Physics Asked on March 20, 2021
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:
(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!
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 $$ tanh(beta^*) = e^{-2beta}.tag{$star$} $$ 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 logcosh(beta^*) + O(N^{-1}) + frac{1}{(2N)^2} log mathcal{Z}_{Lambda_N^*;beta^*}^{rm free} = -2beta + O(N^{-1}) + frac{1}{(2N)^2} log mathcal{Z}_{Lambda_N;beta}^+ . end{multline} Now, we use the fact that $$ lim_{Ntoinfty} frac{1}{(2N)^2} log mathcal{Z}_{Lambda_N^*;beta^*}^{rm free} = phi(beta^*) $$ and $$ lim_{Ntoinfty} frac{1}{(2N)^2} log mathcal{Z}_{Lambda_N;beta}^+ = phi(beta), $$ since the free energy density is independent of the boundary condition in the thermodynamic limit.
We thus obtain begin{align} phi(beta) &= 2beta - log(2) - 2logcosh(beta^*) + phi(beta^*) &= phi(beta^*) - log sinh(2beta^*), 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 $$ tanh(beta) = e^{-2beta}. $$ Indeed, if $beta_{rm c} neq beta_{rm sd}$, then the above identity implies that $phi$ must have another singularity at $$ beta_{rm c}^* = mathrm{atanh}(e^{-2beta_{rm c}}), $$ which would contradict the assumption that $phi$ has only one singularity.
Finally, it is easy to check that $beta_{rm sd} = frac12log(1+sqrt{2})$, 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.
Correct answer by Yvan Velenik on March 20, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP