3.3.3: Only one state and parallel measurement

For example, consider the following situation:
$(a):$ There are two cups $A_1$ and $A_2$ in which water is filled. Assume that the temperature of the water in the cup $A_k$ $(k=1,2)$ is $\omega_k$ °C $(0 {{\; \leqq \;}}\omega_k {{\; \leqq \;}}100)$. Consider two questions "Is the water in the cup $A_1$ cold or hot?" and "How many degrees(°C) is roughly the water in the cup $A_2$?". This implies that we take two measurements such that
 $\quad$ $\left\{\begin{array}{ll} \mbox{$(\sharp_1)$:${\mathsf M}_{L^\infty ( \Omega )} ( {\mathsf O}_{{{{{c}}}}{{{{h}}}}} {{=}} (\{{{{{c}}}},{{{{h}}}}\} , 2^{\{{{{{c}}}},{{{{h}}}}\}}, F_{{{{{c}}}}{{{{h}}}}} ), S_{[\omega_1]} )$in Example 2.31} \\ \\ \mbox{$(\sharp_2)$:${\mathsf M}_{L^\infty ( \Omega )} ({\mathsf O}^{\triangle}  {{=}} ({\mathbb N}_{10}^{100} , 2^{{\mathbb N}_{10}^{100} }, G^{\triangle} ), S_{[\omega_2]} ) $in Example2.32 } \end{array}\right.$
However, as mentioned in the above, \begin{align} \mbox{ "only one state" must be demanded. } \end{align}
Thus, we have the following problem.
Problem 3.21 Represent two measurements ${\mathsf M}_{L^\infty ( \Omega )} ( {\mathsf O}_{{{{{c}}}}{{{{h}}}}} {{=}} (\{{{{{c}}}},{{{{h}}}}\} ,$ $2^{\{{{{{c}}}},{{{{h}}}}\}}, F_{{{{{c}}}}{{{{h}}}}} ), S_{[\omega_1]} )$ and ${\mathsf M}_{L^\infty ( \Omega )} ({\mathsf O}^{\triangle}$ ${{=}} ({\mathbb N}_{10}^{100} ,$ $2^{{\mathbb N}_{10}^{100} }, G^{\triangle} ),$ $S_{[\omega_2]} )$ by only one measurement.

This will be answered in what follows.

Definition 3.22 [ Parallel observable] For each $k = 1,2,\ldots,n$, consider a basic structure $[{\mathcal A}_k \subseteq \overline{\mathcal A}_k \subseteq B(H_k)]$, and an observable ${\mathsf O}_k$ $=$ $(X_k , {\cal F}_k , F_k{})$ in $\overline{\mathcal A}_k$. Define the observable $\widetilde{\mathsf O}= ( {\times_{k=1}^n X_k }, \boxtimes_{k=1}^n {\cal F}_k , \widetilde{F}{})$ in $\bigotimes_{k=1}^n \overline{\mathcal A}_k$ such that

\begin{align} & {\widetilde F}(\Xi_1 \times \Xi_2 \times \cdots \times \Xi_n{}) = F_1 (\Xi_1{}) \otimes F_2 (\Xi_2{}) \otimes \cdots \otimes F_n (\Xi_n{}) \tag{3.15} \\ & \qquad \;\; \forall \Xi_k \in {\cal F}_k \; (k=1,2,\ldots,n ) \nonumber \end{align}

