= Solution
Write the spin-one-half <Heisenberg antiferromagnet> as
$$
H_z=J\sum_{\langle ij\rangle}\mathbf S_i\mathbin\cdot\mathbf S_j,
\qquad J>0,
$$
on a bipartite lattice of <coordination number of a lattice>[coordination number] $z$. A one-spin <density operator> is $\rho=(I+\mathbf r\mathbin\cdot\boldsymbol\sigma)/2$, where $\mathbf r$ is its <Bloch vector> and $|\mathbf r|\leq1$. In a two-sublattice <product state>,
$$
\langle\mathbf S_i\mathbin\cdot\mathbf S_j\rangle
=\frac14\mathbf r_A\mathbin\cdot\mathbf r_B.
$$
The minimum is $-1/4$, attained by pure antiparallel Bloch vectors, so the minimizing product state is a <Néel state>. In the $z\to\infty$ limit this <mean-field approximation> becomes exact and the ground-state energy per bond is therefore
$$
e_{\rm bond}=-\frac J4.
$$
The phrase “energy density” requires a coupling convention. For the unscaled Hamiltonian above, every site belongs to $z/2$ bonds and
$$
\frac{E_0}{N}=-\frac{Jz}{8},
$$
which diverges as $z\to\infty$. With the standard <coordination number of a lattice>[Kac normalization]
$$
H_z^{\rm Kac}=\frac Jz\sum_{\langle ij\rangle}\mathbf S_i\mathbin\cdot\mathbf S_j,
$$
the finite energy density is
$$
\lim_{z\to\infty}\frac{E_0}{N}=-\frac J8.
$$
If the convention divides by the spatial dimension $d=z/2$ instead, the answer is $-J/4$ per site. A Hamiltonian written with $\boldsymbol\sigma_i\mathbin\cdot\boldsymbol\sigma_j$ rather than $\mathbf S_i\mathbin\cdot\mathbf S_j$ multiplies all these energies by four.
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