9.2: Simple examples in mixed measurement theory

Recall the following wise sayings:

 $\quad$ experience is the best teacher, or custom makes all things

Thus, in what follows I do it Let us start fron the following review in $\S$5.2:

Review 9.4 [Answer 5.7 to Problem 5.2 by Fisher's maximum likelihood method]

You do not know which the urn behind the curtain is, $U_1$ or $U_2$. Assume that you pick up a white ball from the urn. The urn is $U_1$ or $U_2$? $\quad$ Which do you think? Figure 9.1(= Figure 5.6):Pure measurement (Fisher's maximum likelihood method)

Answer: Consider the state space $\Omega=\{\omega_1, \omega_2\}$ with the discrete topology and the measure $\nu$ such that

\begin{align} \nu(\{ \omega_1 \})=1, \qquad \nu(\{ \omega_2 \})=1 \tag{9.3} \end{align}

In the classical basic structure $[C_0(\Omega) \subseteq L^\infty (\Omega, \nu ) \subseteq B(L^2 (\Omega, \nu ))]$, consider the measurement ${\mathsf M}_{{L^\infty (\Omega) }} ({\mathsf O} {{=}}$ $( \{ W,$ $B \},$ $2^{\{ W, B \} } ,$ $F_{WB}) , S_{ [{}{\ast}]})$, where the observable ${\mathsf O}_{WB} = ( \{ W,B \}, 2^{\{ W,B \} } , F_{WB})$ in $L^\infty (\Omega)$ is defined by

\begin{align} & [F_{WB}(\{ W \})](\omega_1)= 0.8, & \quad & [ F_{WB}(\{ B \})](\omega_1)= 0.2 \nonumber \\ & [F_{WB}(\{ W \})](\omega_2)= 0.4, & \quad & [F_{WB}(\{ B \})] (\omega_2)= 0.6 \tag{9.4} \end{align}

Here, we see:

\begin{align} & \max \{[F_{WB}(\{W\})](\omega_1), [F_{WB}(\{W\})](\omega_2) \} \\ = & \max \{0.8, 0.4\} = 0.8 = F_{WB}(\{W\})](\omega_1) \end{align}

Then, Fisher's maximum likelihood method (Theorem 5.6) says that

\begin{align} [\ast ] = \omega_1 \end{align}

Therefore, there is a reason to infer that the urn behind the curtain is $U_1$.

${\square} \quad$

Thus, we exercise the following problem.
Problem 9.5 [mixed measurement ${\mathsf M}_{L^\infty (\Omega, \nu )} ({\mathsf O}=(X,{\mathcal F}, F ) , S_{[\ast]} (w) )$] Figure 9.2[ Mixed measurement (Urn problem)
 $(\sharp_1):$ Assume an unfair coin-tossing $(T_{p,1-p})$ such that $( 0 {{\; \leqq \;}}p {{\; \leqq \;}}1 )$: That is, $\quad$ $\left\{\begin{array}{ll} \mbox{the possibility that "head" appears is} \mbox{$100p \% $} \\ \mbox{the possibility that "tail" appears is} \mbox{$100(1-p) \% $} \end{array}\right.$ If "head" [resp. "tail"]appears, put an urn $U_1({\approx} \omega_1)$ [resp. $U_2({\approx} \omega_2)$] behind the curtain. Assume that you do not know which urn is behind the curtain, $U_1$ or $U_2$). The unknown urn is denoted by $[*{}](\in \{\omega_1, \omega_2\})$. This situation is represented by $w \in L^1_{+1}(\Omega, \nu )$ (with the counting measure $\nu$), that is, \begin{align} w(\omega ) = \left\{\begin{array}{ll} p \qquad & \mbox{( if $\omega=\omega_1$ )} \\ 1-p \qquad & \mbox{( if $\omega=\omega_2$ )} \end{array}\right. \end{align} $(\sharp_2):$ Consider the "measurement" such that a ball is picked out from the unknown urn. This "measurement" is denoted by ${\mathsf M}_{L^\infty (\Omega,\nu)}({\mathsf O}, {\overline S}_{[*]}(w))$, and called a mixed measurement.

