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Commutation of observables

  1. Apr 13, 2008 #1
    [SOLVED] commutation of observables

    1. The problem statement, all variables and given/known data

    Prove: If the observables (operators) Q1 and Q2 are both constant of the motion for some Hamiltonian H, then the commutator [Q1, Q2] is also a constant of the motion.

    okay, first question.. am i being asked to prove that [[Q1, Q2], H] = 0? If so, then consider the following.. if not.. well, that sucks.

    Given that they are 'constant of the motion', that means they commute with the Hamiltonian, right..? So that gives us

    \[[H,Q_1 ] = 0\]
    \[[H,Q_2 ] = 0\]

    Then maybe I can say..

    \[HQ_1 - Q_1 H = HQ_2 - Q_2 H\]

    I can multiply through the left side by Q2 to get

    \[Q_2 HQ_1 - Q_2 Q_1 H = 0\]

    Since Q2 is an observable, it's a hermitan operator, which allows me to move Q2 to the other side in that first term, like this..

    \[HQ_1 Q_2 - Q_2 Q_1 H = 0\]

    which is equivalent to saying

    \[[H,Q_1 Q_2 ] = 0\]

    Going back to what I was asked to prove (hopefully), which was:
    \[[[Q_1 Q_2 - Q_2 Q_1 ],H] = 0\]

    Expanding that out, we have
    \[Q_1 Q_2 H - Q_2 Q_1 H - HQ_1 Q_2 + HQ_2 Q_1 = 0\]

    but I've already proved that [H, Q1Q2] = 0, so we can rewrite this as:


    ({Q_1 Q_2 H - HQ_1 Q_2 }) - ({Q_2 Q_1 H - HQ_2 Q_1 }) = 0


    which we know is equivalent to
    \[[H,Q_1 Q_2 ] - [H,Q_1 Q_2 ] = 0\]

    Since [H, Q1Q2] = 0, both sides are equal to zero, and the proof is... complete? Have I done anything wrong?
    Last edited: Apr 13, 2008
  2. jcsd
  3. Apr 13, 2008 #2
    There might be a breakdown of logic here. It seems to me that you are also assuming that Q1 and Q2 commute, which is an unnecessary assumption. I think the proof should be easier than you are making it.
  4. Apr 13, 2008 #3
    Why is it unnecessary? Is what I've said incorrect?
  5. Apr 13, 2008 #4
    It is only incorrect if Q1 and Q2 don't commute. In general, they don't have to for the statement that you need to prove to be correct.
    You can only move the Q2 through the Q1 (toward the right) if they commute. But nothing in the problem says that you should assume they commute. In fact, you can prove what you need to without this assumption.
  6. Apr 13, 2008 #5


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    Like Pacopag said, you are assuming that Q1 Q2 = Q2 Q1 which is not necessarily the case.

    It's easy. Simply expand [[Q1,Q2],H] completely. Then use that fact that you can move H though Q1 and Q2.
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