This section is extracted from the following:

 [1]: S. Ishikawa, gA New Interpretation of Quantum Mechanics,Journal of Quantum Information Science,h Vol. 1 No. 2, 2011, pp. 35-42. doi: 10.4236/jqis.2011.12005 ( download free)

4.5.1: Bell's inequality is violated in both classical and quantum systems

Firstly, let us mention Bell's inequality in mathematics.

Theorem 4.17 [Bell's inequality] Let $(Y,{\cal G}, \mu)$ be a probability space. Consider measurable functions $f_k :Y \to \{-1,1\}$, $(k=1,2,3,4)$, and define the correlations: $\qquad C_{13}=\int_Y f_1(y) \cdot f_3(y) \mu(dy)$, $\qquad C_{14}=\int_Y f_1(y) \cdot f_4(y) \mu(dy)$, $\qquad C_{23}=\int_Y f_2(y) \cdot f_3(y) \mu(dy)$, $\qquad C_{24}=\int_Y f_2(y) \cdot f_4(y) \mu(dy)$ Then, we have Bell's inequality such that \begin{align} |C_{13}-C_{14}|+|C_{23}+C_{24}|{{\; \leqq \;}}2 \tag{4.48} \end{align} Proof. It is easy as follows. \begin{align} & |C_{13}-C_{14}|+|C_{23}+C_{24}| \\ \le & \int_Y f_1(y) \cdot |f_3(y)-f_4(y)| \mu(dy) + \int_Y f_2(y) \cdot |f_3(y)+f_4(y)| \mu(dy)=2 \end{align}
In this section we are not concerned with the relation between "the hidden variables" and Bell's inequality, But we devote ourselves to the Aspects' experiment in the classical systems.
Here, let us prepare three steps (I$\sim$III) as follows.

[Step I]: Consider the basic structure: \begin{align} [{\mathcal A} \subseteq \overline{\mathcal A} \subseteq B(H)] \end{align}

Define the measured value space $X^2=\{-1,1\}^2$ such that $X^2=\{-1,1\}^2 = \{ (1,1), (1,-1),$ $(-1,1),$ $(-1,-1) \}$. Consider two complex numbers $a= {\alpha}_1 +{\alpha}_2{}{\sqrt{-1}}$ and $b= {}{\beta}_1 +{\beta}_2{}{\sqrt{-1}}$ such that $| a |$ $\equiv$ $\sqrt{{}| \alpha_1 |^2 + | {\alpha}_2 |^2{}} =1$ and $| b |$ $\equiv$ $\sqrt{{}| \beta_1 |^2 + | {\beta}_2 |^2{}} =1$. Define the probability space $( X^2, {\cal P}(X^2), \nu_{ab})$ such that

\begin{align} & \nu_{ab} ( \{(1,1)\} ) \! = \nu_{ab}( \{(-1,-1)\} ) \! =(1-\alpha_1 \beta_1 - \alpha_2 \beta_2 )/4 \nonumber \\ & \nu_{ab}( \{(1,-1)\} ) \! = \nu_{ab}( \{(-1,1)\} ) \! =(1+\alpha_1 \beta_1 + \alpha_2 \beta_2 )/4. \tag{4.49} \end{align} The correlation function $P(a,b)$ is calculated as \begin{align} P(a,b) \equiv \!\!\!\!\!\!\!\! \sum_{(x_1, x_2) \in X\times X } \!\!\!\!\!\!\!\! x_1 \cdot x_2 \nu_{ab} ( \{(x_1,x_2)\} ) = -\alpha_1 \beta_1 - \alpha_2 \beta_2 \tag{4.50} \end{align} Our present problem is as follows.
(D):Problem
 ${}$ Find the measurement ${\mathsf M}_{\overline {\mathcal A} } ({\mathsf O}_{ab}:=( X^2,$ ${\cal P}(X^2),$ $F_{ab}) ,$ $S_{[\rho_0]})$ that satisfies \begin{align} \nu_{ab} (\Xi)= \rho_0( F_{ab} (\Xi)) \;\; \quad (\forall \Xi \in {\cal P}(X^2)) \end{align}
This will be answered in the following step [II].
[Step: II]: Consider the problem in the two cases. That is,
 ${}$ $\left\{\begin{array}{ll} \mbox{(i):quantum case: [${\cal A}=B({\mathbb C}^2 \otimes {\mathbb C}^2)$] } \\ \mbox{(ii):classical case: [${\cal A}=C_0(\Omega \times \Omega)$] } \end{array}\right.$
(i): quantum case [${\cal A}=B({\mathbb C}^2) \otimes B({\mathbb C}^2) =B({\mathbb C}^2 \otimes {\mathbb C}^2)$ ] Put \begin{align} e_1= \left[\begin{array}{l} 1 \\ 0 \end{array}\right], \quad e_2= \left[\begin{array}{l} 0 \\ 1 \end{array}\right] \quad (\in {\mathbb C}^2 ). \end{align}

For each $c \in \{a,b\}$, define the observable ${\mathsf O}_c$ $\equiv$ $\bigl(X ,{\cal P}(X) , G_c \bigl)$ in $B({\mathbb C}^2)$ such that

\begin{align} & G_{c}(\{1\}) = \frac{1}{2} \left[\begin{array}{ll} 1 & {\bar c}\; \\ c & 1 \end{array}\right], \quad G_{c}(\{-1\}) = \frac{1}{2} \left[\begin{array}{ll} 1 & -{\bar c} \;\\ - c & 1 \end{array}\right]. \end{align}

