# Expectation value in zero and nonzero temperature

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?

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 $$Z\rightarrow e^{-\beta E_0}$$.
• 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. Apr 4 '20 at 5:27