Contents:

### 4.2.1: The sample space of infinite parallel measurement $\bigotimes_{k=1}^\infty{\mathsf M}_{\overline{\mathcal A}} ({\mathsf O}=(X,{\mathcal F},{ F} ), S_{[\rho]})$

Consider the basic structure

\begin{align} & \qquad \qquad \qquad \qquad [{\mathcal A} \subseteq \overline{\mathcal A} \subseteq B(H)] \\ & \Big( \mbox{that is, } [{\mathcal C}(H) \subseteq B(H) \subseteq B(H)], \mbox{ or } [C_0(\Omega) \subseteq L^\infty (\Omega, \nu) \subseteq B( L^2 (\Omega, \nu) )] \Big) \end{align}

and measurement ${\mathsf M}_{\overline{\mathcal A}} ({\mathsf O}=(X,{\mathcal F},{ F} ), S_{[\rho]})$, which has the sample probability space $(X,{\mathcal F}, P_\rho )$ Note that the existence of the infinite parallel observable $\widetilde{\mathsf O}$ $(=\bigotimes_{k=1}^\infty {\mathsf O} )$ $=(X^{\mathbb N} ,$ $\boxtimes_{k=1}^\infty {\cal F} ,$ ${\widetilde F} ({{=}} \bigotimes_{k=1}^\infty F ) )$ in an infinite tensor $W^*$-algebra $\bigotimes_{k=1}^\infty \overline{\mathcal A}$ is assured by Kolmogorov's extension theorem (Corollary 4.2). For completeness, let us calculate the sample probability space of the parallel measurement ${\mathsf M}_{\bigotimes_{k=1}^\infty \overline{\mathcal A}} (\widetilde{\mathsf O} , S_{[\bigotimes_{k=1}^\infty \rho]})$ in both cases (i.e., quantum case and classical case):

Preparation 4.4 [I]:quantum system: The quantum infinite tensor basic structure is defined by \begin{align} [ {\mathcal C}(\otimes_{k=1}^\infty H) \subseteq B(\otimes_{k=1}^\infty H) \subseteq B(\otimes_{k=1}^\infty H)] \end{align} Therefore,infinite tensor state space is characterized by \begin{align} {\frak S}^p({\mathcal Tr}(\otimes_{k=1}^\infty H)) \subset {\frak S}^m({\mathcal Tr}(\otimes_{k=1}^\infty H))= \overline{\frak S}^m({\mathcal Tr}(\otimes_{k=1}^\infty H)) \tag{4.5} \end{align} Since Definition 2.17 says that ${\mathcal F} = {\mathcal F}_\rho$ $( \forall \rho \in {\frak S}^p({\mathcal Tr}( H)) )$, the sample probability space $(X^{{\mathbb N}},$ $\boxtimes_{k=1}^\infty{\mathcal F},$ $P_{\bigotimes_{k=1}^\infty \rho})$ of the infinite parallel measurement ${\mathsf M}_{\bigotimes_{k=1}^\infty B(H)} (\otimes_{k=1}^\infty {\mathsf O}= (X^{{\mathbb N}},$ $\boxtimes_{k=1}^\infty{\mathcal F},$ $\otimes{k=1}^\infty { F} ),$ $S_{[\bigotimes_{k=1}^\infty \rho]})$ is characterized by \begin{align} & P_{\bigotimes_{k=1}^\infty \rho} (\Xi_1 \times \Xi_2 \times \cdots \times \Xi_n{} \times(\times_{k=n+1}^\infty X)) = \times_{k=1}^n {}_{{}_{{\mathcal Tr}(H)}} \Big( \rho, F(\Xi_k ) \Big){}_{{}_{B(H)}} \tag{4.6} \\ & \quad ( \; \forall \Xi_k \in {\cal F}={\cal F}_\rho, (\; k=1,2,\ldots, n), n=1,2,3 \cdots ) \end{align} which is equal to the infinite product probability measure $\bigotimes_{k=1}^n P_\rho$.
[II]:classical system: Without loss of generality, we assume that the state space $\Omega$ is compact, and $\nu(\Omega)=1$ (cf. Note 2.1). Then, the classical infinite tensor basic structure is defined by \begin{align} [ C_0(\times_{k=1}^\infty \Omega ) \subseteq L^\infty (\times_{k=1}^\infty \Omega, \otimes_{k=1}^\infty \nu) \subseteq B( L^2 ( \times_{k=1}^\infty \Omega, \otimes_{k=1}^\infty \nu ) ) ] \tag{4.7} \end{align} Therefore, the infinite tensor state space is characterized by \begin{align} {\frak S}^p(C_0(\times_{k=1}^\infty \Omega)^*)\Big( \approx \times_{k=1}^\infty \Omega \Big) \tag{4.8} \end{align} Put $\rho=\delta_\omega$. the sample probability space $(X^{{\mathbb N}},$ $\boxtimes_{k=1}^\infty{\mathcal F},$ $P_{\bigotimes_{k=1}^\infty \rho})$ of the infinite parallel measurement ${\mathsf M}_{ L^\infty (\times_{k=1}^\infty \Omega, \otimes_{k=1}^\infty \nu) } (\otimes_{k=1}^\infty {\mathsf O}= (X^{{\mathbb N}},$ $\boxtimes_{k=1}^\infty{\mathcal F},$ $\otimes{k=1}^\infty { F} ),$ $S_{[\bigotimes_{k=1}^\infty \rho]})$ is characterized by \begin{align} & P_{\bigotimes_{k=1}^\infty \rho} (\Xi_1 \times \Xi_2 \times \cdots \times \Xi_n{} \times(\times_{k=n+1}^\infty X)) = \times_{k=1}^n [ F(\Xi_k )](\omega ) \tag{4.9} \\ & \quad ( \; \forall \Xi_k \in {\cal F}={\cal F}_\rho, (\; k=1,2,\ldots, n), n=1,2,3 \cdots ) \end{align} which is equal to the infinite product probability measure $\bigotimes_{k=1}^n P_\rho$.
[III]: Conclusion: Therefore, we can conclude
 $(\sharp):$ in both cases, the sample probability space $(X^{{\mathbb N}},$ $\boxtimes_{k=1}^\infty {\mathcal F},$ $P_{\bigotimes_{k=1}^\infty \rho})$ is defined by the infinite product probability space $(X^{{\mathbb N}},$ $\boxtimes_{k=1}^\infty {\mathcal F},$ ${\bigotimes_{k=1}^\infty P_\rho})$
Summing up, we have the following theorem ( the law of large numbers ).
Theorem 4.5 [The law of large numbers]

