3.3.1: Only one observable (⇐ Only one measurement is permitted ) In this section, we examine the linguistic interpretation $\S$3.1, i.e., "Only one measurement is permitted".
"Only one measurement" implies that "only one observable" and "only one state". That is, we see: \begin{align} \qquad \mbox{[only one measurement]} \Longrightarrow \left\{\begin{array}{ll} \mbox{only one observable (=measuring instrument)} \\ \\ \mbox{only one state} \end{array}\right. \tag{3.10} \end{align}

$\fbox{Note 3.3}$ Although@there may be several opinions,@I believe that@the standard Copenhagen interpretation ( in physics ) also says@"only one measurement is permitted".@Thus, some think that@this spirit is inherited@to quantum language.@However,@our assertion is reverse, namely,@the Copenhagen standard interpretation@is an outcome of the linguistics interpretation.@That is, we assert that

 ${}$ @not "$\fbox{Copenhagen interpretation}\Longrightarrow \fbox{Linguistic interpretation}$" but "$\fbox{Linguistic interpretation} \Longrightarrow\fbox{Copenhagen interpretation}$"

Recall the measurement in Example 2.31 (Cold or hot?) and Example 2.32 (Approximate temperature), and consider the following situation:
$(a):$ There is a cup in which water is filled. Assume that the temperature is $\omega$ °C $(0 {{\; \leqq \;}}\omega {{\; \leqq \;}}100)$. Consider two questions:
 ${}$ $\left\{\begin{array}{ll} \mbox{ "Is this water cold or hot?" (cf. Figure 1.2 below ) } \\ \\ \mbox{ "How many degrees(°C) is roughly the water?" (cf. Figure 2.3 below )} \end{array}\right.$

This implies that we take two measurements such that
 ${}$ $\left\{\begin{array}{ll} \mbox{$(\sharp_1)$:${\mathsf M}_{L^\infty ( \Omega )} ( {\mathsf O}_{{{{{c}}}}{{{{h}}}}} {{=}} (\{{{{{c}}}},{{{{h}}}}\} , 2^{\{{{{{c}}}},{{{{h}}}}\}}, F_{{{{{c}}}}{{{{h}}}}} ), S_{[\omega]} )$in Example 2.31 } \\ \\ \mbox{$(\sharp_2)$:${\mathsf M}_{L^\infty ( \Omega )} ({\mathsf O}^{\triangle}  {{=}} ({\mathbb N}_{10}^{100} , 2^{{\mathbb N}_{10}^{100} }, G^{\triangle} ), S_{[\omega]} ) $in Example 2.32 } \end{array}\right.$
However, as mentioned in the linguistic interpretation,

"only one measurement"$\Longrightarrow$"only one observable"

Thus, we have the following problem.
Problem 3.10 Represent two measurements ${\mathsf M}_{L^\infty ( \Omega )} ( {\mathsf O}_{{{{{c}}}}{{{{h}}}}} {{=}} (\{{{{{c}}}},{{{{h}}}}\} ,$ $2^{\{{{{{c}}}},{{{{h}}}}\}}, F_{{{{{c}}}}{{{{h}}}}} ), S_{[\omega]} )$ and ${\mathsf M}_{L^\infty ( \Omega )} ({\mathsf O}^{\triangle} {{=}} ({\mathbb N}_{10}^{100} ,$ $2^{{\mathbb N}_{10}^{100} }, G^{\triangle} ),$ $S_{[\omega]} )$ by only one measurement.
This will be answered in what follows.

Definition 3.11 [Product measurable space] For each $k =1,2,\ldots,n$, consider a measurable $(X_k ,$ ${\cal F}_k )$. The product space $\times_{k=1}^n X_k$ of $X_k$ $(k=1,2,\ldots,n )$ is defined by

\begin{align*} \times_{k=1}^n X_k = \{ (x_1, x_2,\ldots, x_n ) \;|\; x_k \in X_k \;\; (k=1,2,\ldots,n )\} \end{align*}

Similarly, define the product $\times_{k=1}^n \Xi_k$ of $\Xi_k ( \in {\cal F}_k )$ $(k=1,2,\ldots,n )$ by

\begin{align*} \times_{k=1}^n \Xi_k = \{ (x_1, x_2,\ldots, x_n ) \;|\; x_k \in \Xi_k \;\; (k=1,2,\ldots,n )\} \end{align*}

