= Solution
Taking the sum once over each unordered pair, write the <All-to-all Heisenberg model> as
$$
H=\sum_{i<j}\boldsymbol\sigma_i\mathbin{\cdot}\boldsymbol\sigma_j
=2\mathbf S^2-\frac{3N}{2},
\qquad
\mathbf S=\frac12\sum_i\boldsymbol\sigma_i.
$$
For even $N$, the minimum total spin is $S=0$, hence
$$
\boxed{E_0=-\frac{3N}{2}}.
$$
For odd $N$, $S=1/2$ and $E_0=3/2-3N/2$. If the paper's $\sum_{ij}$ counts ordered pairs or includes $i=j$, the corresponding harmless factors and additive constant change, but the minimizing total-spin sector is the same.
Now partition an even number of spins into disjoint pairs and put every pair in a <spin-one-half singlet state>. Each pair has total spin zero, so their tensor product also has $S=0$ and is itself a ground state. Its internal pair contributes $-3$, while correlations between different singlets vanish, giving $-3N/2$ in total. Therefore
$$
\boxed{E_{\rm paired}-E_0=0}.
$$
The large degeneracy is special to equal all-to-all coupling: the Hamiltonian sees only total spin and cannot distinguish different singlet coverings.
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