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Mar 14, 2023 at 14:47 comment added user105620 $Q$ is a quantum operator, so it can't be taken out of the trace
Mar 13, 2023 at 17:51 comment added Ján Lalinský Why not $Z = \text{Tr}~ e^{-\frac{H - \mu Q}{kT}} = e^{\frac{\mu Q}{kT}} ~\text{Tr}~e^{-\frac{H}{kT}}$ ?
Mar 13, 2023 at 14:40 history asked user105620 CC BY-SA 4.0