# Weiss mean field theory for the Ising model

Let us now study the mean field theory of an Ising model, called Weiss mean field theory. As usual, we start from the Hamiltonian of a nearest-neighbour interaction Ising model, so that the partition function of the system is:

$Z_{N}=\operatorname {Tr} e^{-\beta {\mathcal {H}}}=\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}e^{\beta \left({\frac {1}{2}}\sum _{\left\langle \right\rangle }J_{ij}S_{i}S_{j}+H\sum _{i}S_{i}\right)}$ We define our order parameter as the single-spin magnetization:
$m=\left\langle S_{i}\right\rangle =\left\langle S\right\rangle$ and we introduce it in the partition function through the identity:
$S_{i}S_{j}=(S_{i}-m+m)(S_{j}-m+m)=-m^{2}+m(S_{i}+S_{j})+(S_{i}-m)(S_{j}-m)$ Note that the last term is $(S_{i}-m)(S_{j}-m)=(S_{i}-\left\langle S_{i}\right\rangle )(S_{j}-\left\langle S_{j}\right\rangle )$ , which exactly measures the fluctuations of the spin variables from their mean value. We now neglect it, so that in the partition function we can approximate:
${\frac {1}{2}}\sum _{\left\langle i,j\right\rangle }J_{ij}S_{i}S_{j}\approx {\frac {1}{2}}\sum _{\left\langle i,j\right\rangle }J_{ij}\left[-m^{2}+m(S_{i}+S_{j})\right]$ If our system is isotropic then $J_{ij}=J$ , so that we can write $\sum _{\left\langle i,j\right\rangle }J_{ij}=Jz\sum _{i}$ where $z$ is the coordination number of the lattice. This means that:
${\frac {1}{2}}\sum _{\left\langle i,j\right\rangle }J_{ij}m(S_{i}+S_{j})=Jzm\sum _{i}S_{i}$ and the partition function reduces to:
{\begin{aligned}Z_{N}=\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}e^{\beta \left(-Jzm^{2}{\frac {1}{2}}\sum _{i}1+Jzm\sum _{i}S_{i}+H\sum _{i}S_{i}\right)}=\\\\=e^{-\beta {\frac {Jzm^{2}}{2}}N}\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}e^{\beta (Jzm+H)\sum _{i}S_{i}}=\\=e^{-\beta {\frac {Jzm^{2}}{2}}N}\sum _{S_{1}=\pm 1}\cdots \sum _{S_{N}=\pm 1}\prod _{i=1}^{N}e^{\beta (Jzm+H)S_{i}}=\\\\=e^{-\beta {\frac {Jzm^{2}}{2}}N}\left(\sum _{S=\pm 1}e^{\beta (Jzm+H)S}\right)^{N}=e^{-\beta {\frac {Jzm^{2}}{2}}N}\left[2\cosh \left(\beta (Jzm+H)\right)\right]^{N}\end{aligned}} Therefore the free energy of the system in the thermodynamic limit is:
$f=-\lim _{N\to \infty }{\frac {k_{B}T}{N}}\ln Z_{N}={\frac {1}{2}}Jzm^{2}-k_{B}T\ln \left[\cosh \left(\beta (Jzm+H)\right)\right]-k_{B}T\ln 2$ and the magnetization is:
$m=-{\frac {\partial f}{\partial H}}_{|T}=\tanh \left[\beta (Jzm+H)\right]$ We have thus found the so called self-consistency equation for the magnetization:
$m=\tanh \left[\beta (Jzm+H)\right]$ which is almost the same as the one we have encountered in The role of interaction range. We now proceed exactly as before. Since we are interested in studying the possible occurrence of phase transitions and in particular the existence of spontaneous magnetization we focus on the case $H=0$ :
$m=\tanh(\beta Jzm)$ which is just like the self-consistency equation that we found in The role of interaction range, but taking into account also the coordination number of the lattice. In this case we will have that the phase transition occurs when $\beta Jz=1$ , i.e. at the temperature:
$\beta _{c}Jz=1\quad \Rightarrow \quad T_{c}={\frac {z}{k_{B}}}J$ Now, in order to better understand the physical properties of our system at different temperatures let us expand the free energy $f$ around $m=0$ , still in the case $H=0$ . Using the Taylor series for $\cosh$ and $\ln$ we have $\ln(\cosh x)\sim x^{2}/2-x^{4}/12+\cdots$ , so:

$f(m,H=0){\stackrel {m\approx 0}{\sim }}-k_{B}T\ln 2+{\frac {Jz}{2}}(1-\beta Jz)m^{2}+{\frac {\beta ^{3}}{12}}J^{4}z^{4}m^{4}$ Therefore, the behaviour of $f$ near $m=0$ depends strongly on the sign of the coefficient of $m^{2}$ ; in particular:

• for $\beta Jz<1$ , namely for $T>T_{c}$ the free energy $f$ has only one minimum for $m=0$ : this is the paramagnetic (disordered) phase
• for $\beta Jz>1$ , namely $T the free energy $h$ has two stable minima at:
$m=\pm {\overline {m}}:=\pm {\sqrt {\frac {3(\beta Jz-1)}{(\beta Jz)^{3}}}}$ This is the magnetized (ordered) phase. Trend of $f$ for $m\approx 0$ in the two cases considered

Note that in the computations we have just made we have never imposed a particular value for the dimensionality of the system. This means that the results of this approximation should be valid also for $d=1$ , but we know that in one dimension the Ising model does not exhibit a phase transition. This is an expression of the fact that in the one-dimensional case mean field theory is not a good approximation (again, the dimensionality of the system is still too low).

Let us finally note a fact. As we have previously stated mean field theories are characterized by the fact that the fluctuations of the order parameter are neglected, and we can see that the self-consistency equation can be indeed obtained by neglecting the fluctuations of the single spins. In other words we can find it if we consider all the spins "freezed" to the value ${\overline {m}}$ ; in fact if we consider a single spin $S$ we will have:

$\left\langle S_{0}\right\rangle ={\overline {m}}={\frac {\sum _{S=\pm 1}e^{\beta Jz{\overline {m}}S}S}{\sum _{S=\pm 1}e^{\beta Jz{\overline {m}}S}}}={\frac {e^{\beta Jz{\overline {m}}}-e^{-\beta Jz{\overline {m}}}}{e^{\beta Jz{\overline {m}}}+e^{-\beta Jz{\overline {m}}}}}=\tanh(\beta Jz{\overline {m}})$ (where $\left\langle S\right\rangle ={\overline {m}}$ because we are requiring that the mean values of all the spins are equal).
1. The $1/2$ factor has been introduced for later convenience (it can be reabsorbed in the definition of $J_{ij}$ ).