12.1: Finite realized causal observable

In dualism (i.e., quantum language), Axiom 2 (Causality) is not used independently. Axiom 2 (Causality) is always used with Axiom 1 (measurement). That is because

 $(A_1):$ To be is to be perceived (by George Berkeley(A.D. 1685- A.D. 1753))

$\fbox{Note 12.1}$ Note that Berkeley's words is opposite to Einstein's words:
 $(\sharp):$ The moon is there whether one looks at it or not.
in Einstein and Tagore's conversation.

In this chapter, we devote ourselves to {finite} realized causal observable. ( For the infinite realized causal observable, see Chapter 14.) The readers should understand:

 $\bullet$ "realized causal observable" is a direct consequence of the linguistic interpretation, that is, \begin{align} \mbox{ only one measurement is permitted } \end{align}

Now we will review the following theorem:

Theorem 12.1 [=Theorem 11.1Theorem:Causal operator and observable] Consider the basic structure: \begin{align} [ {\mathcal A}_k \subseteq \overline{\mathcal A}_k \subseteq {B(H_k)}] \qquad (k=1,2) \end{align}

Let $\Phi_{1,2}:\overline{\mathcal A}_2 \to \overline{\mathcal A}_1$ be a causal operator, and let ${\mathsf O}_2$ $=$ $(X , {\cal F} , F_2)$ be an observable in ${\overline{\mathcal A}_2}$. Then, $\Phi_{1,2} {\mathsf O}_2$ $=$ $(X , {\cal F} , \Phi_{1,2} F_2)$ is an observable in ${\overline{\mathcal A}_1}$.

Proof See the proof of Theorem 11.1.
$\square \quad$

In this section,we consider the case that the {tree} ordered set $T(t_0)$ is finite. Thus, putting $T(t_0) = \{ t_0, t_1,\ldots , t_N \}$, consider the finite tree $(T(t_0), {{\; \leqq \;}})$ with the root $t_0$, which is represented by $(T{{=}} \{ t_0,t_1,\ldots, t_N\} , \pi: T \setminus \{ t_0 \} \to T)$ with the the parent map $\pi$.

Definition 12.2 [(finite)sequential causal observable] Consider the basic structure: \begin{align} [ {\mathcal A}_k \subseteq \overline{\mathcal A}_k \subseteq {B(H_k)}] \qquad (t \in T(t_0)=\{t_0, t_1, \cdots, t_n \}) \end{align}

in which, we have a sequential causal operator $\{ \Phi_{t_1,t_2}{}:$ $\overline{\mathcal A}_{t_2} \to \overline{\mathcal A}_{t_1} \}_{(t_1,t_2) \in T^2_{\leqq}}$ (cf. Definition 10.10 ) such that

 $(i):$ for each $(t_1,t_2) \in T^2_{\leqq}$, a causal operator $\Phi_{t_1,t_2}{}:$ ${ \overline{\mathcal A}_{t_2}} \to {\overline{\mathcal A}_{t_1}}$ satisfies that $\Phi_{t_1,t_2} \Phi_{t_2,t_3} = \Phi_{t_1,t_3}$ $(\forall (t_1,t_2)$, $\forall (t_2,t_3) \in T^2_{\leqq})$. Here, $\Phi_{t,t} : {\overline{\mathcal A}_{t}} \to {\overline{\mathcal A}_{t}}$ is the identity.

For each $t \in T$, consider an observable ${\mathsf O }_t {{=}} (X_t, {\cal F}_t , F_t)$ in $\overline{\mathcal A}_{t}$. The pair $[{}\{ {\mathsf O}_t \}_{ t \in T} , \{ \Phi_{t_1,t_2}{}:$ $\overline{\mathcal A}_{t_2} \to \overline{\mathcal A}_{t_1} \}_{(t_1,t_2) \in T^2_{\leqq}}$ $]$ is called a sequential causal observable , denoted by $[{}{\mathsf O}_T{}]$ or $[{}{\mathsf O}_{T(t_0)}{}]$. That is, $[{}{\mathsf O}_T{}]$ $=$ $[{}\{ {\mathsf O}_t \}_{ t \in T} , \{ \Phi_{t_1,t_2}{}:$ $\overline{\mathcal A}_{t_2} \to \overline{\mathcal A}_{t_1} \}_{(t_1,t_2) \in T^2_{\leqq}}$ $]$. Using the parent map $\pi: T\setminus \{t_0\} \to T$, $[{}{\mathsf O}_T{}]$ is also denoted by $[{}{\mathsf O}_T{}]$ $=$ $[{}\{ {\mathsf O}_t \}_{ t \in T} ,$ $\{ \overline{\mathcal A}_{t} \xrightarrow[]{{\Phi_{ \pi(t), t } }} \overline{\mathcal A}_{\pi(t)} \}_{t \in T \setminus \{ t_0 \}) {}}]$.

