Let $\beta=1/T$ and $A(\tau)=e^{H\tau}Ae^{-H\tau}$. For $0<\tau<\beta$, cyclicity of the trace gives
$$
\operatorname{Tr}\!\left(e^{-\beta H}A(\tau)A(0)\right)
=\operatorname{Tr}\!\left(e^{-\beta H}A(0)A(\tau-\beta)\right).
$$
This is the bosonic <Kubo--Martin--Schwinger condition>. It identifies the two time orderings across the end of the thermal interval, so
$$
\boxed{C(\tau+\beta)=C(\tau)}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 337 / 1 / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-337/1/iv/solution