Consider the measurement ${\mathsf M}_{\overline{\mathcal A}} ({\mathsf O}=(X,{\mathcal F},{ F} ), S_{[\rho]})$ with the sample probability space $(X,{\mathcal F}, P_\rho )$. Then, by Kolmogorov's extension theorem (Corollary 4.2Corollary), we have the infinite parallel measurement:

\begin{align} {\mathsf M}_{\bigotimes_{k=1}^\infty \overline{\mathcal A}} (\otimes_{k=1}^\infty {\mathsf O}= (X^{{\mathbb N}},\boxtimes_{k=1}^\infty{\mathcal F},\otimes_{k=1}^\infty { F} ), S_{[\bigotimes_{k=1}^\infty \rho]}) \end{align}

The sample probability space $(X^{{\mathbb N}},\boxtimes_{k=1}^\infty{\mathcal F},$ $P_{\bigotimes_{k=1}^\infty \rho})$ is characterized by the infinite probability space $(X^{{\mathbb N}},$ $\boxtimes_{k=1}^\infty {\mathcal F},$ ${\bigotimes_{k=1}^\infty P_\rho})$. Furthermore, we see

 $(A):$ for any $f \in L^1( X, P_{\rho})$, put \begin{align} & D_f=\Big\{ (x_1,x_2, \ldots ) \in X^{\mathbb N} \;|\; \lim_{n \to \infty } \frac{f(x_1) + f(x_2) + \cdots + f({x_n})}{n} = E(f) \Big\} \\ & \qquad \mbox{ ( where, $E(f) = \int_X f(x) P_\rho ( dx )$ ) } \end{align} Then, it holds that \begin{align} P_{\bigotimes_{k=1}^\infty \rho} (D_f)=1 \tag{4.11} \end{align}
That is,we see, almost surely, \begin{align} \underset{\mbox{ (population mean)}}{\fbox{$\int_X f(x) P_\rho ( dx )$}} = \underset{\mbox{ (sample mean)}}{\fbox{$\lim_{n \to \infty } \frac{f(x_1) + f(x_2) + \cdots + f({x_n})}{n}$}} \tag{4.12} \end{align} Remark 4.6 [Frequency probability] In the above, consider the case that \begin{align} f(x)=\chi_{{}_{\Xi}}(x)= \left\{\begin{array}{ll} 1 \;\; & (x \in \Xi) \\ 0 \;\; & (x \notin \Xi) \end{array}\right. \quad (\Xi \in {\mathcal F} ) \end{align} Then, put \begin{align} & D_{\chi_{{}_{\Xi}}}=\Big\{ (x_1,x_2, \ldots ) \in X^{\mathbb N} \;|\; \lim_{n \to \infty } \frac{ \sharp [ \{ k \;|\; x_k \in \Xi, 1\le k \le n \} }{n} = P_\rho ( \Xi ) \Big\} \tag{4.13} \\ & \qquad \mbox{ (where, $\sharp[A]$ is the number of the elements of the set $A$) } \end{align} Then, it holds that \begin{align} P_{\bigotimes_{k=1}^\infty \rho} (D_{\chi_{{}_{\Xi}}})=1 \tag{4.14} \end{align} Therefore, the law of large numbers (Theorem 4.5) says that
 $(\sharp):$ the probability in Axiom 1 ( $\S$2.7) can be regarded as "frequency probability"
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### 4.2.2: Mean,variance,unbiased variance

