equation leads to $$U=\frac{\displaystyle
-e^\beta+e^{-\frac{1}{2}\beta}}{\displaystyle
e^\beta+2e^{-\frac{1}{2}\beta}}. \label{U}$$ From Eqs. (\[N1\]) and (\[U\]), we obtain a quantitative relation between the negativity and the internal energy $${\cal N(\rho)}=\frac{1}{3}\max[0,-1-2U].\label{U1}$$ The above equation builds a connection between the microscopic entanglement and the macroscopic thermodynamical function, the internal energy. The internal energy completely determine the thermal entanglement. From the equation, we can also read that the thermal state becomes entangled if and only if the internal energy $U<-1/2$. Since $U=\langle H\rangle=\langle \bf s_1\cdot\bf
S_1\rangle$, we have $${\cal N(\rho)}=\frac{1}{3}\max[0,-1-2\langle\bf s_1\cdot\bf
S_1\rangle],\label{U2}$$ which is consistent with the result obtained in Ref. [@Schliemann] by the group-theoretical technique.

### Three-spin case

We now consider the three-spin case and the schematic representation of the corresponding Hamiltonian is given by Fig. 1. In this situation, there are two types of pairwise entanglement, the entanglement between spin 1/2 and spin 1 and the entanglement between two spin halves.

The eigenvalue problem can be solved analytically, and after tracing out the third spin-half system the reduced density matrix $\rho_{12}$ is still of the same form as in Eq. (\[rho\]) with matrix elements given by $$\begin{aligned}
a_1&=&\left(\frac{5}{4}e^{-\frac{5}{4}\beta}+\frac{3}{4}e^{\frac{3}{4}\beta}\right)/Z,\nonumber\\
a_2&=&\left(\frac{5}{6}e^{-\frac{5}{4}\beta}+e^{\frac{3}{4}\beta}+\frac{1}{6}e^{\frac{7}{4}\beta}\right)/Z,\nonumber\\
a_3&=&\left(\frac{5}{12}e^{-\frac{5}{4}\beta}+\frac{5}{4}e^{\frac{3}{4}\beta}+\frac{1}{3}e^{\frac{7}{4}\beta}\right)/Z,\nonumber\\
b_1&=&\left(\frac{5\sqrt{2}}{12}e^{-\frac{5}{4}\beta}-\frac{\sqrt{2}}{4}e^{\frac{3}{4}\beta}-\frac{\sqrt{2}}{6}e^{\frac{7}{4}\beta}\right)/Z,\nonumber\\
Z&=&5e^{-\frac{5}{4}\beta}+6e^{\frac{3}{4}\beta}+e^{\frac{7}{4}\beta}.\end{aligned}$$ Substituting the above equations to Eq. (\[N\]) leads to the negativity $${\cal N}(\rho_{12})=\max\Big[0,
