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Are super-operators always writable in a basis independent form?

  1. Nov 16, 2011 #1
    in particular, i wonder if the trasposition super operator is basis independent or not.

    We can always write an operator W as
    [itex]\hat{W}=\sum_{i,j} c_{i,j} |i\rangle\langle j| [/itex]
    and for the transposed we obtain
    [itex]\hat{W}^T=\sum_{i,j} c_{j,i} |i\rangle\langle j| [/itex]
    we obtain a relation true for each \psi,\phi
    [itex]\langle\psi|\hat{W}^T|\phi\rangle=\langle\phi|\hat{W}|\psi\rangle [/itex] (1)
    Now let's write the transposition super-operator as
    [itex]\Lambda[\hat{W}]=\sum_{i,j} |i\rangle\langle j| \hat{W} |i\rangle\langle j| = \hat{W}^T[/itex]

    Now seems that the transposed matrix does not depend from the basis,
    I have to admit it because the formula (1) is true for each vector and don't depend from the basis.
    But that's sound strange to me, because if I have an operator, it can always be diagonal in a certain basis, and so it is equal to its transposition.
    We can always write W diagonal in a certain basis [itex]\{|\tilde{k}>\}[/itex] not neccesary equal to [itex]\{|i>\}[/itex]:
    [itex]
    \hat{W}=\sum_{k} c_k |\tilde{k}\rangle\langle\tilde{k}|[/itex]
    and from here do i get
    [itex]
    \hat{W}^T=(\sum_{k} c_k |\tilde{k}\rangle\langle\tilde{k}|)^T=(\sum_{k} c_k |\tilde{k}\rangle\langle\tilde{k}|=\hat{W}[/itex]

    Where i'm wrong?
    Can I write every super operator in a representation that is independent from the basis?
     
    Last edited: Nov 16, 2011
  2. jcsd
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