# Lagrangian for Goldstone mode + topological excitation

The XY-model Hamiltonian is the following,

$${\cal H}~=~-J\sum_{\langle i,j\rangle} \cos (\theta_i -\theta_j).$$

The Goldstone mode corresponds to term $(\nabla \theta)^2$ in the effective Lagrangian.

Then what's the form of effective Lagrangian that produces both Goldstone mode and topological excitation (vortices)?

And how to derive the effective Lagrangian from the XY-Hamiltonian?

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I read the original KT-transition paper, but it is not very clear. – ChenChao Dec 6 '12 at 9:52

Consider the change in angle produced as we perform a closed loop around some point. Naively we obtain \begin{align} \Delta \theta = \oint d \vec r \cdot \nabla \theta(\vec r) = \int d^2 r \, \vec e_z \cdot \vec \nabla \times \nabla \theta(\vec r) = 0, \end{align} where Stokes theorem was used in the second step and the fact that $\vec \nabla \times \nabla f = 0$ for any scalar function f. However, we know that due to the compact nature of our angle variable $\theta(\vec r)$, we don't have to get back exactly to the same angle, but have more freedom and are allowed to have $\Delta \theta = 2\pi n$ with $n \in \mathbb{Z}$. When Taylor-expanding the original cosine, we have lost the information about the compact nature of the $\theta$-variables and thus have to reinclude it manually.
To include this additional freedom, we have to generalize the phase difference field $\vec u$ in the form $\vec u = \nabla \theta(\vec r) + \vec \nabla \times \vec A(\vec r)$, where the vector field $\vec A(\vec r) = \psi(\vec r) \vec e_z$ can have only an $z$-component in order for $\vec u$ to lie within the xy-plane. Then \begin{align} 2 \pi n = \oint d \vec r \cdot \vec u(\vec r) = \int d^2 r \, \vec e_z \cdot \vec \nabla \times \vec u(\vec r) = -\int d^2 r \, \vec \nabla^2 \psi(\vec r) \end{align} where $\vec \nabla^2 = (\vec \nabla \cdot \vec \nabla)$ is the Laplace-operator and the vector identity $\vec \nabla \times (\vec \nabla \times \vec A) = \nabla (\vec \nabla \cdot \vec A) - (\vec \nabla \cdot \vec \nabla) \vec A$ was used whose first term vanishes for $\vec A = \psi(\vec r) \vec e_z$ since $\psi(\vec r)$ has no $z$-dependence. Consequently, one finds \begin{align} \vec \nabla^2 \psi(\vec r) = -2\pi \sum_i n_i \delta(\vec r - \vec r_i) \end{align} and we interpret this by saying that the field $\psi(\vec r)$ describes vortices with winding numbers $n_i$ centered at positions $\vec r_i$. The above equation for $\psi(\vec r)$ is the defining equation of linear combinations of fundamental solutions of the Laplace equation (in 2D) which are given by \begin{align} \psi(\vec r) = - 2 \pi \sum_i n_i \log(\vert \vec r - \vec r_i\vert). \end{align} Plugging the phase difference field into the effective Hamiltonian, one finds \begin{align} H = -\frac{J}{2} \int d^2 r \, \vec u^2 = -\frac{J}{2} \int d^2 r \, (\nabla \theta(\vec r))^2 + (\vec \nabla \times \psi(\vec r) \vec e_z)^2, \end{align} where the mixed term involving both the $\theta$ and the $\psi$ field vanished after a partial integration. The first part corresponds to the spin-wave excitations you mentioned, while the second encodes the vortices. It is straight-forward to show that $(\vec \nabla \times \psi(\vec r) \vec e_z)^2 = (\nabla \psi)^2 = -\psi (\vec \nabla \cdot \vec \nabla) \psi$ after a partial integration, such that the topological part one ends up with is given by \begin{align} H_\text{top} = \frac{J}{2} \sum_{ij} n_i n_j \frac{\log(\vert \vec r_i - \vec r_j\vert)}{2\pi}. \end{align} The passage to the Lagrangian formulation of the problem is trivial since no canonical momenta are involved.