To have a matrix representation of the Hamiltonian, we choose the following basis $$\Big\{ |-\frac{1}{2},-1\rangle,  |\frac{1}{2}, 0\rangle,
|-\frac{1}{2}, 1\rangle, |\frac{1}{2} , -1\rangle, |-\frac{1}{2},
0\rangle, |\frac{1}{2} ,1\rangle\Big\},$$ where $|m,M\rangle$ is the eigenstate of ${s}_z$ and $S_z$ with the corresponding eigenvalues given by $m$ and $M$, respectively.

![Schematic representation of the mixed spin Hamiltonian for the number of spins from 2 to 5.](mixfig1.eps){width="35.00000%"}

In the above basis, the Hamiltonian can be written as a block-diagonal form with the dimension of each block being at most $2\times 2$. Thus, the density matrix $\rho_{12}$ for the thermal state is obtained as $$\rho_{12}=\left(
\begin{array}{llllll}
a_1&0 & 0 &0 & 0 & 0\\
0 &a_2&b_1&0 & 0 & 0 \\
0& b_1&a_3&0 & 0 & 0 \\
0& 0  &0  &a_4&b_2&0\\
0& 0 & 0 & b_2&a_5&0\\
0& 0 & 0 & 0 & 0 & a_6
\end{array}
\right) , \label{rho}$$ with the partition function and the matrix elements given by $$\begin{aligned}
Z&=&2 e^\beta+4e^{-\frac{1}{2}\beta},\label{para1}\\
a_1&=&a_6=e^{-\frac{1}{2}\beta}/Z,\label{para22}\\
a_2&=&a_5=\frac{1}{6},\label{para222}\\
a_3&=&a_4=\frac{1}3\left(2e^\beta+e^{-\frac{1}{2}\beta}\right)/Z,\\
b_1&=&b_2=\frac{\sqrt{2}}{3}\left(e^{-\frac{1}{2}\beta}-e^\beta\right)/Z,\nonumber\\
&=&\sqrt{2}(a_1-a_2). \label{para2}\end{aligned}$$ After the partial transpose with respect to the first spin-half subsystem, we can get $\rho_{12}^{T_1}$ $$\rho_{12}^{T_1}=\left(
\begin{array}{llllll}
a_1&b_2&0  &0  &0  &0\\
b_2&a_2&0  &0  &0  &0\\
0  &0  &a_3&0  &0  &0\\
0  &0  &0  &a_4&0  &0\\
0  