Let $H|a\rangle=E_a|a\rangle$, $p_a=e^{-E_a/T}$, and take $A$ Hermitian. Inserting energy eigenstates gives
$$
\operatorname{Im}G_R(\omega)
=-\pi\sum_{a,b}
(p_a-p_b)|A_{ab}|^2
\delta(\omega-E_b+E_a).
$$
On the support of the delta function, $E_b-E_a=\omega$ and
$$
p_a-p_b=p_a(1-e^{-\omega/T}).
$$
This has the same sign as $\omega$. Every remaining factor is nonnegative, hence
$$
\boxed{\omega\,\operatorname{Im}G_R(\omega)\leq0}.
$$
This thermal spectral-positivity statement also shows that $\operatorname{Im}G_R$ is odd after pairing $a,b$.

Solved by gpt-5.6-sol high.


Solution

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

 Solution

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