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Mathematical Quantum Statistics: Why is A*rho of trace class?

  1. Aug 11, 2012 #1
    Hi,

    as we know a density operator [itex]\rho[/itex] is defined to be a non-negative definite operator of trace class (with trace 1).
    We also know that for a given observable A, which is a (possibly unbounded) self-adjoint operator, the expectation value can be calculated as [itex]\operatorname{tr}(A \cdot \rho)[/itex].

    But that's where I just paused: In my lecture about functional analysis I learned that calculating traces in infinite dimensional spaces is a tricky thing. For a given operator O, you must first check that it is of trace class, only then do you know that [itex]\sum \langle \psi_n | O | \psi_n \rangle[/itex] makes sense (i.e. gives the same value for all complete orthogonal sets [itex]\{ \psi_n \}[/itex]).
    In that lecture, we did prove that [itex]A \cdot \rho[/itex] is of trace class if [itex]\rho[/itex] is of trace class and A is an arbitrary bounded operator. But what happens if A is an unbounded operator?

    Thank you for reading and I hope you can give me a hint to the solution :smile:

    (I wasn't sure if I should post this here or in the Calculus & Analysis forum... please tell me if my choice was wrong :wink:)
     
  2. jcsd
  3. Aug 11, 2012 #2

    dextercioby

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    [itex]A\rho[/itex] is generally not trace class. For unbounded operators the expectation value needn't be finite (as their spectrum needn't be finite), so there's no problem if [itex] A\rho [/itex] is unbounded.
     
    Last edited: Aug 11, 2012
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