2.8.1: Linguistic world-view ---Wonder of man's linguistic competence The applied scope of physics physics (realistic world-description method) is rather clear. But the applied scope of mea urement theory is ambiguous. What we can do in measurement theory (= quantum language) is

 (a): $\left\{\begin{array}{ll} \mbox{(a$_1$): Use the language defined by Axiom 1 ($\S$2.7) } \\ \\ \mbox{(a$_2$): Trust in man's linguistic competence } \end{array}\right.$
Thus, some readers may doubt that
 (b): $\qquad \qquad$ Is it science?
However, it should be noted that the spirit of measurement theory is different from that of physics.
2.8.2 Elementary examples---urn problem, etc.
Since measurement theory is a language, we can not master it without exercise. Thus, we present simple examples in what follows.
Example 2.32 [ The measurement of the approximate temperature of water in a cup (continued from Example 2.22: triangle observable)] Consider the classical basic structure: \begin{align*} \mbox{ $[C_0(\Omega ) \subseteq L^\infty (\Omega, \nu ) \subseteq B(L^2 (\Omega, \nu ))]$ } \end{align*}

where $\Omega =$"the closed interval $[0, 100]$" with the Lebesgue measure $\nu$.

Let testees drink water with various temperature $\omega$ °C $(0 {{\; \leqq \;}}\omega {{\; \leqq \;}}100)$. And you ask them@"How many degrees(°C) is roughly this water?"@Gather the data,@( for example,@$h_{n}(\omega)$@persons say@$n$ °C@${(}n=0,10,20,\ldots,90,100)$.@and@normalize them,@that is,@get@the polygonal lines.@For example,@define the state space@$\Omega$@by@the closed interval@$[0,100]$@$(@\subseteq {\mathbb R})$@with the Lebesgue measure.@For each@$n \in {\mathbb N}_{10}^{100} = \{0,10,20,\ldots,100\}$,@define the (triangle)@continuous function@$g_{n}:\Omega \to [0,1]$@by

\begin{align*} g_{n} (\omega) = \left\{\begin{array}{ll} 0 & \quad (0 {{\; \leqq \;}}\omega {{\; \leqq \;}}n-10 ) \\ {\displaystyle \frac{\omega - n -10}{10} } & \quad (n-10 {{\; \leqq \;}}\omega {{\; \leqq \;}}n ) \\ {\displaystyle - \frac{\omega - n + 10}{10} } & \quad (n {{\; \leqq \;}}\omega {{\; \leqq \;}}n+10 ) \\ 0 & \quad (n+10 {{\; \leqq \;}}\omega {{\; \leqq \;}}100 ) \end{array}\right. \end{align*}
 (a): You choose one person from the testees,@and@you ask him/her@"How many degrees(°C) is roughly this water?".@Then the probability that@he/she@says $\left[\begin{array}{ll} {} \mbox{"about 40°C"} \\ {} \mbox{"about 50°C"} \end{array}\right]$@is@given by@$\left[\begin{array}{ll} {}g_{\text 40}(47)=0.25 \\ {} f_{\text 50}(47)=0.75 \end{array}\right]$

This is described in terms of@Axiom 1 ( $\S$2.7)@in what follows. Putting@$Y={\mathbb N}_{10}^{100}$,@define the triangle observable@${\mathsf O}^{\triangle}= (Y , 2^Y, G^{\triangle} )$@in $L^\infty (\Omega )$@such that

\begin{align*} & [G^{\triangle}(\emptyset )](\omega ) = 0, \qquad [G^{\triangle}(Y )](\omega ) = 1 \\ & [G^{\triangle} (\Gamma )](\omega ) = \sum\limits_{n \in \Gamma } g_n (\omega ) \quad \qquad (\forall \Gamma \in 2^{{\mathbb N}_{10}^{100} }, \forall \omega \in \Omega =[0, 100]) \end{align*}

Then, we have the triangle observable ${\mathsf O}^{\triangle}= (Y (= {{\mathbb N}_{10}^{100} } ), 2^Y, G^{\triangle} )$ in $L^\infty ([0,100])$. And we get a measurement ${\mathsf M}_{L^\infty ( \Omega )} ( {\mathsf O}^{\triangle}, S_{ [\delta_\omega]} )$. For example, putting $\omega$=47°C, we see,@by@Axiom 1 ( $\S$2.7),@that

 $(b):$ @the probability@that@a measured value@obtained by the measurement@${\mathsf M}_{L^\infty ( \Omega )} ( {\mathsf O}^{\triangle},$ $S_{[ \omega(=47)]} )$ is $\left[\begin{array}{ll} {\mbox{about 40°C}} \\ \mbox{about 50°C} \end{array}\right]$ is given by $\left[\begin{array}{ll} {[G^{\triangle}( \{ 40 \})](47)=0.3} \\ {[G^{\triangle}( \{ 50 \})](47)=0.7} \end{array}\right]$

Therefore, we see:

\begin{align} \underset{\mbox{ (ordinary language)}}{\fbox{statement (a)}} \xrightarrow[{\mbox{ translation}}]{} \underset{\mbox{ (quantum language)}}{\fbox{statement (b)}} \tag{2.69} \end{align}
Example 2.33 [Exact measurement] 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 ${\mathcal B}_\Omega$ be the Borel field. Then, define the exact observable ${\mathsf O}^{{ (exa)}} = (X(=\Omega) , {\mathcal F}(={\mathcal B}_\Omega), F^{\rm (exa)}{})$ in $L^\infty (\Omega, \nu )$ such that

\begin{align*} [F^{\rm (exa)}(\Xi)](\omega) = \chi_{{}_\Xi} (\omega) = \left\{\begin{array}{ll} 1 \quad & ( \omega \in \Xi )\\ \\ 0 & ( \omega \notin \Xi ) \end{array}\right. \qquad (\forall \Xi \in {\mathcal B}_\Omega{}) \end{align*}

Let $\delta_{\omega_0 } \approx \omega_0 (\in \Omega )$. Consider the exact measurement ${\mathsf M}_{L^\infty ( \Omega, \nu )} ( {\mathsf O}^{{ (exa)}}, S_{[\delta_{\omega_0} ]})$. Here, Axiom 1 ($\S$2.7) says:

 (a): Let $D (\subseteq \Omega)$ be arbitrary open set such that $\omega_0 \in D$. Then, the probability that a measured value obtained by the exact measurement ${\mathsf M}_{L^\infty ( \Omega, \nu )} ({\mathsf O}^{ (exa)}, S_{[\delta_{\omega_0 }]})$ belongs to $D$ is given by \begin{align*} {{}_{C_0(\Omega)^*}} \Big( \delta_{{\omega_0}}, \chi_{{}_D} \Big){{}_{L^\infty(\Omega, \nu )}} = 1 \end{align*}
From the arbitrariness of $D$, we conclude that
 $(b):$ a measured value $\omega_0$ is, with the probability $1$, obtained by the exact measurement ${\mathsf M}_{L^\infty ( \Omega, \nu )}$ $( {\mathsf O}^{ (exa)},$ $S_{[\delta_{\omega_0 }]})$.
Furthermore, put \begin{align*} {\mathcal F}_{\omega_0}=\{ \Xi \in {\mathcal F} \;:\; \omega_0 \notin \mbox{"the closure of $\Xi$"$\setminus$ "the interior of $\Xi$"} \} \end{align*}

Then, when $\Xi \in {\mathcal F}_{\omega_0}$, $F(\Xi)$ is continuous at $\omega_0$. And, ${\mathcal F}$ is the smallest $\sigma$-field that contains ${\mathcal F}_{\omega_0}$.{\;\;} Therefore, we have the probability space $(X, {\mathcal F}, P_{\delta_{\omega_0}})$ such that

\begin{align*} P_{\delta_{\omega_0}}( \Xi ) = [F(\Xi )](\omega_0 ) \qquad (\forall \Xi \in {\mathcal F}_{\omega_0}) \end{align*} that is,
 $(c):$ the exact measurement ${\mathsf M}_{L^\infty ( \Omega, \nu )} (@{\mathsf O}^{ (exa)}, S_{[\delta_{\omega_0 }]})$@has the sample space@$(X, {\mathcal F}, P_{\delta_{\omega_0}})$@$@(=@(\Omega,$@${\mathcal B_{\Omega}} ,$@$P_{\delta_{\omega_0}}@))@$
Example 2.34@uUrn problem]@There are two urns ${U}_1$@and@${U}_2$.@The urn ${U}_1$ [resp. ${U}_2$]@contains@8 white and 2 black balls@[resp.@4 white and 6 black balls]@(cf.@Table 2.2, Figure 2.7).
Here, consider the following statement (a):
 (a): When one ball is picked up from the urn $U_2$, the probability that the ball is white is $0.4$.
In measurement theory, the statement (a) is formulated as follows: Assuming \begin{align*} U_1 \quad \cdots \quad & \mbox{"the urn with the state $\omega_1$"} \\ U_2 \quad \cdots \quad & \mbox{"the urn with the state $\omega_2$"} \end{align*}

define the state space $\Omega$ by $\Omega = \{ {\omega}_1 , {\omega}_2 \}$ with the discrete metric and the {counting measure} $\nu$ (i.e., $\nu(\{\omega_1\})=\nu(\{\omega_2\})=1$). That is, we assume the identification;

\begin{align*} U_1 \approx \omega_1, \quad U_2 \approx \omega_2, \quad \end{align*} Thus, consider the classical basic structure: \begin{align*} \mbox{ $[C_0(\Omega ) \subseteq L^\infty (\Omega, \nu ) \subseteq B(L^2 (\Omega, \nu ))]$ } \end{align*}

