Discussion
Which one of the following is correct for the phase velocity \(v_{p}\) and group velocity \(v_{g}\)? (\(c\) is the speed of light in vacuum)
Answer: D
The correct statement is that for matter waves in free space, \(v_{p} \ne v_{g}\).
For a free relativistic particle \(E^{2} = p^{2}c^{2} + m^{2}c^{4}\). The group velocity \(\frac{dE}{dp} = \frac{pc^{2}}{E}\) equals the particle speed \(v\), while the phase velocity \(\frac{E}{p} = \frac{c^{2}}{v}\) exceeds \(c\). They differ for any massive particle.
The other statements fail. The same relations give \(v_{p}v_{g} = c^{2}\) exactly, not greater. Energy in a dispersive medium travels at the group velocity (in normal dispersion), not the phase velocity. And while \(v_{p}\) can exceed \(c\) in a medium, the group velocity in the normal case does not, so 'both can be more than \(c\)' is not the accepted statement.
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