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Commutator relations for the Ehrenfest Theorem

  1. May 21, 2008 #1
    Hi there,...

    For a derivation of the Ehrenfestequations i found the following commutator relations for the Hamilton-Operator in a book:
    [tex]H = \frac{p_{op}^2}{2m} + V(r,t)[/tex]
    and the momentum-operator [tex]p_{op} = - i \hbar \nabla[/tex] respectively the position-operator [tex]r[/tex] in position space:

    [tex][H,p_{op}] = -i \hbar (V(r,t) \nabla - \nabla V(r,t)) = i \hbar \nabla V(r,t)[/tex]


    [tex][H,r] = \frac{\hbar^2}{2m} (\Delta r - r \Delta) = - \frac{\hbar^2}{m} \nabla[/tex]

    so i don't understand how to get these results. it looks nearly like the use of the product rule but the signs don't match and especially in equation two: how does the laplace-operator become a nabla?

    sorry if this problem is obviously but i dont't see it.

    thanks and greetings.
  2. jcsd
  3. May 21, 2008 #2


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    Homework Helper

    Working in one dimension for simplicity:

    [tex] [H,p] = [V,p]=-i\hbar(V \frac{d}{dx} - \frac{d}{dx} V)[/tex]

    Now, the operator [itex] d/dx V [/itex] corresponds to first applying the operator V (multiplying the wavefunction by V) and then the operator d/dx. It might help to use a test function:

    [tex] \frac{d}{dx} V f = \frac{d}{dx} (Vf) = \frac{dV}{dx} f + V \frac{df}{dx} [/tex]

    So that, in terms of operators:

    [tex] \frac{d}{dx} V = \frac{dV}{dx} + V \frac{d}{dx} [/tex]

    Note the first term on the RHS is different from the LHS: it just says to multiply the wavefunction by the function dV/dx.

    Plugging this in above:

    [tex] [H,p] = -i\hbar(V \frac{d}{dx} - \frac{dV}{dx} - V\frac{d}{dx}) = i \hbar \frac{dV}{dx}[/tex]

    You can work out [H,x] similarly. I would recommend using a test function first.
  4. May 21, 2008 #3
    thanks a lot,... i was blind ;)
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