Golden–Thompson inequality

In physics and mathematics, the Golden–Thompson inequality is a trace inequality between exponentials of symmetric and Hermitian matrices proved independently by and. It has been developed in the context of statistical mechanics, where it has come to have a particular significance.

Statement
The Golden–Thompson inequality states that for (real) symmetric or (complex) Hermitian matrices A and B, the following trace inequality holds:


 * $$ \operatorname{tr}\, e^{A+B} \le \operatorname{tr} \left(e^A e^B\right).$$

This inequality is well defined, since the quantities on either side are real numbers. For the expression on the right hand side of the inequality, this can be seen by rewriting it as $$\operatorname{tr}(e^{A/2}e^B e^{A/2})$$ using the cyclic property of the trace.

Motivation
The Golden–Thompson inequality can be viewed as a generalization of a stronger statement for real numbers. If a and b are two real numbers, then the exponential of a+b is the product of the exponential of a with the exponential of b:
 * $$ e^{a+b} = e^a e^b .$$

If we replace a and b with commuting matrices A and B, then the same inequality $$ e^{A+B} = e^A e^B$$ holds.

This relationship is not true if A and B do not commute. In fact, proved that if A and B are two Hermitian matrices for which the Golden–Thompson inequality is verified as an equality, then the two matrices commute. The Golden–Thompson inequality shows that, even though $$e^{A+B}$$ and $$e^Ae^B$$ are not equal, they are still related by an inequality.

Generalizations
The Golden–Thompson inequality generalizes to any unitarily invariant norm. If A and B are Hermitian matrices and $$\|\cdot\|$$ is a unitarily invariant norm, then
 * $$\|e^{A+B}\| \leq \|e^{A/2}e^Be^{A/2}\| .$$

The standard Golden–Thompson inequality is a special case of the above inequality, where the norm is the Schatten norm with $$p=1$$. Since $$e^{A+B}$$ and $$e^{A/2}e^Be^{A/2}$$ are both positive semidefinite matrices, $$\operatorname{tr}(e^{A+B}) = \|e^{A+B}\|_1$$ and $$\operatorname{tr}(e^{A/2}e^Be^{A/2}) = \|e^{A/2}e^Be^{A/2}\|_1$$.

The inequality has been generalized to three matrices by and furthermore to any arbitrary number of Hermitian matrices by. A naive attempt at generalization does not work: the inequality
 * $$\operatorname{tr}(e^{A+B+C}) \leq |\operatorname{tr}(e^Ae^Be^C)|$$

is false. For three matrices, the correct generalization takes the following form:


 * $$ \operatorname{tr}\, e^{A+B+C} \le \operatorname{tr} \left(e^A \mathcal{T}_{e^{-B}} e^C\right),

$$

where the operator $$\mathcal{T}_f$$ is the derivative of the matrix logarithm given by $$ \mathcal{T}_f(g) = \int_0^\infty \operatorname{d}t \, (f+t)^{-1} g (f+t)^{-1} $$. Note that, if $$f$$ and $$g$$ commute, then $$ \mathcal{T}_f(g) = gf^{-1}$$, and the inequality for three matrices reduces to the original from Golden and Thompson.

used the Kostant convexity theorem to generalize the Golden–Thompson inequality to all compact Lie groups.