# Is the phase velocity of plane wave solutions of the Klein-Gordon equation larger than $c$?

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?