Relativistic matter-wave velocities ($v_{\mathrm{ph}}=E/p,v_{\mathrm{g}}=dE/dp$)

For a one-dimensional free particle with $m>0$ and $p>0$, take the positive total energy from the relativistic energy-momentum relation:A fixed plane wave quantum phase obeys $kx-\omega t=\text{constant}$. Its phase velocity is thereforeFor a narrow wave packet centered at $p_0$, expand its dispersion relation as $E(p)\simeq E(p_0)+E^{\prime }(p_0)(p-p_0)$. In the Fourier inversion integral building the wave packet, the envelope then depends on $x-E^{\prime }(p_0)t$. Thus its group velocity isThe superluminal phase velocity follows a repeating phase marker; it does not describe a localized signal or superluminal transport of energy. Higher derivatives of the dispersion relation produce wave packet spreading, so the translated-envelope argument is a narrow-band approximation. For $v=0.6c$, the Lorentz factor is $1.25$, $p=0.75mc$ and $E=1.25m{c}^2$: the group velocity is $0.6c$ while the phase velocity is $5c/3$.

The total-energy convention matters. Define $\psi =e^{im{c}^{2}t/\hslash }\Psi$ to remove the common rest energy from a wavefunction $\Psi$ with total-energy time dependence. ThenThis is exactly the free-particle momentum convention used by the Schrödinger equation here. Removing a common time-dependent global phase changes the reported phase velocity, while leaving the de Broglie wavelength, relative spatial quantum phases and group velocity unchanged. Adding $m{c}^2$ again to a total energy would count the rest energy twice.