# Quantum Mapping of a Generalized 2D XY Model

I have been reading this paper, which discusses the deconfinement transition in the 2D generalized XY model. The classical Hamiltonian is given by the following. See page 2 of the paper.

$$H = - J\sum_{\langle i,j \rangle}\bigg[(1 - \Delta)\cos\big(\theta_i - \theta_j \big) + \Delta \cos\big(2\theta_i - 2\theta_j \big) \bigg]$$

Directly beneath this equation the paper mentions that

In an associated quantum chain, the two terms ... correspond respectively to the hopping of single bosons and of boson pairs.

In order to see what the corresponding quantum model is, I considered the partition function of the classical system for small $$K$$. We set $$-\beta J$$ to $$K$$.

For a 2D square lattice with periodic boundary conditions on the rows and columns, the above Hamiltonian can be written as the following. Note that $$\theta_{(i,j)}$$ is the rotor at the $$i$$-th column of the $$j$$-th row

$$-\beta H = K \sum_\text{row j} \sum_\text{site i in j}\bigg[(1 - \Delta)\cos\big(\theta_{(i,j)} - \theta_{(i+1,j)} \big) + (1 - \Delta)\cos\big(\theta_{(i,j)} - \theta_{(i,j+1)} \big) + \Delta\cos\big(2\theta_{(i,j)} - 2\theta_{(i+1,j)} \big) + \Delta\cos\big(2\theta_{(i,j)} - 2\theta_{(i,j+1)} \big) \bigg]$$

Therefore

$$e^{-\beta H } = \prod_i \prod_j e^{K(1 - \Delta)\cos\big(\theta_{(i,j)} - \theta_{(i+1,j)} \big)} e^{K(1 - \Delta)\cos\big(\theta_{(i,j)} - \theta_{(i,j+1)} \big)} e^{K\Delta\cos\big(2\theta_{(i,j)} - 2\theta_{(i+1,j)} \big)} e^{K\Delta\cos\big(2\theta_{(i,j)} - 2\theta_{(i,j+1)} \big)}$$ $$e^{-\beta H }\approx \prod_i \prod_j \big(1 + K(1 - \Delta)\cos\big(\theta_{(i,j)} - \theta_{(i+1,j)} \big) \big)\big(1 + K(1 - \Delta)\cos\big(\theta_{(i,j)} - \theta_{(i,j+1)} \big)\big)\big(1 + K\Delta\cos\big(2\theta_{(i,j)} - 2\theta_{(i+1,j)} \big)\big)\big(1 + K\Delta\cos\big(2\theta_{(i,j)} - 2\theta_{(i,j+1)} \big)\big)$$ $$e^{-\beta H }\approx \prod_i \prod_j \bigg[ 1 + K(1 - \Delta)\cos\big(\theta_{(i,j)} - \theta_{(i+1,j)} \big) + K(1 - \Delta)\cos\big(\theta_{(i,j)} - \theta_{(i,j+1)} \big) + K\Delta\cos\big(2\theta_{(i,j)} - 2\theta_{(i+1,j)} \big) + K\Delta\cos\big(2\theta_{(i,j)} - 2\theta_{(i,j+1)} \big)\bigg]$$ $$e^{-\beta H }\approx 1 + \sum_\text{row j} \sum_\text{site i in j} K(1 - \Delta)\cos\big(\theta_{(i,j)} - \theta_{(i+1,j)} \big) + \sum_\text{row j} \sum_\text{site i in j}K(1 - \Delta)\cos\big(\theta_{(i,j)} - \theta_{(i,j+1)} \big) + \sum_\text{row j} \sum_\text{site i in j}K\Delta\cos\big(2\theta_{(i,j)} - 2\theta_{(i+1,j)} \big) + \sum_\text{row j} \sum_\text{site i in j} K\Delta\cos\big(2\theta_{(i,j)} - 2\theta_{(i,j+1)} \big)$$

I am not sure how to proceed from here. How does the quantum mapping work?

$$\hat{a}=\sqrt{n}e^{i\theta}$$, $$\hat{a}^\dagger=\sqrt{n}e^{-i\theta}$$
Single-photon hopping means terms like $$\hat{a_i}^\dagger\hat{a_j}+h.c.$$ and two-photon hopping $${\hat{a_i}^\dagger}^2\hat{a_j}^2+h.c.$$ . Substitute the above and assume $$n$$ to be constant