How can I find the expectation value of any operator at zero temperature in quantum statistical mechanics formalism? Expectation value of any operator $\hat{O}$ is given as $$\langle \hat{O}\rangle = Z^{-1}\mathop{\mathrm{Tr}}[\hat{O} e^{-\beta \hat{K}}]$$

where $Z = \mathop{\mathrm{Tr}}[e^{-\beta \hat{K}}]$, $\hat{K} = \hat{H}-\mu\hat{N}$ for grand canonical and $\hat{K}=\hat{H}$ for canonical ensemble. And $\beta=1/T$.

At $T=0\implies \beta=\infty$, which means $$e^{-\beta\hat{K}}=0 \Longrightarrow Z=0\quad\text{and}\quad \langle\hat{O}\rangle=0/0=\infty$$

My question not about explaining the procedure for finding the expectation value at $T = 0$. I am only asking how does quantum statistical formalism work at $T = 0$ because for the above given reason, I am confused.

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    $\begingroup$ Have you tried directly taking the limit with some common examples? $\endgroup$ – probably_someone May 2 at 12:31
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    $\begingroup$ The system is in its ground state at $T = 0$. Your density operator simply turns into $\rho = | \psi_0 \rangle \langle \psi_0 |$, where $| \psi_0 \rangle$ is the lowest energy quantum state: $\hat{H} | \psi_0 \rangle = E_{\text{min}} | \psi_0 \rangle$. $\endgroup$ – Cham May 2 at 12:59
  • $\begingroup$ @probably_someone Thank you for suggesting this. I found some examples for canonical ensembles. In limit $\beta \to \infty$, only state with lowest energy survive i.e. $H |\psi_0\rangle = E_{0} |\psi_0\rangle$. What happens in grand canonical case? Does chemical potential have no effect in $T\to 0$ limit? $\endgroup$ – Luqman Saleem May 3 at 12:33
  • $\begingroup$ @Cham thank you. How does chemical potential effect the ground state in zero temperature limit? $\endgroup$ – Luqman Saleem May 3 at 12:39
  • $\begingroup$ @LuqmanSaleem, actually, I'm not sure for the chemical potential. I believe it is 0 when $T \rightarrow 0$. If not, then the quantum state may be a eigenvector of the operator $\hat{K} = \hat{H} - \mu \hat{N}$. $\endgroup$ – Cham May 3 at 12:52

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