As $T\to0$, the Matsubara sum becomes $\int d\omega/(2\pi)$. Put $p_0=\omega/v$ and $M=m/v$. The factors of $v$ cancel from both sides, leaving
$$
\frac1g
=\int^\Lambda\frac{d^4p}{(2\pi)^4}
\frac1{p^2+M^2}.
$$
Using four-dimensional spherical coordinates,
$$
\begin{aligned}
\frac1g
&=\frac{2\pi^2}{(2\pi)^4}
\int_0^\Lambda\frac{p^3\,dp}{p^2+M^2}\\
&=\boxed{
\frac1{16\pi^2}
\left[
\Lambda^2-M^2
\log\left(1+\frac{\Lambda^2}{M^2}\right)
\right]}.
\end{aligned}
$$
At the onset of <spontaneous symmetry breaking>, the symmetric-phase gap closes. Thus
$$
\boxed{
\frac1{g_c}=\frac{\Lambda^2}{16\pi^2}},
\qquad
\boxed{
g_c=\frac{16\pi^2}{\Lambda^2}}.
$$
For $g<g_c$ the constraint is instead satisfied by an ordered condensate; a positive symmetric gap exists for $g>g_c$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 337 / 2 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-337/2/a/ii/solution