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Hermitian conjugate of operators

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the hermitian conjugates, where A and B are operators.

    a.) AB-BA
    b.) AB+BA
    c.) i(AB+BA)
    d.) [itex] A^\dagger A [/itex]


    2. Relevant equations

    [itex] (AB)^\dagger =B^\dagger A^\dagger [/itex]



    3. The attempt at a solution

    Are they correct and can I simplify them more?

    a.) [itex] (AB-BA)^\dagger = (AB)^\dagger - (BA)^\dagger= B^\dagger A^\dagger - A^\dagger B^\dagger = BA-AB = -(AB-BA) [/itex]

    b.) [itex] (AB+BA)^\dagger = (AB)^\dagger + (BA)^\dagger= B^\dagger A^\dagger + A^\dagger B^\dagger = BA+AB [/itex]

    c.) [itex] [i(AB+BA)]^\dagger = -i[(AB)^\dagger + (BA)^\dagger]= -i(B^\dagger A^\dagger + A^\dagger B^\dagger) = BA+AB [/itex]

    d.) [itex] [A^\dagger A]^\dagger = A^\dagger (A^\dagger)^\dagger=A^\dagger A [/itex]
     
  2. jcsd
  3. Nov 17, 2011 #2

    dextercioby

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    You might review point c.)
     
  4. Nov 17, 2011 #3
    For point C.) [itex] -i(B^\dagger A^\dagger + A^\dagger B^\dagger) [/itex]
     
  5. Nov 17, 2011 #4

    dextercioby

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    Alright. The last equality is not true. That's what I meant.
     
  6. Nov 17, 2011 #5
    Is it ok until [itex] -i(B^\dagger A^\dagger + A^\dagger B^\dagger) [/itex]?
     
  7. Nov 17, 2011 #6

    dextercioby

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    Yes, it is.
     
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