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Hamiltonian/Angular Momentum Commuter

  1. Apr 1, 2006 #1

    eep

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    I'm having trouble proving that the Hamiltonian commutes with each component of angular momentum as long as the potential only depends on r.

    I have gotten to the following step:

    [tex]
    [H,L_x] = [\frac{p^2}{2m} + V(r), L_x] = [V(r), L_x]
    [/tex]

    [tex]
    [V(r), L_x] = [V(r), yp_z - zp_y]
    [/tex]

    [tex]
    = V(r)yp_z - V(r)zp_y - yp_zV(r) + zp_yV(r)
    [/tex]

    [tex]
    = y[V(r), p_z] - z[V(r), p_y]
    [/tex]


    I'm not sure where to go from here... the problem states that V depends only on r but I'm not sure if I should interperet that as V being linear in terms or r or if there can be higher powers. Help, please!
     
    Last edited: Apr 1, 2006
  2. jcsd
  3. Apr 1, 2006 #2

    Physics Monkey

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    [tex] V(r) [/tex] is meant to be a general function of [tex] r = \sqrt{x^2 + y^2 + x^2} [/tex]. The simplest way to evaluate those commutators is to use the position representation where [tex] \vec{p} = - i \hbar \vec{\nabla} [/tex]. You will find that the key feature is that [tex] V [/tex] depends on [tex] x [/tex], [tex] y[/tex], [tex] z [/tex] only through [tex] r [/tex].
     
  4. Apr 1, 2006 #3

    eep

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    Thank you!
     
  5. Apr 1, 2006 #4

    nrqed

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    Physics Monkey already gave you the answer..Let me just add that it will be useful to use

    [tex] {\partial V \over \partial z} = {\partial V \over \partial r} {\partial r \over \partial z} [/tex]

    and so on.

    Patrick
     
  6. Apr 3, 2006 #5

    dextercioby

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    The fact that [itex] \hat{H} [/itex] is rotationally invariant follows simply from the fact that [itex] V [/itex] depends only on [itex] |\vec{r}| [/itex]. Since the rotation group is [itex] SO(3) [/itex] and its generators are the angular momentum operators, it follows by definition that

    [tex] [\hat{L}_{i},\hat{H}]_{-} =\hat{0} , \forall \ i=1,2,3. [/tex].


    Daniel.
     
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