Consider the measurement ${\mathsf M}_{\overline{\mathcal A}} ({\mathsf O}=({\mathbb R}, {\mathcal B}_{\mathbb R}, F), S_{[{}\rho{}] }{})$. Let $( {\mathbb R}, {\mathcal B}_{\mathbb R}, P_\rho )$ be its sample probability space. That is, consider the case that a measured value space $X={\mathbb R}$. Here, define: \begin{align} &{\mbox{ population mean}}(\mu_{\mathsf O}^{\rho}):E[{\mathsf M}_{\overline{\mathcal A}} ({\mathsf O}=({\mathbb R}, {\mathcal B}_{\mathbb R} F), S_{[{}\rho{}] }{})] = \int_{\mathbb R} x P_\rho (dx)(=\mu) \tag{4.15} \\ &{\mbox{ population variance}}((\sigma_{\mathsf O}^{\rho})^2):\;\; V[{\mathsf M}_{\overline{\mathcal A}} ({\mathsf O}=({\mathbb R}, {\mathcal B}_{\mathbb R} F), S_{[{}\rho{}] }{})] = \int_{\mathbb R} (x- \mu )^2 P_\rho (dx) \tag{4.16} \end{align} Assume that a measured value $(x_1, x_2, x_3,..., x_n ) (\in {\mathbb R}^n)$ is obtained by the parallel measurement $\otimes_{k=1}^n {\mathsf M}_{\overline{\mathcal A}} ({\mathsf O}, S_{[{}\rho{}] }{})$. Put \begin{align} & {\mbox{ sample distribution}}(\nu_n): \nu_n =\frac{ \delta_{x_1}+\delta_{x_2}+ \cdots +\delta_{x_n}}{n} \in {\mathcal M}_{+1}(X) \\ & {\mbox{ sample mean}}(\overline{\mu}_n):{\overline E}[\otimes_{k=1}^n{\mathsf M}_{\overline{\mathcal A}} ({\mathsf O}, S_{[{}\rho{}] }{})] =\frac{ {x_1}+{x_2}+ \cdots +{x_n}}{n} (={\overline{\mu}}) \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\int_{\mathbb R} x \nu_n ( dx ) \\ & {\mbox{ sample variance}}(s^2_n):{\overline V}[\otimes_{k=1}^n{\mathsf M}_{\overline{\mathcal A}} ({\mathsf O}, S_{[{}\rho{}] }{})] =\frac{(x_1 - \overline{ \mu})^2+(x_2 - \overline{ \mu})^2+ \cdot +(x_2 - \overline{ \mu})^2}{n} \\ & \qquad \qquad \qquad \qquad =\int_{\mathbb R} (x- {\overline \mu })^2 \nu_n ( dx ) \\ & {\mbox{ unbiased variance}}(u^2_n):{\overline U}[\otimes_{k=1}^n{\mathsf M}_{\overline{\mathcal A}} ({\mathsf O}, S_{[{}\rho{}] }{})] =\frac{(x_1 - \overline{ \mu})^2+(x_2 - \overline{ \mu})^2+ \cdot +(x_2 - \overline{ \mu})^2}{n-1} \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\frac{n}{n-1 }\int_{\mathbb R} (x- {\overline \mu })^2 \nu_n ( dx ) \end{align} Under the above preparation, we have:
