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. 