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I am reading the section on the 2D Ising model Krammer-Wannier duality in the book Exactly Solved Models in Statistical Mechanics (pg. ~76) by R.J. Baxter. I have two questions:

  1. What was the motivation behind studying this problem in the dual lattice? It seems very artificial to me.
  2. I understand how to get to the "low temperature" expression of the partition function $$ \mathcal{Z}_L(N,L,K) = 2 \exp(M(K+L)) \sum_P \exp(-2(Lr+Ks)) $$ and to the "high temperature" expression $$ \mathcal{Z}_H(N,\tilde{L},\tilde{K}) = 2^N(\cosh(\tilde{L})\cosh(\tilde{K}))^M\sum_P v^rw^s $$ where $N$ is the no. of spins and $K \&L$ are the coupling constants with $\beta$ included inside them. As far as I understood, these expansions are exact in the thermodynamic limit, $N=M, N\rightarrow \infty$, (as the difference at the boundaries, if any, would vanish) so where does the low and high temperature regimes come in to play?

(Edited)

Following @YvanVelenik great help (as always!) if I let $v \equiv \tanh(\tilde{K})=e^{-2L}$ and $(K \leftrightarrow L)$ in $\mathcal{Z}_H$ hen one yields the following suggestive expression $$ \mathcal{Z}_H(N,L,K) = 2^N(\cosh(\tanh^{-1}(e^{-2L}))\cosh(\tanh^{-1}(e^{-2K})))^M \sum_P \exp(-2(Lr+Ks)). $$ If indeed we solve for $\sum_P \exp(-2(Lr+Ks))$ in $\mathcal{Z}_H$ and $\mathcal{Z}_L$ one yields $$ \mathcal{Z}_H = g(M,N,K,L) \mathcal{Z}_L, $$ where $$ g(M,N,K,L) = \frac{2^N(\cosh(\tanh^{-1}(e^{-2L}))\cosh(\tanh^{-1}(e^{-2K})))^M}{2 \exp(M(K+L)) } $$ is the claim that $g(M,N,K,L) \rightarrow 1$ in the high/low limits?

Thanks in advance!

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1 Answer 1

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These expressions are valid in finite systems (in any simply connected box), but the duality transformation maps a model with $+$ boundary condition to a model with free boundary condition, which affects the finite-volume free energy. Of course, as you say, in the thermodynamic limit, the boundary condition becomes irrelevant and one obtains a nontrivial identity for the free energy density of the planar Ising model (equ. (6.2.15) in the book).

The reason they are called low-temperature and high-temperature is that $\mathcal{Z}_L$ can be seen as a polynomial in $e^{-\beta}$, which is a small quantity when $\beta$ is large, that is, at low temperatures, while $\mathcal{Z}_H$ can be seen as a polynomial in $\tanh(\beta)$, which is a small quantity when $\beta$ is small, that is, at high temperatures.

In particular, these two representations of the model lead to an expansion of the infinite-volume free energy density that is convergent at low, respectively high temperatures. (More information on these aspects can be found in Chapter 5 of this book; see Sections 5.7.3 and 5.7.4. The Kramers-Wannier duality itself is discussed in Chapter 3, Section 3.10.1.)

Moreover, the duality transformation sends the model at inverse temperature $\beta>\beta_{\rm c}$ to the model at inverse temperature $\beta<\beta_{\rm c}$ (and vice versa); the critical point $\beta_{\rm c}$ coincides with the fixed point of the transformation. In particular, this duality interchanges the low-temperature region and the high-temperature region.


In a comment, you ask why these expansions are necessary to obtain the value of the critical point. Let me briefly address this here (more information can be found in the book I mention above). For simplicity, I assume that all coupling constants are equal to $1$, so that I only have the inverse temperature $\beta$ to deal with, which simplifies a bit the expressions.

Let $\Lambda_N = \{-N, \dots, N\}^2$ and $\Lambda_N^* = \{-N-\frac12,-N+\frac12,-N+\frac32,\dots,N+\frac12\}^2$ be the dual box.

