One-Dimensional Non-Relativistic Case

Louis de Broglie^9.1 made an analogy between matter waves and light waves, pointing out that wave packets of light change their velocity as the result of spatial variations in the index of refraction of the medium in which they are travelling. This behavior comes about because the dispersion relation for light traveling through a medium with index of refraction $n$ is $\omega = kc/n$, so that the group velocity, $u_g = d \omega /dk = c/n$. Thus, when $n$ increases, $u_g$ decreases, and vice versa.^9.2

The problem for matter waves is to determine how analogous modifications can be made to the free particle dispersion relation so as to produce accelerations of wave packets consistent with Newtonian dynamics in the geometrical optics limit. In this section we make a simple guess as to how to modify the free particle dispersion relation for matter waves, limiting consideration initially to the one-dimensional, non-relativistic case. As in many situations in physics, the simple guess turns out to be correct! This leads to a connection between wave dynamics and Newtonian mechanics.

The dispersion relation for free matter waves is $\omega = (k^2 c^2 + \mu^2 )^{1/2}$. In the non-relativistic limit $k^2 c^2 \ll \mu^2$ and this equation can be approximated as

$$\omega = \mu (1 + k^2 c^2 / \mu^2 )^{1/2} \approx \mu + k^2 c^2 /(2 \mu ) .$$ (9.2)

Let us modify this equation by adding a term $S'$ which depends on $x$, and which plays a role analogous to the index of refraction for light:

$$\omega = S' (x) + \mu + k^2 c^2 /(2 \mu ) = S(x) + k^2 c^2 /(2 \mu ) .$$ (9.3)

The rest frequency has been made to disappear on the right side of the above equation by defining $S = S' + \mu$. This is done to simplify the notation.

Let us now imagine that all parts of the wave governed by this dispersion relation oscillate in phase. The only way this can happen is if $\omega$ is constant, i. e., it takes on the same value in all parts of the wave. It turns out we can do this if $S$ is not a function of time. (The reasons for this will be discussed in the next chapter.)

If $\omega$ is constant, the only way $S$ can vary with $x$ in equation (8.3) is if the wavenumber varies in a compensating way. Thus, constant frequency and spatially varying $S$ together imply that $k = k(x)$. Solving equation (8.3) for $k$ yields

$$k(x) = \left[ \frac{2 \mu [ \omega - S(x) ]}{c^2} \right]^{1/2} .$$ (9.4)

Since $\omega$ is constant, the wavenumber becomes smaller and the wavelength larger as the wave moves into a region of increased $S$.

In the geometrical optics limit, we assume that $S$ doesn't change much over one wavelength so that the wave remains reasonably sinusoidal in shape with approximately constant wavenumber over a few wavelengths. However, over distances of many wavelengths the wavenumber and amplitude of the wave are allowed to vary considerably.

As yet we have no idea what causes $S$ or where it comes from. For now we simply explore the consequences of its presence, especially in the geometric optics limit in which quantum dynamics gives way to Newtonian dynamics.

The group velocity calculated from the dispersion relation given by equation (8.3) is

$$u_g = \frac{d \omega}{dk} = \frac{kc^2}{\mu} = \left( \frac{2c^2 ( \omega - S )}{\mu} \right)^{1/2}$$ (9.5)

where $k$ is eliminated in the last step with the help of equation (8.4). The resulting equation tells us how the group velocity varies as a matter wave traverses a region of slowly varying $S$. Thus, as $S$ increases, $u_g$ decreases and vice versa.

We can now calculate the acceleration of a wave packet resulting from the spatial variation in $S$. We assume that $x(t)$ represents the position of the wave packet, so that $u_g = dx/dt$. Using the chain rule $d u_g /dt = (d u_g /dx)(dx/dt) = (d u_g /dx)u_g$, we find

$$a = \frac{d u_g}{dt} = \frac{d u_g}{dx} u_g = \frac{d u_g^2... ...c{c^2}{\mu} \frac{dS}{dx} = - \frac{\hbar}{m} \frac{dS}{dx} .$$ (9.6)

The group velocity is eliminated in favor of $S$ by squaring equation (8.5) and substituting the result into equation (8.6).

Using Newton's second law to infer the force from the acceleration, we find

$$F = ma = - \frac{dU}{dx} ~~~ \mbox{(force from potential energy)} ,$$ (9.7)

where we have defined $U \equiv \hbar S$. The quantity $U$ is called the potential energy. We have thus established a relationship between quantum mechanical and Newtonian dynamics, in that $U = \hbar S$ dictates the form of the force in Newton's second law, while $S$ governs the refraction of matter waves. A force which equals minus the derivative of some potential energy $U(x)$ is called conservative. Certain forces such as friction are not conservative. At present we will deal mainly with conservative forces.