Put "$w$" = "white"$\!\!,\;$ "$b$" = "black"$\!\!\;$, and put $X=\{w,b\}$. And define the observable ${\mathsf O} \big(\equiv (X \equiv \{w,b\}, 2^{\{w,b\}}, F) \big)$ in $L^\infty (\Omega)$ by

Thus, we get the measurement ${\mathsf M}_{L^\infty ({}\Omega{})} ({}{\mathsf O} , S_{[{} \delta_{\omega_2}]})$. Here, Axiom 1 ( $\S$2.7) says that

 (b): the probability that a measured value ${w}$ is obtained by ${\mathsf M}_{L^\infty ({}\Omega{})} ({}{\mathsf O} , S_{[{} \delta_{\omega_2}]})$ is given by \begin{align*} F(\{b\})(\omega_2) = 0.4 \end{align*}
Therefore, we see: \begin{align} \underset{\mbox{ (ordinary language)}}{\fbox{statement (a)}} \xrightarrow[{\mbox{ translation}}]{} \underset{\mbox{ (quantum language)}}{\fbox{statement (b)}} \tag{2.70} \end{align}
 $\fbox{Note 2.5}$ [$L^\infty (\Omega, \nu )$, or in short, $L^\infty (\Omega)$] In the above example, the counting measure $\nu$ (i.e., $\nu(\{\omega_1\})=\nu(\{\omega_2\})=1$) is not absolutely indispensable. For example,even if we assume that@$\nu(\{\omega_1\})=2$@and@$\nu(\{\omega_2\})=1/3$,@we can assert the same conclusion.@Thus,@in this note, \begin{align*} \mbox{ $L^\infty (\Omega, \nu )$ is often abbreviated to $L^\infty (\Omega)$. } \end{align*}
$\fbox{Note 2.6}$@The statement@(a)@in Example 2.34@is not necessarily guaranteed,@that is,
 $(\flat)$ @When one ball is picked up from the urn@$U_2$,@the probability that@the ball is white@is $0.4$.
is not guaranteed.@What we say is that
 $(\flat)$ @the statement (a) in ordinary language@should be written by@the measurement theoretical statement@(b)

It is a matter of course that@"probability"@can not be derived from mathematics itself.@For example,@the following@$(\sharp_1)$@and@$(\sharp_2)$@are not guaranteed.

 $(\sharp_1):$ @From the set $\{1,2,3,4,5\}$,@choose one number.@Then, the probability that the number is even@is@given by@$2/5$ $(\sharp_2)$: @From the closed interval $[0,1]$,@choose one number $x$.@Then, the probability@that $x \in [a, b ]\subseteq [0,1]$@is@given by@$|b-a|$

The common sense@---@"probability"@can not be derived from mathematics itself@---@is well known as Bertrand's paradox@(cf. $\S$9.12).@Thus,@it is usual to add the term@"at random"@to@the above $(\sharp_1)$@and@$(\sharp_2)$.@In this note,@this term "at random"@is usually omitted.

Example 2.35@[Blood type system]@

The ABO blood group system is the most important blood type system (or blood group system) in human blood transfusion. @Let ${U}_1$@be the whole Japanese's@set@and let@${U}_2$@be the whole Indian's set.@Also, assume that@the distribution of@the ABO blood group system [O:A:B:AB]@concerning Japanese and Indians@is determined in@Table 2.3.

