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Lie Derivative and acceleration

  1. Sep 15, 2011 #1

    This thread looks like GR/SR, but it has grounds QM and maybe only stays in that realm, which is what I'm asking

    I was looking at some everyday non-relativistic quantum mechanics and I spotted something I thought was interesting. Consider the time evolution of an observable
    \frac{d \hat{A}}{dt}=\frac{i}{\hbar}[H,\hat{A}]
    and next consider a form of that for the position operator
    \frac{d^2 \hat{x}}{dt^2}=\frac{i}{\hbar}[H,\hat{v}]
    This seems to give an expression for the acceleration. My question is whether there is anything deeper going on here. Is this expression trying to say that "the extent to which the velocity field doesn't commute with the Hamiltonian of the system is the acceleration"? More specifically, can this be generalized to
    \frac{d^2 \mathbf{x} }{dt^2}=\mathcal{L}_{H}\,\mathbf{u}?

  2. jcsd
  3. Sep 17, 2011 #2


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    depends... what's \mathcal{L}_H and what's u?
  4. Sep 17, 2011 #3
    Well, in my mind I was thinking that [itex]\mathcal{L}_{H}[/itex] was the lie derivative with respect to the Hamiltonian, and that [itex]\mathbf{u}[/itex] was the velocity vector. I wasn't sure how to parametrize the second derivative since this is coming from my guess, but I figured that the Hamiltonian has to do with time translation so I wrote it wrt time, I wasn't sure a) if it even is true, or b) if it should be proper time...
  5. Sep 17, 2011 #4


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    In [itex]\mathcal{L}_{X}[/itex], shouldn't the X be a vector? I.e., in coordinates, shouldn't it be something like [itex]\mathcal{L}_{X} = X^k \partial_k[/itex] ?
    (But a Hamiltonian by itself is not a vector.)

    See also http://en.wikipedia.org/wiki/Lie_derivative
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