# Absence of phase transitions for finite systems

We now analyse some symmetry properties of the Ising model, which will allow us to show that for finite systems no phase transition can occur at all.

To begin with, let us note that for any function ${\displaystyle \varphi }$ of the spin configurations ${\displaystyle \lbrace S_{i}\rbrace }$ we have:

${\displaystyle \sum _{\lbrace S_{i}=\pm 1\rbrace }\varphi (\lbrace S_{i}\rbrace )=\sum _{\lbrace S_{i}=\pm 1\rbrace }\varphi (\lbrace -S_{i}\rbrace )\quad \Rightarrow \quad \operatorname {Tr} \varphi (\lbrace S_{i}\rbrace )=\operatorname {Tr} \varphi (\lbrace -S_{i}\rbrace )}$
which can be "proved" by explicitly writing all the terms[1]. Then, we can also see that ${\displaystyle f}$ is an even function of ${\displaystyle H}$. In fact, from the definition of the Hamiltonian:
${\displaystyle -{\mathcal {H}}=J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}+H\sum _{i}S_{i}}$
it is immediate to see that:
${\displaystyle {\mathcal {H}}(H,J,\lbrace S_{i}\rbrace )={\mathcal {H}}(-H,J,\lbrace -S_{i}\rbrace )}$
Therefore:
{\displaystyle {\begin{aligned}Z(-H,J,T)=\operatorname {Tr} e^{-\beta {\mathcal {H}}(-H,J,\lbrace S_{i}\rbrace )}=\operatorname {Tr} e^{-\beta {\mathcal {H}}(-H,J,\lbrace -S_{i}\rbrace )}=\\=\operatorname {Tr} e^{-\beta {\mathcal {H}}(H,J,\lbrace S_{i}\rbrace )}=Z(H,J,T)\end{aligned}}}
If we now take the logarithm on both sides and multiply by ${\displaystyle -k_{B}T}$, we have:
${\displaystyle F(H,J,T)=F(-H,J,T)}$
namely, the free energy is an even function of ${\displaystyle H}$. This implies that the system (note that we have never taken the thermodynamic limit, so its size is still finite) can never exhibit a spontaneous magnetization when ${\displaystyle H=0}$, and so there are no phase transitions at all (the system will always remain in its paramagnetic phase). In fact, we have:
${\displaystyle M(H)=-{\frac {\partial }{\partial H}}F(H)=-{\frac {\partial }{\partial H}}F(-H)={\frac {\partial }{\partial (-H)}}F(-H)=-M(-H)}$
and if ${\displaystyle H=0}$:
${\displaystyle M(0)=-M(0)\quad \Rightarrow \quad M(0)=0}$
i.e. the spontaneous magnetization is always null.

Note that this result has been obtained only with the use of symmetry properties, and we have never resorted to the "traditional" approach of statistical mechanics, namely the explicit computation of the partition function and subsequently the derivation of the thermodynamics of the system. This will be done later on.

1. Very simply, since ${\displaystyle S_{i}=\pm 1}$, when we sum over all possible values of ${\displaystyle S_{i}}$ we will in both cases cover all the possibilities for the argument ${\displaystyle \lbrace S_{i}\rbrace }$ of ${\displaystyle \varphi }$.