I tried to compare the phase and group velocity of mater waves, say a free electron. I did this once using the de Broglie equation and then using relativity and got different results: $$E = \hbar \omega = \frac{p^2}{2m} = \frac{\hbar^2}{2m}k^2,\ \textrm{where }\ \ p= \hbar k$$ $$\omega = \frac{\hbar}{2m} k^2$$ $$v_{ph} = \frac{\omega}{k} = \frac{\hbar}{2m}k \ \ \ \ \ \ \ \ \ \ \ v_{gr}= \frac{\partial \omega}{\partial k}=\frac{\hbar}{m}k $$ Now using relativity: $$E = \sqrt{p^2 c^2 + m_0^2c^4}$$ $$v_\textrm{ph} = \frac{\hbar \omega}{\hbar k} = \frac{E}{p} = \frac{\gamma m_0 c^2}{\gamma m_0 v} = \frac{c^2}{v} $$ $$v_\textrm{gr} = \frac{\partial E}{\partial p} = \frac{\partial \sqrt{p^2 c^2 + m_0^2c^4}}{\partial p} = \frac{c^2 p}{E}= v$$ So in the first case the phase velocity is half the group velocity and in the second case it is always bigger. If the first case works only for $v\ll c$ than it still should yield a much higher phase velocity. Which of the two is correct and why?

  • $\begingroup$ Hi Rosen, please don't misunderstand this comment (as I am not speaking about this particular post and I am way, way too old to worry about my rep points on this site : ), but many new users either don't know or plain forget that (when they acquire enough points, I don't know the actual number) they can also accept answers, as well as up vote them. I write this for your future questions, but wait a few days before acceptance of any of them, in case you get a better one. thanks. $\endgroup$
    – user108787
    Nov 6, 2016 at 14:42

1 Answer 1


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

  • $\begingroup$ Thanks for your answer. Only question I have now is, why is the dispersion relation not just equal to $\omega_{(k)} = \frac{\hbar}{2m} k^2$ or rather where did the relation you used come from? $\endgroup$ Nov 5, 2016 at 9:45
  • $\begingroup$ That , fingers crossed, should be the relativistic version of E. It came from my olde, oldes notes, the tattiest set of scraps of paper I am embarrassed to say. If you put hbar and c back in, as I am used to setting them to 1, you should be able to reconcile them. Sorry about that, I will google to see if I can get them in your form, but it's a good exercise (for both of us) to try and reconcile them. $\endgroup$
    – user108787
    Nov 5, 2016 at 10:40
  • $\begingroup$ Just to finish off, please ignore my answer if it's too much hassle, there a million versions of this and I don't want you wasting your time on a mistake of mine. I think Schaum's book of exercises covers it. It did get 1 upvote, it's that not you then someone thinks it's OK. $\endgroup$
    – user108787
    Nov 5, 2016 at 11:09
  • $\begingroup$ I tried to upvote your answer because it was helpful but unfortunately I couldn't because I have less than 15 reputation. $\endgroup$ Nov 5, 2016 at 13:29
  • $\begingroup$ Your anwser is probably the correct one because it agrees with the relativistic way of solving the problem and this one is in wikipedias article for phase velocity I think. I don't know though how to derive the dispersion relation and why it is the one we should use as opposed to the parabolic one ($\omega_{(k)} = \frac{\hbar}{2m} k^2$), that I've seen to be correct as well. I will look for the book you mentioned and if I find a solution I'll update the post. $\endgroup$ Nov 5, 2016 at 13:35

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