Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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 ;)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook