8.5: Combined observable ---Only one measurement is permitted

The linguistic interpretation says that \begin{align} & \mbox{"Only one measurement is permitted"} \\ & \Rightarrow \mbox{ "only one observable"} \Rightarrow \mbox{ "the necessity of the combined observable"} \end{align} Thus, we prepare the following theorem.

Theorem8.9 [The existence theorem of classical combined observable] Consider the classical basic structure \begin{align} [C_0(\Omega ) \subseteq L^\infty ( \Omega, \nu ) \subseteq B(L^2 ( \Omega, \nu ))] \end{align} And consider observables ${\mathsf O}_{12}{{=}} (X_1 \times X_2 , {\mathcal F}_1 \boxtimes {\mathcal F}_2, F_{12})$ and ${\mathsf O}_{23}{{=}} $ $(X_2 \times X_3 ,$ $ {\mathcal F}_2 \boxtimes {\mathcal F}_3, F_{23})$ in $L^\infty ( \Omega, \nu )$. Here, for simplicity, assume that $X_i{{=}} \{x^1_i, x^2_i,\ldots, x^{n_i}_i\}$ $(i=1,2,3)$ is finite, Also, assume that ${\mathcal F}_i = 2^{X_i}$. Furthermore assume that \begin{align} {\mathsf O}_{12}^{(2)} = {\mathsf O}_{23}^{(2)} \quad (\mbox{That is, }\;\; F_{12}(X_1 \times \Xi_2 ) = F_{23}(\Xi_2 \times X_3 ) \quad(\forall \Xi_2 \in 2^{X_2})) \end{align} Then, we have the observable ${\mathsf O}_{123}{{=}} (X_1 \times X_2 \times X_3, {\mathcal F}_1 \times {\mathcal F}_2 \times {\mathcal F}_3, F_{123})$ in $L^\infty (\Omega)$ such that \begin{align} {\mathsf O}_{123}^{(12)} = {\mathsf O}_{12}, \quad {\mathsf O}_{123}^{(23)} = {\mathsf O}_{23} \end{align} That is, \begin{align} F_{123}^{(12)}(\Xi_1 \times \Xi_2 \times X_3) & = F_{12}(\Xi_1 \times \Xi_2 ),\;\; F_{123}^{(23)}(X_1 \times \Xi_2 \times \Xi_3 ) = F_{23}(\Xi_2 \times \Xi_3 ) \tag{8.8} \\ & \quad(\forall \Xi_1 \in {\mathcal F}_1, \forall \Xi_2 \in {\mathcal F}_2, \forall \Xi_3 \in {\mathcal F}_3 )) \nonumber \end{align} The ${\mathsf O}_{123}$ is called the {combined observable of ${\mathsf O}_{12}$ and ${\mathsf O}_{23}$.

Proof. ${\mathsf O}_{123}$ $=$ $(X_1 \times X_2 \times X_{3} , $ ${ {\mathcal F}_1 \times {\mathcal F}_{2} \times {\mathcal F}_3}, $ $F_{123})$ is, for example, defined by

\begin{align} & \; \; [{}F_{123} (\{(x_{1} , x_{2} , x_{3}^{}) \}) ] (\omega) \\ & = \left\{\begin{array}{ll} {\displaystyle \frac{ [{}F_{12} (\{(x_{1} , x_{2}) \}){}] (\omega) \cdot [{}F_{23} (\{(x_{2} , x_{3}) \}){}] (\omega) } { [{}F_{12} (X_1 \times \{ x_{2} \}){}] (\omega) } } \\ & \hspace{-2cm} ( [{}F_{12} (X_1 \times \{ x_{2} \}){}] (\omega) \; \not= 0\mbox{ and }) \\ \\ 0 \\ & \hspace{-2.0cm} ( [{}F_{12} (X_1 \times \{ x_{2} \}){}] (\omega) = 0\mbox{ and }) \\ \end{array}\right. \\ & \qquad \qquad \qquad (\forall \omega \in \Omega, \forall (x_1,x_2,x_3) \in X_1 \times X_2 \times X_3) \end{align} This clearly satisfies (8.8). \qed

Counter example8.10 [Counter example in quantum systems] Theorem 8.9Theorem does not hold in the quantum basic structure

\begin{align} [{\mathcal C}(H) \subseteq B(H) \subseteq B(H)] \end{align}

For example, put $H={\mathbb C}^n$, and consider the three Hermitian $(n \times n)$-matrices $T_1$, $T_2$, $T_3$ in $B(H)$ such that

\begin{align} T_1 T_2 =T_2 T_1, \quad T_2 T_3 =T_3 T_2, \quad T_1 T_3 \not= T_3 T_1 \tag{8.9} \end{align}

