# The role of interaction range

In order to see how the range of the interactions between the degrees of freedom of the system affects its properties, let us consider a one-dimensional Ising model with infinite-ranged interactions:

${\displaystyle -{\mathcal {H}}={\frac {J_{0}}{2}}\sum _{i,j}S_{i}S_{j}+H\sum _{i}S_{i}}$
(note that the sum in the interaction term is not restricted, and the ${\displaystyle 1/2}$ factor has been introduced for later convenience). This model can be solved with the technique of Hubbard transformation, also called auxiliary field method. First, we must note that ${\displaystyle J_{0}}$ can't be a constant independent of the dimension of the system because the sum ${\textstyle \sum _{i,j}S_{i}S_{j}}$ contains a number of terms of the order of ${\displaystyle N^{2}}$ and so in the thermodynamic limit it would diverge; we must therefore use the so called Kac prescription, setting ${\displaystyle J_{0}=J/N}$ so that the thermodynamic limit exists. Under these assumptions the partition function of the system is:
${\displaystyle Z_{N}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{{\frac {\beta J}{2N}}\sum _{i,j}S_{i}S_{j}+\beta H\sum _{i}S_{i}}}$
Since the double sum is not restricted, we have:
${\displaystyle \sum _{i,j}S_{i}S_{j}=\left(\sum _{i}S_{i}\right)^{2}}$
If we now call ${\displaystyle x=\sum _{i}S_{i}}$ and ${\displaystyle a=\beta J}$, we can use the Hubbard-Stratonovich identity[1]:
${\displaystyle e^{\frac {ax^{2}}{2N}}={\sqrt {\frac {aN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {aN}{2}}y^{2}+axy}dy}$
where ${\displaystyle \operatorname {Re} a>0}$. The advantage of this approach is that the variable ${\displaystyle x}$, which contains all the degrees of freedom of the system, is linear and not quadratic in the exponential; however we have "paid" the price of having introduced another field, ${\displaystyle y}$ (the auxiliary field from which this method takes its name). The partition function then becomes:
${\displaystyle Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {\beta JN}{2}}y^{2}}\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{\beta (H+Jy)\sum _{i}S_{i}}dy}$
Physically this can be interpreted as the "mean value" of the partition functions of non interacting Ising models subjected to an external field ${\displaystyle H-Jy}$ whose component ${\displaystyle y}$ is distributed along a Gaussian.

We can therefore write:

${\displaystyle Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {\beta JN}{2}}y^{2}}Q_{y}dy\quad \qquad Q_{y}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{\beta (H+Jy)\sum _{i}S_{i}}}$
and ${\displaystyle Q_{y}}$ can be easily computed factorizing the sum, similarly to what we have done for the Ising model in Bulk free energy, thermodynamic limit and absence of phase transitions:
${\displaystyle Q_{y}=\prod _{i}\sum _{S=\pm 1}e^{\beta (H+Jy)S}=2^{N}\cosh ^{N}\left[\beta (H+Jy)\right]}$
Therefore:
${\displaystyle Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{N{\mathcal {L}}}dy\quad \qquad {\mathcal {L}}=\ln \left(2\cosh \left[\beta (H+Jy)\right]\right)-{\frac {\beta J}{2}}y^{2}}$
Now, since the exponent in the integral that defines ${\displaystyle Z_{N}}$ is extensive (${\displaystyle {\mathcal {L}}}$ doesn't depend on ${\displaystyle N}$) and ${\displaystyle N}$ is large, we can compute it using the saddle point approximation (see appendix The saddle point approximation). This consists in approximating the integral with the largest value of the integrand, namely:
${\displaystyle Z_{N}\sim {\sqrt {\frac {\beta JN}{2\pi }}}e^{N{\mathcal {L}}(y_{0})}\sim e^{N{\mathcal {L}}(y_{0})}}$
where ${\displaystyle y_{0}}$ is the maximum of ${\displaystyle {\mathcal {L}}}$, thus given by the condition:
${\displaystyle {\frac {\partial {\mathcal {L}}}{\partial y}}_{|y_{0}}=0}$
which yields:
${\displaystyle \tanh(h+Ky_{0})=y_{0}}$
Since it must be a maximum, we also must have:
${\displaystyle {\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}}<0}$
Now, we can see that the physical meaning of ${\displaystyle y_{0}}$ is the magnetization of the system in the thermodynamic limit. In fact:
{\displaystyle {\begin{aligned}m=\lim _{N\to \infty }-{\frac {1}{N}}{\frac {\partial F}{\partial H}}=\lim _{N\to \infty }{\frac {1}{N}}{\frac {\partial \ln Z}{\partial h}}=\\=\lim _{N\to \infty }{\frac {1}{N}}{\frac {\partial }{\partial h}}\left[N\ln \left(2\cosh(h+Ky_{0})\right)-{\frac {K}{2}}y_{0}^{2}\right]=\\=\lim _{N\to \infty }{\frac {1}{N}}N\tanh(h+Ky_{0})=\tanh(h+Ky_{0})=y_{0}\end{aligned}}}
Since we are interested in determining if the system can exhibit a spontaneous magnetization, we consider the case ${\displaystyle h=0}$; therefore we will have:
${\displaystyle m=\tanh(Km)}$
which is a transcendent equation, so it can't be solved analytically. However, we can solve it graphically:

