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

1. Aug 11, 2012

### Illuvatar

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

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

2. Aug 11, 2012

### dextercioby

$A\rho$ 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.

Last edited: Aug 11, 2012