# 1 dimensional Ising model

How to solve the Ising model in 1D by low temperature, and high temperature expansion, and by change of variable method? Can you please give me some reference links?

• Chapter 7 of Kardar's "Statistical Physics of Fields" explains the low and high temperature expansion of the Ising model. – leongz Jun 1 '13 at 18:46
• Although I've had to do these expansions myself, I've never seen them in the literature. But once you've seen how it works in 2D, it's really simple to generalise to 1D (the counting is much simpler). – Vibert Jun 1 '13 at 21:36

Unfortunately, I don't know of any good references, although I fully expect that this is done in various textbooks. I also don't have time for a detailed answer, so in each case I'll choose boundary conditions simplifying the computations (of course, this has no impact on the limiting free energy density).

## High temperature expansion

I'll consider the 1d Ising model with $+$ boundary condition in the box $\{1,\ldots,L\}$, that is, I consider the Hamiltonian $$H(\sigma) = - \beta \sum_{i=0}^L \sigma_i\sigma_{i+1},$$ with boundary condition $\sigma_0=\sigma_{L+1}=1$. The high temperature expansion amounts to observing that $$e^{\beta\sigma_i\sigma_{i+1}} = \cosh(\beta) \bigl(1+\sigma_i\sigma_{i+1}\tanh(\beta)\bigr).$$ This implies that $$Z_L^{++} = \sum_{\substack{\sigma_i=\pm 1\\i=1,\ldots,L}} \prod_{i=0}^L e^{\beta\sigma_i\sigma_{i+1}} = (\cosh(\beta))^{L+1}\, \sum_{\substack{\sigma_i=\pm 1\\i=1,\ldots,L}} \prod_{i=0}^L \bigl( 1+\sigma_i\sigma_{i+1}\tanh(\beta) \bigr).$$ Expanding the product yields $$\prod_{i=0}^L \bigl( 1+\sigma_i\sigma_{i+1}\tanh(\beta) \bigr) = \sum_E \tanh(\beta)^{|E|}\prod_{(i,i+1)\in E} \sigma_i\sigma_{i+1},$$ where the sum is over collections $E$ of edges $(i,i+1)$ between vertices of $\{0,\ldots,L+1\}$, and $|E|$ denotes the cardinality of $E$.

Interchanging the sums over spins and edges, we obtain $$Z_L^{++} = (\cosh(\beta))^{L+1}\,\sum_E \tanh(\beta)^{|E|} \sum_{\substack{\sigma_i=\pm 1\\i=1,\ldots,L}}\prod_{(i,i+1)\in E} \sigma_i\sigma_{i+1}.$$ The last sum is easily evaluated. Let us denote by $I_E(i)$ the number of edges of $E$ incident at $i$ (so, $I_E(i)\in\{0,1,2\}$). We have $$\sum_{\substack{\sigma_i=\pm 1\\i=1,\ldots,L}}\prod_{(i,i+1)\in E} \sigma_i\sigma_{i+1} = \prod_{i=1}^L \bigl( \sum_{\sigma_i=\pm 1} \sigma_i^{I_E(i)} \bigr).$$ Since $\sum_{\sigma_i=\pm 1} \sigma_i^{I_E(i)}=0$ if $I_E(i)$ is odd, we see that there are only two sets $E$ yielding a nonzero contribution : the empty set and the full set. In both cases, the sum over spin configurations yields a factor $2^L$. Therefore, $$Z_L^{++} = (\cosh(\beta))^{L+1} 2^L \bigl( 1+(\tanh(\beta)^{L+1} \bigr).$$ The free energy is thus given by $$f(\beta) = \lim_{L\to\infty} -\frac1{\beta L} \log Z_L^{++} = - \frac1\beta(\log\cosh(\beta) + \log 2).$$

## Low temperature expansion

This time, I'll consider free boundary condition. The low temperature expansion follows from recording all "contours" separating $+$ and $-$ spins. In dimension 1, this amounts to recording the position of edges $(i,i+1)$ such that $\sigma_i\neq\sigma_{i+1}$.

To make this precise, let us write $$Z_L^{\varnothing\varnothing} = \sum_{\substack{\sigma_i=\pm 1\\i=1,\ldots,L}} \prod_{i=1}^{L-1} e^{\beta\sigma_i\sigma_{i+1}} = e^{\beta(L-1)}\, \sum_{\substack{\sigma_i=\pm 1\\i=1,\ldots,L}} \prod_{i=1}^{L-1} e^{\beta(\sigma_i\sigma_{i+1}-1)}.$$ Now factors in the last product are either equal to $1$ (if the spins agree) or to $e^{-2\beta}$ (if they don't). Therefore, $$Z_L^{\varnothing\varnothing} = 2\, e^{\beta(L-1)}\, \sum_{n=0}^{L-1} \binom{L-1}{n} (e^{-2\beta})^n = 2\, e^{\beta(L-1)}\, (1+e^{-2\beta})^{L-1}.$$ Again, we obtain $$f(\beta) = \lim_{L\to\infty} -\frac1{\beta L} \log Z_L^{\varnothing\varnothing} = - \frac1\beta(\log\cosh(\beta) + \log 2).$$

## Change of variables

The final method you ask for is the "change of variables". I'll consider only the following boundary condition : $\sigma_0=+1$ at the left end, but free boundary condition at the right end.

