Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 337 / 2

A ferromagnetic vacuum breaks the special unitary group $SU(2)$ to the subgroup of rotations about its magnetization, so its vacuum manifold is the two-sphere $S^2$. The massless field is therefore a unit vector $\mathbf{n}(x,t)$ with ${\mathbf{n}}^2=1$.

The only rotational scalar linear in $\dot{\mathbf{n}}$ that can be formed directly from $\mathbf{n}$ is $\mathbf{n}\cdot \dot{\mathbf{n}}=\frac{1}{2}{\partial }_t({\mathbf{n}}^2)=0$. A local one-form $\mathbf{A}(\mathbf{n})\cdot \dot{\mathbf{n}}$ can describe the required Berry phase, but no choice of $\mathbf{A}$ is globally nonsingular and strictly rotationally invariant on $S^2$. One therefore needs coordinate patches, an extension, or a redundant spinor parametrization.

In the CP1 spinor representation,where the ${\sigma }_i$ are Pauli matrices. The transformation $z\mapsto e^{i\alpha (x,t)}z$ leaves $\mathbf{n}$ unchanged. For $U\in SU(2)$,with $R$ in the special orthogonal group $SO(3)$, so $z\mapsto Uz$ induces $n_i\mapsto R_{ij}{n}_j$.

A rotationally invariant first-order term is the Ferromagnetic Wess–Zumino termUnder the phase redundancy it changes as ${\mathcal{L}}_{\mathrm{WZ}}\mapsto {\mathcal{L}}_{\mathrm{WZ}}-\kappa \dot{\alpha }$. The change is a total derivative, so the action has the required invariance.

For $z=(e^{-i\varphi }\cos (\theta /2),\sin (\theta /2){)}^T$,Thus $i\kappa z^{†}\dot{z}=(\kappa /2)(1+\cos \theta )\dot{\varphi }$. Dropping the total derivative $(\kappa /2)\dot{\varphi }$ and writing $s=\kappa /2$ givesThe leading rotationally invariant gradient energy is $({\rho }_s/2)(\nabla \mathbf{n}{)}^2$, so an effective Lagrangian is

Put $q=\delta \theta$, $p=\delta \varphi$, so $\theta =\pi /2-q$ and $\varphi =p$. The quadratic Lagrangian is obtained by takingThe two real fluctuations form a coordinate and its canonical momentum rather than two independent modes. Their Euler-Lagrange equations combine to giveand similarly for $p$. The ferromagnetic magnon therefore has the quadratic dispersion relation

With the convention $G_R(t)=-i\Theta (t)⟨[q(t),q(0)]⟩$, the retarded Green function isDefining the spectral function by $\varrho =-2\mathrm{Im}{G}_R$ givesOther common spectral-weight conventions differ by an overall factor.

The Fluctuation-dissipation theorem in this convention readsIn the classical limit $|\omega |\ll T$, this becomesThe inverse temporal Fourier transform is therefore

For a damped ferromagnetic spin wave, shifting both poles into the lower half-plane preserves causality:Its spectral function isThe classical Fluctuation-dissipation theorem then givesHence the zero-frequency static structure factor iswhich has the stated proportionality after absorbing the normalization $4/s$.

Write $D={\rho }_s/s$, so ${\omega }_k=D{k}^2$, and take ${\Gamma }_k=\gamma |k|$. The small-wavenumber static structure factor behaves asThe spatial Fourier transform of $|k{|}^{-1}$ in $d$ dimensions scales as $|x{|}^{1-d}$. Thusand in three dimensions $C(x)\sim |x{|}^{-2}$. Since this two-point correlation function tends to zero rather than to a nonzero constant at large separation, it is incompatible with true long-range ferromagnetic order and hence with spontaneous symmetry breaking at this higher temperature.