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Graduate Mathematical Quantum Statistics: Why is A*rho of trace class?
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...- Illuvatar
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- Class Mathematical Quantum Quantum statistics Statistics Trace
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- Forum: Quantum Physics