Then, the observable $\widetilde{\mathsf O}$ $=$ $(\times_{k=1}^n X_k , \boxtimes_{k=1}^n{\cal F}_k , \widetilde{F}{})$ is called the parallel observable in $\bigotimes_{k=1}^n \overline{\mathcal A}_k$, and denoted by ${\widetilde F}=\bigotimes_{k=1}^n F_k$, $\widetilde{\mathsf O}$ $=$ $\bigotimes_{k=1}^n {\mathsf O}_k$. the measurement of the parallel observable ${\widetilde{\mathsf O}}=$　$\bigotimes_{k=1}^n {\mathsf O}_k$,　that is,　the measurement　${\mathsf M}_{ \bigotimes_{k=1}^n \overline{\mathcal A}_k}$　$(\widetilde{\mathsf O},$　$S_{ [\bigotimes_{k=1}^n \rho_k] })$　is called a　parallel measurement　and denoted by　${\mathsf M}_{ \bigotimes_{k=1}^n \overline{\mathcal A}_k} ( \bigotimes_{k=1}^n {\mathsf O}_k,$　$S_{[ \bigotimes_{k=1}^n \rho_k]})$　or　$\bigotimes_{k=1}^n {\mathsf M}_{\overline{\mathcal A}_k} ({\mathsf O}_k,$　$S_{[ \rho_k]})$.

The meaning of　the parallel measurement　is as follows.

Our present purpose is

 ${}$ to　take both measurements　${\mathsf M}_{\overline{\mathcal A}_1 }({\mathsf O}_1, S_{[\rho_1]})$　and　 ${\mathsf M}_{\overline{\mathcal A}_2 }({\mathsf O}_2, S_{[\rho_2]})$
Then. image the following:
 $(b):$ $\left\{\begin{array}{ll} \overset{{}}{\underset{ \rho_1 (\in {\frak S}^p({\mathcal A}_1^*) )} {{ \fbox{state}}}} \xrightarrow[]{\qquad \qquad} \overset{{}} { \underset{ {\mathsf O}_1} {{ \fbox{observable}}} } \xrightarrow[{\mathsf M}_{\overline{\mathcal A}_1 }({\mathsf O}_1, S_{[{\rho_1}]})]{\qquad \qquad} \overset{{}} { \underset{ x_1 (\in X_1 )} {{ \fbox{measured value}}} } \\ \\ \overset{{}}{\underset{ \rho_2 (\in {\frak S}^p({\mathcal A}_2^*))} {{ \fbox{state}}}} \xrightarrow[]{\qquad \qquad} \overset{{}} { \underset{ {\mathsf O}_2} {{ \fbox{observable}}} } \xrightarrow[{\mathsf M}_{\overline{\mathcal A}_2}({\mathsf O}_2, S_{[{\rho_2}]})]{\qquad \qquad} \overset{{}} { \underset{ x_2 (\in X_2 )} {{ \fbox{measured value}}} } \end{array}\right.$
However,　according to the linguistic interpretation ($\S$3.1),　two measurements can not be taken. 　Hence,
The (b) is impossible
Thus, two states $\rho_1$ and $\rho_1$ are regarded as one state $\rho_1 \otimes \rho_2$, and further, combining two observables ${\mathsf O}_1$ and ${\mathsf O}_2$, we construct the parallel observable ${\mathsf O}_1\otimes {\mathsf O}_2$, and take the parallel measurement ${\mathsf M}_{ \overline{\mathcal A}_1 \otimes \overline{\mathcal A}_2 }({\mathsf O}_1 \otimes {\mathsf O}_2 , S_{[ \rho_1 \otimes \rho_2 ]})$ in what follows.
 $(c):$ $\overset{{}}{\underset{ \rho_1 \otimes \rho_2 (\in {\frak S}^p({\mathcal A}_1^*) \otimes {\frak S}^p({\mathcal A}_2^*) ) } {{ \fbox{state}}}} \xrightarrow[]{} { \underset{ {\mathsf O}_1 \otimes {\mathsf O}_2} {{ \fbox{parallel observable}}} } \xrightarrow[ {\mathsf M}_{ \overline{\mathcal A}_1 \otimes \overline{\mathcal A}_2 }({\mathsf O}_1 \otimes {\mathsf O}_2 , S_{[ \rho_1 \otimes \rho_2 ]})] {\qquad \qquad} { \underset{ (x_1,x_2) (\in X_1 \times X_2 )} {{ \fbox{measured value}}} }$
The (c) is always possible

