obtained as [@Vidal] $${\cal N}=\left[\big(\sum_n c_n\big)^2-1\right]/2.$$ Here, $C_n$ are the Schmidt coefficients, and $\{|e_n\rangle\}$ and $\{|f_n\rangle\}$ are bases for subsystems 1 and 2, respectively. Applying this formula to the ground state, we immediately have ${\cal N}=\sqrt{2}/3$. When $B=B_\text{th}=3/2$, the ground state is two-fold degenerate, and when $B>B_\text{th}$ the ground state becomes non-degenerate and the corresponding wave function is given by $$|\psi_0\rangle=|-\frac{1}2,-1\rangle,$$ which is obviously of no entanglement. Then, the ground-state negativity forms a platform when $0<B<B_\text{th}$. The jump of negativity at $B=B_\text{th}$ is due to the level crossing. For $N>2$, the effects of magnetic fields on entanglement can be also explained by level crossing. For instance, for $N=4$, there are two level crossing, and the entanglement displays two jumps.

Effects of next-nearest-neighbor interactions on entanglement
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We have studied the effects of finite temperature and magnetic fields on entanglement, and now consider the model containing two kinds of spins, spin $\frac{1}{2}$ and $1$, alternating on a ring with antiferromagnetic exchange coupling between both the NN spins and the NNN spins. The Hamiltonian can be expressed as $$H=J_1\sum^{N/2}_{i=1}\big({\bf s}_i\cdot {\bf  S}_{i}+ {\bf
S}_{i}\cdot {\bf s}_{i+1}\big)+J_2\sum^{N/2}_{i=1}\big({\bf
s}_i\cdot {\bf  s}_{i+1}+ {\bf S}_{i}\cdot {\bf S}_{i+1}\big),$$ where the ${\bf s}_i$ and ${\bf S}_i$ are spin-1/2 