Consider the two particles quantum system in ${ B }({\mathbb C}^2 \otimes {\mathbb C}^2)$. Consider two states $\rho_s$ $= | {\psi_s} \rangle \langle {\psi_s} |$ and $\rho_0$ $= | {\psi_0} \rangle \langle {\psi_0} |$ $\bigl(\in {\frak S}^p( { B }({\mathbb C}^2 \otimes {\mathbb C}^2)^*) \bigl)$. Here, put $\psi_s=(e_1\otimes e_2 -e_2\otimes e_1)/{\sqrt{2}}$ and $\psi_0=e_1\otimes e_1$. Consider the unitary operator $U$ $( \in B({\mathbb C}^2 \otimes {\mathbb C}^2)$ such that $U\psi_0 =\psi_s$. Consider an observable ${\mathsf O}_{ab}$ $=$ $(X^2 , {\cal P}(X^2) , F_{ab}:= U^* (G_a \otimes G_b)U )$ in $B({\mathbb C}^2 \otimes {\mathbb C}^2)$, and get the measurement ${\mathsf M}_{B({\mathbb C}^2 \otimes {\mathbb C}^2)} ({\mathsf O}_{ab} ,S_{[{}\rho_0{}]})$. This clearly satisfies (D). That is because we see that, for each $(x_1,x_2) \in X^2$,

\begin{align} & \rho_0 (F_{ab} (\{(x_1 , x_2) \}) ) = \langle \psi_0, F_{ab} (\{(x_1 , x_2) \}) \psi_0 \rangle \\ = & \langle \psi_s, (G_a (\{{}x_1 \}) \otimes {}G_b (\{{}x_2 \}) ) \psi_s \rangle = \nu_{ab} (\{(x_1,x_2)\}). \end{align}

(ii): classical case: [${\cal A}=C_0(\Omega) \otimes C_0(\Omega)=C_0(\Omega \times \Omega)$] Put $\omega_0 (=(\omega_0', \omega''_0)) \in \Omega \times \Omega$, and $\rho_0 = \delta_{\omega_0}$ ($\in {\frak S}^p ({C_0(\Omega \times \Omega)}^*)$). Define the observable ${\mathsf O}_{ab}:=( X^2, {\cal P}(X^2), F_{ab})$ in $L^\infty (\Omega \times \Omega)$ such that

\begin{align} [ F_{ab}( \{(x_1,x_2)\} )](\omega_0 ) = \nu_{ab} ( \{(x_1,x_2)\} ) \end{align} Therefore, we get the measurement ${\mathsf M}_{L^\infty (\Omega\times \Omega)} ({\mathsf O}_{ab} ,S_{[{}\delta_{\omega_0}{}]})$, which clearly satisfies (D).
[Step III]:

For each $k=1,2$, consider two complex numbers $a^k (=\alpha_1^k+\alpha_2^k{\sqrt{-1}} )$ and $b^k(=\beta_1^k+\beta_2^k{\sqrt{-1}})$ such that $| a^k |=| b^k |=1$. Consider the tensor parallel measurement $\otimes_{i,j=1,2}$ ${\mathsf M}_{ {\cal A} } ({\mathsf O}_{a^ib^j}:=( X^2, {\cal P}(X^2), F_{a^ib^j}) ,$ $S_{[\rho_0]})$ in the tensor $W^*$-algebra $\otimes_{{}_{i,j=1,2}} \overline{\mathcal A}$. Assume the measured value $x (\in X^8)$. That is,

\begin{align} x & = \big( (x_{1}^{11} , x_{2}^{11}), (x_{1}^{12} , x_{2}^{12}), (x_{1}^{21} , x_{2}^{21}), (x_{1}^{22} , x_{2}^{22}) \big) \\ & \in \times_{i,j=1,2} X^2 \end{align} Here, we see, by (4.50), that, for any $i,j=1,2$, \begin{align} P({a^{i},b^{j}}) & = \!\!\! \sum_{(x^{ij}_1, x^{ij}_2) \in X\times X } \!\!\! x^{ij}_1 \cdot x^{ij}_2 \rho_0 (F_{a^ib^j}( \{(x^{ij}_1,x^{ij}_2)\} )) \\ & = -\alpha^{i}_1 \beta^{j}_1 - \alpha^{i}_2 \beta^{j}_2 \end{align} Putting \begin{align} a^1 ={\sqrt{-1}}, \; b^1 = \frac{1+{{\sqrt{-1}}}}{ \sqrt{2} }, \; a^2 = 1, \; b^2 = {}\frac{1-{{\sqrt{-1}}}}{ \sqrt{2} }, \end{align} we get the following equality: \begin{align} |P(a^1 , b^1) - P(a^1 , b^2)| \; + \; |P(a^2 , b^1) + P(a^2 , b^2 )| = 2 \sqrt{2} \tag{4.51} \end{align} Thus, in both cases ( i.e., quantum case [${\cal A}=B({\mathbb C}^2 \otimes {\mathbb C}^2)$] and classical case [${\cal A}=C_0(\Omega\times \Omega)$]), the formula (4.51) holds.

This fact is often said that
Bell's inequality is violated
though we do not mention the reason to compare the equality (4.51) and Bell's inequality (4.48) in this section.

4.5.2: Stop being bothered

Remark 4.18 [Shut up and calculate]. The above argument may suggest that there is something faster than light. However, when faster-than-light appears, our standing point is
Stop being bothered
This is not only our opinion but also most physicists'. In fact, in Mermin's book, he said
 $(a):$ "Most physicists, I think it is fair to say, are not bothered." (b): If I were forced to sum up in one sentence what the Copenhagen interpretation says to me, it would be "Shut up and calculate"
If it is so, we want to assert that the linguistic interpretation ($\S$3.1) is the true colors of "the Copenhagen interpretation". That is because I also say
 $(c):$ If I were forced to sum up in one sentence what the linguistic interpretation says to me, it would be "Shut up and calculate"