Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

The order parameter of Ising model can be defined as $m=\frac{N_1-N_2}{N}$, if $N$ is the total number of lattice points, $N_1$ and $N_2$ is the number of lattice points spin up and down respectively, $N=N_1+N_2$.

But I am not able to write down the order parameter of the q-state Potts model. Any help will be appreciated!

share|cite|improve this question
Can you give references for further study on this please ? – cosmicraga Apr 1 '13 at 15:08 – hlew Apr 2 '13 at 7:10

1 Answer 1

up vote 4 down vote accepted

First one needs to gain a deeper understanding of the order parameter of the Ising model. The magnetization $m$ can be written as $m=(+1)n_1+(-1)n_2$, where $n_p=N_p/N$ ($p=1,2$) is the number density of each spin. So what do the coefficients $(\pm1)$ stand for? They are two possible magnetizations of a single Ising spin, which form the representation of the $\mathbb{Z}_2$ group. So the Ising spin is also known as the $\mathbb{Z}_2$ spin.

Now it is straight forward to extend the formulation to the $q$-state Potts model. Simply replace the number 2 by $q$, i.e. to consider the Potts spin as the $\mathbb{Z}_q$ spin, whose "magnetization" must be taken from the representation of the $\mathbb{Z}_q$ group, which are $q$th roots of unity $e^{2\pi ip/q}$ ($p=1,2,\cdots,q$). Given the representation of the Potts spin, it is easy to write down the order parameter $$m=\sum_{p=1}^qe^{2\pi i\frac{p}{q}}n_p,$$ with $n_p=N_p/N$ (let $N=\sum_pN_p$) being the number density of the $p$th type of the Potts spin. Note that the order parameter is complex in general. In the disordered phase, all types of Potts spin appear with equal probability, i.e. $n_1=n_2=\cdots=n_q$, and in this case, we do have $m=0$ due to the cancellation of the phase, which is consistent with the idea of the order parameter: "a quantity that is zero in the disordered phase and non-zero in the ordered phase".

share|cite|improve this answer
So the dimensionality of the order parameter is 2. Am I right? We can define the universality class of the phase transition from it. – hlew Apr 2 '13 at 7:24
@hlew The dimensionality of the order parameter depends on the specific model. – Everett You Apr 2 '13 at 10:00
The order parameter is complex when $q>2$, so it has two components(the real and imaginary parts). Isn't the dimensionality 2? Thank you! – hlew Apr 2 '13 at 12:32
@hlew Ok, I though you were talking about the scaling dimension. If you are concerning the degree of freedom, then you are right. – Everett You Apr 2 '13 at 17:07
This is a very interesting order parameter, however, it is not the one usually found in the literature. Could you please provide some references for this? – Asaf Nov 13 '14 at 15:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.