For each $k=1,2,3$, define the spectrum decomposition ${\mathsf O}_k=(X_k, {\mathcal F}_k, F_k )$ in $H$ (which is regarded as a projective observable) such that

\begin{align} T_k = \int_{X_k} x_k F_k ( d x_k ) \tag{8.10} \end{align}

where $X_k ={\mathbb R}, {\mathcal F}_k={\mathcal B}_{\mathbb R}$. From the commutativity, we have the simultaneous observables

\begin{align} & {\mathsf O}_{12}{{=}} {\mathsf O}_1 \times {\mathsf O}_2 =(X_1 \times X_2 , {\mathcal F}_1 \boxtimes {\mathcal F}_2, F_{12}=F_1 \times F_2) \nonumber \end{align} \begin{align} & {\mathsf O}_{23}{{=}} {\mathsf O}_2 \times {\mathsf O}_3= (X_2 \times X_3 , {\mathcal F}_2 \boxtimes {\mathcal F}_3, F_{23}=F_2 \times F_3) \end{align}

It is clear that \begin{align} {\mathsf O}_{12}^{(2)} = {\mathsf O}_{23}^{(2)} \quad (\mbox{that is,}\;\; F_{12}(X_1 \times \Xi_2 ) =F_2(\Xi_2 ) = F_{23}(\Xi_2 \times X_3 ) \quad(\forall \Xi_2 \in {\mathcal F}_2)) \end{align}

However, it should be noted that there does not exist the observable ${\mathsf O}_{123}{{=}} (X_1 \times X_2 \times X_3, {\mathcal F}_1 \boxtimes {\mathcal F}_2 \boxtimes {\mathcal F}_3, F_{123})$ in $B(H)$ such that

\begin{align} {\mathsf O}_{123}^{(12)} = {\mathsf O}_{12}, \quad {\mathsf O}_{123}^{(23)} = {\mathsf O}_{23} \end{align}

That is because, if ${\mathsf O}_{123}$ exists, Theorem 8.3Theorem says that ${\mathsf O}_1$ and ${\mathsf O}_3$ commute, and it is in contradiction with the (8.9). Therefore, the combined observable ${\mathsf O}_{123}$ of ${\mathsf O}_{12}$ and ${\mathsf O}_{23}$ does not exist.

8.5.2: Combined observable and Bell's inequality

Now we consider the following problem:

Problem 8.11 [combined observable and Bell's inequality] Consider the basic structure \begin{align} [ {\mathcal A} \subseteq \overline{\mathcal A} \subseteq B(H)] \end{align} Put $X_1=X_2=X_3=X_4=\{-1, 1\}$. Let ${\mathsf O}_{13}{{=}} (X_1 \times X_3 , 2^{X_1} \times 2^{X_3} , F_{13})$, ${\mathsf O}_{14}{{=}} (X_1 \times X_4 , 2^{X_1} \times 2^{X_4} , F_{14})$, ${\mathsf O}_{23}{{=}} $ $(X_2 \times X_3 ,$ $ 2^{X_2} \times 2^{X_3} , F_{23})$ and ${\mathsf O}_{24}{{=}} $ $(X_2 \times X_3 ,$ $ 2^{X_2} \times 2^{X_4} , F_{24})$ be observables in $L^\infty (\Omega)$ such that \begin{align} {\mathsf O}_{13}^{(1)}={\mathsf O}_{14}^{(1)}, \;\; {\mathsf O}_{23}^{(2)}={\mathsf O}_{24}^{(2)}, \;\; {\mathsf O}_{13}^{(3)}={\mathsf O}_{23}^{(3)}, \;\; {\mathsf O}_{14}^{(4)}={\mathsf O}_{24}^{(4)} \end{align}

Define the probability measure $\nu_{ab}$ on $\{-1, 1 \}^2$ by the formula (4.49). Assume that there exists a state $\rho_0 \in {\frak S}^p({\mathcal A}^*) $ such that

\begin{align} & {}_{{\mathcal A}^*} \big(\rho_0, F_{13}(\{(x_1,x_3)\}) \big) {}_{\overline{\mathcal A} } =\nu_{a^1b^1}(\{(x_1,x_3)\}, \\ & {}_{{\mathcal A}^*} \big(\rho_0, F_{14}(\{(x_1,x_4)\}) \big) {}_{\overline{\mathcal A} } =\nu_{a^1b^2}(\{(x_1,x_4)\} \\ & {}_{{\mathcal A}^*} \big(\rho_0, F_{23}(\{(x_2,x_3)\}) \big) {}_{\overline{\mathcal A} } =\nu_{a^2b^1}(\{(x_2,x_3)\}, \\ & {}_{{\mathcal A}^*} \big(\rho_0, F_{24}(\{(x_2,x_4)\}) \big) {}_{\overline{\mathcal A} } =\nu_{a^2b^2}(\{(x_2,x_4)\} \end{align}
Now we have the following problem:
$(a):$ Does the observable ${\mathsf O}_{1234}{{=}} (\times_{k=1}^4 X_k , \times_{k=1}^4{\mathcal F}_k , F_{1234})$ in $\overline{\mathcal A}$ satisfying the following ($\sharp$) exist?
$(\sharp):$ $ \displaystyle {\mathsf O}_{1234}^{(13)} = {\mathsf O}_{13}, \;\; {\mathsf O}_{1234}^{(14)} = {\mathsf O}_{14}, \;\; {\mathsf O}_{1234}^{(23)} = {\mathsf O}_{23}, \;\; {\mathsf O}_{1234}^{(24)} = {\mathsf O}_{24} \;\; $
In what follows, we show that the above observable ${\mathsf O}_{1234}$ does not exist.


