Group velocity, quantum mechanics

1. Feb 10, 2009

raul_l

1. The problem statement, all variables and given/known data

A particle in classical mechanics with a velocity v becomes a wave packet with a group velocity v_g in quantum mechanics.
I have to show that $$\vec{v} = \vec{v}_g$$

2. Relevant equations

$$\vec{v}_g = \frac{\partial \omega(\vec{k})}{\partial \vec{k}}$$

$$\omega(\vec{k}) = \hbar^{-1} \sqrt{m_{0}{2} c^4+\hbar^2 k^2 c^2}$$

3. The attempt at a solution

$$\vec{v}_g = \frac{\partial}{\partial \vec{k}} \hbar^{-1} \sqrt{m_{0}{2} c^4+\hbar^2 \vec{k} \vec{k} c^2} = \frac{\vec{k}}{\hbar \sqrt{m_{0}{2} c^4+\hbar^2 k^2 c^2}} = \frac{\vec{k}}{\hbar \sqrt{m_{0}{2} c^4+\hbar^2 k^2 c^2}} = \frac{m\vec{v}}{\hbar^2 E} = \frac{m\vec{v}}{\hbar^2 m c^2} = \frac{\vec{v}}{\hbar^2 c^2}$$

2. Feb 10, 2009

raul_l

In Wikipedia it's done like this:
$$v_g = \frac{\partial \omega}{\partial k} = \frac{\partial (E/\hbar)}{\partial (p/\hbar)} = \frac{\partial E}{\partial p} = ...$$

From there on it's fairly easy.

But I'm wondering what's wrong with my approach.

3. Feb 10, 2009

Dick

How come you don't get an hbar^2*c^2 from the chain rule in the numerator when you do your differentiation with respect to k?

4. Feb 10, 2009

raul_l

How could I not see this.
I've never felt more stupid

Thanks