Since we want to analyse the Ising model in order to determine if and how it exhibits phase transitions, and that phase transitions are characterized by singularities in the thermodynamic potentials, it is important to study the analytic properties of the Ising model's free energy so that we can find possible non-analytic behaviours and therefore understand when such phenomena can occur.

Let us therefore consider an Ising model where the external field $H$ is constant and the interaction $J_{ij}$ occurs only between nearest neighbouring spins, namely:

$J_{ij}={\begin{cases}J&S_{i}{\text{ and }}S_{j}{\text{ are nearest neighbours}}\\0&{\text{otherwise}}\end{cases}}$

This way the Hamiltonian of the system is:

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

where the notation

$\left\langle i,j\right\rangle$ means that

$i$ and

$j$ are nearest neighbours

^{[1]}.

We now state the most important analytic properties of the bulk free energy density $f$ (which hold in general).

**Theorem **

Let $f$ be the free energy density of a system at temperature $T$ subject to an external (generic) field $H$. Then:

- $f$ is negative, $f<0$
- $f$ is a continuous function of its arguments

- The derivatives of $f$ exist almost everywhere; furthermore, right and left derivatives of $f$ exist everywhere and are equal almost everywhere

- $-\partial f/\partial T\geq 0$, namely the entropy per site is non-negative

- $\partial f/\partial T$ is monotonically non-increasing with $T$, namely $\partial ^{2}f/\partial T^{2}\leq 0$; this implies, from its definition, that $C_{H}\geq 0$
- $\partial f/\partial H$ is monotonically non-increasing with $H$, namely $\partial ^{2}f/\partial H^{2}\leq 0$; this implies, from its definition, that $\chi _{T}\geq 0$

We shall only prove the fourth point; the general strategy is to start from a finite system and then take the thermodynamic limit.

*Proof *

If we assume the existence of the derivatives, then:

${\begin{aligned}-{\frac {\partial F}{\partial T}}={\frac {\partial }{\partial T}}\left[k_{B}T\ln \left(\operatorname {Tr} e^{-{\frac {\mathcal {H}}{k_{B}T}}}\right)\right]=\\{}\\=k_{B}\ln \left(\operatorname {Tr} e^{-{\frac {\mathcal {H}}{k_{B}T}}}\right)+k_{B}T{\frac {1}{\operatorname {Tr} e^{-{\frac {\mathcal {H}}{k_{B}T}}}}}\operatorname {Tr} \left(e^{-{\frac {\mathcal {H}}{k_{B}T}}}{\frac {\mathcal {H}}{k_{B}T^{2}}}\right)=\\{}\\=k_{B}\ln \left(\operatorname {Tr} e^{-\beta {\mathcal {H}}}\right)+k_{B}{\frac {\operatorname {Tr} \left(\beta {\mathcal {H}}e^{-\beta {\mathcal {H}}}\right)}{Z}}=k_{B}\left[\ln Z+{\frac {\operatorname {Tr} \left(\beta {\mathcal {H}}e^{-\beta {\mathcal {H}}}\right)}{Z}}\right]\end{aligned}}$

If we now define:

$\rho ={\frac {e^{-\beta {\mathcal {H}}}}{Z}}={\frac {e^{-\beta {\mathcal {H}}}}{\operatorname {Tr} e^{-\beta {\mathcal {H}}}}}$

we have:

$-{\frac {\partial F}{\partial T}}=-k_{B}\operatorname {Tr} (\rho \ln \rho )$

Now, since by definition

$0<\rho <1$,

$\ln \rho <0$ and thus

$-\partial F/\partial T$ is a sum of positive terms; therefore

$-\partial F/\partial T$ is positive. Dividing by

$N$ and taking the thermodynamic limit we get

$-\partial f/\partial T\geq 0$.

We can even prove something more (we will do it in the particular case of the Ising model):

**Theorem **

The free energy density $f$ is a concave function of $H$.

*Proof *

The proof is based upon the *Hölder inequality*, which we now recall.
If $\lbrace g_{k}\rbrace$ and $\lbrace h_{k}\rbrace$ are two sequences with $g_{k},h_{k}\geq 0$ for all $k$ and $\alpha _{1},\alpha _{2}\in \mathbb {R}$ are such that $\alpha _{1}+\alpha _{2}=1$, then:

$\sum _{k}g_{k}^{\alpha _{1}}h_{k}^{\alpha _{2}}\leq \left(\sum _{k}g_{k}\right)^{\alpha _{1}}\left(\sum _{k}h_{k}\right)^{\alpha _{2}}$

Let us now consider

$Z(H)$ (for simplicity we will write only

$H$ as its argument). By definition:

$Z(H)=\operatorname {Tr} e^{\beta H\sum _{i}S_{i}+\beta J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}}=\operatorname {Tr} \left(e^{\beta H\sum _{i}S_{i}}G[S]\right)$

where we have defined

$G[S]:=e^{\beta J\sum _{\left\langle i,j\right\rangle }S_{i}S_{j}}$. Then we have:

${\begin{aligned}Z(\alpha _{1}H_{1}+\alpha _{2}H_{2})=\operatorname {Tr} \left(e^{\beta \alpha _{1}H_{1}\sum _{i}S_{i}+\beta \alpha _{2}H_{2}\sum _{i}S_{i}}G[S]\right)=\\{}\\=\operatorname {Tr} \left[\left(e^{\beta H_{1}\sum _{i}S_{i}}G[S]\right)^{\alpha _{1}}\left(e^{\beta H_{2}\sum _{i}S_{i}}G[S]\right)^{\alpha _{2}}\right]\leq \\\leq \left(\operatorname {Tr} e^{\beta H_{1}\sum _{i}S_{i}}\right)^{\alpha _{1}}\left(\operatorname {Tr} e^{\beta H_{1}2\sum _{i}S_{i}}\right)^{\alpha _{2}}\end{aligned}}$

where we have also used the fact that

$\alpha _{1}+\alpha _{2}=1$, and thus:

$Z(\alpha _{1}H_{1}+\alpha _{2}H_{2})\leq Z(H_{1})^{\alpha _{1}}Z(H_{2})^{\alpha _{2}}$

Therefore, taking the logarithm at both sides, multiplying for

$-k_{B}T$ and taking the thermodynamic limit we get:

$f(\alpha _{1}H_{1}+\alpha _{2}H_{2})\geq \alpha _{1}f(H_{1})+\alpha _{2}f(H_{2})$

and so

$f$ is indeed a concave function of

$H$.

- ↑ With this notation we are also implicitly assuming that we are not counting twice terms that are equal; to be explicit, considering the unidmensional case if we sum over all the possible nearest neighbours we would have terms like : every term (except eventually for the boundaries, but only if we don't take periodic conditions) is counted twice (), so we should multiply everything by to correct this exceeding number of terms. With our notation we are implicitly doing that, and so the sum can be explicitly written as ; however, sometimes in the future it will be convenient to use the other convention, so we will count every possible term and then divide by 2.