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?
 A: 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,
$$
f(\beta) = \lim_{L\to\infty} -\frac1{\beta L} \log Z_L^{+\varnothing} = - \frac1\beta(\log\cosh(\beta) + \log 2).
$$
A: 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 .
