At zero temperature, the <Matsubara sum> becomes a frequency integral:
$$
\frac1{vg}
=\int_{|\mathbf k|<\Lambda}\frac{d^2k}{(2\pi)^2}
\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}
\frac1{\omega^2+v^2k^2+m^2}.
$$
The frequency integral is $1/(2\sqrt{v^2k^2+m^2})$, and radial momentum integration gives
$$
\frac1{vg}
=\frac{\sqrt{v^2\Lambda^2+m^2}-m}{4\pi v^2}.
$$
Define the <critical coupling> by the massless equation
$$
\frac1{vg_c}=\frac{\Lambda}{4\pi v}.
$$
Taking the cutoff to infinity in the difference gives
$$
\frac1g-\frac1{g_c}=-\frac{m}{4\pi v},
$$
and hence
$$
\boxed{m=\frac{4\pi v(g-g_c)}{gg_c}.}
$$
A positive mass solution exists only for $g>g_c$. For $g<g_c$, the symmetric saddle cannot enforce the constraint with $m^2>0$; instead the $O(N)$ symmetry is <spontaneous symmetry breaking>[spontaneously broken] to $O(N-1)$, the field acquires <Néel order>, and the ordered phase contains massless <Goldstone bosons>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 337 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-337/1/c/solution