Effects of Varying Potential Energy

The problem of determining the wave function in the case in which potential energy varies arbitrarily in all three dimensions is mathematically complex and beyond the scope of this course. However, in the special case in which the potential energy depends only on one dimension (say $x$) and furthermore varies only slowly with $x$, the wave function takes the form

$$\psi = k^{-1/2} \exp [i( \int k dx - \omega t)]$$ (10.12)

where the wavenumber $k = k(x)$ is no longer constant, but depends on position.

In case you haven't seen the skinny ``S'' before in the above equation, it indicates the integral of the function $k(x)$. What does this mean? If we have

$$s(x) = \int k dx ,$$ (10.13)

then all we are saying is that

$$k = \frac{ds}{dx} .$$ (10.14)

Thus, for example, if $k(x) = a x^2$ where $a$ is a constant, then $s(x) = a x^3 /3 + C$, where $C$ is an arbitrary constant. You can verify this by plugging $s(x)$ into the above equation. The arbitrary constant is there because the derivative of a constant is zero, and it vanishes when computing $k$. Most of the time we can simply omit it in this application.

We can obtain the dependence of $k$ on position from energy conservation. If the particle energy is $E = \hbar \omega = K + U(x)$ where $K = \Pi^2 /(2M) = \hbar^2 k^2 /(2M)$ is the kinetic energy and $M$ is the particle mass, then we can get $k(x)$ in terms of the potential energy $U(x)$ and the total energy $E$, which is assumed to be constant:

$$k(x) = [2M (E - U)]^{1/2} / \hbar .$$ (10.15)

We stated above that this method works when the potential energy varies ``slowly'' with $x$. We can now state this condition more quantitatively: $U$ must change slowly enough so that the fractional change in $k$ over a distance of one wavelength is small. However, the wavelength itself is related to $k$: $\lambda = 2 \pi /k$. It is natural to ask which value of $k$ should be used to determine the wavelength if $k$ itself is changing? The answer is that if the change in $k$ over a wavelength is small, it doesn't matter, since in this case the change in the wavelength itself is small! On the other hand, if the fractional change in $k$ is large for any plausible computation of wavelength, then this method of computing the wave function is invalid.

Why is there a factor $k^{-1/2}$ in front of the complex exponential in equation (9.12)? Recall that if $k$ is larger, then so is the momentum, which means that the particle is moving faster. Think of traffic flowing down a highway. If a steady stream of cars moves from a region where the speed limit is $100 \mbox{ km} \mbox{ h}^{-1}$ to a region where it is $50 \mbox{ km} \mbox{ h}^{-1}$, then the cars will be half the distance apart in the region of lower speed limit. In other words, the probability of finding a car at any given point on the road will be twice as high in this region. The factor of $k^{-1/2} = \hbar^{1/2} \Pi^{-1/2} = [\hbar / (M v)]^{1/2}$ accounts for this effect. When the probability is computed from the wave function, we find that $P = \vert\psi \vert^2 \propto 1/v$ in agreement with the above example.