Example 3.23 [The answer to Problem 3.21] Put $\Omega_1 = \Omega_2 = [0,100]$, and define the state space $\Omega_1 \times \Omega_2$. And consider two observables, that is, the [C-H]-observable ${\mathsf O}_{{{{{c}}}}{{{{h}}}}}= (X {{=}} \{ {{{{c}}}} , {{{{h}}}} \}, 2^X, F_{{{{{c}}}}{{{{h}}}}} )$ in $L^\infty (\Omega_1)$ (in Example 2.31) and triangle-observable ${\mathsf O}^{\triangle}= (Y( {{=}} {\mathbb N}_{10}^{100}) , 2^Y, G^{\triangle} )$ in $L^\infty(\Omega_2)$ (in Example 2.32). Thus, we get the parallel observable ${\mathsf O}_{{{{{c}}}}{{{{h}}}}} \otimes {\mathsf O}^{\triangle}$ $=$ $(\{ {{{{c}}}} , {{{{h}}}} \}\times {\mathbb N}_{10}^{100}, 2^{\{ {{{{c}}}} , {{{{h}}}} \}\times {\mathbb N}_{10}^{100}}, F_{{{{{c}}}}{{{{h}}}}} \otimes G^{\triangle} )$ in $L^\infty (\Omega_1 \times \Omega_2 )$, take the parallel measurement ${\mathsf M}_{L^\infty (\Omega_1 \times \Omega_2 )}({\mathsf O}_{{{{{c}}}}{{{{h}}}}} \otimes {\mathsf O}^{\triangle}, S_{[(\omega_1,\omega_2)]})$. Here, note that

\begin{align} \delta_{\omega_1} \otimes \delta_{\omega_2} = \delta_{( \omega_1, \omega_2)} \approx {( \omega_1, \omega_2)}. \end{align} For example, putting $(\omega_1,\omega_2 ) =(25, 55)$, we see the following.
 $(d):$ When the parallel measurement ${\mathsf M}_{L^\infty (\Omega_1 \times \Omega_2 )}({\mathsf O}_{{{{{c}}}}{{{{h}}}}} \otimes {\mathsf O}^{\triangle}, S_{[(25,55)]})$ is taken, the probability \begin{align} & \mbox{that } \mbox{the measured value } \left[\begin{array}{ll} (\mbox{c}, \mbox{about 50°C}) \\ (\mbox{c}, \mbox{about 60°C}) \\ (\mbox{h}, \mbox{about 50°C}) \\ (\mbox{h}, \mbox{about 60°C}) \end{array}\right] \mbox{is obtained is given by} \left[\begin{array}{ll} 0.375 \\ 0.375 \\ 0.125 \\ 0.125 \end{array}\right] \end{align}
That is because \begin{align} & [(F_{{{{{c}}}}{{{{h}}}}} \otimes G^{\triangle})( \{ (\mbox{c}, \mbox{about 50°C} ) \} )] (25,55) \\ = & [F_{{{{{c}}}}{{{{h}}}}}(\{ \mbox{c} \} )](25) \cdot [G^{\triangle}(\{ \mbox{about 50°C}\})](55) =0.75 \cdot 0.5=0.375 \end{align} Thus, similarly, \begin{align} & [(F_{{{{{c}}}}{{{{h}}}}} \otimes G^{\triangle})( \{ (\mbox{c}, \mbox{about 60°C} ) \} )] (25,55) =0.75 \cdot 0.5=0.375 \\ & [(F_{{{{{c}}}}{{{{h}}}}} \otimes G^{\triangle})( \{ (\mbox{h}, \mbox{about 50°C} ) \} )] (25,55) =0.25 \cdot 0.5=0.125 \\ & [(F_{{{{{c}}}}{{{{h}}}}} \otimes G^{\triangle})( \{ (\mbox{h}, \mbox{about 60°C} ) \} )] (25,55) =0.25 \cdot 0.5=0.125 \end{align}

Remark 3.24 Also, for example, putting $(\omega_1,\omega_2 ) =(55, 55)$, we see:

