Wang-Landau Analysis of the q state Potts Model


The Potts model is a generalization of the Ising model in which each site or "spin" can have one of q states. The energy between two sites is 0 if their states are different, and -K if the states are the same. The q = 2 Potts model is equivalent to the Ising model, with half the critical temperature, because the energy difference between two like and two unlike states is half what it is when the spins in the Ising model take on values of ±1.

    One of the important features of the Potts model is that the nature of the phase transition from a disordered to ordered state changes with q. For small q the transition is continuous as it is in the Ising model (in zero magnetic field). For q > 4 on the square lattice the transition becomes first order, with a jump in the energy at the transition temperature.

    We use the Wang-Landau algorithm to generate an approximate density of states g(E). Because the energy is a discrete variable in the Potts model, g(E) becomes the number of microstates of energy E. The probability that the system has energy E is given by P(E) = g(E)exp(-ßE), where ß = 1/kT. The program uses units such that k = 1 and K = 1.

    The program plots ln [(g(E)/g(E0)] versus E, where E0 = -2N is the lowest possible energy for a system of N sites on a square lattice. The program also plots the unnormalized value of ln P(E) for the value of the temperature T specified by the user and the histogram of energies for each pass of the Wang Landau algorithm. Also shown is the specific heat as a function of temperature, which can be computed from P(E).


  1. Run the simulation with q = 2 at a temperature that is close to the transition temperature T ≅ 1.135. Run the simulation until the parameter f is near 1.0. What is the shape of ln P(E)? Under the Views menu you can change the vertical scale so that you can see more of the peak structure. Compare the plots of ln P(E) for T = 0.5, T = 1, and T = 2.
  2. Run the simulation with q = 10 at a temperature which is close to the transition temperature T ≅ 0.70. What is the shape of ln P(E)? To obtain more insight set T = 0.5 and describe the shape of ln (P(E). Then repeat for T = 1. Compare the plots of ln P(E) for different temperatures. How does the shape of P(E) compare to what you found for q = 2? Explain why P(E) has two peaks near the transition for q = 10.

Java Classes

Updated 28 December 2009.