For a nearest-neighbor bond in the positive coordinate direction $\mu$, smoothness and orthogonality give
$$
\mathbf n_i\mathbin\cdot\mathbf n_{i+\hat\mu}
=-1+2|\mathbf m|^2
+\frac{a^2}{2}|\partial_\mu\widetilde{\mathbf n}|^2
+\text{higher derivatives and powers of }\mathbf m.
$$
There are two such bonds per site on the square lattice. Omitting the constant ground-state energy,
$$
S^2J\sum_{\langle ij\rangle}
\mathbf n_i\mathbin\cdot\mathbf n_j
\longrightarrow
\int d^2x\left[
\frac{JS^2}{2}|\nabla\widetilde{\mathbf n}|^2
+\frac{4JS^2}{a^2}|\mathbf m|^2
\right].
$$
The staggered part of the field coupling cancels between the two sublattices, whereas
$$
S\sum_i\mathbf B\mathbin\cdot\mathbf n_i
\longrightarrow
\frac S{a^2}\int d^2x\,
\mathbf B\mathbin\cdot\mathbf m.
$$
Including the overall minus sign of the Hamiltonian contribution to the real-time action, the continuum Lagrangian density is therefore
$$
\boxed{
\mathcal L
=\frac S{a^2}\mathbf m\mathbin\cdot
(\widetilde{\mathbf n}\times\partial_t\widetilde{\mathbf n})
-\frac{4JS^2}{a^2}|\mathbf m|^2
-\frac{JS^2}{2}|\nabla\widetilde{\mathbf n}|^2
-\frac S{a^2}\mathbf B\mathbin\cdot\mathbf m.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 337 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-337/2/d/solution