Golden–Thompson lemma

Trace exponential inequality: for self-adjoint A,B, Tr e^{A+B} ≤ Tr(e^{A}e^{B}).
Golden–Thompson lemma

Definitions and notation

Statement

Let AA and BB be self-adjoint operators on a finite-dimensional Hilbert space (or assume eA+Be^{A+B} is trace-class so the trace is finite). Then

Tr(eA+B)Tr(eAeB). \operatorname{Tr}\big(e^{A+B}\big) \le \operatorname{Tr}\big(e^{A}e^{B}\big).

If AA and BB commute, then eA+B=eAeBe^{A+B}=e^{A}e^{B} and equality holds.

Key hypotheses and conclusions

Hypotheses

  • AA and BB are self-adjoint.
  • The traces involved are finite (automatic in finite dimension).

Conclusions

  • A fundamental bound for trace exponentials: Tr(eA+B)Tr(eAeB)\operatorname{Tr}(e^{A+B}) \le \operatorname{Tr}(e^{A}e^{B}).
  • Equality when [A,B]=0[A,B]=0.