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Time derivative of an observable

  1. Jan 24, 2012 #1
    According to my book:

    [tex]\frac{d}{dt} \langle Q \rangle = \frac{i}{\hbar} \langle [\hat{H}, \hat{Q}] \rangle + \langle \frac{\partial \hat{Q}}{\partial t} \rangle [/tex].

    No derivation for this is given. How can derive you this?
  2. jcsd
  3. Jan 25, 2012 #2


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    This is nothing else but the quantum mechanical translation (Poisson bracket of phase space function replaced by operator commutation relation) of the equation of motion in the Hamiltonian formulation of classical mechanics.
  4. Jan 25, 2012 #3


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    You can use the product rule for the derivative of <Q(t)>=<ψ(t)|*Q(t)*|ψ(t)> and then apply Schrödinger's equation two times.
  5. Jan 25, 2012 #4
  6. Jan 25, 2012 #5


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    oh, sorry, I missed the <.>; yes, it's Ehrenfest´s theorem, not the operator equation itself
  7. Jan 25, 2012 #6


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    To be fair, the equation for the time derivative of a Heisenberg operator looks pretty much exactly like that. lol.
  8. Jan 25, 2012 #7


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    The full derivation has been given in 2009 by 2 German guys. You can find it by searching arxiv.org for the name <Ehrenfest>.

    8. arXiv:1003.3372 [pdf, ps, other]
    Title: A sharp version of Ehrenfest's theorem for general self-adjoint operators
  9. Jan 25, 2012 #8
    Thanks guys.
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