# Bulk free energy, thermodynamic limit and absence of phase transitions

Let us therefore consider a one-dimensional Ising model with $N$ spins and free boundary conditions, i.e. the first and the last spin can assume any value. Using the nearest neighbour interaction Hamiltonian we have:

$-{\mathcal {H}}=H\sum _{i}S_{i}+J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}$ and defining $h=\beta H$ and $K=\beta J$ for the sake of simplicity, the partition function of the system will be:
$Z_{N}=\operatorname {Tr} e^{-\beta {\mathcal {H}}}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{h\sum _{i}S_{i}+K\sum _{i}S_{i}S_{i+1}}$ If we now set $h=0$ , namely if the system is not subjected to any external field, then:
$Z_{N}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{K\sum _{i}S_{i}S_{i+1}}=\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}e^{KS_{1}S_{2}+\cdots KS_{N-1}S_{N}}$ In order to compute $Z_{N}$ we use the so called recursion method: from the expression of $Z_{N}$ we can deduce the expression of the partition function $Z_{N+1}$ of the same system with one additional spin added to the lattice:
$Z_{N+1}=\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}\sum _{S_{N+1}=\pm 1}e^{K(S_{1}S_{2}+\cdots S_{N-1}S_{N})}e^{KS_{N}S_{N+1}}$ However, the sum over $S_{N+1}$ gives:
$\sum _{S_{N+1}=\pm 1}e^{KS_{N}S_{N+1}}=e^{KS_{N}}+e^{-KS_{N}}=2\cosh(KS_{N})=2\cosh K$ where we have used the evenness of the hyperbolic cosine, namely the fact that $\cosh(\pm x)=\cosh x$ . Therefore we have:
$Z_{N+1}=Z_{N}\cdot 2\cosh K$ and iterating the relation $N$ times we get:
$Z_{N+1}=Z_{1}(2\cosh K)^{N}\quad \Rightarrow \quad Z_{N}=Z_{1}(2\cosh K)^{N-1}$ Since:
$Z_{1}=\sum _{S_{1}=\pm 1}1=1+1=2$ we get the final result:
$Z_{N}=2(2\cosh K)^{N-1}$ Now, following all the prescriptions we know, we get to:

$F(T)=-k_{B}T\left[\ln 2+(N-1)\ln \left(2\cosh {\frac {J}{k_{B}T}}\right)\right]$ and in the thermodynamic limit:
$f(T)=-k_{B}T\ln \left(2\cosh {\frac {J}{k_{B}T}}\right)$ Let us note that $f$ does indeed respect the properties we have seen in Analytic properties of the Ising model. Furthermore, since the logarithm and the hyperbolic cosine are analytic functions we see that $f(T)$ is itself an analytic function of $T$ : it is therefore impossible that the system will exhibit any kind of phase transition (at least for $T\neq 0$ ) even in the thermodynamic limit.

1. This can be justified as follows: $Z_{1}$ is the partition function of a single-spin Ising model and we are considering two-spin interactions, so this single spin will not interact with anything. Therefore the Hamiltonian of the system in this case is null, and so $e^{-\beta {\mathcal {H}}}=e^{0}=1$ .
2. In this case, in fact, there can be problems. From this fact we can state that the only "phase transition" that can happen in a one-dimensional Ising model occurs at $T=0$ (which is of course an unphysical case), where all the spins are aligned.