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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 arbitraryboundedoperator. 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

(I wasn't sure if I should post this here or in the Calculus & Analysis forum... please tell me if my choice was wrong )

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

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