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Commutator of a density matrix and a real symmetric matix

  1. Oct 18, 2006 #1
    Let p1,p2 be two density matrices and M be a real, symmetric matrix.
    Now,
    <<p1|[M,p2]>>=
    <<p1|M*p2>>-<<p1|p2*M>>=
    Tr{p1*M*p2}-Tr{p1*p2*M}=
    2i*Tr{(Im(p1|M*p2))}.
    Why is it that this works out as simply as (x+iy)-(x-iy)?
    How is Tr{p1*p2*m}=conjugate(Tr{p1*M*p2})? I can't seem to figure
    out why this works out so cleanly. Thanks for any comments.
     
  2. jcsd
  3. Oct 18, 2006 #2
    For the diagonal terms of an operator A, in some matrix representation,

    [tex](A^\dag)_{ii}=(A_{ii})^*[/tex]

    and so

    [tex]Tr(A^\dag) = Tr(A)^*[/tex].


    Of course, we mean [tex]A\equiv M^\dag \rho_2[/tex]. Since [tex]Tr(\rho_1)^*=Tr(\rho_1)[/tex] by self-adjointness,

    [tex]Tr(\rho_1^\dag \rho_2^\dag M)=Tr(\rho_1^\dag) \dot Tr (M^\dag \rho_2)^* = Tr(\rho_1 M^\dag \rho_2)^*[/tex]
     
    Last edited: Oct 18, 2006
  4. Oct 18, 2006 #3
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