Theorem 4.7 [Population mean,population variance,sample mean,sample variance] Assume that a measured value $(x_1, x_2, x_3, \cdots ) (\in {\mathbb R}^{\mathbb N})$ is obtained by the infinite parallel measurement $\bigotimes_{k=1}^\infty {\mathsf M}_{\overline{\mathcal A}} ({\mathsf O}=({\mathbb R}, {\mathcal B}_{\mathbb R}, F), S_{[{}\rho{}] }{})$. Then, the law of large numbers (Theorem 4.5 ) says that \begin{align} & {{}{(4.15)}}= {\mbox{population mean}}(\mu_{\mathsf O}^{\rho})=\lim_{n \to \infty } \frac{x_1 + x_2 + \cdots + x_n }{n} =:\overline{\mu}=\mbox{sample mean} \\ & {{}{(4.16)}}= {\mbox{population variance}}(\sigma_{\mathsf O}^{\rho})=\lim_{n \to \infty } \frac{(x_1-\mu_{\mathsf O}^{\rho})^2 + (x_2-\mu_{\mathsf O}^{\rho})^2 + \cdots + (x_n-\mu_{\mathsf O}^{\rho})^2 }{n} \nonumber \\ & \qquad \qquad = \lim_{n \to \infty } \frac{(x_1-\overline{\mu})^2 + (x_2-\overline{\mu})^2 + \cdots + (x_n-\overline{\mu})^2 }{n} =:\mbox{sample variance} \end{align}
Example 4.8 [Spectrum decomposition] Consider the quantum basic structure \begin{align} [{\mathcal C}(H ) \subseteq B(H ) \subseteq B(H )] \end{align} Let $A$ be a self-adjoint operator on $H$, which has the spectrum decomposition (i.e., projective observable) ${\mathsf O}_A=({\mathbb R}, {\mathcal B}_{\mathbb R}, F_A)$ such that \begin{align} A= \int_{\mathbb R} \lambda F_A (d \lambda ) \end{align} That is, under the identification: \begin{align} \mbox{ self-adjoint operator: $A$ } \underset{\mbox{ identification}}{\longleftrightarrow} \mbox{ spectrum decomposition:${\mathsf O}_A=({\mathbb R}, {\mathcal B}_{\mathbb R}, F_A)$ } \end{align} the self-adjoint operator $A$ is regarded as the projective observable ${\mathsf O}_A=({\mathbb R}, {\mathcal B}_{\mathbb R}, F_A)$. Fix the state $\rho_u = |u \rangle \langle u | \in {\frak S}^p({\mathcal Tr}(H))$. Consider the measurement ${\mathsf M}_{B(H)} ({\mathsf O}_A , S_{[{}|u\rangle \langle u |{}] }{})$. Then, we see \begin{align} &{\mbox{ population mean}}(\mu_{{\mathsf O}_A}^{\rho_u}):E[{\mathsf M}_{B(H)} ({\mathsf O}_A , S_{[{}|u\rangle \langle u |{}] }{})] = \int_{\mathbb R} \lambda \langle u, F_A(d \lambda ) u \rangle =\langle u, Au \rangle \tag{4.17} \\ &{\mbox{ population variance}}((\sigma_{{\mathsf O}_A}^{\rho_u})^2):\;\;V[{\mathsf M}_{B(H)} ({\mathsf O}_A , S_{[{}|u\rangle \langle u |{}] }{})] = \int_{\mathbb R} (\lambda - \langle u, Au \rangle )^2 \langle u, F_A(d \lambda ) u \rangle \nonumber \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \| (A - \langle u, Au \rangle)u \|^2 \tag{4.18} \end{align}

