Ising Mean Field

The results of mean-field theory require the numerical solution of the self-consistent equation $$m = \tanh \big[(Jqm + H)/kT\big].\label{m}$$ To understand the solutions it is helpful to also calculate the Helmholtz function (per spin) $$f = -kT \ln \bigg[2\cosh \big[(Jqm + H)/kT\big] \bigg]. \label{f}$$

The program plots $m$ versus $m$ and $\tanh \big[(Jqm + H)/kT\big]$ versus $m$ in one window and $f$ versus $m$ in another window. From the first window we can pick out the possible equilibrium values of $m$. From the minima of $f(m)$ we can determine the actual equilibrium value(s) of $m$. Units are chosen so that $k = 1$ and $J = 1$.

Problem: Numerical solutions of \eqref{m}

Program IsingMeanField uses Newton's method to find the zeros of the function $g(m) = m - \tanh \beta (Jqm + H)$.

  1. Set $H = 0$ and $q=4$ and confirm the mean-field approximation for the critical temperature $T_c$ of the Ising model on a square lattice. Start with $kT/Jq=10$ and then proceed to lower temperatures. The equilibrium value of $m$ is the solution with the lowest value of the Helmholtz function.
  2. Plot the temperature dependence of $m$ for $T < T_c$.
  3. Determine $m(T)$ for the one-dimensional Ising model ($q=2$) and $H =0$ and $H=1$. Compare your values for $m(T)$ with the exact solution in one dimension.
Problem: Minima of the Helmholtz function

It can be shown that $f(m)$ near the critical point can be written as \begin{equation} \label{eq:5/freemf} f(m) = a + b(1-\beta q J)m^2 + cm^4, \end{equation} where $a$, $b$, and $c$ are positive.

  1. Show that the minimum value of $f$ for $T>T_c$ is at $m=0$, and that $m =\pm m_0$ corresponds to a lower value of $f$ for $T \lt T_c$.
  2. Use Program IsingMeanField to plot $f(m)$ as a function of $m$ for $T>T_c$. For what value of $m$ does $f(m)$ have a minimum?
  3. Plot $f(m)$ for $T=1$. Where are the minima of $f(m)$? Do they have the same depth? If so, what is the meaning of this result?
  4. If $H$ is nonzero but small, there is an additional term $-mH$ in \eqref{eq:5/freemf}. Choose $T=1$ and $H=0.5$. Do the two minima have the same depth? The global minimum of $f$ corresponds to the stable phase. If we quickly “flip” the field and let $H \to -0.5$, the minimum at $m \approx 1$ will become a local minimum. The system will remain in this local minimum for some time before it switches to the global minimum.


Problems 5.15 and 5.17 in Statistical and Thermal Physics: With Computer Applications, 2nd ed., Harvey Gould and Jan Tobochnik, Princeton University Press (2021).