I have the following problem that I was unable to solve for class, but I had a couple first steps that I started with that I am unable to finish. I know I can't get this since it's already been turned in, but I would like to see if this was even a viable option for doing this proof in the first place.

``Given an orthogonal set of states, $\{\phi_{n}\}$, and a Hamiltonian, $\hat{H}$, show that the partition function, $Q_{\beta}$, satisfies the following $$ Q_\beta \geq \sum_{n}\exp\{-\beta \langle \phi_n|\hat{H}|\phi_n\rangle \}$$ with equality holding when the $\phi_n$ states are eigenstates of the Hamiltonian.''

I started by dropping in the identity in the exponential (as eigenstates of the Hamiltonian) $$ \sum_n \exp\{ -\beta\sum_k \langle \phi_n|\psi_k\rangle\langle \psi_k |\hat{H}|\phi_n\rangle\}=\sum_n \exp\{ -\beta \sum_k E_k |c_{nk}|^2\}$$ Then I am left with showing that $$\sum_n \exp\{ -\beta \sum_k E_k |c_{nk}|^2\} \leq \sum_k \exp\{ -\beta E_k\}$$ with equality showing up again the same way, with a Kronecker delta $\delta_{nk}$ collapsing the sum in the exponential.

I realize I didn't make it very far, so this might not be the best way to show this, but it seems manifestly true just by looking at it, but I can't actually show it. Does anyone have any hints about how to continue with this?


1 Answer 1


Here is a sketched proof of the inequality. The problem is to show that

$$ \sum_n\langle \phi_n|e^{-\beta \hat{H}}|\phi_n\rangle ~\stackrel{?}{\geq}~ \sum_n e^{-\beta\langle \phi_n|\hat{H}|\phi_n\rangle} ,\qquad\qquad (1)$$

where the Hamiltonian $\hat{H}$ is a selfadjoint operator, and $|\phi_n\rangle $ denote orthonormal basis vectors in the Hilbert space of states. The lhs. of eq. (1) is the partition function ${\rm Tr}(e^{-\beta \hat{H}})$. By scaling $\hat{H}$, we may assume that $\beta=-1$. The inequality (1) would follow if we can show the inequality for each and every summand

$$ \langle \phi_n|e^{ \hat{H}}|\phi_n\rangle ~\stackrel{?}{\geq}~ e^{ \langle \phi_n|\hat{H}|\phi_n\rangle} ,\qquad\qquad (2)$$

or equivalently, in a simplified notation for fixed $n$,

$$ \langle e^{ \hat{H}}\rangle ~\stackrel{?}{\geq}~ e^{ \langle \hat{H} \rangle} .\qquad\qquad (3)$$

But eq. (3) is just Jensen's inequality for a convex function (with the exponential function playing the role of the convex function).


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