Now we can show our present problem.

Problem 12.3 We want to formulate the measurement of a sequential causal observable$[{}{\mathsf O}_T{}]$ $=$ $[{}\{ {\mathsf O}_t \}_{ t \in T} , \{ \Phi_{t_1,t_2}{}:$ $\overline{\mathcal A}_{t_2} \to \overline{\mathcal A}_{t_1} \}_{(t_1,t_2) \in T^2_{\leqq}}$ $]$ for a system $S$ with an initial state $\rho_{t_0} (\in {\frak S}^p({\mathcal A}_{t_0}^*))$.

\begin{align} \color{magenta}{ \mbox{ How do we formulate this measurement? } } \end{align}

Now let us solve this problem as follows. Note that the linguistic interpretation says that

\begin{align} \mbox{ only one measurement (and thus, only one observable) is permitted} \end{align}

Thus, we have to combine many observables in a sequential causal observable$[{}{\mathsf O}_T{}]$ $=$ $[{}\{ {\mathsf O}_t \}_{ t \in T} ,$ $\{ \Phi_{t_1,t_2}{}:$ $\overline{\mathcal A}_{t_2} \to \overline{\mathcal A}_{t_1} \}_{(t_1,t_2) \in T^2_{\leqq}}$ $]$. This is realized as follows.

Definition 12.4 [Realized causal observable]

Let $T(t_0) = \{ t_0, t_1,\ldots , t_N \}$ be a {finite} tree. Let $[{\mathsf O}_{T(t_0)}]$ $=$ $[{}\{ {\mathsf O}_t \}_{ t \in T}, \{ \Phi_{\pi(t), t }{}:$ $\overline{\mathcal A}_{t}$ $\xrightarrow[]{{\Phi_{ \pi(t), t } }}$ $\overline{\mathcal A}_{\pi(t)} \}_{ t \in T\setminus \{t_0\} }$ $]$ be a sequential causal observable.

For each $s$ $(\in T)$, put $T_s = \{ t \in T \;|\; t {\; \geqq \;}s \}$. Define the observable $\widehat{\mathsf O}_s {{=}} (\times_{t \in T_s } X_t,$ $\boxtimes_{t \in T_s } {\cal F}_t, {\widehat F}_s)$ in $\overline{\mathcal A}_{s}$ such that

\begin{align} \widehat{\mathsf O}_s = \left\{\begin{array}{ll} {\mathsf O}_s \quad & \mbox{( if $s \in T \setminus \pi (T) \;$)} \\ \\ {\mathsf O}_s {\times} (\times_{t \in \pi^{-1} (\{ s \})} \Phi_{ \pi(t), t} \widehat {\mathsf O}_t) \quad & \mbox{( if $s \in \pi (T) \;$)} \end{array}\right. \tag{12.1} \end{align}

(In quantum case,the existence of $\widehat{\mathsf O}_s$ is not always guaranteed). And further, iteratively, we get the observable $\widehat{\mathsf O}_{t_0{}}$ $=$ $(\times_{t \in T } X_t,$ $\boxtimes_{t \in T } {\cal F}_t,$ ${\widehat F}_{t_0})$ in $\overline{\mathcal A}_{t_0}$. Put $\widehat{\mathsf O}_{t_0{}}$ $=$ $\widehat{\mathsf O}_{T(t_0){}}$.

The observable $\widehat{\mathsf O}_{T(t_0){}}$ $=$ $(\times_{t \in T } X_t,$ $\boxtimes_{t \in T } {\cal F}_t,$ ${\widehat F}_{t_0})$ is called the (finite) realized causal observable of the sequential causal observable$[{}{\mathsf O}_{T(t_0)}{}]$ $=$ $[{}\{ {\mathsf O}_t \}_{ t \in T} , \{ \Phi_{\pi(t), t }{}:$ $\overline{\mathcal A}_{t} \to \overline{\mathcal A}_{\pi(t)} \}_{ t \in T\setminus \{t_0\} }$ $]$.

