Theoretical question about waves and interference

1. Yes, the group velocity of a non-relativistic matter wave is twice the phase velocity.

2. Yes, the phase velocity of a relativistic matter wave is always faster than light.

For more details, see e.g. my Phys.SE answer here.

Take a plane wave:

$$ A(x, t) = Ae^{iphi(x, t)} = Ae^{i(kx-omega t)}$$

where

$$ phi(x, t) = kx-omega t$$

is the phase. The time derivative of the phase,

$$ frac{partial phi}{partial t} = -omega $$

gives the frequency. Meanwhile, the spatial derivative

$$ frac{partial phi}{partial x} = k $$

is the wavenumber:

$$ k = frac{2pi}{lambda} $$

The phase velocity is their ratio:

$$ v_{ph} = frac{omega} k $$

So that answer to (1) is yes.

In the above wave:

$$ omega(k) = v_{ph} k$$

This is called the dispersion relation, or: how does the frequency depend on wavenumber?

The linear relation is called "dispersionless": all frequencies propagate at the same speed.

For a non-linear relation, such as:

$$ omega(k) = sqrt{(ck)^2 + (mc^2)^2} $$

then:

$$ v_{ph} = frac{sqrt{(ck)^2 + (mc^2)^2}} k = csqrt{1+frac{m^2c^4}{k^2}} > c$$

which is larger than $c$. So the answer to (2) is "yes".

You should convince yourself that the phase $phi(x, t)$ is local, that is, as it changes and say, the peak, or the zero crossing, moves: no information is being transferred (really: nothing more than an apparent position is "moving").

Energy (or information) travels at the group velocity:

$$ v_{gp} = frac{domega}{dk} $$

which in the example given, is:

$$ v_{gp} = cfrac k {omega} < c$$