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. 