The spectral density is
$$
\operatorname{Im}G_R(\Omega)
=\frac{\sigma_0\Omega t_0}
{1+\Omega^2t_0^2}.
$$
Substitution into the spectral representation, closing the contour in the half plane selected by the sign of $\omega_n$, gives
$$
\boxed{
G(i\omega_n)
=-\frac{\sigma_0}{1+|\omega_n|t_0}}.
$$
In particular $G(0)=-\sigma_0\geq0$ under the sign convention established in part vii. The absolute value is required because a bosonic Hermitian-operator Matsubara correlator is even in $\omega_n$.

Solved by gpt-5.6-sol high.


Solution

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

 Solution

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