The change of averages with time
Now we want to show you something else. How do averages change with time? Suppose for the moment that we have an operator $\Aop$, which does not itself have time in it in any obvious way. We mean an operator like $\xop$ or $\pop$. (We exclude things like, say, the operator of some external potential that was being varied with time, such as $V(x,t)$.)  Now suppose we calculate $\av{A}$, in some state $\ket{\psi}$, which is \begin{equation} \label{Eq:III:20:76} \av{A}=\bracket{\psi}{\Aop}{\psi}. \end{equation} How will $\av{A}$, depend on time? Why should it? One reason might be that the operator itself depended explicitly on time—for instance, if it had to do with a time-varying potential like $V(x,t)$. But even if the operator does not depend on $t$, say, for example, the operator $\Aop=\xop$, the corresponding average may depend on time. Certainly the average position of a particle could be moving. How does such a motion come out of Eq. (20.76) if $\Aop$ has no time dependence? Well, the state $\ket{\psi}$ might be changing with time. For nonstationary states we have often shown a time dependence explicitly by writing a state as $\ket{\psi(t)}$. We want to show that the rate of change of $\av{A}$, is given by a new operator we will call $\Adotop$. Remember that $\Aop$ is an operator, so that putting a dot over the $A$ does not here mean taking the time derivative, but is just a way of writing a new operator $\Adotop$ which is defined by \begin{equation} \label{Eq:III:20:77} \ddt{}{t}\,\av{A}=\bracket{\psi}{\Adotop}{\psi}. \end{equation} Our problem is to find the operator $\Adotop$.  First, we know that the rate of change of a state is given by the Hamiltonian. Specifically, \begin{equation} \label{Eq:III:20:78} i\hbar\,\ddt{}{t}\,\ket{\psi(t)}=\Hop\,\ket{\psi(t)}. \end{equation} This is just the abstract way of writing our original definition of the Hamiltonian: \begin{equation} \label{Eq:III:20:79} i\hbar\,\ddt{C_i}{t}=\sum_jH_{ij}C_j. \end{equation} If we take the complex conjugate of Eq. (20.78), it is equivalent to \begin{equation} \label{Eq:III:20:80} -i\hbar\,\ddt{}{t}\,\bra{\psi(t)}=\bra{\psi(t)}\,\Hop. \end{equation} Next, see what happens if we take the derivatives with respect to $t$ of Eq. (20.76). Since each $\psi$ depends on $t$, we have  \begin{equation} \label{Eq:III:20:81} \ddt{}{t}\av{A}=\!\biggl(\!\ddt{}{t}\bra{\psi}\!\biggr)\Aop\ket{\psi}\!+\! \bra{\psi}\Aop\biggl(\!\ddt{}{t}\ket{\psi}\!\biggr). \end{equation}   Finally, using the two equations in (20.78) and (20.80) to replace the derivatives, we get \begin{equation*} \ddt{}{t}\,\av{A}=\frac{i}{\hbar}\,\{ \bracket{\psi}{\Hop\Aop}{\psi}- \bracket{\psi}{\Aop\Hop}{\psi}\}. \end{equation*} This equation is the same as \begin{equation*} \ddt{}{t}\,\av{A}=\frac{i}{\hbar}\, \bracket{\psi}{\Hop\Aop-\Aop\Hop}{\psi}. \end{equation*} Comparing this equation with Eq. (20.77), you see that \begin{equation} \label{Eq:III:20:82} \Adotop=\frac{i}{\hbar}\,(\Hop\Aop-\Aop\Hop). \end{equation} That is our interesting proposition, and it is true for any operator $\Aop$.  Incidentally, if the operator $\Aop$ should itself be time dependent, we would have had \begin{equation} \label{Eq:III:20:83} \Adotop=\frac{i}{\hbar}\,(\Hop\Aop-\Aop\Hop) +\ddp{\Aop}{t}. \end{equation}  Let us try out Eq. (20.82) on some example to see whether it really makes sense. For instance, what operator corresponds to $\xdotop$? We say it should be \begin{equation} \label{Eq:III:20:84} \xdotop=\frac{i}{\hbar}\,(\Hop\xop-\xop\Hop). \end{equation} What is this? One way to find out is to work it through in the coordinate representation using the algebraic operator for $\Hcalop$. In this representation the commutator is  \begin{equation*} \Hcalop x-x\Hcalop=\biggl\{-\frac{\hbar^2}{2m}\,\frac{d^2}{dx^2}+ V(x)\biggr\}x-x\biggl\{-\frac{\hbar^2}{2m}\,\frac{d^2}{dx^2}+ V(x)\biggr\}. \end{equation*} 
 \begin{align*} \Hcalop x&-x\Hcalop\,=\\[1.5ex] \!\biggl\{\!-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\!+\! V(x)\!\biggr\}x&-x\biggl\{\!-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\!+\! V(x)\!\biggr\}. \end{align*}  If you operate with this on any wave function $\psi(x)$ and work out all of the derivatives where you can, you end up after a little work with \begin{equation} -\frac{\hbar^2}{m}\,\ddt{\psi}{x}.\notag \end{equation} But this is just the same as \begin{equation} -i\,\frac{\hbar}{m}\,\Pcalop_x\psi,\notag \end{equation} so we find that \begin{equation} \label{Eq:III:20:85} \Hop\xop-\xop\Hop=-i\,\frac{\hbar}{m}\,\pop_x \end{equation} or that \begin{equation} \label{Eq:III:20:86} \xdotop=\frac{\pop_x}{m}. \end{equation} A pretty result. It means that if the mean value of $x$ is changing with time the drift of the center of gravity is the same as the mean momentum divided by $m$. Exactly like classical mechanics.  Another example. What is the rate of change of the average momentum of a state? Same game. Its operator is \begin{equation} \label{Eq:III:20:87} \pdotop=\frac{i}{\hbar}\,(\Hop\pop-\pop\Hop). \end{equation} Again you can work it out in the $x$ representation. Remember that $\pop$ becomes $d/dx$, and this means that you will be taking the derivative of the potential energy $V$ (in the $\Hcalop$)—but only in the second term. It turns out that it is the only term which does not cancel, and you find that \begin{equation} \Hcalop\Pcalop-\Pcalop\Hcalop=i\hbar\,\ddt{V}{x}\notag \end{equation} or that \begin{equation} \label{Eq:III:20:88} \pdotop=-\ddt{V}{x}. \end{equation} Again the classical result. The right-hand side is the force, so we have derived Newton’s law! But remember—these are the laws for the operators which give the average quantities. They do not describe what goes on in detail inside an atom.  Quantum mechanics has the essential difference that $\pop\xop$ is not equal to $\xop\pop$. They differ by a little bit—by the small number $i\hbar$. But the whole wondrous complications of interference, waves, and all, result from the little fact that $\xop\pop-\pop\xop$ is not quite zero.  The history of this idea is also interesting. Within a period of a few months in 1926, Heisenberg and Schrödinger independently found correct laws to describe atomic mechanics. Schrödinger invented his wave function $\psi(x)$ and found his equation. Heisenberg, on the other hand, found that nature could be described by classical equations, except that $xp-px$ should be equal to $i\hbar$, which he could make happen by defining them in terms of special kinds of matrices. In our language he was using the energy-representation, with its matrices. Both Heisenberg’s matrix algebra and Schrödinger’s differential equation explained the hydrogen atom. A few months later Schrödinger was able to show that the two theories were equivalent—as we have seen here. But the two different mathematical forms of quantum mechanics were discovered independently. 