# Do we know the exact ground state of the Antiferromagnetic Heisenberg Model?

I've been reading about the Heisenbeg Model:

\begin{equation} H=-J\sum_{<i, j>} \hat{\vec{S}}_i .\hat{\vec{S}}_j = -J\sum_{<i, j>}\Big[\hat{S^z}_i \hat{S^z}_j + \frac{1}{2}(\hat{S^+}_i \hat{S^-}_j+\hat{S^-}_i \hat{S^+}_j)\Big] \end{equation}

Where $\hat{\vec{S}}_i$ is the Spin Vector Operator in the site i. In a bipartite lattice (where you can divide the lattice in two sublattices such that every site is surrounded by sites of the other sublattice) the classical antiferromagnetic ($J<0$) ground state is every spin of sublattice A pointing at one direction while every spin in sublattice B pointing in the opposite direction. In the Heisenberg Model this is not an eigenstate because acting with$\hat{S^+}_i$ and $\hat{S^-}_i$ modify neighbouring spins.

My question is: Is this problem solved? Do we know the exact ground state of this model? I'm familiar with the Holstein-Primakoff transformation and the ondulatory solutions known as magnons but I'm wondering if there is an exact known ground state.

• We “know” the ground state in the one-dimensions case. In fact the eigenstates were somehow found by Bethe in 1927. However this expressions are really hard to work with and computing expectation value of operators is hard. For local operators (on neighboring sites) we know everything. But the exact form of correlations has been found (in the long-distance limit) only recently and with very indirect methods – lcv Dec 26 '17 at 21:02