# Problem in de-broglie relation!

In the starting of quantum mechanics we encounter Plank's relation $E=\hbar \omega$, and from special relativity we have $Ev_p=pc^2$ , where $v_p$ is the velocity of the particle (we are assuming momentum is a particle aspect). Putting the two together, we get $$\hbar \omega v_p = pc^2 \, .$$ Now we assume that $\omega$ is a wave concept and use $\omega / k =v_w$ where $v_w$ is the wave velocity. For now we assume the wave velocity to be different from that of the particle for we don't know if the wave is the particle itself or an excitement in the probability field of the universe. That's a different topic. Here I just want to say that $v_p$ and $v_w$ might be different, or even the same. We don't know so why take risks?

Plugging the expression, we have $$p = \hbar v_p v_w k /c^2 \, .$$

Then, plugging it into the relativity relation $E^2 = p^2c^2 + m^2 c^4$ and finding out $d \omega /dk$, we get $$d\omega / dp = v_p^2v_w/c^2 \, .$$ This is the group velocity which has to be equal to the particle velocity if it has to have some meaning! Therefore, $$v_p = v_p^2v_w/c^2$$ or $$v_p v_w = c^2$$ or $$v_w = c^2 / v_p$$ which automatically implies $v_w>c$ (assuming $v_p<c$ , which is the very core of special relativity).

What is going on here? The wave velocity has come out more than the speed of light!

• @Willy Billy Williams Care to explain why? – Yuzuriha Inori May 13 '17 at 14:39
• I erased it, that was correct. I believe the issue is that both $v_w=c$ and $v_p=c$ in vacuum – user126422 May 13 '17 at 14:45
• De Broglie wave travel with a group velocity which is less than c – Shashaank May 13 '17 at 14:53
• $E v_p \neq pc^2$ – Ofek Gillon May 13 '17 at 14:53
• @Shashaank That's the thing... Then why does calculations suggest otherwise? – Yuzuriha Inori May 13 '17 at 14:55