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Hermitian conjugation identity

  1. Jul 12, 2017 #1
    1. The problem statement, all variables and given/known data
    ##(\hat A \times \hat B)^*=-\hat B^* \times \hat A^*##
    Note that ##*## signifies the dagger symbol.

    2. Relevant equations
    ##(\hat A \times \hat B)=-(\hat B \times \hat A)+ \epsilon_{ijk} [a_j,b_k]##

    3. The attempt at a solution
    Using as example ##R## and ##P## operators:
    ##(\hat R \times \hat P)^*_i=-(\hat P \times \hat R)^*_i+ \epsilon_{ijk} [Y,P_z]##
    ##(\hat R \times \hat P)^*_i=-(\hat P \times \hat R)^*_i##
    ##(\hat R \times \hat P)^*=-\hat P^* \times \hat R^*##
     
  2. jcsd
  3. Jul 12, 2017 #2

    strangerep

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    Start instead from a component-wise definition:
    $$(A \times B)_i ~=~ \epsilon_{ijk} A_j B_k $$ (where the usual summation convention applies to repeated indices).
     
  4. Jul 13, 2017 #3
    Since ##\epsilon_{ijk}## is antisymmetric then we have
    ##\epsilon_{ijk}A_jB_k=A_jB_k-A_kB_j##
    ##A_jB_k-A_kB_j=-(B_jA_k-B_kA_j)##
    ##(A \times B)_i=-(B \times A)_i##
    Since A and B are Hermitian the same equlity holds for their self-adjoint counterparts.
     
    Last edited: Jul 13, 2017
  5. Jul 13, 2017 #4

    strangerep

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    That equation does not make sense. On the LHS, ##i## is a free index, but ##j,k## are dummy summation indices. However, on your RHS both ##j## and ##k## are free indices, and there's no ##i## at all. Both left and right hand sides of such an equation must have exactly the same free indices.

    The LHS uses a version of the summation convention. It is short for $$\sum_{j,k} \epsilon_{ijk}A_jB_k $$
    This is wrong if ##B_j## and ##A_k## don't commute (which is presumably the case here, since the problem statement didn't specify commutativity). So you can't blithely interchange ##A## and ##B## like that.

    Your original problem statement doesn't say that ##A,B## are Hermitian.

    But that doesn't solve the problem as stated.

    Start with this: $$\left( \sum_{j,k} \epsilon_{ijk}A_jB_k \right)^\dagger ~=~ \dots\,? \dots $$ Hint: for arbitrary operators ##X,Y##, what is ##(XY)^\dagger## ?
     
  6. Jul 14, 2017 #5
    Thanks for the replies.
    The title mentions that I am solving the identity for Hermitian operators.
    I know that ##\epsilon_{ijk}A_jB_k## is a sum over j and k, with ##j,k=1,2,3##.
    ##(\epsilon_{ijk}A_jB_k)^\dagger=(\epsilon_{ijk}B^\dagger_kA^\dagger_j)=-(B^\dagger \times A^\dagger)_i##
    Therefore:
    ##(A \times B)^\dagger_i=-(B^\dagger \times A^\dagger)_i##
     
  7. Jul 14, 2017 #6

    strangerep

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    No it doesn't -- it mentions "Hermitian conjugation", which is an operation which can be performed on any operator.

    Anyway, I take it you're now happy with the solution.
     
  8. Jul 15, 2017 #7
    Bad wording I guess then.
    Thanks for the help, anyway!
     
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