Furthermore, the $\sigma$-{{field}} $\boxtimes_{k=1}^n{\cal F}_k$ on the {product space} $\times_{k=1}^n X_k$ is defined by

 ${}:$ $\boxtimes_{k=1}^n{\cal F}_k$ is the smallest field including $\{ \times_{k=1}^n \Xi_k \;|\; \Xi_k \in {\cal F}_k \;\; (k=1,2,\ldots,n ) \}$

$({}\times_{k=1}^n X_k , \boxtimes_{k=1}^n{\cal F}_k)$ is called the product measurable space. Also, in the case that $(X,{\cal F})=(X_k,{\cal F}_k)$ $(k=1,2,\ldots,n)$, the product space $\times_{k=1}^n X_k$ is denoted by $X^n$, and the product measurable space $({}\times_{k=1}^n X_k ,$ $\boxtimes_{k=1}^n{\cal F}_k)$ is denoted by $(X^n,{\cal F}^n)$.

Definition 3.12 [Simultaneous observable, simultaneous measurement]Consider the basic structure $[{\mathcal A} \subseteq \overline{\mathcal A} \subseteq B(H)]$. Let $\rho \in {\frak S}^p({\mathcal A}^*)$. For each $k =1,2,\ldots,n$, consider a measurement ${\mathsf M}_{\overline{\mathcal A}}$ $({\mathsf O}_ =(X_k , {\cal F}_k , F_k{}), S_{[\rho]} )$ in $\overline{\mathcal A}$. Let $({}\times_{k=1}^n X_k , \boxtimes_{k=1}^n{\cal F}_k)$ be the product measurable space. An observable $\widehat{\mathsf O}$ $=$ $({}\times_{k\in K } X_k ,$ $\boxtimes_{k=1}^n{\cal F}_k ,$ $\widehat{F}{})$ in $\overline{\mathcal A}$ is called the simultaneous observable of $\{ {\mathsf O}_k \;: \; k=1,2,...,n \}$, if it satisfies the following condition:

\begin{align} & {\widehat F}({}\Xi_1 \times \Xi_2 \times \cdots \times \Xi_{n}{}) = F_1 ({}\Xi_1{}) \cdot F_2 ({}\Xi_2{}) \cdots F_n ({}\Xi_{n}{}) \tag{3.11} \\ & \qquad \qquad ( \;\; \forall \Xi_k \in {\cal F}_k \; (k=1,2,\ldots,n )) \nonumber \end{align}

$\widehat{\mathsf O}$ is also denoted by $\times_{k=1}^n{\mathsf O}_k$, $\widehat{F}$ $=$ $\times_{k=1}^n{F}_k$. Also, the measurement ${\mathsf M}_{\overline{\mathcal A}} (\times_{k=1}^n{\mathsf O}_k, S_{[\rho]})$ is called the simultaneous measurement Here, it should be noted that

 $\bullet$ the existence of the simultaneous observable $\times_{k=1}^n {\mathsf O}_k$ is not always guaranteed.

though it always exists in the case that $\overline{\mathcal A}$ is commutative (this is, $\overline{\mathcal A}=L^\infty (\Omega)$).

In what follows, we shall explain the meaning of "simultaneous observable".
Let us explain the simultaneous measurement. We want to take two measurements ${\mathsf M}_{\overline{\mathcal A}}({\mathsf O}_1,$ $S_{[\rho]})$ and measurement ${\mathsf M}_{\overline{\mathcal A}}({\mathsf O}_2, S_{[\rho]})$. That is, it suffices to image the following:

However, according to the linguistic interpretation ($\S$3.1), two measurements ${\mathsf M}_{\overline{\mathcal A}}({\mathsf O}_1,$ $S_{[\rho]})$ and ${\mathsf M}_{\overline{\mathcal A}}({\mathsf O}_2, S_{[\rho ]})$ can not be taken. That is,

