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

T \to 0

, the Matsubara sum becomes

\int d ω / (2 π)

. Put

p_{0} = ω / v

and

M = m / v

. The factors of

v

cancel from both sides, leaving

\frac{1}{g} = \int^{Λ} \frac{d ^{4} p}{( 2 π ) ^{4}} \frac{1}{p ^{2} + M ^{2}} .

Using four-dimensional spherical coordinates,

\frac{1}{g} = \frac{2 π ^{2}}{( 2 π ) ^{4}} \int_{0}^{Λ} \frac{p ^{3} d p}{p ^{2} + M ^{2}} = \frac{1}{16 π ^{2}} [Λ^{2} - M^{2} lo g (1 + \frac{Λ ^{2}}{M ^{2}})] .

At the onset of spontaneous symmetry breaking, the symmetric-phase gap closes. Thus

\frac{1}{g _{c}} = \frac{Λ ^{2}}{16 π ^{2}}, g_{c} = \frac{16 π ^{2}}{Λ ^{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.

 Ancestors (11)

There are no discussions about this article yet.

There are currently no matching articles.