Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 337 (Applications of Quantum Field Theory)

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 $\eta =0$.

Let ${\eta }_i=+1$ and $-1$ on the two sublattices of the square lattice. A unit-vector decomposition that separates staggered and uniform magnetization iswhereThe alternating part is the Néel order parameter, while $\mathbf{m}$ is the slowly varying uniform magnetization generated by canting the two sublattices.

For a smooth configuration, neighboring sites lie on opposite sublattices and contribute Spin coherent-state Berry phases with opposite orientations. The terms $S\Gamma [\tilde{\mathbf{n}}]$ and $S\Gamma [-\tilde{\mathbf{n}}]$ therefore cancel pairwise modulo the quantized solid-angle ambiguity. Their smooth bulk contribution vanishes in the continuum limit.

Singular spacetime configurations can leave lattice Berry phases attached to hedgehog events, but those are outside the smooth sector used in this continuum derivation. Thus the order-parameter-only Wess–Zumino term may be dropped here.

At a site with background ${\eta }_i\tilde{\mathbf{n}}$, the linear variation is $\delta {\mathbf{n}}_i=\mathbf{m}$. The supplied variation formula givesReplacing the lattice sum by $a^{-2}\int d^{2}x$ yieldsThis Berry-phase term makes the uniform canting field the momentum conjugate to rotations of the Néel order parameter.

For a nearest-neighbor bond in the positive coordinate direction $\mu$, smoothness and orthogonality giveThere are two such bonds per site on the square lattice. Omitting the constant ground-state energy,The staggered part of the field coupling cancels between the two sublattices, whereasIncluding the overall minus sign of the Hamiltonian contribution to the real-time action, the continuum Lagrangian density is therefore

SetAfter adding $\lambda \mathbf{m}\cdot \tilde{\mathbf{n}}$, the terms involving the massive canting field areCompleting the square givesThe Gaussian functional integral over $\mathbf{m}$ contributes only a field-independent determinant. Hence

Because $\tilde{\mathbf{n}}\cdot (\tilde{\mathbf{n}}\times {\partial }_t\tilde{\mathbf{n}})=0$, the component of $\mathbf{q}$ parallel to $\tilde{\mathbf{n}}$ is $-\mathbf{B}\cdot \tilde{\mathbf{n}}$. The Gaussian integral over $\lambda$ removes precisely this longitudinal component. ThusUsing the vector triple-product identity and $|\tilde{\mathbf{n}}|=1$, this becomes the $O(3)$ nonlinear sigma model in a background field:The magnetic field acts as the temporal component of an $O(3)$ background gauge connection.

Choose the orientation of spherical coordinates aswhich differs from the opposite azimuth convention only by $\varphi \mapsto -\varphi$. With $\mathbf{B}=B\hat{\mathbf{x}}$, $\theta =\pi /2-\delta \theta$, and $\varphi =\delta \varphi$,Substitution into the effective action giveswhere

Let $\psi =\delta \theta +i\delta \varphi$ and define the zero-field spin-wave velocityThe linearized equation isFor a plane wave, the two circular polarizations therefore obeyor, with signed-frequency branches, $\omega =ck\pm B$ and their negative-frequency partners. At $B=0$ these are the two degenerate, linearly dispersing antiferromagnetic spin waves. The field Zeeman-splits the two opposite circular polarizations by shifting their frequencies in opposite directions.