### 4.2.3: Robertson uncertainty relation

Now we can introduce Robertson's uncertainty principle
Theorem 4.9 [Robertson's uncertainty principle (parallel measurement)] Consider the quantum basic structure $[{\mathcal C}(H) \subseteq B(H) \subseteq B(H)]$. Let ${A_1}$ and ${{A_2}}$ be unbounded self-adjoint operators on a Hilbert space $H$, which respectively has the spectrum decomposition: \begin{align} {\mathsf O}_{A_1}=({\mathbb R}, {\mathcal B}_{\mathbb R}, F_{A_1}) \;\; \mbox{ to } \;\; {\mathsf O}_{A_1}=({\mathbb R}, {\mathcal B}_{\mathbb R}, F_{A_1}) \end{align} Thus, we have two measurements ${\mathsf M}_{B(H)}({\mathsf O}_{A_1}, S_{[{\rho_u}]})$ and ${\mathsf M}_{B(H)}({\mathsf O}_{A_2}, S_{[{\rho_u}]})$, where $\rho_u= |u\rangle \langle u |$ $\in {\frak S}^p({\mathcal C}(H)^*)$. To take two measurements means to take the parallel measurement ${\mathsf M}_{B({\mathbb C}^n)}({\mathsf O}_{A_1}, S_{[{\rho_u}]})$ $\otimes$ ${\mathsf M}_{B({\mathbb C}^n)}({\mathsf O}_{A_2}, S_{[{\rho_u}]})$, namely, \begin{align} {\mathsf M}_{B(H)\otimes B(H)}({\mathsf O}_{A_1}\otimes {\mathsf O}_{A_2}, S_{[{\rho_u} \otimes {\rho_u}]}) \end{align} Then, the following inequality (i.e., Robertson's uncertainty principle ) holds that \begin{align} \sigma_{A_1}^{\rho_u} \cdot \sigma_{A_2}^{\rho_u} {\; \geqq \;} \frac{1}{2} | \langle u , ({A_1}{A_2}-{A_2}{A_1}) u \rangle | \qquad (\forall |u \rangle \langle u |= {\rho_u}, \;\; \| u \|_H=1 ) \end{align} where $\sigma_{A_1}^{\rho_u}$ and $\sigma_{A_2}^{\rho_u}$ are shown in (4.18), namely, \begin{align} \left\{\begin{array}{ll} \sigma_{A_1}^{\rho_u} = \left[ \langle {A_1} u , {A_1} u \rangle - |\langle u , {A_1} u \rangle|^2 \right]^{1/2} =\| (A_1 - \langle u, A_1 u \rangle)u \| \\ \sigma_{A_2}^{\rho_u} = \left[ \langle {A_2} u , {A_2} u \rangle - |\langle u , {A_2} u \rangle|^2 \right]^{1/2} = \| (A_2 - \langle u, A_2 u \rangle)u \| \end{array}\right. \end{align} Therefore, putting $[A_1, A_2 ] \equiv A_1 A_2 - A_2 A_1$, we rewrite Robertson's uncertainty principle as follows: \begin{align} \| A_1 u \| \cdot \| A_2 u \| \ge \| (A_1 - \langle u, A_1 u \rangle)u \| \cdot \| (A_2 - \langle u, A_2 u \rangle)u \| \ge | \langle u, [A_1, A_2 ] u \rangle |/2 \tag{4.19} \end{align} For example, when $A_1(=Q)$ [resp. $A_2(=P)$ ] is the position observable [resp. momentum observable ] (i.e., $QP-PQ=\hbar {\sqrt{-1}}$), it holds that \begin{align} \sigma_Q^{\rho_u} \cdot \sigma_P^{\rho_u} {\; \geqq \;} \frac{1}{2} \hbar \end{align} Proof.: Robertson's uncertainty principle (4.19) is essentially the same as Schwarz inequality, that is, \begin{align} & |\langle u, [A_1, A_2]u \rangle | = | \langle u, (A_1 A_2- A_2 A_1)u \rangle | \\ = & \Big| \Big\langle u, \Big( (A_1 -\langle u , A_1 u \rangle )(A_2 -\langle u , A_2 u \rangle ) - (A_2 -\langle u , A_2 u \rangle )(A_1 -\langle u , A_1 u \rangle ) \Big) u \Big\rangle \Big| \\ \le & 2 \| (A_1 - \langle u, A_1 u \rangle)u \| \cdot \| (A_2 - \langle u, A_2 u \rangle)u \| \end{align}