Golden–Thompson inequality in statistical mechanics

Trace bound Tr(e^{A+B}) ≤ Tr(e^A e^B) and its standard application to quantum partition functions and free-energy bounds.

Golden–Thompson inequality in statistical mechanics

Statement

Let $A$ and $B$ be self-adjoint operators on a finite-dimensional Hilbert space (equivalently, Hermitian matrices). Then

\operatorname{Tr}\!\left(e^{A+B}\right)\le \operatorname{Tr}\!\left(e^{A}e^{B}\right).

This is the Golden–Thompson inequality , emphasized here in the form most commonly used in quantum statistical mechanics.

If $H_1$ and $H_2$ are Hamiltonians and $H=H_1+H_2$ , then for $\beta>0$ ,

Z(\beta)=\operatorname{Tr}\!\left(e^{-\beta(H_1+H_2)}\right) \le \operatorname{Tr}\!\left(e^{-\beta H_1}e^{-\beta H_2}\right),

$A=A^\ast$ and $B=B^\ast$ (Hermitian).
Finite-dimensional setting (or more generally: $A,B$ bounded with trace-class exponentials; the finite-dimensional case is the cleanest for stat-mech applications).

The trace of the exponential of a sum is bounded by the trace of the product of exponentials:
$\operatorname{Tr}(e^{A+B}) \le \operatorname{Tr}(e^A e^B).$
For decomposed Hamiltonians $H=H_1+H_2$ , this yields an upper bound on $Z(\beta)$ and hence a lower bound on the corresponding free energy (via $F=-(1/\beta)\log Z$ as in statistical free energy ).

A standard proof uses the Lie–Trotter product formula

e^{A+B}=\lim_{n\to\infty}\left(e^{A/n}e^{B/n}\right)^n

together with trace Hölder inequalities to compare $\operatorname{Tr}(e^{A+B})$ to $\operatorname{Tr}(e^{A}e^{B})$ .

In statistical mechanics, the inequality is a core tool for:

bounding interacting partition functions by splitting a Hamiltonian into “solvable + perturbation” pieces (useful alongside Peierls–Bogoliubov bounds);
establishing convexity/monotonicity properties of $\log Z$ and related thermodynamic potentials;
controlling approximations where noncommuting terms obstruct naive factorization.