Quantum correlation

In quantum mechanics, quantum correlation is the expected value of the product of the alternative outcomes. (Wikipedia).

In Bell’s 1964 paper, assume two outcomes $O_1$ and $O_2$, and each has two outcomes $+1$, $-1$. Then by definition of quantum correlation, first we need “the product of the alternative outcomes”: \begin{equation*} (+1)\times(+1) = +1,\ (+1)\times(-1) = -1,\ (-1)\times(+1) = -1,\ (-1)\times(-1) = +1 \end{equation*}

Then the quantum correlation between $A$ and $B$ is \begin{equation*} \rho_{O_1O_2} = P_{++}-P_{+-}-P_{-+}+P_{–} \end{equation*}

In an experiment, the estimated quantum correlation is \begin{equation*} \hat\rho_{O_1O_2} = \frac{N_{++}-N_{+-}-N_{-+}+N_{–}}{N_{total}} \end{equation*} where $N_{\cdot\cdot}$ is the observed number of specific combination of $O_1$ and $O_2$ outcomes.

Bell’s inequality “in a nutshell”

Assume hidden-variable theory (Wikipedia) holds, assume one particle is split into two (denote as $A$ and $B$) and flying to opposite direction. Measure the spin of $A$ and $B$ on $xyz$ axes, denote as $(A_x, A_y, A_z)$ and $(B_x, B_y, B_z)$. Then since the conservation of spin on a certain axis, we have $A_i+B_i=0$. Futhermore, since we assume the hidden-variable theory, then we have the following 8 combinations of $(A_x, A_y, A_z, B_x, B_y, B_z)$:

	Ax	Ay	Az	Bx	By	Bz
case 1	+	+	+	-	-	-
case 2	+	+	-	-	-	+
case 3	+	-	+	-	+	-
case 4	+	-	-	-	+	+
case 5	-	+	+	+	-	-
case 6	-	+	-	+	-	+
case 7	-	-	+	+	+	-
case 8	-	-	-	+	+	+

Assume the probability of “case $i$” is $P_i$, then by the definitoin of quantum correlation, we have \begin{equation*} \rho_{A_xB_y}=-P_1-P_2+P_3+P_4+P_5+P_6-P_7-P_8 \end{equation*} \begin{equation*} \rho_{A_xB_z}=-P_1+P_2-P_3+P_4+P_5-P_6+P_7-P_8 \end{equation*} \begin{equation*} \rho_{A_yB_z}=-P_1+P_2+P_3-P_4-P_5+P_6+P_7-P_8 \end{equation*}

Notice that \begin{equation*} |\rho_{A_xB_z}-\rho_{A_yB_z}|=2|-P3+P_4+P_5-P_6|=2|(P_4+P_5)-(P_3+P_6)|\leq2(|P_4+P_5|+|P_3+P_6|) \end{equation*} also notice that all $P_i$ are positive, and $\sum_iP_i=1$ then \begin{equation*} 2(|P_4+P_5|+|P_3+P_6|)=2(P_3+P_4+P_5+P_6) \end{equation*} \begin{equation*} =1+(-P_1-P_2+P_3+P_4+P_5+P_6-P_7-P_8)=1+\rho_{A_xB_y} \end{equation*}

In summary, we have \begin{equation*} |\rho_{A_xB_z}-\rho_{A_yB_z}|\leq1+\rho_{A_xB_y} \end{equation*}

Or write it in a more commonly seen way: \begin{equation*} |P_{xz}-P_{zy}|\leq1+P_{xy} \end{equation*}