How to derive the theorem about CHSH inequality in the $\mathbb{C}^2\otimes\mathbb{C}^2$ space in (Horodecki, 1995)?

Question

In the $\mathbb{C}^2\otimes\mathbb{C}^2$ space, a state can be represented as: \begin{equation} \rho=\frac{1}{4}(\mathbb{I}\otimes\mathbb{I}+\mathbf{r}\cdot\mathbf{\sigma}\otimes\mathbb{I}+\mathbb{I}\otimes\mathbf{s}\cdot\mathbf{\sigma}+\sum^3_{n,m=1}t_{nm}\sigma_n\otimes\sigma_m) \end{equation} where $\mathbf{r}$, $\mathbf{s} \in\mathbb{R}^3$

And the Bell operator associated with the CHSH inequality is given by: \begin{equation} \mathcal{B}_{CHSH}=\mathbf{\hat a}\cdot\sigma\otimes(\mathbf{\hat b}+\mathbf{\hat b}')\cdot\sigma+\mathbf{\hat a}'\cdot\sigma\otimes(\mathbf{\hat b}-\mathbf{\hat b}')\cdot\sigma \end{equation} where $\mathbf{\hat a}$,$\mathbf{\hat a}'$,$\mathbf{\hat b}$,$\mathbf{\hat b}'$are unit vectors in $\mathbb{R}^3$

and the CHSH inequality states that: \begin{equation} \mathrm{tr}(\rho\mathcal{B}_{CHSH})\leq2 \end{equation}

and this paper* states that by some calculation

\begin{equation} \mathrm{tr}(\rho\mathcal{B}_{CHSH})=\langle\mathbf{\hat a},T_{\mathscr{l}}(\mathbf{\hat b}+\mathbf{\hat b}')\rangle+\langle\mathbf{\hat a'},T_{\mathscr{l}}(\mathbf{\hat b}-\mathbf{\hat b}')\rangle \end{equation} where $T_{\mathscr{l}}$ is a $3\times 3$ matrix formed by $t_{nm}$.

I've tried to expand every term in its matrix form, but doing so is simply not feasible. What is the proper way to arrive this result?

*Ryzard Horodecki, Paweł Horodecki, Michał Horodecki, (1995). Violating Bell Inequality by Mixed Spin- 1/2 States: necessary and sufficient condition, Physics Letter A, 200, 340-344

Answer 1

Huh, the calculation is actually quite simple. I am surprised I didn't notice that earlier.

By noticing that the trace of $\mathbf{e}\cdot\sigma$, $\forall \mathbf{e}\in \mathbb{R}^3$ is actually zero , and the following property: \begin{equation} \mathrm{tr}(A\otimes B)=\mathrm{tr}(A)\mathrm{tr}(B) \end{equation}

We can see that most of the terms in $\mathrm{tr}(\rho\mathcal{B}_{CHSH})$ is actually zero except for the last term, which gives:

\begin{align} \mathrm{tr}(\rho\mathcal{B}_{CHSH}) &=\sum^3_{n,m=1}t_{nm}[a_n(b_m+b_m')+a_n'(b_m-b_m')]\\ &=\sum^3_{n=1}a_n(T_{\mathscr{l}}(\mathbf{\hat b}+\mathbf{\hat b}'))_n+a_n'(T_{\mathscr{l}}(\mathbf{\hat b}-\mathbf{\hat b}'))_n\\ &=\langle\mathbf{\hat a},T_{\mathscr{l}}(\mathbf{\hat b}+\mathbf{\hat b}')\rangle+\langle\mathbf{\hat a}',T_{\mathscr{l}}(\mathbf{\hat b}-\mathbf{\hat b}')\rangle \end{align}