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

After integration by parts, define the Euclidean quadratic operatorThe action can then be written asEach component of $\mathbf{n}$ gives the same bosonic Gaussian functional integral, soDiscarding a $\lambda$-independent normalization, the remaining functional determinant gives

Every term in $\tilde{S}$ is proportional to $N$. The Large-N expansion therefore suppresses fluctuations of the auxiliary field by powers of $1/N$, and the leading partition function comes from a saddle-point approximation.

Vary with respect to $i\lambda$ and use $\delta \mathrm{Tr}\log K=\mathrm{Tr}(K^{-1}\delta K)$. At a translation-invariant saddle $i\lambda =m^2$, the gap equation isThe frequencies are the bosonic Matsubara frequencies imposed by periodicity around the imaginary-time thermal circle.

At zero temperature, the Matsubara sum becomes a frequency integral:The frequency integral is $1/(2\sqrt{v^2{k}^2+m^2})$, and radial momentum integration givesDefine the critical coupling by the massless equationTaking the cutoff to infinity in the difference givesand henceA 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 spontaneously broken to $O(N-1)$, the field acquires Néel order, and the ordered phase contains massless Goldstone bosons.

For fixed momentum putIn the contour formula, $F(z)=1/(E_k^2-z^2)$ has poles at $z=\pm E_k$. Deforming the contour onto those poles and using the oddness of $\coth (z/(2T))$ gives the standard Bosonic Matsubara sumEquivalently,where $n_B(E)=1/(e^{E/T}-1)$ is the Bose-Einstein distribution. The first term is the zero-point fluctuation and the second is its thermal occupation.

The finite-temperature gap equation isWith $E=\sqrt{v^2{k}^2+m^2}$, the radial measure obeys $kdk=EdE/v^2$, soAs $\Lambda \to \infty$, subtraction of the massless, zero-temperature equation at $g_c$ leavesThe zero-temperature relation with $m=\Delta$ isEquating the finite parts givesand thereforeThe subtraction is a renormalization condition: it trades the cutoff-dependent bare coupling for the physical zero-temperature gap.

When $T\ll \Delta$, the argument of the inverse hyperbolic sine is large. Using $\mathrm{arsinh}z=\log (2z)+1/(4z^2)+O(z^{-4})$ givesThe mass remains the zero-temperature gap, with an exponentially small correction from thermally activated excitations.

When $T\gg \Delta$, expand around $z=1/2$. Ifis the golden ratio, then $\mathrm{arsinh}(1/2)=\log \phi$, andThe leading value $m/T=2\log \phi$ is universal. This is the quantum-critical regime: temperature is the only leading energy scale and the correlation length is of order $v/T$.

The analytic continuation from bosonic imaginary frequency to a retarded frequency is $i{\omega }_n\mapsto \omega +i{0}^{+}$. ThusWriting $E_k=\sqrt{v^2{k}^2+m(T{)}^2}$ and using the Sokhotski–Plemelj formula gives, in this sign convention,The opposite overall convention for the retarded Green function reverses this sign; the corresponding spectral function is conventionally chosen positive at positive frequency.

Near the quantum critical point, $m/T$ is a function only of $\Delta /T$. The Green function has the scaling formwithThis is quantum critical scaling with dynamical critical exponent $z=1$ and leading large-$N$ anomalous dimension $\eta =0$.