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What is a Gibbs state and what does it differ from a pure state? Say I have a two-level atom and it is described by a Gibbs state $\rho_G = \dfrac{e^{- \frac{H}{kT}}}{Z}$. I know $Z$ is a partition function. How can I express $\rho_G$ explicitly in the diagonalized form?

Let the eigenstates of $H$ be the ground state $\left|g\right>$ and the excited state $\left|e\right>$ of the atom. Is $\rho_G = \left|g\right>\left<g\right|$ in this case with near zero temperature?

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The expression you give, $$\rho_G=\frac 1Z e^{-H/kT},$$ is already explicit. If you want an explicitly diagonalized expression, then you can use the fact that a function $f:\mathbb C\to\mathbb C$ is defined as acting on Hilbert space operators $A$, using the eigenvector route, as giving $f(A)|a⟩=f(a)|a⟩$ whenever $A|a⟩=a|a⟩$. Thus if $$ H=\sum_n E_n|E_n⟩⟨E_n| $$ (taking a point spectrum for simplicity), with possible degeneracies and so on, then $$ \rho_G=\frac 1Z e^{-H/kT}=\sum_n \frac 1Z e^{-E_n/kT}|E_n⟩⟨E_n|. $$

As always, $Z$ is the partition function, chosen to make $\rho_G$ normalized to $\mathrm{Tr}(\rho_G)=1$, so $$ Z =\mathrm{Tr}\mathopen{}\left(e^{-H/kT}\right)\mathclose{} =\sum_n e^{-E_n/kT}. $$

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  • $\begingroup$ How to define a Gibbs state if the Hamiltonian is non-Hermitian (e.g. $H=H_S +i\Gamma$ for effective system-bath coupling); where the density matrix become non-Hermitian? $\endgroup$ – donnydm Mar 22 '18 at 14:56
  • $\begingroup$ @donnydm You should ask that as a separate question. $\endgroup$ – Emilio Pisanty Mar 22 '18 at 15:26

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