Group velocity, quantum mechanics

[raul_l]

Homework Statement

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

Homework 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}

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}

 

[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.

 

[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?

 

[raul_l]

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

Thanks