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  Boundary correlations in the 2D Ising model: a 50-year-old open problem

+ 7 like - 0 dislike

The 2D Ising model is heralded as one of the few "exactly-soluble" systems in statistical mechanics that exhibits complex behavior. However, not much is known about the boundary correlation functions of the 2D Ising model. In particular, for the simplest case of an Ising model on the upper-half plane, the limit of the correlation function

\[M=\lim_{N\to\infty}\langle \sigma_{00}\sigma_{N0}\rangle\]

Is known. However, the rate of approach is completely unknown. 

There is numerical evidence that the rate is exponential with the exponential being the same correlation length as the correlation length in the bulk. I do not know of a proof of this (You would think that after 50 years someone would have figured out how to compute it). Is there anything quantitative known about the rate of convergence?

asked Nov 6, 2016 in Open problems by David B Roberts (135 points) [ no revision ]
recategorized Nov 7, 2016 by dimension10

1 Answer

+ 3 like - 0 dislike

Note that the question only makes sense when $T\geq T_{\rm c}$, since the 2-point function does not decay when $T<T_{\rm c}$ (neither in the half-plane, nor in the full plane).

I don't know how to prove the result when $T=T_{\rm c}$, although it is quite plausible that the technology developed around SLE makes it possible nowadays. The decay will not be exponential in this case, of course.

On the other hand, it is possible to prove the result when $T>T_{\rm c}$. I will write $\langle\cdot\rangle_{\rm full}$ for the expectation in the full plane. You can first use Griffiths' inequality to prove that

$$\langle \sigma_{00} \sigma_{N0} \rangle \leq \langle \sigma_{00} \sigma_{N0} \rangle _{\rm full}.$$

To obtain a lower bound, use again Griffiths' inequality to show that

$$\langle \sigma_{00} \sigma_{N0} \rangle \geq \langle \sigma_{00} \sigma_{0M} \rangle  \langle \sigma_{0M} \sigma_{NM} \rangle \langle \sigma_{NM} \sigma_{N0} \rangle \geq e^{-c M} \langle \sigma_{0M} \sigma_{NM} \rangle.$$

Choose $M=N^\alpha$ for some arbitrary $\tfrac12<\alpha<1$. Then, it follows from the analysis developed here that

$$\langle \sigma_{0M} \sigma_{NM} \rangle = (1+o(1)) \langle \sigma_{0M} \sigma_{NM} \rangle_{\rm full} = (1+o(1)) \langle \sigma_{00} \sigma_{N0} \rangle_{\rm full},$$ where the term $o(1)$ vanishes as $N\to\infty$. We thus have

$$\langle \sigma_{00} \sigma_{N0} \rangle \geq e^{-c M} \langle \sigma_{00} \sigma_{N0} \rangle_{\rm full}.$$

The conclusion follows, since the decay is exponential in the full plane and $\lim_{N\to\infty}M/N = 0$. (Note that the factor $e^{-cM}$ is very poor; with enough work, it should however be possible to compute precisely the order of the prefactor, but this is not necessary for our purpose here.)

To conclude, I have no idea whether the decay of this 2-point function can be computed explicitly directly, but this problem is easily solved using the many techniques that mathematical physicists have developed in the last decades to analyze nonperturbatively the Ising model with full mathematical rigor. Actually, the above proof of the equality of the rate of decay remains true when $d\geq 3$ (and you work in a half-space), since it has been proved that the decay in full space is exponential when $T>T_{\rm c}$ and the Ornstein-Zernike result I use above also remains valid in higher dimensions. So one can go much further than what is doable with exact computations.

answered Nov 21, 2016 by Yvan Velenik (1,110 points) [ revision history ]
edited Nov 21, 2016 by Yvan Velenik

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