Ising model 2-dimensional - ground state configuration I have to prove something about the 2-dimensional ising model. 
The problem is the following: Prove that every nearest-neighbour and next-nearest-neighbour interaction on the square lattice $\mathbb{Z}^2$ which is symmetric has at least one periodic ground state configuration. 
Actually, I don't know how to procede.. the only tools that I know and maybe I can use them are: the definition of ground state configuration i.e. the configuration of minimal energy, the fact that the square lattice is symmetric hence it depends only on types of particles and distances between them.
It will be great if someone could give me a hint !
Thank you very much!
 A: Suppose the spins on the sites of the square lattice are coupled nearest neighbor so that the Hamiltonian is defined as
$$H=-J\sum_{<ij>}S_iS_j$$
where $<ij>$ denotes nearest neighbors $i$ and $j$ (each site has 4 nearest neighbors). The ground state of the Hamiltonian will be given by finding a configuration of spins $\{S_i\}$ for which the pairwise interaction terms in the sum are +1 or -1 depending on the sign of $J$, the coupling (if $J < 0$, then we want the terms in the sum to individually be 1, and if $J > 0$, we want the terms to be -1). This means that there are two ground states in the nearest-neighbor Hamiltonian, because if, suppose, I set $S_1 = +1$, then we know that all of its nearest neighbor spins must also be $+1$. Then we just repeat the reasoning on the nearest neighbors of those sites whose spins we just set. We could get the other ground state by picking $S_1 = -1$, which by this reasoning gives that the other sites are also $-1$. Now, if $J < 0$, then the same logic can be used: we start by labeling $S_1 = +1$ and then all of its nearest neighbors must be $S_i=-1$, but since this is a square lattice, if we then look at the nearest neighbors of one of these $-1$ sites, we can label all of its nearest neighbors $+1$ without having to undo the first spin we set (this property is referred to as "frustration-free"). Note also that starting from either of these two states, if you flip any of the spins, the energy increases (hence we know that these are truly the ground states).
These two states (all $S_i=+1$ or all $S_i=-1$) have a discrete translational symmetry in both the $x$ and $y$ directions (by 1 lattice site). I am not fully sure if this what you mean by "periodic ground state configuration", but I am unable to comment and ask, but hopefully this gives you enough information to figure it out.
We can repeat this logic for the next-nearest-neighbor model, too. In this scenario, the spin at one site is coupled to eight next-nearest neighbors instead of 4 nearest neighbors. Suppose again $J > 0$ and we choose one site and set $S_1=+1$ and mark all of its next-nearest neighbors also $S_i=+1$ so that our interaction terms in the sum are each +1. We can then move to one of these sites that we just marked and again mark all of its next-nearest neighbors also $S_i=+1$, and repeat this algorithm. But now we have left a subset of the total set of spins that are not yet marked. Since the graph (whose edges correspond to the couplings between next-nearest neighbors) of these spins is disjoint from the graph of the spins we just set to $+1$, we can set all of the spins in this grid to either $+1$ or $-1$ and still minimize $H$. So in total, you have four ground states (all $S_i=+1$, half $S_i=+1$/half $S_i=-1$, half $S_i=-1$/half $S_i=+1$, all $S_i=-1$) and the first and last ones have the same discrete translational symmetry described before by one lattice spacing while the other two have translational symmetries by two lattice spacings. If $J < 0$, the situation is a bit complicated as you cannot create a frustration-free configuration. I hope this helps you work out this problem! I recommend sketching some graphs.
