The classical 2d Ising model has a Hamiltonian of the form:

\begin{equation} H = -\sum_{m,n = 0}^{M,N} J_1 x_{m,n}x_{m+1,n} + J_2 x_{m,n}x_{m,n+1} \end{equation}

The partition function for the model can be written as the sum over all configuration sof the spins $x_{ij}$ times the boltzman factor. Up to an over all multiplicative constant we can re-write this as:

\begin{equation} Z = \sum_{x_{ij}}\prod_m (1+t_1 x_{mn} x_{m+1,n})(1+t_2 x_{mn} x_{m,n+1}) \end{equation}

Then we can perform the sum by which ever means we like. My current favourite is by following this reference: http://link.springer.com/article/10.1007%2FBF01017042?LI=true, or see http://link.springer.com/article/10.1007/BF02896231. They re-write the partition sum as an integral over Grassmann variables. The computation is technical at points but to my understanding, they essentially sum over all loops in the plane with appropriate boltzmann weights by a convenient choice of integral. This can be seen by expanding the exponential and recognizing that the only terms left after performing the integration will be loops that describe domain walls.

They successfully perform the integration on a finite lattice with the topology of a torus and obtain the free energy. When I have more time I will detail some of these calculations. But, for now I have a question: What if we want to consider vortices in the system?

One way to do introduce two vortices in the model is to require that some finite number of neighbouring rows satisfy different boundary conditions from all others.

Another may be to consider the system Hamiltonian on a cylinder and specify antiperiodic boundary conditions on the lower half and periodic on the upper half.

In particular, I would be interested in knowing what happens to the free energy relative to the case of no vortices in the thermodynamic limit (this may be the only tractable part of the problem).

  • $\begingroup$ I am not sure what you call vortices in the context of the Ising model (you can have them in continuous spin systems of course). What you can do (and maybe what you mean) is to force the presence of interfaces in the system. Forcing the presence of finitely many will not affect the (limiting) free energy density. $\endgroup$ – Yvan Velenik Apr 26 '14 at 8:36
  • $\begingroup$ @YvanVelenik I should have been more clear, by vortex I mean a string negative couplings dragged in from infinite and terminating at some point. So any Wilson loop which contains this point will be negative while all others positive. I think the free energy should change to order 1/N - do you agree? $\endgroup$ – Stackexchange_user23 Apr 26 '14 at 16:19
  • $\begingroup$ Yes, I agree, it will affect the finite-volume free energy density by a term of order $1/N$ (that is, the energy associated to the defect lines, of order $N$, divided by the total volume, of order $N^2$). $\endgroup$ – Yvan Velenik Apr 26 '14 at 18:50

Let me write up the (trivial) argument, as it is too long for comments.

Let us consider the Ising model in a finite set $\Lambda$ with an arbitrary, but fixed, boundary condition. We actually consider two versions of the model:

  • The standard ferromagnetic model, with Hamiltonian $$ H_\Lambda (\sigma) = -J\sum_{(i,j)\cap\Lambda\neq\emptyset} \sigma_i\sigma_j \,, $$ where the first sum is over all pairs of nearest-neighbours at least one of which belongs to $\Lambda$.
  • A model in which the sign of the coupling constants has been changed for a set $\mathcal{E}$ of nearest-neighbours, that is, with Hamiltonian \begin{align} H_\Lambda^\mathcal{E} (\sigma) &= -J\sum_{\substack{(i,j)\cap\Lambda\neq\emptyset\\(i,j)\not\in\mathcal{E}}} \sigma_i\sigma_j +J\sum_{\substack{(i,j)\cap\Lambda\neq\emptyset\\(i,j)\in\mathcal{E}}} \sigma_i\sigma_j\\ &= H_\Lambda(\sigma) + 2 J\sum_{\substack{(i,j)\cap\Lambda\neq\emptyset\\(i,j)\in\mathcal{E}}} \sigma_i\sigma_j \,. \end{align}

We immediately deduce that, for any configuration $\sigma$, $$ |H_\Lambda(\sigma) - H_\Lambda^\mathcal{E} (\sigma)| \leq 2 J |\mathcal{E}|\,, $$ where $|\mathcal{E}| = \#\{(i,j)\cap\Lambda\neq\emptyset\,:\, (i,j)\in\mathcal{E}\}$ is the number of affected pairs of neighbours in $\Lambda$. Consequently, denoting by $Z_\Lambda$ and $Z_\Lambda^{\mathcal{E}}$ the corresponding partition functions, $$ Z_\Lambda e^{-2\beta J |\mathcal{E}|} \leq Z_\Lambda^{\mathcal{E}} \leq Z_\Lambda e^{+2\beta J |\mathcal{E}|}\,, $$ and, therefore, $$ \bigl| \frac1{\beta|\Lambda|}\log Z_\Lambda - \frac1{\beta|\Lambda|}\log Z_\Lambda^{\mathcal{E}} \bigr| \leq 2J\frac{|\mathcal{E}|}{|\mathcal{\Lambda}|}\,, $$ where $|\Lambda|$ is the number of vertices in $\Lambda$.

In your case, $|\Lambda| = N^2$ (say, for a square box of sidelength $N$) and $|\mathcal{E}| \leq CN$ for some fixed constant $C$ (if you introduce a finite number of defects). The change in the finite-volume free energy density is thus at most of order $1/N$. To prove that it is indeed of this order (and not much smaller) is more difficult (and only true at low temperatures).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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