= Relativistic matter-wave velocities
{title2=$v_{\mathrm{ph}}=E/p,\quad v_{\mathrm g}=dE/dp$}
For a one-dimensional <free particle> with $m>0$ and $p>0$, take the positive total <energy> from the <relativistic energy-momentum relation>:
$$
E(p)=\sqrt{m^2c^4+p^2c^2},\qquad
\omega=E/\hbar,\qquad k=p/\hbar.
$$
A fixed <plane wave> <quantum phase> obeys $kx-\omega t=\text{constant}$. Its <phase velocity> is therefore
$$
v_{\mathrm{ph}}=\frac{\omega}{k}=\frac{E}{p}=\frac{c^2}{v}>c.
$$
For a narrow <wave packet> centered at $p_0$, expand its <dispersion relation> as $E(p)\simeq E(p_0)+E'(p_0)(p-p_0)$. In the <Fourier inversion> <integral> building the <wave packet>, the envelope then depends on $x-E'(p_0)t$. Thus its <group velocity> is
$$
v_{\mathrm g}=\frac{d\omega}{dk}=\frac{dE}{dp}=\frac{pc^2}{E}=v<c,\qquad
v_{\mathrm{ph}}v_{\mathrm g}=c^2.
$$
The 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 $E=1.25mc^2$: the <group velocity> is $0.6c$ while the <phase velocity> is $5c/3$.
\Image[/matter-wave-phase-and-group-velocity.png]
{title=A <matter wave>'s <phase velocity> and <group velocity> are different slopes}
{description=Original one-dimensional <Fourier inversion> model with $m=c=\hbar=1$, central <momentum> $p_0=0.75$ and <Gaussian function> momentum amplitude $a(p)\propto\exp[-(p-p_0)^2/(4\sigma_p^2)]$, $\sigma_p=0.055$. The blue curve is the real part of $\Psi(x,t)=\int a(p)e^{i(px-E(p)t)}\,dp$ and the orange curves bound its amplitude. Dashed lines mark $v_{\mathrm g}t$; dotted lines track one phase of the central <plane wave>, $x=Et/p$. The right panel compares the secant and tangent slopes. This scalar model illustrates the <dispersion relation>, without representing <Electron> <spin> or measured particle paths.}
{source=matter-wave-phase-and-group-velocity.py}
The total-<energy> convention matters. Define $\psi=e^{imc^2t/\hbar}\Psi$ to remove the common <rest energy> from a <wavefunction> $\Psi$ with total-<energy> time dependence. Then
$$
K=E-mc^2\simeq\frac{p^2}{2m},\qquad
\omega_{\mathrm{NR}}=K/\hbar,\qquad
\frac{\omega_{\mathrm{NR}}}{k}\simeq\frac{p}{2m},\qquad
\frac{d\omega_{\mathrm{NR}}}{dk}\simeq\frac{p}{m}.
$$
This 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 $mc^2$ again to a total <energy> would count the <rest energy> twice.
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