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Silly adjoint operator proof

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data

    For operators A, B, prove that (AB)^t = (B)^t(A)^t where ^t is representing the Hermitian adjoint.

    I know that this should be similar to proofs I did about matrix transposes in linear algebra, but I'm not sure how to do it without seeing the operators as matrices with indices. I've been trying to do it with dirac notation but that's been confusing...

    2. Relevant equations



    3. The attempt at a solution

    Well... it's a long shot. I don't think this works:

    <(AB)^t psi1| psi2> = ((AB)^t)* < psi1|psi2> = B^t A^t < psi1|psi2> = <psi1| B^tA^t psi2> ==> (AB)^t = B^t A^t.

    I think I kind of made it up at the part with the complex conjugate, so yeah, basically I'm confused, even though this is supposed to be the easiest proof ever... :grumpy:
     
  2. jcsd
  3. Sep 14, 2008 #2

    Dick

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    <(AB)^T(psi1)|psi2>=<psi1|(AB)(psi2)>=<psi1|A(B(psi2))>=<A^T(psi1)|B(psi2)>. Can you continue?
     
  4. Sep 14, 2008 #3
    aha... I wasn't sure if I was allowed to "split them up", so to speak, at the point where the A moves back to the rhs of the statement. If I'm allowed to do that I'm done! :biggrin:
     
  5. Sep 14, 2008 #4
    oooh wait, it's just composition of functions!!! hehehe
     
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