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If the eigenstate $| i\rangle$ of the Hamiltonian $\hat H$ has energy $E_i$ the relative probability of the system being in that state is $e^{-\beta E_i}$ where $\beta = 1/\left(k_BT\right)$.

The density matrix is then:

$$\hat{\rho} = \frac{1}{Z} \sum_i e^{- \beta E_i} | i \rangle \langle i| = \frac{e^{-\beta \hat H}}{Z} \tag{1}$$

where

$$ Z = \sum_i e^{- \beta E_i} = \mathrm{Tr}(e^{- \beta \hat H}) . \tag{2}$$

I can easily prove to myself the validity of (2) however I am unsure about (1), namely why it is that:

$$\sum_i e^{- \beta E_i} | i \rangle \langle i| = e^{-\beta \hat H}$$

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  • $\begingroup$ Equation one is the eigendecomposition of the function of operators $e^{\beta \hat{H}}$. It’s important to note that the $H$ on the right hand side of your equation is an operator and not a function of c-numbers. It follows from the spectral theorem essentially en.wikipedia.org/wiki/Spectral_theorem $\endgroup$
    – gabe
    Commented Dec 25, 2020 at 2:25

2 Answers 2

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Perhaps it's easiest to work backwards. Let the right hand side of the equation act on an arbitrary vector $|A\rangle$. $$e^{-\beta \hat H}|A\rangle$$ Next, let's label the eigenvectors of the operator $\hat H$ with an index $i$ so that $$\hat H|i\rangle = E_i|i\rangle .$$ This also implies that $$e^{-\beta\hat H}|i\rangle = e^{-\beta E_i}|i\rangle .$$ These eigenvectors form a complete basis so we can write $$|A\rangle = \sum_i |i\rangle \langle i|A\rangle .$$ Now we have $$e^{-\beta \hat H}|A\rangle = \sum_i e^{-\beta \hat H}|i\rangle \langle i|A\rangle = \sum_i e^{-\beta E_i}|i\rangle \langle i|A\rangle . $$ But $|A\rangle$ was arbitrary so we have the result you wanted as a general matrix identity.

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The easiest way to justify that is to evaluate the elements of $\hat{\rho}$. Then

$$ \langle j|\left(\sum_{i} e^{-\beta E_{i}}|i\rangle\langle i|\right)|m\rangle=\langle j|e^{-\beta H}|m\rangle $$ which, by using the fact that $|i\rangle$ is an eigenstate of $\hat{H}$ (the basis generated by the eigenstates of $\hat{H}$ can always be chosen to be an orthonormal basis) implies that $$ \sum_{i} e^{-\beta E_{i}}\delta_{ji}\delta_{im}=e^{-\beta E_m}\delta_{jm}, $$ where in the left-hand side we used the fact that a function of an operator evaluated at an eigenstate, is the function evaluated at the corresponding eigenvalue times the eigenstate (e.g. https://en.wikipedia.org/wiki/Matrix_function). By summing over $i$ (it needs to be at least equal to $j$) we obtain $$ e^{-\beta E_{j}}\delta_{jm}=e^{-\beta E_j}\delta_{jm}, $$ indicating that all elements of both operators are the same, hence they are equal.

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