# Critical temperature difference between Ising and XY model

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?

• 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. – Yvan Velenik Jun 28 '14 at 17:37
• No, the loop model is an approximation of the O(n) spin model. You can look, for example, at the explanations in Nienhuis' Les Houches lectures on Loop models, which can be downloaded from his homepage. – Yvan Velenik Jul 1 '14 at 8:30
• ps: In the Ising case, I managed to very precisely check the critical coupling value given by the formula above. That's why it's even more puzzling for me. – Learning is a mess Jul 1 '14 at 15:41
• The value of the critical points of the spin and loop models should not be related at all. However, the critical behaviors are expected to be the same (they should belong to the same universality class). – Yvan Velenik Jul 1 '14 at 16:53
• I have no idea what the value of the coupling constant of the spin O(2) model on the triangular lattice is. There are very efficient cluster methods for this model (as far as I know, I am not a specialist in the numerical aspects) which should provide precise estimates. – Yvan Velenik Jul 1 '14 at 16:56