# 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:

$-{\mathcal {H}}={\frac {J_{0}}{2}}\sum _{i,j}S_{i}S_{j}+H\sum _{i}S_{i}$ (note that the sum in the interaction term is not restricted, and the $1/2$ 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 $J_{0}$ 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 $N^{2}$ and so in the thermodynamic limit it would diverge; we must therefore use the so called Kac prescription, setting $J_{0}=J/N$ so that the thermodynamic limit exists. Under these assumptions the partition function of the system is:
$Z_{N}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{{\frac {\beta J}{2N}}\sum _{i,j}S_{i}S_{j}+\beta H\sum _{i}S_{i}}$ Since the double sum is not restricted, we have:
$\sum _{i,j}S_{i}S_{j}=\left(\sum _{i}S_{i}\right)^{2}$ If we now call $x=\sum _{i}S_{i}$ and $a=\beta J$ , we can use the Hubbard-Stratonovich identity:
$e^{\frac {ax^{2}}{2N}}={\sqrt {\frac {aN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {aN}{2}}y^{2}+axy}dy$ where $\operatorname {Re} a>0$ . The advantage of this approach is that the variable $x$ , 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, $y$ (the auxiliary field from which this method takes its name). The partition function then becomes:
$Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {\beta JN}{2}}y^{2}}\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{\beta (H+Jy)\sum _{i}S_{i}}dy$ Physically this can be interpreted as the "mean value" of the partition functions of non interacting Ising models subjected to an external field $H-Jy$ whose component $y$ is distributed along a Gaussian.

We can therefore write:

$Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {\beta JN}{2}}y^{2}}Q_{y}dy\quad \qquad Q_{y}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{\beta (H+Jy)\sum _{i}S_{i}}$ and $Q_{y}$ 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:
$Q_{y}=\prod _{i}\sum _{S=\pm 1}e^{\beta (H+Jy)S}=2^{N}\cosh ^{N}\left[\beta (H+Jy)\right]$ Therefore:
$Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{N{\mathcal {L}}}dy\quad \qquad {\mathcal {L}}=\ln \left(2\cosh \left[\beta (H+Jy)\right]\right)-{\frac {\beta J}{2}}y^{2}$ Now, since the exponent in the integral that defines $Z_{N}$ is extensive (${\mathcal {L}}$ doesn't depend on $N$ ) and $N$ 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:
$Z_{N}\sim {\sqrt {\frac {\beta JN}{2\pi }}}e^{N{\mathcal {L}}(y_{0})}\sim e^{N{\mathcal {L}}(y_{0})}$ where $y_{0}$ is the maximum of ${\mathcal {L}}$ , thus given by the condition:
${\frac {\partial {\mathcal {L}}}{\partial y}}_{|y_{0}}=0$ which yields:
$\tanh(h+Ky_{0})=y_{0}$ Since it must be a maximum, we also must have:
${\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}}<0$ Now, we can see that the physical meaning of $y_{0}$ is the magnetization of the system in the thermodynamic limit. In fact:
{\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 $h=0$ ; therefore we will have:
$m=\tanh(Km)$ 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 $K=\beta J$ ):

• for $\beta J>1$ there are three solutions, one for $m=0$ and two at $\pm {\overline {m}}$ • for $\beta J=1$ these three solutions coincide
• for $\beta J<1$ there is only one solution: $m=0$ This means that two possible non null solutions appear when:

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

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

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

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

Therefore, if $T>T_{c}=J/k_{B}$ the only acceptable value for the magnetization of the system is $m={\overline {m}}=0$ , while if $T then $m=\pm {\overline {m}}\neq 0$ : 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:
$-{\frac {aN}{2}}y^{2}+axy=-{\frac {aN}{2}}\left(y-{\frac {x}{2}}\right)^{2}+{\frac {ax^{2}}{2N}}$ and computing the Gaussian integral.