$(a):$

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. $

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

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

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

$({}\times_{k=1}^n X_k , \boxtimes_{k=1}^n{\cal F}_k)$ 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:

$\widehat{\mathsf O}$ 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

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

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

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

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

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

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:

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

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 $y={\widetilde g}(\omega)$.