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This question is based on this question asked in physics SE. In this question, they say that phase velocity greater than velocity of light. Ok, greater than velocity of light for phase velocity is acceptable. But group velocity must be $\leq{c}$ . But in my knowledge, for a quantum free particle, phase velocity $$V_p=\frac{\omega}{k}=\frac{p}{2m}$$ and group velocity $$V_g=\frac{d\omega}{dk}=\frac{p}{m} .$$ Obviously $$V_g=2V_p.$$ But how this is possible? This means group velocity also greater than velocity of light, and is pure nonsense. Where is the mistake happened to me?

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  • $\begingroup$ what is the equation you are using for what you call a "quantum free particle". If it is a simple plane wave there is no d(omega) or d(k) $\endgroup$ – anna v Nov 18 '15 at 4:50
  • $\begingroup$ @annaV I can't understand what you are asking about, but, I consider the particle as a wave packet. ( Because a quantum 'free' particle can't be expressed as a simple plane wave, because such wave functions are non normalisable). So, it is a combination of plane waves of different wavelength $d\lambda$ around a central wave. Thus $d\omega$ and $dk$ came. $\endgroup$ – Muhsin Ibn Al Azeez Nov 18 '15 at 5:54
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The main issue is that one cannot use a non-relativistic dispersion formula $E=\frac{p^2}{2m}$ to deduce what happens at relativistic velocities. This Phys.SE post deals with the same theme.

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