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

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