It seems that the question OP asked can be paraphrased as: What is the ground state of an antiferromagnetic many-body system?
Let me explain why this is the case.
Prelude:
Quantum magnetism starts by finding the eigenstates and eigenenergies of the Heisenberg Hamiltonian:
$$ H = J \sum_{i,j} \mathbf{S}_{i}.\mathbf{S}_{j} $$
Where the exchange integral $J$ is (positive) negative for the (anti) ferromagnetic case. The archetypical way of solving the eigenproblem for $\mathcal{H}$ is to place spins on some lattice $\Lambda$. The eigen system of $H$ depends on the dimensionality and the geometry of the underlying lattice geometry.
In the case of the one-dimensional ferromagnet, Hans Bathe 1931 first calculated the exact eigensystem of the Heisenberg Hamiltonian using an ansatz, which now goes by his name. Later, Bethe's result was extended by Hulthen for the anti-ferromagnet (AFM). Although Bethe's solution is exact, we don't know how to generalize it in higher dimensions. We even don't know precisely why it works (That's why it's called ansatz).
The ground state of the one-dimensional AFM is a mess due to entanglement.
No analytical ground state for the AFM Heisenberg system is known in dimensions higher than one. But one can guess, just like 1D, the higher dimensional AFM ground state will also be highly entangled.
Variational ground states:
The absence of an analytical solution of $H$ in higher dimensions leads people to consider variational wavefunctions based on some educated guess. In the variational approach, the wave function is minimized with respect to a set of "chosen" parameters to understand the properties of the GS.
Let $\Psi_{\mathrm{var}}$ be the variational wavefunction for some AFM. Obviously we expect that $\Psi_{\mathrm{var}}$ is the eigenstate of a "Variational Hamiltonian" $H_{\mathrm{var}}$ with eigenvalue $E_{\mathrm{var}}$.
$$ H_{\mathrm{var}}\Psi_{\mathrm{var}} = E_{\mathrm{var}}\Psi_{\mathrm{var}} $$
Note that $H_{\mathrm{var}}$ may depend on additional parameters that are absent for the original Heisenberg Hamiltonian $H$. These additional interaction parameters help us to understand the GS properties of the original Hamiltonian $H$. We call $H_{\mathrm{var}}$ as parent Hamiltonian. One example of the $\Psi_{\mathrm{var}}$ is the semiclassical large $N$ wavefunction.
In some sense, the word parent is misleading because it does not correspond to the original Heisenberg Hamiltonian; instead, it is a new one with additional coupling parameters.
Constructing the parent Hamiltonians using the spin projectors, which are discussed below.
Example:
The classical Neel state $|\uparrow\downarrow\uparrow\downarrow\cdots\rangle$ is not the GS of the Heisenberg Hamiltonian due to the quantum fluctuations. However, it is possible to construct a singlet between a pair of neighboring spins $i$ and $i+1$ of an AFM, known as the Valnce Bond State or VBS:
$$ |\mathrm{VBS}\rangle = \frac{1}{\sqrt{2}}|(|\uparrow_{i}\rangle | \downarrow_{i+1} \rangle - | \downarrow_{i} \rangle | \uparrow_{i+1} \rangle) $$
which has lower energy than the $\mathrm{Neel}$ state. One can ask whether it is possible to construct a Hamiltonian for which $|\mathrm{VBS}\rangle$ is its eigenstate.
One way to motivate is by introducing an AFM interaction between NN (nearest neighbor) and NNN (next to nearest neighbor) terms, leading to the well-known $J_{1}-J_{2}$ Hamiltonian.
$$ H_{J_{1}J_{2}} = J_{1} \sum_{\langle ij \rangle} \mathbf{S}_{i}.\mathbf{S}_{j} + J_{2} \sum_{\langle\langle ij \rangle\rangle} \mathbf{S}_{i}.\mathbf{S}_{j} $$
Where $\langle\rangle$ and $\langle\langle\rangle\rangle$ are NN and NNN pairs.
In 1D, if we take $J_{2} = J_{1}/2$ that gives the famous Majumdar-Ghosh Hamiltonian
$$ H_{\mathrm{MG}} = \sum_{i = 1}^{\mathcal{N}} \Bigl(\mathbf{S_{i}} . \mathbf{S_{i+1}} + \frac{1}{2} \mathbf{S_{i}} . \mathbf{S_{i+2}} \Bigr) $$
Whose exact ground state is the VBS states
$$ |\mathrm{MG}_{\pm}\rangle = \prod_{n = 1}^{\mathcal{N}/2} \left( |\uparrow\rangle_{2n} \otimes |\downarrow\rangle_{2n \pm 1} - |\downarrow\rangle_{2n} \otimes |\uparrow\rangle_{2n \pm 1} \right) / \sqrt{2} $$
Which is written for the even number of sites and periodic boundary conditions. The eigenenergy of MG chain is 3/4. One can see quite clearly that $\mathcal{H}_{\mathrm{MG}}$ is an example of the variational Hamiltonian with VBS as its variational eigenstates.
Note: The $|\mathrm{MG}_{\pm}\rangle$ is the first example of Matrix Product States. Which later generalized to the spin-1 case, leading to the AKLT chain.
Frustration freeness:
The competing terms in the Hamiltonian lead to geometrical frustrations. However, it is possible to find a type of spin system that has competing terms without frustration. Surprisingly, such Hamiltonians have exact yet highly nontrivial ground states, violating the entanglement entropy's area law. Examples of such spin chains are Fredkin Spin Chain and Shor-Movassagh chain.