Bulk free energy, thermodynamic limit and absence of phase transitions

Let us therefore consider a one-dimensional Ising model with $N$$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:

$$-{\mathcal {H}}=H\sum _{i}S_{i}+J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}$$

$$-{\mathcal {H}}=H\sum _{i}S_{i}+J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}$$

and defining $h=\beta H$$h=\beta H$ and $K=\beta J$$K=\beta J$ for the sake of simplicity, the partition function of the system will be:

$$Z_{N}=\operatorname {Tr} e^{-\beta {\mathcal {H}}}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{h\sum _{i}S_{i}+K\sum _{i}S_{i}S_{i+1}}$$

$$Z_{N}=\operatorname {Tr} e^{-\beta {\mathcal {H}}}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{h\sum _{i}S_{i}+K\sum _{i}S_{i}S_{i+1}}$$

If we now set $h=0$$h=0$, namely if the system is not subjected to any external field, then:

$$Z_{N}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{K\sum _{i}S_{i}S_{i+1}}=\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}e^{KS_{1}S_{2}+\cdots KS_{N-1}S_{N}}$$

$$Z_{N}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{K\sum _{i}S_{i}S_{i+1}}=\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}e^{KS_{1}S_{2}+\cdots KS_{N-1}S_{N}}$$

In order to compute $Z_{N}$$Z_{N}$ we use the so called recursion method: from the expression of $Z_{N}$$Z_{N}$ we can deduce the expression of the partition function $Z_{N+1}$$Z_{N+1}$ of the same system with one additional spin added to the lattice:

$$Z_{N+1}=\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}\sum _{S_{N+1}=\pm 1}e^{K(S_{1}S_{2}+\cdots S_{N-1}S_{N})}e^{KS_{N}S_{N+1}}$$

$$Z_{N+1}=\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}\sum _{S_{N+1}=\pm 1}e^{K(S_{1}S_{2}+\cdots S_{N-1}S_{N})}e^{KS_{N}S_{N+1}}$$

However, the sum over $S_{N+1}$$S_{N+1}$ gives:

$$\sum _{S_{N+1}=\pm 1}e^{KS_{N}S_{N+1}}=e^{KS_{N}}+e^{-KS_{N}}=2\cosh(KS_{N})=2\cosh K$$

$$\sum _{S_{N+1}=\pm 1}e^{KS_{N}S_{N+1}}=e^{KS_{N}}+e^{-KS_{N}}=2\cosh(KS_{N})=2\cosh K$$

where we have used the evenness of the hyperbolic cosine, namely the fact that $\cosh(\pm x)=\cosh x$$\cosh(\pm x)=\cosh x$. Therefore we have:

$$Z_{N+1}=Z_{N}\cdot 2\cosh K$$

$$Z_{N+1}=Z_{N}\cdot 2\cosh K$$

and iterating the relation $N$$N$ times we get:

$$Z_{N+1}=Z_{1}(2\cosh K)^{N}\quad \Rightarrow \quad Z_{N}=Z_{1}(2\cosh K)^{N-1}$$

$$Z_{N+1}=Z_{1}(2\cosh K)^{N}\quad \Rightarrow \quad Z_{N}=Z_{1}(2\cosh K)^{N-1}$$

Since:

$$Z_{1}=\sum _{S_{1}=\pm 1}1=1+1=2$$

$$Z_{1}=\sum _{S_{1}=\pm 1}1=1+1=2$$

we get the final result:

$$Z_{N}=2(2\cosh K)^{N-1}$$

$$Z_{N}=2(2\cosh K)^{N-1}$$

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

$$F(T)=-k_{B}T\left[\ln 2+(N-1)\ln \left(2\cosh {\frac {J}{k_{B}T}}\right)\right]$$

$$F(T)=-k_{B}T\left[\ln 2+(N-1)\ln \left(2\cosh {\frac {J}{k_{B}T}}\right)\right]$$

and in the thermodynamic limit:

$$f(T)=-k_{B}T\ln \left(2\cosh {\frac {J}{k_{B}T}}\right)$$

$$f(T)=-k_{B}T\ln \left(2\cosh {\frac {J}{k_{B}T}}\right)$$

Let us note that $f$$f$ does indeed respect the properties we have seen in Analytic properties of the Ising model. Furthermore, since the logarithm and the hyperbolic cosine are analytic functions we see that $f(T)$$f(T)$ is itself an analytic function of $T$$T$: it is therefore impossible that the system will exhibit any kind of phase transition (at least for $T\neq 0$$T\neq 0$) even in the thermodynamic limit.