# Absence of spontaneous magnetization

Let us now come back to free boundary conditions and compute the mean magnetization $\left\langle S_{j}\right\rangle$ of a given site $j$ . By definition we have:

$Z\left\langle S_{j}\right\rangle =\operatorname {Tr} \left(S_{j}e^{-\beta {\mathcal {H}}}\right)=\sum _{\lbrace S_{i}=\pm 1\rbrace }S_{j}e^{K\sum _{i=1}^{N-1}S_{i}S_{i+1}}=\sum _{\lbrace S_{i}=\pm 1\rbrace }\prod _{i=1}^{N-1}S_{j}e^{KS_{i}S_{i+1}}$ We now use the fact that $e^{x}=\cosh x+\sinh x$ , so that in this case:
$e^{KS_{i}S_{i+1}}=\cosh K+S_{i}S_{i+1}\sinh K$ where we have used (like we have done previously) the evenness of $\cosh$ , and now also the oddness of $\sinh$ , i.e. $\sinh(\pm x)=\pm \sinh x$ . This way:
{\begin{aligned}Z\left\langle S_{j}\right\rangle =\sum _{\lbrace S_{i}=\pm 1\rbrace }\prod _{i=1}^{N-1}S_{j}\left(\cosh K+S_{i}S_{i+1}\sinh K\right)=\\=(\cosh K)^{N-1}\sum _{\lbrace S_{i}=\pm 1\rbrace }\prod _{i=1}^{N-1}S_{j}\left(1+S_{i}S_{i+1}\tanh K\right)=\\{}\\=(\cosh K)^{N-1}\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}S_{j}(1+S_{1}S_{2}\tanh K)(1+S_{2}S_{3}\tanh K)\cdots \\\cdots (1+S_{N-1}S_{N}\tanh K)=\\{}\\=(\cosh K)^{N-1}\cdot \sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}\left[S_{j}+\sum _{M=1}^{N-1}S_{j}S_{i_{1}}S_{i_{1}+1}S_{i_{2}}S_{i_{2}+1}\cdots \right.\\\left.\cdots S_{i_{M}}S_{i_{M}+1}(\tanh K)^{M}\right]\end{aligned}} where $S_{i_{1}}$ etc. are the spin variables appropriately rearranged. Let us now consider the two different contributions to $\left\langle S_{j}\right\rangle$ . As of the first:
$\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}S_{j}=2^{N-1}\sum _{S_{j}=\pm 1}S_{j}=2^{N-1}(1-1)=0$ Considering now the second one:
$\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}\sum _{M=1}^{N-1}S_{j}S_{i_{1}}S_{i_{1}+1}S_{i_{2}}S_{i_{2}+1}\cdots S_{i_{M}}S_{i_{M}+1}(\tanh K)^{M}$ we have that for every fixed $M$ this term vanishes; for example, if we consider the contribution relative to a fixed value $M^{*}$ of $M$ and sum (for example) over $S_{i_{1}}$ we have:
{\begin{aligned}S_{j}\cdot (+1)\cdot S_{i_{1}+1}S_{i_{2}}S_{i_{2}+1}\cdots S_{i_{M^{*}}}S_{i_{M^{*}}+1}(\tanh K)^{M^{*}}+\qquad \qquad \qquad \\{}\\\qquad \qquad +S_{j}\cdot (-1)\cdot S_{i_{1}+1}S_{i_{2}}S_{i_{2}+1}\cdots S_{i_{M^{*}}}S_{i_{M^{*}}+1}(\tanh K)^{M^{*}}=0\end{aligned}} Therefore, also the second term vanishes and in the end:
$\left\langle S_{j}\right\rangle =0$ This result perfectly agrees with what we have already seen in Absence of phase transitions for finite systems, but now it has been deduced from a direct computation.

1. Note that the sum on nearest neighbour is done without counting the same terms twice (as we have already stressed). In fact, in our case every spin $S_{i}$ interacts with its nearest neighbours $S_{i-1}$ and $S_{i+1}$ , but the sum in the trace involves every two-spin interaction only once.
2. The form of this last term can be understood more easily doing an explicit computation with a simple example.