\begin{align*} \mbox{The (b) is impossible} \end{align*} Therefore, combining two observables ${\mathsf O}_1$ and ${\mathsf O}_2$, we construct the simultaneous observable ${\mathsf O}_1\times {\mathsf O}_2$, and take the simultaneous measurement ${\mathsf M}_{\overline{\mathcal A}} ( {\mathsf O}_1 \times {\mathsf O}_2, S_{[\rho]})$ in what follows.
 $(c):$ $\overset{{}}{\underset{ \rho (\in {\frak S}^p({\mathcal A}^*) ) } {{ \fbox{state}}}} \xrightarrow[]{\qquad \qquad} \overset{{}} { \underset{ {\mathsf O}_1 \times {\mathsf O}_2} {{ \fbox{simultaneous observable }}} } \xrightarrow[ {\mathsf M}_{\overline{\mathcal A}} ( {\mathsf O}_1 \times {\mathsf O}_2, S_{[\rho]})]{} \overset{{}} { \underset{ (x_1,x_2) (\in X_1 \times X_2 )} {{ \fbox{measured value }}} }$
\begin{align*} \mbox{The (c) is possible if ${\mathsf O}_1 \times {\mathsf O}_2$ exists} \end{align*}

Answer 3.13 [The answer to Problem 3.10] Consider the state space $\Omega$ such that $\Omega =$ $[0,100]$, the closed interval. And consider two observables, that is, [C-H]-observable ${\mathsf O}_{{{{{c}}}}{{{{h}}}}}= (X {{=}} \{ {{{{c}}}} , {{{{h}}}} \}, 2^X, F_{{{{{c}}}}{{{{h}}}}} )$ (in Example 2.31) and triangle observable ${\mathsf O}^{\triangle}= (Y( {{=}} {\mathbb N}_{10}^{100}) , 2^Y, G^{\triangle} )$ (in Example 2.32). Thus, we get the simultaneous observable ${\mathsf O}_{{{{{c}}}}{{{{h}}}}} \times {\mathsf O}^{\triangle}$ $=$ $(\{ {{{{c}}}} , {{{{h}}}} \}\times {\mathbb N}_{10}^{100}, 2^{\{ {{{{c}}}} , {{{{h}}}} \}\times {\mathbb N}_{10}^{100}}, F_{{{{{c}}}}{{{{h}}}}} \times G^{\triangle} )$, and we can take the simultaneous measurement ${\mathsf M}_{L^\infty (\Omega)}({\mathsf O}_{{{{{c}}}}{{{{h}}}}} \times {\mathsf O}^{\triangle}, S_{[\omega]})$. For example, putting@$\omega=55$,@we see

 $(d):$ @when@the simultaneous measurement@${\mathsf M}_{L^\infty (\Omega)}({\mathsf O}_{{{{{c}}}}{{{{h}}}}} \times {\mathsf O}^{\triangle}, S_{[55]})$@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} 0.125 \\ 0.125 \\ 0.375 \\ 0.375 \end{array}\right] \end{align}
That is because \begin{align*} & [(F_{{{{{c}}}}{{{{h}}}}} \times G^{\triangle})( \{ (\mbox{c}, \mbox{about 50°C} ) \} )] (55) \\ = & [F_{{{{{c}}}}{{{{h}}}}}(\{ \mbox{c} \} )](55) \cdot [G^{\triangle}(\{ \mbox{about 50°C}\})](55) =0.25 \cdot 0.5=0.125 \end{align*} and similarly, \begin{align*} & [(F_{{{{{c}}}}{{{{h}}}}} \times G^{\triangle})( \{ (\mbox{c}, \mbox{about 60°C} ) \} )] (55) =0.25 \cdot 0.5=0.125 \\ & [(F_{{{{{c}}}}{{{{h}}}}} \times G^{\triangle})( \{ (\mbox{h}, \mbox{about 50°C} ) \} )] (55) =0.75 \cdot 0.5=0.375 \\ & [(F_{{{{{c}}}}{{{{h}}}}} \times G^{\triangle})( \{ (\mbox{h}, \mbox{about 60°C} ) \} )] (55) =0.75 \cdot 0.5=0.375 \end{align*}
 $\fbox{Note 3.4}$ @The above argument@is not always possible.@In@quantum mechanics,@a simultaneous observable ${\mathsf O}_1\times {\mathsf O}_2$@does not always exist@(See the following Example 3.14@and@Heisenberg's {uncertainty principle in Sec.@4.4j.