Consider the following phenomenon:
 (a): Choose one person from the the whole Indian's set $U_2$ at random. Then the probability that the person's blood type is $\left[\begin{array}{ll} O \\ A \\ B \\ AB \end{array}\right]$ is given by $\left[\begin{array}{ll} 0.3 \\ 0.2 \\ 0.4 \\ 0.1 \end{array}\right]$

In what follows, we shall translate the statement (a) described in ordinary language to quantum language. Put $\Omega = \{ {\omega}_1 , {\omega}_2 \}$ and consider the discrete metric $(\Omega, d_D)$. We get consider the classical basic structure:

\begin{align*} \mbox{{ $[C_0(\Omega ) \subseteq L^\infty (\Omega, \nu ) \subseteq B(L^2 (\Omega, \nu ))]$} } \end{align*} Therefore, the pure state space is defined by \begin{align*} {\frak S}^p (C_0(\Omega)^*)= \{ \delta_{\omega_1} , \delta_{\omega_2} \} \end{align*} Here, consider \begin{align*} & \delta_{\omega_1} \quad \cdots \quad \mbox{"the state of the whole Japanese's set $U_1$(i.e., population)"} \qquad \\ & \delta_{\omega_2} \quad \cdots \quad \mbox{"the state of the whole India's set $U_1$(i.e., population)"}, \end{align*} Note that "population" = "system" (cf. Table 2.1). That is, we consider the following identification: (Therefore, image Figure 2.9):

Define the blood type observable ${\mathsf O}_{\mbox{ BT}} = ( \{ O, A,B,AB \}, 2^{\{ O, A,B,AB \} } , F_{{{\mbox{ BT}}}}{})$ in $L^\infty (\Omega, \nu )$ such that

\begin{align} & [F_{\mbox{ BT}}(\{ O \}{})](\omega_1{})= 0.3, & \quad & [ F_{\mbox{ BT}}(\{ A \}{})](\omega_1{})= 0.4 \nonumber \\ & [F_{\mbox{ BT}}(\{ B \}{})](\omega_1{})= 0.2, & \quad & [ F_{\mbox{ BT}}(\{ AB \}{})](\omega_1{})= 0.1 \tag{2.71} \end{align} \begin{align} & [F_{\mbox{ BT}}(\{ O \}{})](\omega_2{})= 0.3, & \quad & [F_{\mbox{ BT}}(\{ A \}{})] (\omega_2{})= 0.2 \nonumber \\ & [F_{\mbox{ BT}}(\{ B \}{})](\omega_2{})= 0.4, & \quad & [F_{\mbox{ BT}}(\{ AB \}{})] (\omega_2{})= 0.1 \tag{2.72} \end{align}

Thus we get the measurement ${\mathsf M}_{{L^\infty (\Omega, \nu )}} ({\mathsf O}_{{{\mbox{ BT}}}} , S_{ [{}{\delta_{\omega_2}}]}{})$. Hence, the above (a) is translated to the following statement (in terms of quantum language):

 (b): The probability that a measured value $\left[\begin{array}{ll} O \\ A \\ B \\ AB \end{array}\right]$ is obtained by the measurement ${\mathsf M}_{{L^\infty (\Omega, \nu ) }} ({\mathsf O}_{\mbox{ BT}} , S_{ [{}{\delta_{\omega_2}}]}{})$ is given by $\left[\begin{array}{ll} {{}_{C_0(\Omega)^*}} \Big( \delta_{{\omega_2}}, F_{{{\mbox{ BT}}}}(\{O\}) \Big){{}_{L^\infty(\Omega, \nu )}} = [F_{{{\mbox{ BT}}}}(\{O\})](\omega_2) = 0.3 \\ {{}_{C_0(\Omega)^*}} \Big( \delta_{{\omega_2}}, F_{{{\mbox{ BT}}}}(\{A\}) \Big){{}_{L^\infty(\Omega, \nu )}} = [F_{{{\mbox{ BT}}}}(\{A\})](\omega_2) = 0.2 \\ {{}_{C_0(\Omega)^*}} \Big( \delta_{{\omega_2}}, F_{{{\mbox{ BT}}}}(\{B\}) \Big){{}_{L^\infty(\Omega, \nu )}} = [F_{{{\mbox{ BT}}}}(\{B\})](\omega_2) = 0.4 \\ {{}_{C_0(\Omega)^*}} \Big( \delta_{{\omega_2}}, F_{{{\mbox{ BT}}}}(\{AB\}) \Big){{}_{L^\infty(\Omega, \nu )}} = [F_{{{\mbox{ BT}}}}(\{AB\})](\omega_2) = 0.1 \end{array}\right]$

$\fbox{Note 2.7}$ Readers may feel that Examples 2.32--2.35 are too easy. However, as mentioned in (a) of Sec. 2.8.1, what we can do is
 $\bullet$ $\left\{\begin{array}{ll} \mbox{to be faithful to Axioms ( = to learn Axioms by rote ) } \\ \\ \mbox{ to trust in Man's linguistic competence } \end{array}\right.$

If some find the other language that is more powerful than quantum language, it will be praised

as the greatest discovery

in the history of science. That is because this discovery allows us to go beyond quantum mechanics.