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Qm - Hermitian Commutator [A,B]†

  1. Oct 24, 2011 #1
    I have this review question: If operators A and B are hermitian, prove that their commutator is "anti-hermitian", ie) [A,B]†=-[A,B]

    What has me confused is the placement of the dagger on the commutator. Why [A,B]† and not [A†,B†]? Also, I am using Griffith's Intro to QM as a text. I have failed to find where the author uses this [A,B]† commutator in the book. If anyone has the book and can give me a page number of an example, I'd be grateful.
     
    Last edited: Oct 24, 2011
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  3. Oct 24, 2011 #2

    George Jones

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    What does [itex]\left[ A , B \right] [/itex] equal?
     
  4. Oct 24, 2011 #3

    Dick

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    [A,B]=AB-BA. Just use that. You don't need any book references.
     
  5. Oct 24, 2011 #4
    [A,B] = AB - BA

    So, [A, B]† = (AB - BA)† = (AB)† - (BA)† = B†A† - A†B†

    and [A†,B†] = A†B† - B†A†.
     
  6. Oct 24, 2011 #5
    I think there may be some confusion. I know what [A,B] is; What I'm inquiring on is the placement of the dagger.

    Edit: You guys can see my dagger symbol, right?
     
  7. Oct 24, 2011 #6

    George Jones

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    Also,
     
  8. Oct 25, 2011 #7

    dextercioby

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    Because they are generally NOT the same operator. You have the adjoint of the commutator ([A,B]†) versus the commutator of the adjoints ([A†,B†]). For bounded operators, which can be extended to the whole Hilbert space, they are equal, provided A and B are self-adjoint and bounded, of course.
     
  9. Oct 25, 2011 #8
    That makes sense, thanks!
     
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