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:

$$-{\mathcal {H}}={\frac {J_{0}}{2}}\sum _{i,j}S_{i}S_{j}+H\sum _{i}S_{i}$$

$$-{\mathcal {H}}={\frac {J_{0}}{2}}\sum _{i,j}S_{i}S_{j}+H\sum _{i}S_{i}$$

(note that the sum in the interaction term is not restricted, and the $1/2$$1/2$ factor has been introduced for later convenience). This model can be solved with the technique of Hubbard transformation, also called auxiliary field method. First, we must note that $J_{0}$$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}}$${\textstyle \sum _{i,j}S_{i}S_{j}}$ contains a number of terms of the order of $N^{2}$$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$$J_{0}=J/N$ so that the thermodynamic limit exists. Under these assumptions the partition function of the system is:

$$Z_{N}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{{\frac {\beta J}{2N}}\sum _{i,j}S_{i}S_{j}+\beta H\sum _{i}S_{i}}$$

$$Z_{N}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{{\frac {\beta J}{2N}}\sum _{i,j}S_{i}S_{j}+\beta H\sum _{i}S_{i}}$$

Since the double sum is not restricted, we have:

$$\sum _{i,j}S_{i}S_{j}=\left(\sum _{i}S_{i}\right)^{2}$$

$$\sum _{i,j}S_{i}S_{j}=\left(\sum _{i}S_{i}\right)^{2}$$

If we now call $x=\sum _{i}S_{i}$$x=\sum _{i}S_{i}$ and $a=\beta J$$a=\beta J$, we can use the Hubbard-Stratonovich identity:

$$e^{\frac {ax^{2}}{2N}}={\sqrt {\frac {aN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {aN}{2}}y^{2}+axy}dy$$

$$e^{\frac {ax^{2}}{2N}}={\sqrt {\frac {aN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {aN}{2}}y^{2}+axy}dy$$

where $\operatorname {Re} a>0$$\operatorname {Re} a>0$. The advantage of this approach is that the variable $x$$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$$y$ (the auxiliary field from which this method takes its name). The partition function then becomes:

$$Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {\beta JN}{2}}y^{2}}\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{\beta (H+Jy)\sum _{i}S_{i}}dy$$

$$Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {\beta JN}{2}}y^{2}}\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{\beta (H+Jy)\sum _{i}S_{i}}dy$$

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$$H-Jy$ whose component $y$$y$ is distributed along a Gaussian.

We can therefore write:

$$Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {\beta JN}{2}}y^{2}}Q_{y}dy\quad \qquad Q_{y}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{\beta (H+Jy)\sum _{i}S_{i}}$$

$$Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{-{\frac {\beta JN}{2}}y^{2}}Q_{y}dy\quad \qquad Q_{y}=\sum _{\lbrace S_{i}=\pm 1\rbrace }e^{\beta (H+Jy)\sum _{i}S_{i}}$$

and $Q_{y}$$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:

$$Q_{y}=\prod _{i}\sum _{S=\pm 1}e^{\beta (H+Jy)S}=2^{N}\cosh ^{N}\left[\beta (H+Jy)\right]$$

$$Q_{y}=\prod _{i}\sum _{S=\pm 1}e^{\beta (H+Jy)S}=2^{N}\cosh ^{N}\left[\beta (H+Jy)\right]$$

Therefore:

$$Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{N{\mathcal {L}}}dy\quad \qquad {\mathcal {L}}=\ln \left(2\cosh \left[\beta (H+Jy)\right]\right)-{\frac {\beta J}{2}}y^{2}$$

$$Z_{N}={\sqrt {\frac {\beta JN}{2\pi }}}\int _{-\infty }^{+\infty }e^{N{\mathcal {L}}}dy\quad \qquad {\mathcal {L}}=\ln \left(2\cosh \left[\beta (H+Jy)\right]\right)-{\frac {\beta J}{2}}y^{2}$$

Now, since the exponent in the integral that defines $Z_{N}$$Z_{N}$ is extensive (${\mathcal {L}}$${\mathcal {L}}$ doesn't depend on $N$$N$) and $N$$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:

$$Z_{N}\sim {\sqrt {\frac {\beta JN}{2\pi }}}e^{N{\mathcal {L}}(y_{0})}\sim e^{N{\mathcal {L}}(y_{0})}$$

$$Z_{N}\sim {\sqrt {\frac {\beta JN}{2\pi }}}e^{N{\mathcal {L}}(y_{0})}\sim e^{N{\mathcal {L}}(y_{0})}$$

where $y_{0}$$y_{0}$ is the maximum of ${\mathcal {L}}$${\mathcal {L}}$, thus given by the condition:

$${\frac {\partial {\mathcal {L}}}{\partial y}}_{|y_{0}}=0$$

$${\frac {\partial {\mathcal {L}}}{\partial y}}_{|y_{0}}=0$$

which yields:

$$\tanh(h+Ky_{0})=y_{0}$$

$$\tanh(h+Ky_{0})=y_{0}$$

Since it must be a maximum, we also must have:

