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  Critical temperature difference between Ising and XY model

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The following formula gives the critical coupling (more precisely the ratio of the spin-spin coupling over the temperature) for $O(n)$ models on a triangular lattice:

$$\text{e}^{-2K}=\frac{1}{\sqrt{2+\sqrt{2-n}}}$$

with $K=\beta J$

Numerically, it says that:

Ising model (n = 1) has $K \approx 0.27$

XY model (n=2) has $K \approx 0.17$

Thus, the critical temperature for the XY model is higher than the Ising model. I've been thinking about it but I can't come out with a reason of why allowing the order parameter to take continuous values means that we need to go higher in temperature to destroy order. Is there a (semi) intuitive reason for that?

This post imported from StackExchange Physics at 2014-06-29 09:38 (UCT), posted by SE-user Learning is a mess
asked Jun 28, 2014 in Theoretical Physics by Learning is a mess (75 points) [ no revision ]

1 Answer

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Your intuition is correct. It is not difficult to prove that the 2-point function of the Ising model is always an upper bound on the corresponding quantity for the XY model. In particular, if $\beta_c^{XY}$ denotes the inverse temperature at which the Kosterlitz-Thouless phase transition occurs, then $\beta_c^{XY}\geq 2\beta_c^{I}$, where $\beta_c^I$ is the inverse critical temperature for the Ising model. You can find the proof here. Note that the formula you give above is not for the $O(n)$ spin model, but for the $O(n)$ loop model.This post imported from StackExchange Physics at 2014-06-29 09:38 (UCT), posted by SE-user Yvan Velenik
answered Jun 28, 2014 by Yvan Velenik (1,110 points) [ no revision ]

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