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.
The Golden–Thompson inequality states that for (real) symmetric or (complex) Hermitian matrices A and B, the following trace inequality holds:
\operatorname{tr}eA+B\le\operatorname{tr}\left(eAeB\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}(eA/2eBeA/2)
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:
ea+b=eaeb.
eA+B=eAeB
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
eA+B
eAeB
The Golden–Thompson inequality generalizes to any unitarily invariant norm. If A and B are Hermitian matrices and
\| ⋅ \|
\|eA+B\|\leq\|eA/2eBeA/2\|.
p=1
eA+B
eA/2eBeA/2
\operatorname{tr}(eA+B)=\|eA+B\|1
\operatorname{tr}(eA/2eBeA/2)=\|eA/2eBeA/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}(eA+B+C)\leq|\operatorname{tr}(eAeBeC)|
\operatorname{tr}eA+B+C\le\operatorname{tr}\left(eA
l{T} | |
e-B |
eC\right),
where the operator
l{T}f
l{T}f(g)=
infty | |
\int | |
0 |
\operatorname{d}t(f+t)-1g(f+t)-1
f
g
l{T}f(g)=gf-1
used the Kostant convexity theorem to generalize the Golden–Thompson inequality to all compact Lie groups.