Mean field solution for potts model

[Binvestigator]

Homework Statement

The mean-field equation for the three-state Potts model H= -J∑δ_σiδ_σj can be derived as follows using this:
a) show that H is equivalent to -J∑S_i.S_j where Si=(1 0) , (-1/2 √3/2 ) , (-1/2 -√3/2)
b) putting H0= (H0 H'0) show the mean field equation become
H0/jz = exp(3βH0/2 ) -1 / exp(3β H0/2) +2
Fmf= - N kT Ln( exp(β H0) +2 exp (-β H0/2) + N H0^2 / 2 Jz

it is problem 4.3 statistical mechanics of phase transition, Yeomans

Homework Equations

The Attempt at a Solution

In the first section of this chapter,Author used this method for ising model, I know that should use the Bogoliubov inequality, taking the average in ensemble, minimizing the free energy .

I should use the previous problem , but how? How should give ensemble on the vector spin. Am I should indicate that the free energies are equal? for equality of Hamiltonian?
I don't have any idea at start point to indicate that 2 Hamiltonian are same.

Appreciate any help.

 

[mark726]

Hello fellow scientist,

Thank you for bringing up this interesting problem. I will try to explain the derivation of the mean-field equation for the three-state Potts model and how it relates to the Ising model.

Firstly, let's start with the Hamiltonian of the three-state Potts model, which is given by:

H = -J∑δσiδσj

where J is the coupling constant, δ is the Kronecker delta function, and σi and σj are the spin states at site i and j, respectively. Now, we can rewrite this Hamiltonian in terms of the spin states as:

H = -J∑Si.Sj

where Si and Sj are the spin vectors at site i and j, respectively. Note that Si and Sj are defined as:

Si = (1 0), (-1/2 √3/2), (-1/2 -√3/2)

These vectors correspond to the three possible spin states in the Potts model. Now, if we consider the Ising model, we can see that the spin vectors in this model are defined as:

Si = (1 0), (-1 0)

In other words, the Ising model only has two possible spin states. This means that we can rewrite the Hamiltonian of the Potts model in terms of the Ising model as:

H = -J∑Si.Sj = -J∑Si.Sj + J∑Si.Sj = -J∑Si.Sj + J∑Si.Sj = -J∑Si.Sj + J∑Si.Sj

where the second term represents the sum over the spin states of the Ising model. Now, if we consider the mean-field approximation, we can replace the sum over the spin states of the Ising model by its average value, which is given by:

<H> = (exp(βH) + 2exp(-βH) + exp(βH))/4 = (exp(βH) + 2exp(-βH) + exp(βH))/4

where β = 1/kT is the inverse temperature and H is the external magnetic field.

Now, if we substitute this expression into the Hamiltonian of the Potts model, we get:

H = -J∑Si.Sj + J∑Si.Sj = -J∑Si.Sj +