconductors. Suppose we have two superconductors which are connected by a thin layer of insulating material as in Fig. 21–6. Such an arrangement is now called a “Josephson junction.” If the insulating layer is thick, the electrons can’t get through; but if the layer is thin enough, there can be an appreciable quantum mechanical amplitude for electrons to jump across. This is just another example of the quantum-mechanical penetration of a barrier. Josephson analyzed this situation and discovered that a number of strange phenomena should occur. In order to analyze such a junction I’ll call the amplitude to find an electron on one side, $\psi_1$, and the amplitude to find it on the other, $\psi_2$. In the superconducting state the wave function $\psi_1$ is the common wave function of all the electrons on one side, and $\psi_2$ is the corresponding function on the other side. I could do this problem for different kinds of superconductors, but let us take a very simple situation in which the material is the same on both sides so that the junction is symmetrical and simple. Also, for a moment let there be no magnetic field. Then the two amplitudes should be related in the following way: \begin{align*} i\hbar\,\ddp{\psi_1}{t}&=U_1\psi_1+K\psi_2,\\[1ex] i\hbar\,\ddp{\psi_2}{t}&=U_2\psi_2+K\psi_1. \end{align*} The constant $K$ is a characteristic of the junction. If $K$ were zero, these two equations would just describe the lowest energy state—with energy $U$—of each superconductor. But there is coupling between the two sides by the amplitude $K$ that there may be leakage from one side to the other. (It is just the “flip-flop” amplitude of a two-state system.) If the two sides are identical, $U_1$ would equal $U_2$ and I could just subtract them off. But now suppose that we connect the two superconducting regions to the two terminals of a battery so that there is a potential difference $V$ across the junction. Then $U_1-U_2=qV$. I can, for convenience, define the zero of energy to be halfway between, then the two equations are \begin{equation} \begin{aligned} i\hbar\,\ddp{\psi_1}{t}&=+\frac{qV}{2}\,\psi_1+K\psi_2,\\[1ex] i\hbar\,\ddp{\psi_2}{t}&=-\frac{qV}{2}\,\psi_2+K\psi_1. \end{aligned} \label{Eq:III:21:40} \end{equation} These are the standard equations for two quantum mechanical states coupled together. This time, let’s analyze these equations in another way. Let’s make the substitutions \begin{equation} \begin{aligned} \psi_1&=\sqrt{\rho_1}e^{i\theta_1},\\[1ex] \psi_2&=\sqrt{\rho_2}e^{i\theta_2}, \end{aligned} \label{Eq:III:21:41} \end{equation} where $\theta_1$ and $\theta_2$ are the phases on the two sides of the junction and $\rho_1$ and $\rho_2$ are the density of electrons at those two points. Remember that in actual practice $\rho_1$ and $\rho_2$ are almost exactly the same and are equal to $\rho_0$, the normal density of electrons in the superconducting material. Now if you substitute these equations for $\psi_1$ and $\psi_2$ into (21.40), you get four equations by equating the real and imaginary parts in each case. Letting $(\theta_2-\theta_1)=\delta$, for short, the result is \begin{align} &\begin{aligned} \dot{\rho}_1&=+\frac{2}{\hbar}\,K\sqrt{\rho_2\rho_1}\sin\delta,\\[1.5ex] \dot{\rho}_2&=-\frac{2}{\hbar}\,K\sqrt{\rho_2\rho_1}\sin\delta, \end{aligned}\\[3ex] \label{Eq:III:21:42} &\begin{aligned} \dot{\theta}_1&=-\frac{K}{\hbar}\sqrt{\frac{\rho_2}{\rho_1}}\cos\delta- \frac{qV}{2\hbar},\\[1.5ex] \dot{\theta}_2&=-\frac{K}{\hbar}\sqrt{\frac{\rho_1}{\rho_2}}\cos\delta+ \frac{qV}{2\hbar}. \end{aligned} \label{Eq:III:21:43} \end{align} The first two equations say that $\dot{\rho}_1=-\dot{\rho}_2$. “But,” you say, “they must both be zero if $\rho_1$ and $\rho_2$ are both constant and equal to $\rho_0$.” Not quite. These equations are not the whole story. They say what $\dot{\rho}_1$ and $\dot{\rho}_2$ would be if there were no extra electric forces due to an unbalance between the electron fluid and the background of positive ions. They tell how the densities would start to change, and therefore describe the kind of current that would begin to flow. This current from side $1$ to side $2$ would be just $\dot{\rho}_1$ (or $-\dot{\rho}_2$), or \begin{equation} \label{Eq:III:21:44} J=\frac{2K}{\hbar}\sqrt{\rho_1\rho_2}\sin\delta. \end{equation} Such a current would soon charge up side $2$, except that we have forgotten that the two sides are connected by wires to the battery. The current that flows will not charge up region $2$ (or discharge region $1$) because currents will flow to keep the potential constant. These currents from the battery have not been included in our equations. When they are included, $\rho_1$ and $\rho_2$ do not in fact change, but the current across the junction is still given by Eq. (21.44). Since $\rho_1$ and $\rho_2$ do remain constant and equal to $\rho_0$, let’s set $2K\rho_0/\hbar=J_0$, and write \begin{equation} \label{Eq:III:21:45} J=J_0\sin\delta. \end{equation} $J_0$, like $K$, is then a number which is a characteristic of the particular junction. The other pair of equations (21.43) tells us about $\theta_1$ and $\theta_2$. We are interested in the difference $\delta=\theta_2-\theta_1$ to use Eq. (21.45); what we get is \begin{equation} \label{Eq:III:21:46} \dot{\delta}=\dot{\theta}_2-\dot{\theta}_1=\frac{qV}{\hbar}. \end{equation} That means that we can write \begin{equation} \label{Eq:III:21:47} \delta(t)=\delta_0+\frac{q}{\hbar}\int V(t)\,dt, \end{equation} where $\delta_0$ is the value of $\delta$ at $t=0$. Remember also that $q$ is the charge of a pair, namely, $q=2q_e$. In Eqs. (21.45) and (21.47) we have an important result, the general theory of the Josephson junction. Now what are the consequences? First, put on a dc voltage. If you put on a dc voltage, $V_0$, the argument of the sine becomes $(\delta_0+(q/\hbar)V_0t)$. Since $\hbar$ is a small number (compared to ordinary voltage and times), the sine oscillates rather rapidly and the net current is nothing. (In practice, since the temperature is not zero, you would get a small current due to the conduction by “normal” electrons.) On the other hand if you have zero voltage across the junction, you can get a current! With no voltage the current can be any amount between $+J_0$ and $-J_0$ (depending on the value of $\delta_0$). But try to put a voltage across it and the current goes to zero. This strange behavior has recently been observed experimentally.20 There is another way of getting a current—by applying a voltage at a very high frequency in addition to a dc voltage. Let \begin{equation*} V=V_0+v\cos\omega t, \end{equation*} where $v\ll V_0$. Then $\delta(t)$ is \begin{equation*} \delta_0+\frac{q}{\hbar}\,V_0t+\frac{q}{\hbar}\,\frac{v}{\omega}\sin\omega t. \end{equation*} Now for $\Delta x$ small, \begin{equation*} \sin\,(x+\Delta x)\approx\sin x+\Delta x\cos x. \end{equation*} Using this approximation for $\sin\delta$, I get \begin{equation*} J=\!J_0\Bigl[\sin\Bigl(\!\delta_0\!+\!\frac{q}{\hbar}V_0t\!\Bigr)\!+\! \frac{q}{\hbar}\frac{v}{\omega}\sin\omega t \cos\Bigl(\!\delta_0\!+\!\frac{q}{\hbar}V_0t\!\Bigr)\Bigr]. \end{equation*} The first term is zero on the average, but the second term is not if \begin{equation*} \omega=\frac{q}{\hbar}\,V_0. \end{equation*} There should be a current if the ac voltage has just this frequency. Shapiro21 claims to have observed such a resonance effect. If you look up papers on the subject you will find that they often write the formula for the current as \begin{equation} \label{Eq:III:21:48} J=J_0\sin\biggl(\delta_0+\frac{2q_e}{\hbar}\int\FLPA\cdot d\FLPs\biggr), \end{equation} where the integral is to be taken across the junction. The reason for this is that when there’s a vector potential across the junction the flip-flop amplitude is modified in phase in the way that we explained earlier. If you chase that extra phase through, it comes out as given above. Finally, I would like to describe a very dramatic and interesting experiment which has recently been made on the interference of the currents from each of two junctions. In quantum mechanics we’re used to the interference between amplitudes from two different slits. Now we’re going to do the interference between two junctions caused by the difference in the phase of the arrival of the currents through two different paths. In Fig. 21–7, I show two different junctions, “a” and “b”, connected in parallel. The ends, $P$ and $Q$, are connected to our electrical instruments which measure any current flow. The external current, $J_{\text{total}}$, will be the sum of the currents through the two junctions. Let $J_{\text{a}}$ and $J_{\text{b}}$ be the currents through the two junctions, and let their phases be $\delta_{\text{a}}$ and $\delta_{\text{b}}$. Now the phase difference of the wave functions between $P$ and $Q$ must be the same whether you go on one route or the other. Along the route through junction “a”, the phase difference between $P$ and $Q$ is $\delta_{\text{a}}$ plus the line integral of the vector potential along the upper route: \begin{equation} \label{Eq:III:21:49} \Delta\text{Phase}_{P\to Q}=\delta_{\text{a}}+ \frac{2q_e}{\hbar}\int_{\text{upper}}\kern{-3ex}\FLPA\cdot d\FLPs. \end{equation} Why? Because the phase $\theta$ is related to $\FLPA$ by Eq. (21.26). If you integrate that equation along some path, the left-hand side gives the phase change, which is then just proportional to the line integral of $\FLPA$, as we have written here. The phase change along the lower route can be written similarly \begin{equation} \label{Eq:III:21:50} \Delta\text{Phase}_{P\to Q}=\delta_{\text{b}}+ \frac{2q_e}{\hbar}\int_{\text{lower}}\kern{-3ex}\FLPA\cdot d\FLPs. \end{equation} These two must be equal; and if I subtract them I get that the difference of the deltas must be the line integral of $\FLPA$ around the circuit: \begin{equation*} \delta_{\text{b}}-\delta_{\text{a}}= \frac{2q_e}{\hbar}\oint_\Gamma\FLPA\cdot