Summing up the above arguments, we have the following theorem:

In the classical case, the realized causal observable $\widehat{\mathsf O}_{T(t_0){}}$ $=$ $(\times_{t \in T } X_t,$ $\boxtimes_{t \in T } {\cal F}_t,$ ${\widehat F}_{t_0})$ always exists.

 $\fbox{Note 12.2}$ In the above (12.1), the product "$\times$" may be generalized as the quasi-product "$\overset{qp}{\times}$". However, in this note we are not concerned with such generalization.

Example 12.5 [A simple classical example]

Suppose that a tree $(T \equiv \{ 0, 1, ..., 6, 7 \}, \pi)$ has an ordered structure such that $\pi(1) = \pi(6) = \pi(7) = 0$, $\pi(2) = \pi(5) = 1$, $\pi(3) = \pi(4) = 2$.

Consider a sequential causal observable $[{}{\mathsf O}_T{}]$ $=$ $[{}\{ {\mathsf O}_t \}_{ t \in T} ,$ $\{ {L^\infty (\Omega_{t})} {{\Phi_{ \pi(t), t } }\atop{\rightarrow}}$ $L^\infty (\Omega_{\pi(t)}) \}_{t \in T \setminus \{ 0 \}) {}}]$. Now, we will construct its realized causal observable $\widehat{\mathsf O}_{T(t_0){}}$ $=$ $(\times_{t \in T } X_t,$ $\boxtimes_{t \in T } {\cal F}_t,$ ${\widehat F}_{t_0})$ in what follows.

Put

\begin{align} \widehat{\mathsf O}_t = {\mathsf O}_t \quad \mbox{ and thus} \quad \widehat{F}_t = {F}_t \quad (t = 3,4,5,6,7). \end{align}

First we construct the product observable $\widehat{\mathsf O}_2$ in ${L^\infty (\Omega_2)}$ such as

\begin{align} \widehat{\mathsf O}_2 = (X_2 \times X_3 \times X_4, {\cal F}_2 \boxtimes {\cal F}_3 \boxtimes {\cal F}_4, {\widehat F}_2 ) \quad \mbox{ where } {\widehat F}_2 = F_2 \times (\times_{t=3,4} \Phi_{2,t} {\widehat F}_t) , \end{align}

Iteratively, we construct the following:

That is, we get the product observable $\widehat{\mathsf O}_1 \equiv ({\times}_{t=1}^5 X_t, {{\boxtimes}_{t=1}^5 {\cal F}_t}, {\widehat F}_1)$ of ${\mathsf O}_1$, $\Phi_{1,2} \widehat{\mathsf O}_2$ and $\Phi_{1,5} \widehat{\mathsf O}_5$, and finally, the product observable

\begin{align} \widehat{\mathsf O}_0 \equiv ({\times}_{t=0}^7 X_t, {{\boxtimes}_{t=0}^7 {\cal F}_t}, {\widehat F}_0 (= F_0 \times (\times_{t=1,6,7} \Phi_{0,t} {\widehat F}_t) ) \end{align}

of ${\mathsf O}_0$, $\Phi_{0,1} \widehat{\mathsf O}_1$, $\Phi_{0,6} \widehat{\mathsf O}_6$ and $\Phi_{0,7} \widehat{\mathsf O}_7$.

Then, we get the realization of a sequential causal observable $[{}\{{\mathsf O}_t\}_{t \in T },$ $\{ {L^\infty (\Omega_t)} \overset{\Phi_{\pi(t), t}}\to {L^\infty (\Omega_{\pi(t)})} \}_{ t \in T \setminus \{0\} }{}]$. For completeness, ${\widehat F}_0$ is represented by

\begin{align} & \widehat{F}_0 (\Xi_0 \times \Xi_1 \times \Xi_2 \times \Xi_3 \times \Xi_4 \times \Xi_5 \times \Xi_6 \times \Xi_7)] \nonumber \\ = & F_0(\Xi_0) \times \Phi_{0,1} \biggl( F_1(\Xi_1) \times \Phi_{1,5}F_5(\Xi_5) \times \Phi_{1,2} \Big( F_2(\Xi_2) \times \Phi_{2,3}F_3(\Xi_3) \times \Phi_{2,4}F_4(\Xi_4) \Big) \biggl) \nonumber \\ & \qquad \qquad \qquad \qquad \times \Phi_{0,6}( F_6(\Xi_6)) \times \Phi_{0,7}( F_7(\Xi_7)) \tag{12.2} \end{align}

(In quantum case,the existence of $\widehat{\mathsf O}_0$ in not guaranteed).

