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Suppose that $\hat H$ is an operator (typically a Hamiltonian) and $\beta$ is a positive parameter (typically $\beta=1/k_BT$). Show that $$ \mathbf{Tr}\Big(e^{-\beta\hat H}\Big) \geq \sum_{k}e^{-\beta\langle k\vert\hat H k\rangle} $$ My first step is using a Taylor expansion to yield $$ \sum_{n}\frac{(-\beta)^n}{n!}\sum_{k}\Big(\langle k\vert\hat H^n\vert k\rangle-\big(\langle k\vert\hat H k\rangle\big)^n\Big)\geq 0 $$ How can I move forward?


marked as duplicate by John Rennie, Brandon Enright, Danu, Qmechanic Oct 2 '14 at 8:54

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  • $\begingroup$ What happens if $\{\vert k\rangle\}$ is an eigenbasis of $\hat{H}$? $\endgroup$ – Nephente Oct 2 '14 at 6:09
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    $\begingroup$ Starting from the very first expression in the question, literally write out the definition of the trace... $\endgroup$ – DanielSank Oct 2 '14 at 6:11
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    $\begingroup$ @nephente: If ${\vert k \rangle}$ happen to be eigenbasis of $\hat H$, the "=" holds. $\endgroup$ – Roger209 Oct 2 '14 at 6:27
  • $\begingroup$ What's is the properties of your k's? $\endgroup$ – an offer can't refuse Oct 2 '14 at 7:42
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/21420/2451 $\endgroup$ – Qmechanic Oct 2 '14 at 7:49