The dispersion relation for a free relativistic electron wave is $$ω(k) = c\sqrt {k^2 +k_e^2}$$ where $k_e =\frac {m_ec}{\hbar}$

Writing $E = \hbar ω$ and $p = \hbar k$.

A wave with frequency $f$ and wavelength $λ$ has a phase speed $v_p$, which can be defined as $v_p = fλ$.

From $λ = h/ p$ and $f = 2πω$, we can conclude that the phase speed $v_p$ of a free electron wave is $$v_p = ω(k) /k=c(1+\frac {k_e^2} {k^2})^{1/2} \ge c$$

The group speed $v_g$ of a wave is defined as $v_g = \frac {dω(k)} {dk}$

This implies that the group $v_g$ of a free electron wave is $$v_g=c(1+\frac {k_e^2} {k^2})^{-1/2} \le c $$

Then we can say $$ v_pv_g = c^2$$.

The group speed $v_g$ of a free electron wave is equal to the electron particle speed $v$ defined by $$p =\gamma m_ev$$ $$E =\gamma m_ec^2$$

That is, $$v =\frac {pc^2}{E}=c(1+\frac {k_e^2} {k^2})^{-1/2} =v_g$$

The dispersion relation for a free relativistic electron wave is $$ω(k) = c\sqrt {k^2 +k_e^2}$$ where $k_e =\frac {m_ec}{\hbar}$

Writing $E = \hbar ω$ and $p = \hbar k$.

A wave with frequency $f$ and wavelength $λ$ has a phase speed $v_p$, which can be defined as $v_p = fλ$.

From $λ = h/ p$ and $f = 2πω$, we can conclude that the phase speed $v_p$ of a free electron wave is $$v_p = ω(k) /k=c\left(1+\frac {k_e^2} {k^2}\right)^{1/2} \ge c$$

The group speed $v_g$ of a wave is defined as $v_g = \frac {dω(k)} {dk}$

This implies that the group $v_g$ of a free electron wave is $$v_g=c\left(1+\frac {k_e^2} {k^2}\right)^{-1/2} \le c $$

Then we can say $$ v_pv_g = c^2$$.

The group speed $v_g$ of a free electron wave is equal to the electron particle speed $v$ defined by $$p =\gamma m_ev$$ $$E =\gamma m_ec^2$$

That is, $$v =\frac {pc^2}{E}=c\left(1+\frac {k_e^2} {k^2}\right)^{-1/2} =v_g$$