The trick is to change variables to $\eta_i=\sigma_{i-1}\sigma_i$, $i=1,\ldots,L$. This yields $$Z_L^{+\varnothing} = \sum_{\substack{\sigma_i=\pm 1\\i=1,\ldots,L}} \prod_{i=1}^{L-1} e^{\beta\sigma_i\sigma_{i+1}} = \sum_{\substack{\eta_i=\pm 1\\i=1,\ldots,L}} \prod_{i=1}^{L-1} e^{\beta\eta_i}.$$ The $\eta$ spins are non interacting, so that $$Z_L^{+\varnothing} = \prod_{i=1}^{L-1} \bigl(\sum_{\eta_i=\pm 1} e^{\beta\eta_i}\bigr) = \bigl(e^{\beta}+e^{-\beta}\bigr)^{L-1}.$$ And therefore, a final time, $$f(\beta) = \lim_{L\to\infty} -\frac1{\beta L} \log Z_L^{+\varnothing} = - \frac1\beta(\log\cosh(\beta) + \log 2).$$

• Hopefully there aren't too many computational (or others) mistakes, as I had very little time to write this down... Don't hesitate to make corrections. – Yvan Velenik Jun 1 '13 at 19:01
• :could you please explain why you chose different boundary conditions in different problems?(although i know that system is thermodynamically the same whether you twist and join Lth lattice point with the first or not,as L tends to infinity) – sreeram Jun 3 '13 at 3:59
• @sreeram: in each case, I chose boundary conditions leading to simple computations. FOr example, in the case of low temperature expansion, imposing the $++$ boundary condition would add a constraint that the number of "domain walls" has to be even; of course, this can be done, but it makes the argument slightly less immediate. [TO BE CONTINUED] – Yvan Velenik Jun 3 '13 at 6:48
• For the change of variables, the situation is similar: fixing the leftmost spin to be $+1$ has the advantage that the transformation from the $\sigma$ to the $\eta$ variables is invertible, otherwise we would have to keep the value of the original leftmost spin $\sigma_1$. I chose not to impose anything at the right end of the interval, since otherwise this would induce a constraint on the product of the $\eta$ spins. – Yvan Velenik Jun 3 '13 at 6:48
• @sreeram The calculations are correct. But I think the free energy would not be $f(\beta) = \lim_{L\to\infty} -\frac1{\beta L} \log Z_L^{+\varnothing} = - \frac1\beta(\log\cosh(\beta) + \log 2)$. The expression $\log 2$ is suspected.I think the Velenik would like you observe more carefully the calculations. – MathOverview May 11 '15 at 12:39

If I'm right the expansions you are referring to are relevant only in two dimensions; then allowing you to derive the critical temperature by Kramer's duality. In 1D, you solve it by working out the transfer matrix without "expanding" (only neglecting its lowest eigenvalue). It's well described in http://www.damtp.cam.ac.uk/user/tong/statphys/five.pdf .

• No, these expansions do apply in any dimensions. In dimension 1, they allow a simple computation of the free energy of the model, provided there is no magnetic field (otherwise the computations become more complicated than using the transfer matrix). This is also true of the change of variables method (which I assume refers to the change of variables $\eta_i = \sigma_i\sigma_{i+1}$ mapping the model to independent spins), which is only useful in dimension $1$. – Yvan Velenik Jun 1 '13 at 11:16
• @YvanVelenik please give me references. – sreeram Jun 1 '13 at 15:21
• @sreeram: I don't know any, but I'll sketch an answer later if nobody does. – Yvan Velenik Jun 1 '13 at 15:37
• @Learning is a mess: in $D \neq 2$ dimensions the same expansions exist, but the geometrical picture is different. In 1D the dual lattice consists of sites, in 2D of links, in 3D of plaquettes, et cetera. – Vibert Jun 1 '13 at 21:38
• Sorry, I think I wasn't clear enough. In 3D for example, the loops are replaced by plaquettes (essentially squares surrounded by four links) and so on. So you cannot play the same game as in 2D, where the dual model is the same as the normal model, apart from a prefactor which gives you the critical temperature. I understand that in 3D there is something like a "spin-gauge duality" that maps spin Hamiltonians to gauge theories defined on these plaquettes, but it's not my area of expertise. – Vibert Jun 1 '13 at 22:08