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. 