From these figures we can see that there are three possible cases (remembering that by definition ${\displaystyle K=\beta J}$):

• for ${\displaystyle \beta J>1}$ there are three solutions, one for ${\displaystyle m=0}$ and two at ${\displaystyle \pm {\overline {m}}}$
• for ${\displaystyle \beta J=1}$ these three solutions coincide
• for ${\displaystyle \beta J<1}$ there is only one solution: ${\displaystyle m=0}$

This means that two possible non null solutions appear when:

${\displaystyle T
Let us see which of these solutions are acceptable, i.e. which of these solutions are maxima of ${\displaystyle {\mathcal {L}}}$. Still in the case ${\displaystyle h=0}$, we have:
${\displaystyle {\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}=K\left[K\operatorname {sech} ^{2}(Ky)-1\right]}$
and so:

• ${\displaystyle {\boldsymbol {K\leq 1}}}$:in this case ${\displaystyle y_{0}=0}$, so:

${\displaystyle {\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}=0}=K(K-1)<0}$
so ${\displaystyle y_{0}}$ is indeed a maximum for ${\displaystyle {\mathcal {L}}}$

• ${\displaystyle {\boldsymbol {K>1}}}$:if ${\displaystyle y_{0}=0}$ then:

${\displaystyle {\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}=0}=K(K-1)>0}$
so this is not a maximum for ${\displaystyle {\mathcal {L}}}$, and thus is not an acceptable solution of ${\displaystyle m=\tanh(Km)}$. On the other hand, if ${\displaystyle y_{0}=\pm {\overline {m}}}$ then (also using the evenness of ${\displaystyle \operatorname {sech} }$):
${\displaystyle {\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}=\pm {\overline {m}}}=K\left[K\operatorname {sech} ^{2}(K{\overline {m}})-1\right]}$
and this time this derivative is negative, because:
${\displaystyle K\operatorname {sech} ^{2}(K{\overline {m}})-1<0\quad \Longleftrightarrow \quad \operatorname {sech} (K{\overline {m}})<{\frac {1}{\sqrt {K}}}\quad \Longleftrightarrow \quad \cosh(K{\overline {m}})>{\sqrt {K}}>1}$
which is always true.

Therefore, if ${\displaystyle T>T_{c}=J/k_{B}}$ the only acceptable value for the magnetization of the system is ${\displaystyle m={\overline {m}}=0}$, while if ${\displaystyle T then ${\displaystyle m=\pm {\overline {m}}\neq 0}$: a phase transition has occurred, since now the system can exhibit a net spontaneous magnetization. We thus see explicitly that if we let the interactions to be long-ranged the Ising model can undergo phase transitions already in the one-dimensional case.

1. It can be easily verified completing the square in the exponential:
${\displaystyle -{\frac {aN}{2}}y^{2}+axy=-{\frac {aN}{2}}\left(y-{\frac {x}{2}}\right)^{2}+{\frac {ax^{2}}{2N}}}$
and computing the Gaussian integral.