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Context:

As a fundamental rule, in zero temperature, we know that the expectation value of a time-independent observable at time $t$, namely $\langle \hat{O}\rangle _t = \langle G(t)\mid \hat{O}\mid G(t)\rangle ...(1)$. $\mid G(t)\rangle$ is the ground at time $t$. For nonzero temperature, we have $\langle \hat{G}\rangle_t=Tr(\hat{\rho}(t)\hat{O})...(2)$. $Tr(...)$ means the summation over the complete eigenstates of the Hamiltonian. $\hat\rho(t)$ means the density matrix.

My question:

How can we produce the first expression from the second by lowering the temperature from finite to zero or how can we get the expression of the expectation value of an observable that is equivalent to the first one in zero temperature?

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Yes, we can. Let us work in the eigenbasis of the Hamiltonian, i.e. $$\hat{H}|n\rangle = E_n|n\rangle, \langle m|\hat{\rho}|n\rangle = \langle m|\frac{1}{Z}e^{-\beta \hat{H}}|n\rangle = \delta_{n,m}\frac{1}{Z}e^{-\beta E_n}.$$ Then $$\langle\hat{O}\rangle = Tr\left[\hat{\rho}\hat{O}\right] = \sum_{n,m}\langle n|\hat{\rho}|m\rangle\langle m|\hat{O}|n\rangle = \sum_{n}\frac{1}{Z}e^{-\beta E_n}\langle n|\hat{O}|n\rangle. $$ As temperature goes to zero, $T\rightarrow 0$ ($\beta=\frac{1}{k_B T}\rightarrow +\infty$) the term corresponding to the ground state dominates: $e^{-\beta E_0} \gg e^{-\beta E_n}, n>0$, and we eventually can omit all the terms except this one, so that $$\langle\hat{O}\rangle \rightarrow \frac{1}{Z}e^{-\beta E_0}\langle 0|\hat{O}|0\rangle \rightarrow \langle 0|\hat{O}|0\rangle, $$ where $Z\rightarrow e^{-\beta E_0}$.

Remark. There might be substantial differences between the two types of averaging, where it comes to mathematical developments: e.g., in the context of Feynman-Dyson expansion, Wick's theorem at zero temperature is an exact operator identity, whereas its equivalent for finite temperatures is an identity for averages, valid in thermodynamic limit.

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  • $\begingroup$ Excellent! So to be absolutely exact, the expression in zero temperature is nothing BUT an approximation from the finite $T$ case. And it is reasonable to omit all the summation from excited states. $\endgroup$
    – OneMoreSec
    Apr 4 '20 at 5:27
  • $\begingroup$ I don't know in which context you pose your question, so I added a remark regarding a situation where one has to be careful. $\endgroup$ Apr 4 '20 at 6:06
  • $\begingroup$ I do not really get your remark, sad. I just recalled the textbook content in many-body physics. It tells us what the expectation value is in zero and finite temperature. But I can't remember if there's any relation between them. $\endgroup$
    – OneMoreSec
    Apr 4 '20 at 6:19

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