Then, we have the following problems:
 $(a):$ Calculate the probability that a white ball is picked from the unknown urm behind the curtain!
And further,
 $(b):$ when a white ball is picked, calculate the probability that the unknown urm behind the curtain is $U_1$!

We would like to remark as follows.
 $\bullet$ in the above problem, the term "subject probability" is not used.

Answer: Asssume that the state space$\Omega=\{\omega_1, \omega_2\}$ is defined by the discrete metric with the following measure $\nu$: \begin{align} \nu(\{ \omega_1 \})=1, \qquad \nu(\{ \omega_2 \})=1 \tag{9.5} \end{align} Thus, we start from the clasical basic structure: \begin{align} [C_0(\Omega) \subseteq L^\infty (\Omega, \nu ) \subseteq B(L^2 (\Omega, \nu ))] \tag{9.6} \end{align}

in which we consider the mixed measurement ${\mathsf M}_{{L^\infty (\Omega) }} ({\mathsf O} {{=}}$ $( \{ W,$ $B \},$ $2^{\{ W, B \} } ,$ $F) , S_{ [{}{\ast}]}{(w)})$. Here, the observable ${\mathsf O}_{WB} = ( \{ W, B \}, 2^{\{ W, B \} } , F_{WB})$ in $L^\infty (\Omega)$ is defined by

\begin{align} & [F_{WB}(\{ W \})](\omega_1)= 0.8, & \quad & [ F_{WB}(\{ B \})](\omega_1)= 0.2 \nonumber \\ & [F_{WB}(\{ W \})](\omega_2)= 0.4, & \quad & [F_{WB}(\{ B \})] (\omega_2)= 0.6 \tag{9.7} \end{align}

Also,the mixed state $w_0 \in L^1_{+1}(\Omega , \nu )$ is defined by

\begin{align} w_0(\omega_1 ) = p, \qquad w_0(\omega_2)=1-p \tag{9.8} \end{align}

Then, by Axiom$^{(m)}$1 ( in $\S$9.1), we see

 $(a):$ the probability that a measured value $x$ $(\in \{ W , B \})$ is obtained by ${\mathsf M}_{{L^\infty (\Omega) }} ({\mathsf O} {{=}}$ $( \{ W,$ $B \},$ $2^{\{ W, B \} } ,$ $F) , S_{ [{}{\ast}]}{(w)})$ is given by
\begin{align} P(\{ x \}) &= {}_{L^1(\Omega )} \big( w_0, F(\{x \}) \big)_{L^\infty(\Omega)} = \int_\Omega [F(\{ x \})]( \omega) \cdot w_0 (\omega) \nu(d \omega) \nonumber \\ & = p [F(\{ x \})](\omega_1) + (1-p) [F(\{ x \})](\omega_2) \nonumber \\ &= \left\{\begin{array}{ll} 0.8 p + 0.4 (1-p) \quad & (x=W{}\; \mbox{のとき}) \\ 0.2 p + 0.6 (1-p) \quad & (x=B{}\; \mbox{のとき}) \end{array}\right. \tag{9.9} \end{align}

The question (b) will be answered in Answer9.13.

${\square} \quad$

$\fbox{Note 9.1}$ The following question is natural. That is,
 $(\flat_1):$ In the above ($\sharp_1$) in Problem 9.5Problem, why is "the {possibility} that $[{}\ast{}] = {\omega_1}$ is $100p\%$ $\cdots$" replaced by "the probability that $[{}\ast{}] = {\omega_1}$ is $100p\%$ $\cdots$" ?
However, the linguistic interpretation says that
 $(\flat_2):$ there is no probability without measurements.
This is the reason why the term "probability" is not used in (i). However, from the practical point of view, we are not sensitive to the difference between "probability" and "possibility".