The main observation is that, up to simple explicit functions of the inverse temperature and $N$, the low-temperature expansion of $\mathcal{Z}_{\Lambda_N;\beta}^+$ and the high-temperature expansion of $\mathcal{Z}_{\Lambda_N^*;\beta^*}^{\rm free}$ coincide, provided you choose the dual inverse temperature $\beta^*$ in such a way that $$ \tanh(\beta^*) = e^{-2\beta}.\tag{$\star$} $$ Namely, $$ 2^{-4N^2 -8N -4}\cosh(\beta^*)^{-8N^2-10N-2}\mathcal{Z}_{\Lambda_N^*;\beta^*}^{\rm free} = e^{-\beta(8N^2+10N+2)} \mathcal{Z}_{\Lambda_N;\beta}^+. $$ We are really interested in the free energy density in the thermodynamic limit. Therefore, let us take the logarithm and divide by $4N^2$ on both sides: \begin{multline} -\log(2) - 2 \log\cosh(\beta^*) + O(N^{-1}) + \frac{1}{(2N)^2} \log \mathcal{Z}_{\Lambda_N^*;\beta^*}^{\rm free}\\ = -2\beta + O(N^{-1}) + \frac{1}{(2N)^2} \log \mathcal{Z}_{\Lambda_N;\beta}^+ . \end{multline} Now, we use the fact that $$ \lim_{N\to\infty} \frac{1}{(2N)^2} \log \mathcal{Z}_{\Lambda_N^*;\beta^*}^{\rm free} = \phi(\beta^*) $$ and $$ \lim_{N\to\infty} \frac{1}{(2N)^2} \log \mathcal{Z}_{\Lambda_N;\beta}^+ = \phi(\beta), $$ since the free energy density is independent of the boundary condition in the thermodynamic limit.

We thus obtain \begin{align} \phi(\beta) &= 2\beta - \log(2) - 2\log\cosh(\beta^*) + \phi(\beta^*)\\ &= \phi(\beta^*) - \log \sinh(2\beta^*), \end{align} where we used the duality relation $(\star)$ to for the last identity.

Now, assuming that $\phi$ possesses exactly one singularity, located at the critical point $\beta_{\rm c}$, it follows from the above identity that $\beta_{\rm c} = \beta_{\rm sd}$, where $\beta_{\rm sd}$ is the self-dual point, that is, the unique value of $\beta$ such that $$ \tanh(\beta) = e^{-2\beta}. $$ Indeed, if $\beta_{\rm c} \neq \beta_{\rm sd}$, then the above identity implies that $\phi$ must have another singularity at $$ \beta_{\rm c}^* = \mathrm{atanh}(e^{-2\beta_{\rm c}}), $$ which would contradict the assumption that $\phi$ has only one singularity.

Finally, it is easy to check that $\beta_{\rm sd} = \frac12\log(1+\sqrt{2})$, which provides an explicit expression for the critical inverse temperature of the Ising model on $\mathbb{Z}^2$. This is actually the way the latter was first determined, by Kramers and Wannier, before the explicit computation of $\phi$ by Onsager.

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    $\begingroup$ The duality implies a symmetry of the free energy that guarantees that, assuming the latter has a unique singularity, then this singularity has to occur at the self-dual temperature (were it to occur at a different temperature $T_0$, then there would be another singularity at the dual temperature $T_0^*$, which would contradict the assumption that there is only one singularity). This was the way Kramers and Wannier predicted the critical temperature of the 2d Ising model before the explicit computation of the free energy by Onsager. $\endgroup$ Commented Jan 19, 2021 at 18:19
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    $\begingroup$ I have explained this in more detail here. $\endgroup$ Commented Jan 19, 2021 at 18:21
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    $\begingroup$ The need for these expansions is that they both provide (up to trivial prefactors) the same terms, once you identify $e^{-2\beta}$ and $\tanh(\beta^*)$. This is what gives you the required symmetry of the free energy, from which you then deduce that the critical point is given by the self-dual point. $\endgroup$ Commented Jan 19, 2021 at 18:24
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    $\begingroup$ More precisely, these expansions show that $\mathcal{Z}^+(\beta) = g(\beta)\mathcal{Z}^{\rm free}(\beta^*)$, where $g$ is an explicit (volume-dependent) smooth function of $\beta$ and the dual inverse temperature is defined by $\beta^* = \mathrm{atanh}(e^{-2\beta})$. This follows by a term by term comparison of the two sums that you get by using the low-temperature expansion of $\mathcal{Z}^+(\beta)$ and the high-temperature expansion of $\mathcal{Z}^{\rm free}(\beta^*)$. $\endgroup$ Commented Jan 19, 2021 at 19:58
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    $\begingroup$ This identity also requires that the partition function $\mathcal{Z}^{\rm free}(\beta^*)$ be computed on the graph that is dual to the graph used for $\mathcal{Z}^+(\beta)$. So, the use of the dual lattice, although it might appear artificial, plays a crucial role in this derivation. $\endgroup$ Commented Jan 19, 2021 at 20:00

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