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. 