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Adjoint of a Commutator

  1. Sep 5, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that
    [tex] \left [ A,B \right ]^{\dagger}=-\left [A,B \right ] [/tex]
    2. Relevant equations
    [tex] \left [ A,B \right ] = AB-BA [/tex][tex] \left (AB \right)^{\dagger}= B^{\dagger}A^{\dagger} [/tex]
    3. The attempt at a solution
    [tex] \left [ A,B \right ]^{\dagger}=\left (AB-BA \right )^{\dagger} [/tex][tex]=\left (AB \right )^{\dagger}-\left (BA \right )^{\dagger} [/tex][tex]=B^{\dagger}A^{\dagger}-A^{\dagger}B^{\dagger}[/tex][tex]=-\left (A^{\dagger}B^{\dagger}-B^{\dagger}A^{\dagger} \right )[/tex][tex]=-\left [ A^{\dagger},B^{\dagger} \right ][/tex]
    I can only see this working if the operators are Hermitian but the question did not specify it as such.
     
  2. jcsd
  3. Sep 5, 2016 #2

    andrewkirk

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    A quick way to solve the problem of being given a potentially incorrectly specified question is to look for a counterexample. Keep it as simple as possible. In this case, you can choose two 2 x 2 real matrices that are non-symmetric - hence non-Hermitian. Then do the calc and see if the result holds.

    I used the following R code to test this (the 't' function performs a transpose and %*% does matrix mult)

    A=array(c(1,0,1,2),dim=c(2,2))
    A=array(c(2,1,0,1),dim=c(2,2))
    t(A%*%B-B%*%A)
    -(A%*%B-B%*%A)
    t(A%*%B-B%*%A)-(-(A%*%B-B%*%A))
     
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