Example 3.14 [The non-existence of the simultaneous spin observables] Assume that the electron $P$ has the (spin) state $\rho=|u \rangle \langle u|$ $\in$ ${\frak S}^p(B({\mathbb C}^2))$, where

\begin{align*} u= \left[\begin{array}{l} \alpha_1 \\ \alpha_2 \end{array}\right] \quad (\mbox{where, }|u|= (|\alpha_1|^2+ |\alpha_2|^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@${\mathsf M_{B({\mathbb C}^2)}}({\mathsf O}_z =(X,2^X, F^z ), S_{[\rho]})$. 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@${\mathsf M_{B({\mathbb C}^2)}}({\mathsf O}_x =(X,2^X, F^x ), S_{[\rho]})$ Then we have the following problem:

 $(a):$ @Two measurements@${\mathsf M_{B({\mathbb C}^2)}}({\mathsf O}_z =(X,2^X, F^z ), S_{[\rho ]})$@and@${\mathsf M_{B({\mathbb C}^2)}}({\mathsf O}_x =(X,2^X, F^x ), S_{[\rho ]})$@are taken simultaneously?

This is impossible.@That is because@the two observable ${\mathsf O}_z$@and@${\mathsf O}_x$@do not commute.@For example,@we see

\begin{align*} F^z( \{ \uparrow \})F^x( \{ \uparrow \}) = \left[\begin{array}{lL} 1 & 0 \\ 0 & 0 \end{array}\right] \cdot \left[\begin{array}{ll} 1/2 & 1/2 \\ 1/2 & 1/2 \end{array}\right] = \left[\begin{array}{ll} 1/2 & 1/2 \\ 0 & 0 \end{array}\right] \end{align*} \begin{align*} F^x( \{ \uparrow \})F^z( \{ \uparrow \}) = \left[\begin{array}{ll} 1/2 & 1/2 \\ 1/2 & 1/2 \end{array}\right] \cdot \left[\begin{array} 1 & 0 \\ 0 & 0 \end{array}\right] = \left[\begin{array}{ll} 1/2 & 0 \\ 1/2 & 0 \end{array}\right] \end{align*} And thus, \begin{align*} F^x( \{ \uparrow \})F^z( \{ \uparrow \}) \not= F^z( \{ \uparrow \})F^x( \{ \uparrow \}) \end{align*} The following theorem is clear.@For completeness, we add the proof to it.

Theorem@3.15@[Exact measurement@and system quantity]@Consider the classical basic structure: \begin{align*} \mbox{ $[C_0(\Omega ) \subseteq L^\infty (\Omega, \nu ) \subseteq B(L^2 (\Omega, \nu ))]$ } \end{align*} Let@${\mathsf O}^{\rm{(exa)}}_0=(X,{\cal F}, F^{\rm (exa)})$@(i.e.,@$(X,{\cal F},$@$F^{\rm (exa)})=(\Omega,{\cal B}_{\Omega}, \chi )$)@be the exact observable in@$L^\infty(\Omega,\nu)$.@Let@${\mathsf O}_1=({\mathbb R},{\cal B}_{\mathbb R}, G)$@be the observable@that is induced by@a quantity@${\widetilde g}:\Omega \to {\mathbb R}$@as in Example 2.26@(system quantity).@Consider the simultaneous observable@${\mathsf O}^{\rm{(exa)}}_0 \times@{\mathsf O}_1@$.@Let@$(x,y)$@$(\in X \times {\mathbb R})$@be a measured value obtained by@the simultaneous measurement@${{\mathsf M}}_{L^\infty (\Omega, \nu)}(@{\mathsf O}^{\rm{(exa)}}_0\times@{\mathsf O}_1@,$@$S_{[\delta_\omega]} )$.@Then,@we can surely believe@that@$x=\omega$,@and@$y= {\widetilde g}(\omega)$.

Proof
Let $D_0 (\in {\cal B}_{\Omega} )$ be arbitrary open set such that $\omega (\in D_0 \subseteq \Omega {{=}} X )$. Also, let $D_1 (\in {\cal B}_{\mathbb R})$ be arbitrary open set such that ${\widetilde g}(\omega) \in$ $D_1$. The probability that a measured value $(x,y)$ obtained by the measurement ${{\mathsf M}}_{L^\infty (\Omega, \nu)}( {\mathsf O}^{\rm{(exa)}}_0 \times {\mathsf O}_1 , S_{[\delta_\omega]} )$ belongs to $D_0 \times D_1$ is given by $\chi_{_{D_0}}(\omega) \cdot \chi_{_{{\widetilde g}^{-1}(D_1)}} (\omega )=1$. Since $D_0$ and $D_1$ are arbitrary, we can surely believe that $x=\omega$ and $y={\widetilde g}(\omega)$.

$\square \quad$