Angular momentum
Let’s for fun look at another operation—the operation of orbital angular momentum. In Chapter 17 we defined an operator $\Jop_z$ in terms of $\Rop_z(\phi)$, the operator of a rotation by the angle $\phi$ about the $z$-axis. We consider here a system described simply by a single wave function $\psi(\FLPr)$, which is a function of coordinates only, and does not take into account the fact that the electron may have its spin either up or down. That is, we want for the moment to disregard intrinsic angular momentum and think about only the orbital part. To keep the distinction clear, we’ll call the orbital operator $\Lop_z$, and define it in terms of the operator of a rotation by an infinitesimal angle $\epsilon$ by \begin{equation*} \Rop_z(\epsilon)\,\ket{\psi}= \biggl(1+\frac{i}{\hbar}\,\epsilon\,\Lop_z\biggr)\ket{\psi}. \end{equation*}  (Remember, this definition applies only to a state $\ket{\psi}$ which has no internal spin variables, but depends only on the coordinates $\FLPr=x,y,z$.) If we look at the state $\ket{\psi}$ in a new coordinate system, rotated about the $z$-axis by the small angle $\epsilon$, we see a new state \begin{equation*} \ket{\psi'}=\Rop_z(\epsilon)\,\ket{\psi}. \end{equation*}  If we choose to describe the state $\ket{\psi}$ in the coordinate representation—that is, by its wave function $\psi(\FLPr)$, we would expect to be able to write \begin{equation} \label{Eq:III:20:68} \psi'(\FLPr)= \biggl(1+\frac{i}{\hbar}\,\epsilon\,\Lcalop_z\biggr)\psi(\FLPr). \end{equation} What is $\Lcalop_z$? Well, a point $P$ at $x$ and $y$ in the new coordinate system (really $x'$ and $y'$, but we will drop the primes) was formerly at $x-\epsilon y$ and $y+\epsilon x$, as you can see from Fig. 20–2. Since the amplitude for the electron to be at $P$ isn’t changed by the rotation of the coordinates we can write  \begin{equation*} \psi'(x,y,z)=\psi(x-\epsilon y,y+\epsilon x,z)= \psi(x,y,z)-\epsilon y\,\ddp{\psi}{x}+ \epsilon x\,\ddp{\psi}{y} \end{equation*} 
 \begin{align*} \psi'(x,y,z)&=\psi(x-\epsilon y,y+\epsilon x,z)\\[1ex] &=\psi(x,y,z)-\epsilon y\,\ddp{\psi}{x}+ \epsilon x\,\ddp{\psi}{y} \end{align*}  (remembering that $\epsilon$ is a small angle). This means that \begin{equation} \label{Eq:III:20:69} \Lcalop_z=\frac{\hbar}{i}\biggl(x\,\ddp{}{y}-y\,\ddp{}{x}\biggr). \end{equation} That’s our answer. But notice. It is equivalent to \begin{equation} \label{Eq:III:20:70} \Lcalop_z=x\Pcalop_y-y\Pcalop_x. \end{equation} Returning to our quantum-mechanical operators, we can write \begin{equation} \label{Eq:III:20:71} \Lop_z=x\pop_y-y\pop_x. \end{equation} This formula is easy to remember because it looks like the familiar formula of classical mechanics; it is the $z$-component of \begin{equation} \label{Eq:III:20:72} \FLPL=\FLPr\times\FLPp. \end{equation}  One of the fun parts of this operator business is that many classical equations get carried over into a quantum-mechanical form. Which ones don’t? There had better be some that don’t come out right, because if everything did, then there would be nothing different about quantum mechanics. There would be no new physics. Here is one equation which is different. In classical physics \begin{equation*} xp_x-p_xx=0. \end{equation*} What is it in quantum mechanics? \begin{equation*} \xop\pop_x-\pop_x\xop=? \end{equation*} Let’s work it out in the $x$-representation. So that we’ll know what we are doing we put in some wave function $\psi(x)$. We have \begin{equation*} x\Pcalop_x\psi(x)-\Pcalop_xx\psi(x), \end{equation*} or \begin{equation*} x\,\frac{\hbar}{i}\,\ddp{}{x}\,\psi(x)- \frac{\hbar}{i}\,\ddp{}{x}\,x\psi(x). \end{equation*} Remember now that the derivatives operate on everything to the right. We get \begin{equation} \label{Eq:III:20:73} x\,\frac{\hbar}{i}\,\ddp{\psi}{x}- \frac{\hbar}{i}\,\psi(x)- \frac{\hbar}{i}\,x\,\ddp{\psi}{x}=-\frac{\hbar}{i}\,\psi(x). \end{equation} The answer is not zero. The whole operation is equivalent simply to multiplication by $-\hbar/i$: \begin{equation} \label{Eq:III:20:74} \xop\pop_x-\pop_x\xop=-\frac{\hbar}{i}. \end{equation} If Planck’s constant were zero, the classical and quantum results would be the same, and there would be no quantum mechanics to learn!  Incidentally, if any two operators $\Aop$ and $\Bop$, when taken together like this: \begin{equation*} \Aop\Bop-\Bop\Aop, \end{equation*} do not give zero, we say that “the operators do not commute.” And an equation such as (20.74) is called a “commutation rule.” You can see that the commutation rule for $p_x$ and $y$ is \begin{equation*} \pop_x\yop-\yop\pop_x=0. \end{equation*}  There is another very important commutation rule that has to do with angular momenta. It is \begin{equation} \label{Eq:III:20:75} \Lop_x\Lop_y-\Lop_y\Lop_x=i\hbar\Lop_z. \end{equation} You can get some practice with $\xop$ and $\pop$ operators by proving it for yourself. It is interesting to notice that operators which do not commute can also occur in classical physics. We have already seen this when we have talked about rotation in space. If you rotate something, such as a book, by $90^\circ$ around $x$ and then $90^\circ$ around $y$, you get something different from rotating first by $90^\circ$ around $y$ and then by $90^\circ$ around $x$. It is, in fact, just this property of space that is responsible for Eq. (20.75).