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}}<0$$

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}}<0$$

Now, we can see that the physical meaning of $y_{0}$$y_{0}$ is the magnetization of the system in the thermodynamic limit. In fact:

$${\begin{aligned}m=\lim _{N\to \infty }-{\frac {1}{N}}{\frac {\partial F}{\partial H}}=\lim _{N\to \infty }{\frac {1}{N}}{\frac {\partial \ln Z}{\partial h}}=\\=\lim _{N\to \infty }{\frac {1}{N}}{\frac {\partial }{\partial h}}\left[N\ln \left(2\cosh(h+Ky_{0})\right)-{\frac {K}{2}}y_{0}^{2}\right]=\\=\lim _{N\to \infty }{\frac {1}{N}}N\tanh(h+Ky_{0})=\tanh(h+Ky_{0})=y_{0}\end{aligned}}$$

$${\begin{aligned}m=\lim _{N\to \infty }-{\frac {1}{N}}{\frac {\partial F}{\partial H}}=\lim _{N\to \infty }{\frac {1}{N}}{\frac {\partial \ln Z}{\partial h}}=\\=\lim _{N\to \infty }{\frac {1}{N}}{\frac {\partial }{\partial h}}\left[N\ln \left(2\cosh(h+Ky_{0})\right)-{\frac {K}{2}}y_{0}^{2}\right]=\\=\lim _{N\to \infty }{\frac {1}{N}}N\tanh(h+Ky_{0})=\tanh(h+Ky_{0})=y_{0}\end{aligned}}$$

Since we are interested in determining if the system can exhibit a spontaneous magnetization, we consider the case $h=0$$h=0$; therefore we will have:

$$m=\tanh(Km)$$

$$m=\tanh(Km)$$

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$$K=\beta J$):

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

• for $\beta J<1$$\beta J<1$ there is only one solution: $m=0$$m=0$

This means that two possible non null solutions appear when:

$$T<T_{c}:={\frac {J}{k_{B}}}$$

$$T<T_{c}:={\frac {J}{k_{B}}}$$

Let us see which of these solutions are acceptable, i.e. which of these solutions are maxima of ${\mathcal {L}}$${\mathcal {L}}$. Still in the case $h=0$$h=0$, we have:

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}=K\left[K\operatorname {sech} ^{2}(Ky)-1\right]$$

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}=K\left[K\operatorname {sech} ^{2}(Ky)-1\right]$$

and so:

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

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}=0}=K(K-1)<0$$

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}=0}=K(K-1)<0$$

so $y_{0}$$y_{0}$ is indeed a maximum for ${\mathcal {L}}$${\mathcal {L}}$

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

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}=0}=K(K-1)>0$$

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}=0}=K(K-1)>0$$

so this is not a maximum for ${\mathcal {L}}$${\mathcal {L}}$, and thus is not an acceptable solution of $m=\tanh(Km)$$m=\tanh(Km)$. On the other hand, if $y_{0}=\pm {\overline {m}}$$y_{0}=\pm {\overline {m}}$ then (also using the evenness of $\operatorname {sech}$$\operatorname {sech}$):

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}=\pm {\overline {m}}}=K\left[K\operatorname {sech} ^{2}(K{\overline {m}})-1\right]$$

$${\frac {\partial ^{2}{\mathcal {L}}}{\partial y^{2}}}_{|y_{0}=\pm {\overline {m}}}=K\left[K\operatorname {sech} ^{2}(K{\overline {m}})-1\right]$$

and this time this derivative is negative, because:

$$K\operatorname {sech} ^{2}(K{\overline {m}})-1<0\quad \Longleftrightarrow \quad \operatorname {sech} (K{\overline {m}})<{\frac {1}{\sqrt {K}}}\quad \Longleftrightarrow \quad \cosh(K{\overline {m}})>{\sqrt {K}}>1$$

$$K\operatorname {sech} ^{2}(K{\overline {m}})-1<0\quad \Longleftrightarrow \quad \operatorname {sech} (K{\overline {m}})<{\frac {1}{\sqrt {K}}}\quad \Longleftrightarrow \quad \cosh(K{\overline {m}})>{\sqrt {K}}>1$$

which is always true.

Therefore, if $T>T_{c}=J/k_{B}$$T>T_{c}=J/k_{B}$ the only acceptable value for the magnetization of the system is $m={\overline {m}}=0$$m={\overline {m}}=0$, while if $T<T_{c}$$T<T_{c}$ then $m=\pm {\overline {m}}\neq 0$$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.