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

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

And the Bell operator associated with the CHSH inequality is given by: $$$$\mathcal{B}_{CHSH}=\mathbf{\hat a}\cdot\sigma\otimes(\mathbf{\hat b}+\mathbf{\hat b}')\cdot\sigma+\mathbf{\hat a}'\cdot\sigma\otimes(\mathbf{\hat b}-\mathbf{\hat b}')\cdot\sigma$$$$ where $$\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: $$$$\mathrm{tr}(\rho\mathcal{B}_{CHSH})\leq2$$$$

and this paper* states that by some calculation

$$$$\mathrm{tr}(\rho\mathcal{B}_{CHSH})=\langle\mathbf{\hat a},T_{\mathscr{l}}(\mathbf{\hat b}+\mathbf{\hat b}')\rangle+\langle\mathbf{\hat a'},T_{\mathscr{l}}(\mathbf{\hat b}-\mathbf{\hat b}')\rangle$$$$ where $$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

• Is the last term missing a prime on a? About your problem did you try to use a formal for the product of sigma matrices? Commented Nov 16, 2021 at 8:00
• @lalala ah yes, thank you for noticing. And What do you mean by a "formal"? I'm still familiarizing myself in the mathematics and all that. Commented Nov 16, 2021 at 8:10
• Auto correction.. I meant formula, like here en.m.wikipedia.org/wiki/Pauli_matrices in the section 'relation to dot and cross product'. Commented Nov 16, 2021 at 8:26
• Thank you very much! That is most useful. Commented Nov 16, 2021 at 8:28

1 Answer

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: $$$$\mathrm{tr}(A\otimes B)=\mathrm{tr}(A)\mathrm{tr}(B)$$$$

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}