The phase velocity is given by $$ v= \frac{\omega}{k} \, .$$ Using the usual dispersion relation $$ E^2 = p^2c^2+ m^2c^4 \leftrightarrow \omega^2 \hbar^2= k^2\hbar^2 c^2 + m^2c^4$$ yields $$ v= \frac{\sqrt{k^2c^2 + \frac{m^2c^4 }{\hbar^2} }}{k} \, .$$ If we now assume that $k^2\gg \frac{ m^2c^2}{\hbar^2} $, we can Taylor expand the square root \begin{align} v&= \frac{\sqrt{k^2c^2 + \frac{m^2c^4 }{\hbar^2} }}{k} \\ &=\frac{kc \sqrt{ + \frac{m^2c^2 }{\hbar^2 k^2} }}{k} \\ & \approx c(1+\frac{m^2c^2 }{2\hbar^2 k^2}) \, . \end{align} This seems to suggest that $v> c$. Moreover, the velocity gets smaller the larger the wave vector/momentum $k$ is. Is this correct or (more likely) did I do some stupid mistake in the calculation above?
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1Possible duplicates: https://physics.stackexchange.com/q/6912/2451 , https://physics.stackexchange.com/q/503967/2451 and links therein. – Qmechanic Oct 03 '19 at 16:10
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In case anomalous dispersion one can watch phase velocity $ > c $, but this has nothing to do with information or energy transfer, so this fact is not breaking causality, albeit is very interesting – Agnius Vasiliauskas Oct 03 '19 at 16:26