Example 9.6 [Mixed spin measurement ${\mathsf M}_{B({\mathbb C}^2)} ({\mathsf O}=(X= \{ \uparrow, \downarrow \} ,{2}^X, F^z ) , {\overline S}_{[\ast]} (w) )$]

Consider the quantum basic structure: \begin{align} [{\mathcal C}({\mathbb C}^2) ( = B({\mathbb C}^2) ) \subseteq B({\mathbb C}^2) \subseteq B({\mathbb C}^2)] \end{align}

And consider a particle $P_1$ with spin state $\rho_1=|a \rangle \langle a|$ $\in$ ${\frak S}^p(B({\mathbb C}^2))$, where

\begin{align} a= \left[\begin{array}{l} \alpha_1 \\ \alpha_2 \end{array}\right] \in {\mathbb C}^2 \quad (\mbox{ }\|a \|= (|\alpha_1|^2+ |\alpha_2|^2)^{1/2}=1) \end{align}

And consider another particle $P_2$ with spin state $\rho_2=|b \rangle \langle b|$ $\in$ ${\frak S}^p(B({\mathbb C}^2))$, where

\begin{align} b= \left[\begin{array}{l} \beta_1 \\ \beta_2 \end{array}\right] \in {\mathbb C}^2 \quad (\mbox{ }\|b \|= (|\beta_1|^2+ |\beta_2|^2)^{1/2}=1) \end{align} Here, assume that
 $\bullet$ the "probability" that the "particle" $P$ is $\left\{\begin{array}{ll} \mbox{ a particle }P_1 \\ \mbox{ a particle } P_2 \end{array}\right\}$ is given by $\left\{\begin{array}{ll} p \\ 1-p \end{array}\right\}$
That is, \begin{align} \underset{\mbox{ (Particle $P_1$)}}{\fbox{state $\rho_1$}} \xrightarrow[\mbox{probability }\;p]{} \underset{\mbox{ (Particle $P$)}}{\fbox{unknown state $[\ast]$}} \xleftarrow[\mbox{probability}\;1-p]{} \underset{\mbox{ (Particle $P_2$)}}{\fbox{state $\rho_2$}} \end{align}

Here, the unknown state $[\ast]$ of Particle $P$ is represented by the mixed state$w$ $( \in$ ${\frak S}^m ({\mathcal Tr}({\mathbb C}^2)))$ such that

\begin{align} w=p \rho_1 + (1-p) \rho_2 =p |a \rangle \langle a | + (1-p) |b \rangle \langle b| \end{align}

Therefore, we have the mixed measurement ${\mathsf M_{B({\mathbb C}^2)}}({\mathsf O}_z =(X,2^X, F^z ), {\overline S}_{[\ast]}(w))$ of the $z$-axis spin observable ${\mathsf O}_z =(X,{\mathcal F}, F^z )$, where

\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} And we say that
 $(a):$ the probability that a measured value $\left\{\begin{array}{ll} \uparrow \\ \downarrow \end{array}\right\}$ is obtained by the mixed measurement ${\mathsf M_{B({\mathbb C}^2)}}({\mathsf O}_z =(X,2^X, F^z ), {\overline S}_{[\ast]}(w))$ is given by \begin{align} \left\{\begin{array}{ll} {}_{{\mathcal Tr}({\mathbb C}^2)} \Big(w, F^z( \{ \uparrow \}) \Big) {}_{B({\mathbb C}^2)} = p|\alpha_1|^2 +(1-p) |\beta_1|^2 \\ \\ {}_{{\mathcal Tr}({\mathbb C}^2)} \Big(w, F^z( \{ \downarrow \}) \Big) {}_{B({\mathbb C}^2)} = p |\alpha_2|^2 + (1-p)|\beta_2|^2 \end{array}\right\} \end{align}
Remark 9.7

As seen in the above, we say that

 $(a):$ Pure measurement theory is fundamental. Adding the concept of "mixed state", we can construct mixed measurement theory as follows. \begin{align} \underset{\mbox{ ${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O} , {\overline S}_{[\ast ]}{(w)})$}}{\mbox{$\fbox{ mixed measurement theory }$}} := { { \underset{\mbox{ ${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O} , S_{[\ast ]})$}}{\mbox{ $\fbox{ pure measurement theory }$}}} } + { { \underset{\mbox{ $w$}} {\mbox{$\fbox{ mixed state}$}} } } \end{align}
That is, we devote ourselves to the case that
There is no mixed measurement without pure measurement
(though the other cases (cf. Remark 18.2 ) may not be neglected). Hence, in quantum language, there is no confrontation between "frequency probability" and "subjective probability". The reason that a coin-tossing is used in Problem 9.5 is to emphasize that the naming of "subjective probability" is improper.