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I'm studying the Ising model in 2 dimensions in an approximative way. Now my professor has written this formula that links the dual space and the "normal" space:

$$\sinh(2 K) \sinh(2 K^*) = 1 $$

Do you know how can I demonstrate this equivalence?

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    $\begingroup$ I'm going to suggest that your question would be clearer to many readers if you edited it to include a definition of $K$, and perhaps a bit more background about the context in which this equation has arisen. $\endgroup$ – LonelyProf Dec 11 '18 at 22:27
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I presume that you are referring to the Kramers-Wannier duality. This is a symmetry relation which may be used to relate a high-temperature expansion of the square lattice Ising model to a low-temperature expansion of the same model. Kramers and Wannier used it to obtain an expression for the critical temperature of the 2D Ising model a couple of years before Onsager published his famous exact solution. Although the names may suggest that the expansions are only correct at "high" or "low" temperature, this is only the case if they are truncated. The full expansions are both formally exact.

That Wikipedia page does not really give the details you want. But the duality is quite well explained in Rodney Baxter's book on Exactly solved models in statistical mechanics, which is out of print, but available on the ANU Theoretical Physics web page. The section of interest to you is on pp 73-78, and the relevant equations are (6.2.14ab) and (6.2.15). In the derivation, the horizontal and vertical coupling constants are allowed to be different from each other.

I'm not going to reproduce the derivation here, it is well set out in Baxter's book. I'll just say that the "dual space" you mention is really just a dual lattice: the spins are on the vertices of the original lattice, but in the centres of the faces of the dual lattice. The relation is derived by considering two formally exact expressions for the partition function: one is a sum over closed loops or polygons which can be drawn on the original lattice, and the other is a sum over polygons drawn on the dual lattice.

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  • $\begingroup$ Thanks for the answer, so we put that equivalence because we see that there is a link (duality) between the descriptions. $\endgroup$ – MementoMori Dec 16 '18 at 9:30

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