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

Let us therefore consider a one-dimensional Ising model with ${\displaystyle N}$ 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:

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

Now, following all the prescriptions we know, we get to:

${\displaystyle F(T)=-k_{B}T\left[\ln 2+(N-1)\ln \left(2\cosh {\frac {J}{k_{B}T}}\right)\right]}$
and in the thermodynamic limit:
${\displaystyle f(T)=-k_{B}T\ln \left(2\cosh {\frac {J}{k_{B}T}}\right)}$
Let us note that ${\displaystyle f}$ 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 ${\displaystyle f(T)}$ is itself an analytic function of ${\displaystyle T}$: it is therefore impossible that the system will exhibit any kind of phase transition (at least for ${\displaystyle T\neq 0}$[2]) even in the thermodynamic limit.

1. This can be justified as follows: ${\displaystyle Z_{1}}$ 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 ${\displaystyle e^{-\beta {\mathcal {H}}}=e^{0}=1}$.
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 ${\displaystyle T=0}$ (which is of course an unphysical case), where all the spins are aligned.