Assume that the observable ${\mathsf O}_{1234}{{=}} (\times_{k=1}^4 X_k , $ $ \times_{k=1}^4{\mathcal F}_k ,$ $ F_{1234})$ exists. Then, it suffices to show the contradiction. Define $C_{13}(\rho_0)$, $C_{14}(\rho_0)$, $C_{23}(\rho_0)$ and $C_{24}(\rho_0)$ such that

\begin{align} \left\{\begin{array}{ll} \displaystyle C_{13}(\rho_0) = \int_{\times_{k=1}^4 X_k} x_1 \cdot x_3 \; {}_{{\mathcal A}^*} \big(\rho_0, F_{1234}( \times_{k=1}^4 dx_k ) \big) {}_{\overline{\mathcal A} } \\ \hspace{1.1cm} \bigl( = \int_{X_1 \times X_3} x_1 \cdot x_3 \; \nu_{a^1 b^1} ( dx_1 dx_3) \bigl) \\ \displaystyle C_{14}(\rho_0) = \int_{\times_{k=1}^4 X_k} x_1 \cdot x_4 \; {}_{{\mathcal A}^*} \big(\rho_0, F_{1234}( \times_{k=1}^4 dx_k ) \big) {}_{\overline{\mathcal A} } \\ \hspace{1.1cm} \bigl(= \int_{X_1 \times X_4} x_1 \cdot x_4 \; \nu_{a^1b^2}( dx_1dx_4) \bigl) \\ \displaystyle C_{23}(\rho_0) = \int_{\times_{k=1}^4 X_k} x_2 \cdot x_3 \; {}_{{\mathcal A}^*} \big(\rho_0, F_{1234}( \times_{k=1}^4 dx_k ) \big) {}_{\overline{\mathcal A} } \\ \hspace{1.1cm} \bigl(= \int_{X_2 \times X_3} x_2 \cdot x_3 \; \nu_{a^2b^1}( dx_2dx_3) \bigl) \\ \displaystyle C_{24}(\rho_0) = \int_{\times_{k=1}^4 X_k} x_2 \cdot x_4 \; {}_{{\mathcal A}^*} \big(\rho_0, F_{1234}( \times_{k=1}^4 dx_k ) \big) {}_{\overline{\mathcal A} } \\ \hspace{1.1cm} \bigl(= \int_{X_2 \times X_4} x_2 \cdot x_4 \; \nu_{a^2b^2}( dx_2dx_4) \bigl) \end{array}\right. \end{align}

Then, we can easily get the following Bell's inequality: (cf. Bell's inequality (Thorem 4.17)).

\begin{align} & |C_{13}(\rho_0)-C_{14}(\rho_0)| + |C_{23}(\rho_0)+C_{24}(\rho_0)| \nonumber \\ {{\; \leqq \;}} & \int_{\times_{k=1}^4 X_k} \!\!\! | x_1 | \cdot |x_3-x_4| \;\;+ \!\! |x_2 | \cdot |x_3+x_4| \big[F_{1234}(\times_{k=1}^4 dx_k )\big](\rho_0) \nonumber \\ {{\; \leqq \;}} & 2 \qquad (\mbox{since } x_k \in \{-1, 1\} ) \tag{8.11} \end{align}

However, the formula (4.51) says that this (8.11) must be $2{\sqrt 2}$. Thus, by contradiction, we says that ${\mathsf O}_{1234}$ satisfying (a) does not exist. Thus we can not take a measurement ${\mathsf M}_{\overline{\mathcal A}} ( {\mathsf O}_{1234}, S_{[\rho_0]} )$. However, it should be noted that

$(b):$ instead of ${\mathsf M}_{\overline{\mathcal A}} ( {\mathsf O}_{1234}, S_{[\rho_0]} )$. we can take a parallel measurement ${\mathsf M}_{\otimes_{k=1}^4 \overline{\mathcal A}} ( {\mathsf O}_{13} \otimes {\mathsf O}_{14} \otimes {\mathsf O}_{23} \otimes {\mathsf O}_{24} ,$ $ S_{[\otimes_{k=1}^4 \rho_0]})$. In this case, we easily see that (8.11) = $2{\sqrt 2}$ as the formula (4.51}).
That is,

$(c):$ in the case of a parallel measurement, Bell's inequality is broken in both quantum and classical systems.

$\fbox{Note 8.2}$ In the above argument, Bell's inequality is used in the framework of measurement theory. This is of course true. Although the above argument is not related to the problem of "hidden variable", it is interesting as how to use Bel's inequality.