Assuming $c=1$, $v=\frac{\omega}{k}=\frac{\sqrt{k^2+m^2}}{k}>1$, for $m \neq 0$. Why is it not an issue that this $v$ is greater than the speed of light?
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/6912/2451 , physics.stackexchange.com/q/503967/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jul 22, 2022 at 18:12
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$\begingroup$ See also this very clear paper on the subject: Subluminality of relativistic quantum tunneling, Phys. Rev. A 107, 032209 (2023). $\endgroup$– QuilloCommented May 2 at 6:59
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No. The physical speed is the group velocity $$ v_g= \frac{\partial \omega}{\partial k} = \frac{k}{\sqrt{k^2+m^2}} <1. $$ The phase velocity $$ v_\phi = \frac{\omega}{k} $$ is not relevant.
answered Jul 22, 2022 at 18:10
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$\begingroup$ Why is $v_g$ interpreted as the physical velocity? It has the units of s/m $\endgroup$ Commented Jul 22, 2022 at 18:21
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$\begingroup$ It has the same units as $\omega/k$ i,.e dimensionless. it is the speed at which energy and information travel. $\endgroup$ Commented Jul 22, 2022 at 18:48
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$\begingroup$ But if we didn't set $c=1$, this would have units s/m $\endgroup$ Commented Jul 22, 2022 at 18:58
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1$\begingroup$ I got it. The numerator would have a pc^2 if we don't set natural units. $\endgroup$ Commented Jul 22, 2022 at 19:04