 $(e):$ the probability that a measured value $\left[\begin{array}{ll} (\mbox{c}, \mbox{about 50°C}) \\ (\mbox{c}, \mbox{about 60°C}) \\ (\mbox{h}, \mbox{about 50°C}) \\ (\mbox{h}, \mbox{about 60°C}) \end{array}\right]$ is obtained by parallel measurement ${\mathsf M}_{L^\infty (\Omega_1 \times \Omega_2 )}({\mathsf O}_{{{{{c}}}}{{{{h}}}}} \otimes {\mathsf O}^{\triangle}, S_{[(55,55)]})$ is given by $\left[\begin{array}{ll} 0.125 \\ 0.125 \\ 0.375 \\ 0.375 \end{array}\right]$
That is because, we similarly, see \begin{align} \left\{\begin{array}{ll} & [F_{{{{{c}}}}{{{{h}}}}}(\{ \mbox{c} \} )](55) \cdot [G^{\triangle}(\{ \mbox{about 50°C}\})](55) =0.25 \cdot 0.5=0.125 \\ & [F_{{{{{c}}}}{{{{h}}}}}(\{ \mbox{c} \} )](55) \cdot [G^{\triangle}(\{ \mbox{about 60°C}\})](55) =0.25 \cdot 0.5=0.125 \\ & [F_{{{{{c}}}}{{{{h}}}}}(\{ \mbox{h} \} )](55) \cdot [G^{\triangle}(\{ \mbox{about 50°C}\})](55) =0.75 \cdot 0.5=0.375 \\ & [F_{{{{{c}}}}{{{{h}}}}}(\{ \mbox{h} \} )](55) \cdot [G^{\triangle}(\{ \mbox{about 60°C}\})](55) =0.75 \cdot 0.5=0.375 \end{array}\right. \\ & \tag{3.16} \end{align} Note that this is the same as Answer 3.13 (cf. Note 3.5 later).

The following theorem is clear. But, the assertion is significant.

Theorem 3.25 [Ergodic property] For each $k=1,2, \cdots, n$, consider a measurement ${\mathsf M}_{L^\infty ( \Omega)} ( {\mathsf O}_k(:=(X_k,{\mathcal F}_k, F_k)), S_{[\delta_\omega]})$ with the sample probability space $(X_k,{\mathcal F}_k, P_k^{\omega})$. Then, the sample probability spaces of the simultaneous measurement ${\mathsf M}_{L^\infty ( \Omega)} (\times_{k=1}^n {\mathsf O}_k, S_{[\delta_\omega]})$ and the parallel measurement ${\mathsf M}_{L^\infty ( \Omega^n)}$ $( \bigotimes_{k=1}^n {\mathsf O}_k ,$ $S_{[ \otimes_{k=1}^n \delta_{\omega}]})$ are the same, that is, these are the same as the product probability space

\begin{align} (\times_{k=1}^n X_k,\boxtimes_{k=1}^n{\mathcal F}_k,\bigotimes_{k=1}^n P_k^{\omega}) \tag{3.17} \end{align} Proof. It is clear, and thus we omit the proof.( Also, see Note 3.5 later.)

Example 3.26 [The parallel measurement is always meaningful in both classical and quantum systems] The electron $P_1$ has the (spin) state $\rho_1=|u_1 \rangle \langle u_1|$ $\in$ ${\frak S}^p(B({\mathbb C}^2))$ such that

\begin{align} u_1= \left[\begin{array}{l} \alpha_1 \\ \beta_1 \end{array}\right] \quad (\mbox{where, }\|u_1 \|= (|\alpha_1|^2+ |\beta_1|^2)^{1/2}=1) \end{align}

Let ${\mathsf O}_z =(X(=\{ \uparrow, \downarrow \} ),2^X, F^z )$ be the spin observable concerning the $z$-axis such that

\begin{align} F^z( \{ \uparrow \}) = \left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] , \quad F^z( \{ \downarrow \}) = \left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] \end{align}

