Forces; the classical limit
Suppose that we have a particle moving along and passing through a region where there is a potential that varies at right angles to the motion. Classically, we would describe the situation as sketched in Fig. 7–7. If the particle is moving along the $x$-direction and enters a region where there is a potential that varies with $y$, the particle will get a transverse acceleration from the force $F=-\ddpl{V}{y}$. If the force is present only in a limited region of width $w$, the force will act only for the time $w/v$. The particle will be given the transverse momentum \begin{equation*} p_y=F\,\frac{w}{v}. \end{equation*}  The angle of deflection $\delta\theta$ is then \begin{equation*} \delta\theta=\frac{p_y}{p}=\frac{Fw}{pv}, \end{equation*}  where $p$ is the initial momentum. Using $-\ddpl{V}{y}$ for $F$, we get \begin{equation} \label{Eq:III:7:26} \delta\theta=-\frac{w}{pv}\,\ddp{V}{y}. \end{equation}  It is now up to us to see if our idea that the waves go as (7.20) will explain the same result. We look at the same thing quantum mechanically, assuming that everything is on a very large scale compared with a wavelength of our probability amplitudes. In any small region we can say that the amplitude varies as \begin{equation} \label{Eq:III:7:27} e^{-(i/\hbar)[(W+p^2/2M+V)t-\FLPp\cdot\FLPx]}. \end{equation}  Can we see that this will also give rise to a deflection of the particle when $V$ has a transverse gradient? We have sketched in Fig. 7–8 what the waves of probability amplitude will look like. We have drawn a set of “wave nodes” which you can think of as surfaces where the phase of the amplitude is zero. In every small region, the wavelength—the distance between successive nodes—is \begin{equation*} \lambda=\frac{h}{p}, \end{equation*}  where $p$ is related to $V$ through \begin{equation} \label{Eq:III:7:28} W+\frac{p^2}{2M}+V=\text{const}. \end{equation}  In the region where $V$ is larger, $p$ is smaller, and the wavelength is longer. So the angle of the wave nodes gets changed as shown in the figure.  To find the change in angle of the wave nodes we notice that for the two paths $a$ and $b$ in Fig. 7–8 there is a difference of potential $\Delta V=(\ddpl{V}{y})D$, so there is a difference $\Delta p$ in the momentum along the two tracks which can be obtained from (7.28): \begin{equation} \label{Eq:III:7:29} \Delta\biggl(\frac{p^2}{2M}\biggr)=\frac{p}{M}\,\Delta p= -\Delta V. \end{equation}  The wave number $p/\hbar$ is, therefore, different along the two paths, which means that the phase is advancing at a different rate. The difference in the rate of increase of phase is $\Delta k=\Delta p/\hbar$, so the accumulated phase difference in the total distance $w$ is \begin{equation} \label{Eq:III:7:30} \Delta(\text{phase})=\Delta k\cdot w= \frac{\Delta p}{\hbar}\cdot w= -\frac{M}{p\hbar}\,\Delta V\cdot w. \end{equation}  This is the amount by which the phase on path $b$ is “ahead” of the phase on path $a$ as the wave leaves the strip. But outside the strip, a phase advance of this amount corresponds to the wave node being ahead by the amount \begin{equation} \Delta x=\frac{\lambda}{2\pi}\,\Delta(\text{phase})= \frac{\hbar}{p}\,\Delta(\text{phase})\notag \end{equation}  or \begin{equation} \label{Eq:III:7:31} \Delta x=-\frac{M}{p^2}\,\Delta V\cdot w. \end{equation}  Referring to Fig. 7–8, we see that the new wavefronts will be at the angle $\delta\theta$ given by \begin{equation} \label{Eq:III:7:32} \Delta x=D\,\delta\theta; \end{equation}  so we have \begin{equation} \label{Eq:III:7:33} D\,\delta\theta=-\frac{M}{p^2}\,\Delta V\cdot w. \end{equation}  This is identical to Eq. (7.26) if we replace $p/M$ by $v$ and $\Delta V/D$ by $\ddpl{V}{y}$. The result we have just got is correct only if the potential variations are slow and smooth—in what we call the classical limit. We have shown that under these conditions we will get the same particle motions we get from $F=ma$, provided we assume that a potential contributes a phase to the probability amplitude equal to $Vt/\hbar$. In the classical limit, the quantum mechanics will agree with Newtonian mechanics.