What do physicists mean by solving the Ising model?

To me, an Ising model is a setting of discrete objects, that have attributes (spins) that contribute to energy based on interactions with nearby objects. With the energy function (Hamiltonian) written down, we can write down the partition function.

However, I don't get what physicists mean by "solving" the model. What aspects of this model do they want to get? What do they mean by a solution?

Another commonly used notion of "solving a theory" is to find a procedure to compute, at least in principle, all local observables. Sometimes one might also add non-local observables to the mix, it depends on what people are most interested about.

Local observables are correlation functions $$\langle O_1(x_1)\cdots O_n(x_n)\rangle$$, where $$O_i$$ may be the spin field $$\sigma$$, but may also be any other operator of the theory. Knowing the spectrum of operators is part of finding a solution.

One has a variety of methods for approximating those functions, but there are always some regimes where the approximations break down or sometimes they become impossible to deal with after a certain order. A solution would be a procedure that gives exact results and does not suffer from any limitation to its validity nor its applicability in practice.

I'm going to make a list of some theories that have been solved so you get the flavor. Stop whenever you feel bored.

Free theories

A free theory of a field $$\phi$$ has an exact solution because the spectrum is made of all operators of the form $$\partial^{n_1}\phi \partial^{n_2} \phi \partial^{n_3}\cdots \partial^{n_k}\phi$$ and the correlators are obtained by doing Wick contractions. The two-point function, which is the propagator, is of course known exactly.

Two dimensions

The Ising model in $$2d$$ has an exact solution. All correlation functions $$\langle \sigma_{i_1}\sigma_{i_2}\cdots \sigma_{i_k}\rangle$$ can be obtained by taking suitable derivatives of the partition function with respect to the parameters $$J_1$$ and $$J_2$$. It may be a tedious task, but completely determined.

Other examples of exactly solved models are the minimal models in $$2d$$ CFTs (the Ising model at criticality is actually one of them). There are only a finite number of operators and their $$n$$-point functions are entirely fixed by the representation theory of the Virasoro algebra. So there are well defined algorithms to obtain them explicitly.

Actually, in $$2d$$ CFTs, because of the Operator Product Expansion (OPE), one only needs to compute three-point functions since the higher points can be always reduced to fewer points by using the OPE. Another example of a solved theory is therefore the Liouville CFT where the three-point functions are known in an explicit form by the DOZZ formula.

Three dimensions

Examples of exactly solved theories in higher dimensions are rarer. Sometimes they require some rigid structure. An example is when the theory is topological.

For pure Chern-Simons in $$3d$$ (which is topological) we know how to compute the partition function and also how to compute all non-local observables. There are no interesting local observables, the only things one should look at are correlators of Wilson loops. Both the partition function and the correlators of Wilson loops are related to topological invariants. In particular, the latter are knot invariants (like HOMFLY and Jones polynomials) of the loop configuration. These are defined algorithmically and are easy to compute.

Zero dimensions

Yes, QFT in zero dimensions, namely matrix models. The partition function of a matrix model is simply a multivariate integral over an ensemble of matrices. Clearly this could be regarded as "solved" already, but that's not what we want. We want to send the size of the matrices $$N\to\infty$$. This limit makes the integrals non-trivial but they can be exactly solved. In a few words, it is possible to compute in closed form the probability distribution of the matrix eigenvalues $$\rho(\lambda)$$ (I'm talking about hermitian matrices here). So all correlators (namely moments of the ensemble) can be evaluated as ordinary integrals weighted by $$\rho(\lambda)$$.

Four dimensions

We wish.

• Thanks for your thorough explanation. I'm still wondering what you meant by "exact models". Do we need an exact form of $\lim_{N\to \infty}Z$, or we just need that any correlation functions of any observables are completely determinable? More particularly, in the link to "exact solution", Onsager's solution is given as an analytical expression of the free energy. Why is that "exact"? – Student Apr 25 '20 at 20:29
• Exact means that it's not an approximation. It's the true result and you have a formula that you can put in a computer and get an answer with a reasonable precision in a reasonable amount of time. – MannyC Apr 25 '20 at 20:35
• Thanks for your clarification: reasonable precision in a reasonable amount of time! This was what I felt weird! – Student Apr 25 '20 at 21:33

I think most physicists mean computing the partition function. So, given a lattice $$L$$ with edges $$E(L)$$ and vertices $$V(L)$$, solving the corresponding Ising model would mean to compute $$Z(\beta,h):=\sum_{s\in\{-1,1\}^{V(L)}}e^{-\beta H_h(s)},$$ with $$H_h(s):=J\sum_{\{p,q\}\in E(L)}s(p)s(q)+h\sum_{p\in V(L)}s(p).$$ Computing this partition function then allows to deduce other thermodynamical quantitites by taking derivatives with respect to the thermodinamical variables $$\beta$$ and $$h$$.

• Great! That brought me closer! What do you mean by computing $Z$? As a mathematician, that definition is a very good computation already. I guess you mean by computing $Z$ as some combination of other functions, e.g. cosines? Or do you mean by extracting more attributes of $Z$? – Student Apr 25 '20 at 18:06
• Moreover, I often hear physicists talking about exact solutions. Do those relate to what you meant here? – Student Apr 25 '20 at 18:08
• One typically wants $Z$ in the thermodynamics limit, and this is not easy at all to obtain from the explicit expression. At the very least, the explicit expression is not useful at all to compute quantities like the critical temperature or the critical exponents. To get a flavor of what the solution is like check Onsager's solution for $2d$. – MannyC Apr 25 '20 at 18:45
• I see, so they want a "better" expression of $\lim_{N\to\infty}Z$, which hopefully will be turned into other interesting attributes of the system. That's it? – Student Apr 25 '20 at 20:20
• @Student An elementary analogy is summing a geometric series. Just looking at the sum $\sum_n^N p^n$, it's hard to see how this behaves as a function of $p$ in the limit $N\rightarrow \infty$, but once you know $\sum_n^\infty p^n = 1 /(1 - p)$ you can do a lot with it. – d_b Apr 26 '20 at 1:55

In physics, solving a model are sometimes ambiguous. But if someone is from a physics background, they will immediately understand from solving an Ising model is to solve the Schrodinger's equation for that model. So the Hamiltonian for a general Ising model is $$H(\sigma)=\sum_{i,j}-J_{ij}\sigma_i\sigma_j-\mu\sum_jh_j\sigma_j$$ Now can we solve the eigenequation for this Hamiltonian i.e. $$H\psi=E\psi$$, which is nothing but the Schrodinger's Equation. By solving the eigenequation, we can know Energy eigenvalues of the system. Now using the partition function one can calculate the probability of the system to be in a particular eigenstate at a given temperature.

• In the case of the Ising model, Schrödinger's equation is trivial because all operators commute. Therefore, the solutions are immediately all vectors of the form $\otimes_{p\in V(L)}\psi_p$ with $\psi_p\in\{(1,0),(0,1)\}$. The difficult part is computing the partition function. – Iván Mauricio Burbano Apr 25 '20 at 18:01