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

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- Solution d

Solution

For a nearest-neighbor bond in the positive coordinate direction

μ

, smoothness and orthogonality give

n_{i} \cdot n_{i + \overset{μ}{^}} = - 1 + 2∣ m ∣^{2} + \frac{a ^{2}}{2} ∣ \partial_{μ} n ∣^{2} + higher derivatives and powers of m .

There are two such bonds per site on the square lattice. Omitting the constant ground-state energy,

S^{2} J \sum_{⟨ ij ⟩} n_{i} \cdot n_{j} ⟶ \int d^{2} x [\frac{J S ^{2}}{2} ∣\nabla n ∣^{2} + \frac{4 J S ^{2}}{a ^{2}} ∣ m ∣^{2}] .

The staggered part of the field coupling cancels between the two sublattices, whereas

S \sum_{i} B \cdot n_{i} ⟶ \frac{S}{a ^{2}} \int d^{2} x B \cdot m .

Including the overall minus sign of the Hamiltonian contribution to the real-time action, the continuum Lagrangian density is therefore

L = \frac{S}{a ^{2}} m \cdot (n \times \partial_{t} n) - \frac{4 J S ^{2}}{a ^{2}} ∣ m ∣^{2} - \frac{J S ^{2}}{2} ∣\nabla n ∣^{2} - \frac{S}{a ^{2}} B \cdot m .

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