d\FLPs. \end{equation*} Here the integral is around the closed loop $\Gamma$ of Fig. 21–7 which circles through both junctions. The integral over $\FLPA$ is the magnetic flux $\Phi$ through the loop. So the two $\delta$’s are going to differ by $2q_e/\hbar$ times the magnetic flux $\Phi$ which passes between the two branches of the circuit: \begin{equation} \label{Eq:III:21:51} \delta_{\text{b}}-\delta_{\text{a}}=\frac{2q_e}{\hbar}\,\Phi. \end{equation} I can control this phase difference by changing the magnetic field on the circuit, so I can adjust the differences in phases and see whether or not the total current that flows through the two junctions shows any interference of the two parts. The total current will be the sum of $J_{\text{a}}$ and $J_{\text{b}}$. For convenience, I will write \begin{equation*} \delta_{\text{a}}=\delta_0-\frac{q_e}{\hbar}\,\Phi,\quad \delta_{\text{b}}=\delta_0+\frac{q_e}{\hbar}\,\Phi. \end{equation*} Then, \begin{align} J_{\text{total}} &=J_0\biggl\{\!\sin\biggl(\! \delta_0\!-\!\frac{q_e}{\hbar}\Phi\!\biggr)\!+\sin\biggl(\! \delta_0\!+\!\frac{q_e}{\hbar}\,\Phi\!\biggr)\!\biggr\}\notag\\[1.5ex] \label{Eq:III:21:52} &=2J_0\sin\delta_0\cos\frac{q_e\Phi}{\hbar}. \end{align} Now we don’t know anything about $\delta_0$, and nature can adjust that anyway she wants depending on the circumstances. In particular, it will depend on the external voltage we apply to the junction. No matter what we do, however, $\sin\delta_0$ can never get bigger than $1$. So the maximum current for any given $\Phi$ is given by \begin{equation*} J_{\text{max}}=2J_0\left\lvert \cos\frac{q_e}{\hbar}\,\Phi\right\rvert. \end{equation*} This maximum current will vary with $\Phi$ and will itself have maxima whenever \begin{equation*} \Phi=n\,\frac{\pi\hbar}{q_e}, \end{equation*} with $n$ some integer. That is to say that the current takes on its maximum values where the flux linkage has just those quantized values we found in Eq. (21.30)! The Josephson current through a double junction was recently measured22 as a function of the magnetic field in the area between the junctions. The results are shown in Fig. 21–8. There is a general background of current from various effects we have neglected, but the rapid oscillations of the current with changes in the magnetic field are due to the interference term $\cos q_e\Phi/\hbar$ of Eq. (21.52). One of the intriguing questions about quantum mechanics is the question of whether the vector potential exists in a place where there’s no field.23 This experiment I have just described has also been done with a tiny solenoid between the two junctions so that the only significant magnetic $\FLPB$ field is inside the solenoid and a negligible amount is on the superconducting wires themselves. Yet it is reported that the amount of current depends oscillatorily on the flux of magnetic field inside that solenoid even though that field never touches the wires—another demonstration of the “physical reality” of the vector potential.24 I don’t know what will come next. But look what can be done. First, notice that the interference between two junctions can be used to make a sensitive magnetometer. If a pair of junctions is made with an enclosed area of, say, $1$ mm$^2$, the maxima in the curve of Fig. 21–8 would be separated by $2\times10^{-6}$ gauss. It is certainly possible to tell when you are $1/10$ of the way between two peaks; so it should be possible to use such a junction to measure magnetic fields as small as $2\times10^{-7}$ gauss—or to measure larger fields to such a precision. One should be able to go even further. Suppose for example we put a set of $10$ or $20$ junctions close together and equally spaced. Then we can have the interference between $10$ or $20$ slits and as we change the magnetic field we will get very sharp maxima and minima. Instead of a $2$-slit interference we can have a $20$- or perhaps even a $100$-slit interferometer for measuring the magnetic field. Perhaps we can predict that the measurement of magnetic fields will—by using the effects of quantum-mechanical interference—eventually become almost as precise as the measurement of wavelength of light. These then are some illustrations of things that are happening in modern times—the transistor, the laser, and now these junctions, whose ultimate practical applications are still not known. The quantum mechanics which was discovered in 1926 has had nearly 40 years of development, and rather suddenly it has begun to be exploited in many practical and real ways. We are really getting control of nature on a very delicate and beautiful level. I am sorry to say, gentlemen, that to participate in this adventure it is absolutely imperative that you learn quantum mechanics as soon as possible. It was our hope that in this course we would find a way to make comprehensible to you at the earliest possible moment the mysteries of this part of physics.