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.