Substituting the ground-state energy and ${\cal N}_{12}=1/3$ to Eq. (\[relation\]), we obtain ${\cal N}_{13}=-1/2$, indicating that there exists no entanglement between two spin halves. Next, we consider more general situations, i.e., the case of even $N$ sites.

### The case of even $N$ spins

Except for the SU(2) symmetry in the system, there exists exchange symmetry for the case of even spins. For instance, in the four-spin model, the Hamiltonian is invariant when exchanging two spin halves or two spin ones. Thus, for even-spin model, the entanglements between the two nearest-neighbor spins and the correlation functions $\langle{\bf s}_i\cdot{\bf S}_{i}\rangle$ are independent on index $i$. Therefore, the internal energy per spin is equal to the correlation function $\langle{\bf s}_i\cdot{\bf S}_{i}\rangle$ $$u=U/N=\langle{\bf s}_1\cdot{\bf S}_{1}\rangle. \label{uuu}$$ From Eqs. (\[N12\]) and (\[uuu\]), we have $${\cal N}_{12}=-\frac{1}3-\frac{2}3 u. \label{N1212}$$ This equation indicates that for the case of even spins the entanglement between two nearest neighbors is solely determined by the internal energy per spin. And for the case of zero temperature, the entanglement is determined by the ground-state energy. The less the energy, the more the entanglement.

We now apply the above result to the study of ground-state entanglement. We rewrite the four-spin Hamiltonian as 