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

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Hi,

as we know a density operator \rho 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 \operatorname{tr}(A \cdot \rho).

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 \sum \langle \psi_n | O | \psi_n \rangle makes sense (i.e. gives the same value for all complete orthogonal sets \{ \psi_n \}).
In that lecture, we did prove that A \cdot \rho is of trace class if \rho 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:)
 
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A\rho 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 A\rho is unbounded.
 
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