# Goldstone mode as spin wave in 2D?

I'm trying to understand how Goldstone modes destroy long range order in 1D and 2D spin lattice. I started with a spin chain, using 1D XY-model, which has continuous symmetry. $H=- \sum_{<i j>} J_{ij}\; \cos(\theta_i-\theta_j)$ with no external field.

If there's a spin wave on this chain, half a wavelength will destroy the magnetization. The energy cost will be $\Delta E=(-J\cos(\frac{\pi}{N})+J)N$. As $N\rightarrow \infty$, $\Delta E\rightarrow 0$. So no matter how low the temperature is, this mode will destroy spontaneous magnetization.

But I'm having trouble understanding it in 2D. The easiest construction is simply to put $N$ chains together, and the energy cost will multiply by $N$: $\Delta E=(-J\cos(\frac{\pi}{N})+J)N^2$. But this time as $N\rightarrow \infty$, $\Delta E\rightarrow \frac{\pi^2}{2}J$. So if $k_BT\ll\frac{\pi^2}{2}J$, this mode can't exist and spontaneous magnetization will happen. I tried other possible constructions, like a spin wave going diagonal direction, which doesn't work. I also looked at the vortices in XY-model, the 4 spins in the center of the vortex cost $4J$ and I believe summing over other bonds the total cost will be equal or more than $\frac{\pi^2}{2}J$. I'm thinking if it's possible to destroy the magnetization with energy less than $\frac{\pi^2}{2}J$. But Mermin–Wagner theorem states Goldstone mode with zero energy will destroy the ordered state, so there must be something I'm missing. I tried to find an illustration of this mode, but searching "2D spin wave" or "2D Goldstone mode" only returns calculations or experiments. So I'm wondering what does it look like, what is the alignment of the spins exactly?

• How is the title related to the question? – Norbert Schuch Oct 7 '16 at 6:57
• Just a short comment, but the energy of a system at temperature $T$ is extensive, since we have $\sim kT$ for every mode. So $\Delta E \sim NkT$, which in the thermodynamic limit will always be bigger than any finite (non-extensive) energy gap, like your $\frac{\pi^2}{2}J$. – Ruben Verresen Oct 7 '16 at 21:41
• In a way you are actually indirectly arguing for Coleman's theorem: at zero temperature, you've shown that in the 1D case even the ground state is not stable since any infinitesimally small energy would disrupt it, i.e. any state that is not *literally*/100% the ground state would not have a stable continuous symmetry breaking. And then you discovered that in 2D it seems the ground state is stable against finite (but non-extensive) energy perturbations, allowing for spontaneous symmetry breaking of continuous symmetries at zero temperature. (But not at finite temperature, cf Schuch's answer.) – Ruben Verresen Oct 7 '16 at 21:47
• (Disclaimer: note that my comments are only for hand-wavy purposes, and are trying to address the intuition you were getting at in your question, but the real answer is in Schuch's post.) – Ruben Verresen Oct 7 '16 at 21:51
• Thank you for answers and discussions, that helped a lot. I thought 2D case can be visualized like 1D case, a single zero-energy wave destroys the ordered state. Now I see that the 1D case is really special. It's hard to visualize how 1D and 2D lattice are unstable because of phonons, too. The infrared divergence is the true nature of this phenomenon. (Though it's hard to explain that to people with little math background.) – Yi Jiang Oct 8 '16 at 2:26