According to the Schrödinger equation (SE) for a free particle, we have solutions of the form $Ae^{ikx} + Be^{-ikx}$. The wavenumber $k \in (-\infty, \infty)$. It is standard to point out that these traveling waves do not model particles because 1) they are not normalizable, and 2) their phase velocity is one half that predicted by the de Broglie formula $p=\hbar k$.
The solution is to remark that the classical particle travels with a velocity equal to the group velocity, not the phase velocity. As the dispersion relation given by SE is $\omega(k) = \hbar k^2/2m$, the group velocity is $d\omega/dk$, which is twice the phase velocity. Thus
$v_{\mathrm{classical}} = v_{\mathrm{group}} = 2v_{\mathrm{phase}}$.
However, it would appear that, since $k$ is unbounded, this means the predicted speed for a classical particle has no upper limit, and it can go much faster than the speed of light.
The non-relativistic nature of SE and QM is often remarked. Am I right in assuming that this is one area where the non-relativistic-ness comes to the fore? I'm not familiar enough with Dirac's equation to know how it changes the situation here, which is why I ask.
Best Answer
Yes, definitely.
In order to write the Schrodinger equation, one starts from the non relativistic hamiltonian $H = \frac{p^2}{2m}$, hence the dispersion relation $\hbar \omega = {(\hbar k)^2 \over 2m}$, that can be obtained by substituting $H$ with $\hbar \omega$ and $p$ with $\hbar k$.
Klein-Gordon's and Dirac's equations are instead based on the relativistic hamiltonian $H^2 = (pc)^2 + (mc^2)^2$, therefore the relativistic dispersion relation will be
$$(\hbar \omega)^2 = (\hbar k c)^2 + (mc^2)^2$$
which gives a group velocity that behaves properly for a relativistic theory.
$${d \omega \over dk} = \frac{c}{\sqrt{1+({mc \over \hbar k})^2}}$$