Golden–Thompson inequality

Remarks

The right side is a positive real number, because $\tr(e^A e^B) = \tr(e^{A/2} e^B e^{A/2})$ and $e^{A/2} e^B e^{A/2}$ is positive definite. One proof uses the inequality $\tr\big((XY)^{2^k}\big) \le \tr\big(X^{2^k} Y^{2^k}\big)$ for positive definite $X$ and $Y$, which follows from the Cauchy–Schwarz inequality for the trace by induction on $k$. Put $X = e^{A/2^k}$ and $Y = e^{B/2^k}$ and let $k \to \infty$; the Lie product formula turns the left side into $\tr e^{A + B}$. If $AB = BA$, then $e^{A + B} = e^A e^B$ and equality holds. The naive extension to three Hermitian matrices, with $\abs{\tr(e^A e^B e^C)}$ on the right, is false in general.

Going deeper

For a student

In statistical mechanics, $\tr e^{-\beta H}$ is the partition function of a quantum system with Hamiltonian $H$, so the inequality bounds a partition function by a simpler expression. In probability, it is used in the approach of Ahlswede and Winter to matrix concentration inequalities. There one must control the trace of the exponential of a sum of random Hermitian matrices. Lieb found a version for three matrices, with an integral in place of a simple product.