Is the group velocity times the phase velocity always the velocity of light squared? Can there be cases where it is different from $c$ squared? If so in what situations would it differ?
 A: For a classical relativistic scalar field, we have
$$E(p) = \sqrt{(pc)^2 + \left( mc^2 \right)^2},$$
and using $E = \hbar \omega$ and $p = \hbar k$ we find that (in units with $\hbar = c = 1$)
$$\omega(k) = \sqrt{k^2 + m^2},$$
so a wave with wave number $k$ has phase velocity $$\frac{\omega}{k} = \sqrt{1 + \left(\frac{m}{k} \right)^2}$$ and group velocity $$\frac{d\omega}{dk} = \frac{1}{\sqrt{1 + \left(\frac{m}{k}\right)^2}}.$$
So indeed, such a wave has that the product of the group and phase velocities equals $c^2$ for any wave number $k$.
But this only holds for relativistic scalar fields; in general waves can have pretty much any dispersion relation $\omega(k)$ and so any relation between the group and phase velocities. (For example, waves like sounds waves or water waves obviously don't have any direct connection to the speed of light.)
A: $$c^2~\stackrel{?}{=}~v_pv_g~=~\frac{E}{p}\frac{d E}{d p}~=~\frac{d (E^2)}{d (p^2)}\qquad\Leftrightarrow \qquad E^2-c^2p^2 ~=~{\rm const} $$
is true for relativistic scalar matter, but is violated e.g. for non-relativistic matter waves
$$ E~=~\frac{p^2}{2m}\qquad\Rightarrow \qquad v_g~=~2v_p .$$
A: No.
$v_g v_p=c^2$ is true for waves in waveguides.  But that's just one area where $v_g$ and $v_p$ are different, and that happens whenever the velocity depends on the wavelength (or frequency).   In other areas - matter waves, and even the basic dispersion of colours by a prism - it does not apply.
This is quite a common trap for students. If they first meet group velocity when learning about waveguides they assume that $v_g v_p=c^2$ is fundamental. It isn't: it happens to be true in this case but not in general.
