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Quantum/math question

  1. Jun 16, 2008 #1
    Quantum/math question :)

    1. The problem statement, all variables and given/known data
    I need to expand the commutator of the 3 operators A,B,C: [A,BC] in terms of [A,B], [A,C], [B,C].

    2. Relevant equations
    [A,B] is defined to be AB - BA
    also, [A,BC] = B[A,C] +[A,B]C.
    There are some other identities but none that I see relevant.



    3. The attempt at a solution

    Tried lots of opening and additing/substructing of elements using the above formulas but I could never get rid of "operators". Is the a way to write "B" and "C" in terms of the above?
    I'm working on this for over an hour! :(

    Thank you very much!
    Tomer.
     
  2. jcsd
  3. Jun 16, 2008 #2
    Are you sure you're supposed to get rid of the operators? I don't think it's possible to express it otherwise. That would require reducing a sum of 3 operators to a sum of 2 operators.
     
  4. Jun 16, 2008 #3
    Thanks for responding.

    The question is:
    Let A;B;C; be operators
    1. Expand [A;BC] in terms of [A;B]; [A;C]; [B;C]

    If you can understand anything different from the exact question you're most welcomed to explain to me :)
    What do you mean by "That wuold require reducing a sum of 3 operators to a sum of 2"?
    Why is that a consequence?
     
  5. Jun 16, 2008 #4
    well an expansion of [A,BC] is ABC-BCA which is a sum of 3 operators in a row. I don't see how you could express that in the form aAB + bBA +cAC +dCA + eBC + fCB (where a...f are constants) which is the form it would need to take if you wanted it expressed solely in terms of [A,B], [A,C] and [B,C]
     
  6. Jun 16, 2008 #5
    I still don't understand the "paradox" here, nor do I see how the sum you wrote (with the consts a...f) is a sun of two operators... what I see is 6 operators.

    Ah well.... I thank you for trying anyway!
     
  7. Jun 16, 2008 #6
    Your answer will involve a sum of the products of the original operators with the two-operator commutators, as in:

    X*[Y,Z] or [X,Y]*Z

    Where X, Y, and Z are each one of A, B, and C. (I wrote it this way to avoid giving the answer away :)
     
  8. Jun 16, 2008 #7
    But I need an expression consisting only of [A,B], [A,C] and [B,C]
    I already know that [A,BC] = B[A,C] + [A,B]C....
     
  9. Jun 16, 2008 #8
    But clearly that's not possible. For example, let

    [tex]A=p_{x}, B=x, C=y [/tex]

    Then

    [tex] [A,BC] = [p_{x},x]*y = -i \hbar *y [/tex]

    and

    [tex] [A,B] = -i\hbar, [A,C] = [B,C] = 0 [/tex]

    There just isn't any way to do it. I think they mean to write it as an expression where the commutators involve only two of A, B, and C, like the one you gave above.
     
  10. Jun 16, 2008 #9
    *sigh*... I'm really gonna hate them if you're right... :)
    I'll send a mail to my tutor, all the student are too confused with this subject I guess none have noticed yet.

    Thank you very much!
    I'll post here if it turns out otherwise...
     
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