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General QM question

  1. Oct 29, 2007 #1
    Hmm well this question isn't general except in the sense that I have no idea about how to solve it and hence have no idea what physics it involves!

    Two particles

    particle one obeys H_1 = (p_1)^2/(2m_1) + V1(r_1)
    particle two obeys H_2 = (p_2)^2/(2m_2) + V2(r_2)

    The system obeys H = H_1 + H_2 + V(|r1 - r2|)

    (the latter is the relative distance between the two particles' positions)
    r_1 and r_2 are scalars, r1 and r2 are the respective vectors

    1.) Find an operator that commutes with H_1 and state its equation of motion
    2.) Find an operator that commutes with H_2 and state its equation of motion
    3.) Find an operator that commutes with H (and prove that it does)

    How do I go about doing this? I can only think of trivial examples, like H_2 commutes with H_1 and vice-versa, and nothing for H.

    There is no other information given, so I can't assume this is the helium atom or something.
  2. jcsd
  3. Oct 29, 2007 #2
    One non-trivial example of operator satisfying condition 1.) is angular momentum
    [itex]\mathbf{L}_1 = [\mathbf{r}_1 \times \mathbf{p}_1] [/itex]. This may give you a clue on how to approach questions 2.) and 3.).

  4. Oct 29, 2007 #3
    Thanks! Ugh! In fact I had considered angular momentum but I thought that its form was particular to the hydrogen atom. But I guess since there is no phi-dependence on the potential it should work fine. So L1 and L2 work for 1.) and 2.), but how about #3? Will the total angular momentum be conserved? If so, how is it defined? Is it just L = L1 + L2? I tried writing out the Hamiltonians in terms of reduced mass but that didn't seem to do anything.

    Is there any classical mechanics reasoning I can use? e.g. with Hamilton's equations?
    Last edited: Oct 29, 2007
  5. Oct 29, 2007 #4
    Yes, I think L = L1+L2 should work. You can try to calculate [L,H]=0 explicitly. There is, however, a nicer approach. Try to prove the following equivalent condition

    [tex] H = \exp(\frac{i}{\hbar} (\mathbf{L} \cdot \vec{\phi})) H \exp(-\frac{i}{\hbar} (\mathbf{L} \cdot \vec{\phi})) [/tex]

    by noticing that for any vector operator [itex] \mathbf{a} [/itex]

    [tex] \exp(\frac{i}{\hbar} (\mathbf{L} \cdot \vec{\phi})) \mathbf{a} \exp(-\frac{i}{\hbar} (\mathbf{L} \cdot \vec{\phi})) [/tex]

    is the result of rotation of this vector around axis [itex] \vec{\phi} [/itex].

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