Thus, we have the measurement ${\mathsf M_{B({\mathbb C}^2)}}({\mathsf O}_z =(X,2^X, F^z ), S_{[\rho_1]})$. The electron $P_2$ has the (spin) state $\rho_2=|u_2 \rangle \langle u_2|$ $\in$ ${\frak S}^p(B({\mathbb C}^2))$ such that

\begin{align} u= \left[\begin{array}{ll} \alpha_2 \\ \beta_2 \end{array}\right] \quad (\mbox{where, }\|u_2 \|= (|\alpha_2|^2+ |\beta_2|^2)^{1/2}=1) \end{align}

Let ${\mathsf O}_x =(X,2^X, F^x )$ be the spin observable concerning the $x$-axis such that

\begin{align} F^x( \{ \uparrow \}) = \left[\begin{array}{ll} 1/2 & 1/2 \\ 1/2 & 1/2 \end{array}\right] , \quad F^x( \{ \downarrow \}) = \left[\begin{array}{ll} 1/2 & -1/2 \\ -1/2 & 1/2 \end{array}\right] \end{align}

Thus, we have the measurement ${\mathsf M_{B({\mathbb C}^2)}}({\mathsf O}_x =(X,2^X, F^x ), S_{[\rho_2]})$ Then we have the following problem:

 $(a):$ Two measurements ${\mathsf M_{B({\mathbb C}^2)}}({\mathsf O}_z =(X,2^X, F^z ), S_{[\rho_1 ]})$ and ${\mathsf M_{B({\mathbb C}^2)}}({\mathsf O}_x =(X,2^X, F^x ), S_{[\rho_2 ]})$ are taken simultaneously?
This is possible. It can be realized by the parallel measurement \begin{align} {\mathsf M_{B({\mathbb C}^2) \otimes B({\mathbb C}^2)}}({\mathsf O}_z \otimes {\mathsf O}_z=(X \times X,2^{ X \times X }, F^z \otimes F^x), S_{[\rho \otimes \rho ]}) \end{align} That is,
 $(b):$ The probability that a $\mbox{measured value} \left[\begin{array}{ll} ( \uparrow, \uparrow ) \\ ( \uparrow, \downarrow) \\ (\downarrow, \uparrow ) \\ (\downarrow, \downarrow) \end{array}\right]$ is obtained by the parallel measurement ${\mathsf M_{B({\mathbb C}^2) \otimes B({\mathbb C}^2)}}({\mathsf O}_z \otimes {\mathsf O}_z, S_{[\rho \otimes \rho ]})$ is given by $\qquad \qquad \left[\begin{array}{ll} \langle u, F^z(\{\uparrow \})u \rangle \langle u, F^x(\{\uparrow \})u \rangle =p_1 p_2 \\ \langle u, F^z(\{\uparrow \})u \rangle \langle u, F^x(\{\downarrow \})u \rangle =p_1(1- p_2) \\ \langle u, F^z(\{\downarrow \})u \rangle \langle u, F^x(\{\uparrow \})u \rangle =(1-p_1) p_2 \\ \langle u, F^z(\{\downarrow \})u \rangle \langle u, F^x(\{\downarrow \})u \rangle =(1-p_1) (1- p_2) \end{array}\right]$ where $p_1=|\alpha_1|^2, \quad p_2=\frac{1}{2}(|\alpha_1|^2 +\widehat{\alpha}_1 \alpha_2 +\alpha_1\widehat{\alpha}_2 + |\alpha_2|^2)$
 $\fbox{Note 3.5}$ Theorem 3.25 is rather deep in the following sense. For example, "To toss a coin 10 times" is a simultaneous measurement. On the other hand, "To toss 10 coins once" is characterized as a parallel measurement. The two have the same sample space. That is, \begin{align} \mbox{ "spatial average" = "time average" } \end{align} which is called the ergodic property. This means that the two are not distinguished by the sample space and not the measurements (i.e., a simultaneous measurement and a parallel measurement). However, this is peculiar to classical pure measurements. It does not hold in classical mixed measurements and quantum measurement.
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