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Group velocity, quantum mechanics

  1. Feb 10, 2009 #1
    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 [tex] \vec{v} = \vec{v}_g [/tex]

    2. Relevant equations

    [tex] \vec{v}_g = \frac{\partial \omega(\vec{k})}{\partial \vec{k}} [/tex]

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

    3. The attempt at a solution

    [tex] \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} [/tex]
  2. jcsd
  3. Feb 10, 2009 #2
    In Wikipedia it's done like this:
    [tex] v_g = \frac{\partial \omega}{\partial k} = \frac{\partial (E/\hbar)}{\partial (p/\hbar)} = \frac{\partial E}{\partial p} = ... [/tex]

    From there on it's fairly easy.

    But I'm wondering what's wrong with my approach.
  4. Feb 10, 2009 #3


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    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?
  5. Feb 10, 2009 #4
    How could I not see this.
    I've never felt more stupid :redface:

    Thanks :smile:
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