$\square \quad$

Remark 12.6 In the above example, consider the case that ${\mathsf O}_t$ ($t=2,6,7$) is not determined. In this case,it suffices to define ${\mathsf O}_t$ by the {{}}{existence observable } ${\mathsf O}^{\rm{(exi)}}_t {{=}} (X_t , \{ \emptyset, X_t \}, F^{\rm{(exi)}}_t)$. Then, we see that

\begin{align} & \widehat{F}_0 (\Xi_0 \times \Xi_1 \times X_2 \times \Xi_3 \times \Xi_4 \times \Xi_5 \times X_6 \times X_7 ) \nonumber \\ =& F_0(\Xi_0) \times \Phi_{0,1} \biggl( F_1(\Xi_1) \times \Phi_{1,5}F_5(\Xi_5) \times \Phi_{1,2} \Big( \Phi_{2,3}F_3(\Xi_3) \times \Phi_{2,4}F_4(\Xi_4) \Big) \biggl) \tag{12.3} \end{align}

This is true. However, the following is not wrong. Putting $T'=\{0,1,3,4,5 \}$, consider the $[{}{\mathsf O}_{T'}{}]=$ $[{}\{ {\mathsf O}_t \}_{ t \in {T'}} , \{ \Phi_{t_1,t_2}{}:$ ${L^\infty (\Omega_{t_2})} \to {L^\infty (\Omega_{t_1})} \}_{(t_1,t_2) \in (T')^2_{{\; \leqq \;}}}]$.

Then, the realized causal observable $\widehat{\mathsf O}_{T'(0){}}$ $=$ $(\times_{t \in T' } X_t,$ $\boxtimes_{t \in T' } {\cal F}_t,$ ${\widehat F}'_{0})$ is defined by

\begin{align} & \widehat{F}'_0 (\Xi_0 \times \Xi_1 \times \Xi_3 \times \Xi_4 \times \Xi_5 ) = F_0(\Xi_0) \nonumber \\ & \times \Phi_{0,1} \Big( F_1(\Xi_1) \times \Phi_{1,5}F_5(\Xi_5) \times \Phi_{1,4}F_4(\Xi_4)\times \Phi_{1,3}F_3(\Xi_3) \times \Phi_{1,4}F_4(\Xi_4) \Big) \tag{12.4} \end{align}

which is different from the true (12.2). We may sometimes omit "existence observable". However, if we do so, we omit it on the basis of careful cautions.

$\square \quad$

Thus,we can answer Problem 12.3 as follows.

Problem 12.7 [=Problem 12.3](written again) We want to formulate the measurement of a sequential causal observable$[{}{\mathsf O}_T{}]$ $=$ $[{}\{ {\mathsf O}_t \}_{ t \in T} , \{ \Phi_{t_1,t_2}{}:$ $\overline{\mathcal A}_{t_2} \to \overline{\mathcal A}_{t_1} \}_{(t_1,t_2) \in T^2_{\leqq}}$ $]$ for a system $S$ with an initial state $\rho_{t_0} (\in {\frak S}^p({\mathcal A}_{t_0}^*))$.

\begin{align} \mbox{ How do we formulate the measurement? } \end{align}

Answer If the realized causal observable $\widehat{\mathsf O}_{t_0}$ exists, the measurement is formulated by

\begin{align} \mbox{ measurement } {\mathsf M}_{\overline{\mathcal A}_{t_0}}(\widehat{\mathsf O}_{t_0}, S_{[\rho_{t_0}]}) \end{align}

Thus, according to Axiom 1 ( measurement: $\S$2.7), we see that

 $(B):$ The probability that a measured value $(x_t)_{t \in T}$ obtained by the measurement ${\mathsf M}_{\overline{\mathcal A}_{t_0}}(\widehat{\mathsf O}_{T{}}, S_{[\rho_{t_0}]})$ belongs to ${\widehat \Xi}( \in \boxtimes_{t\in T}{\cal F}_t )$ is given by \begin{align} {}_{{\mathcal A}_0^*} \Big( \rho_{t_0}, {\widehat F}_{t_0} ({\widehat \Xi} ) \Big) {}_{\overline{\mathcal A}_{t_0}} \tag{12.5} \end{align}

The following theorem, which holds in classical systems, is frequently used.

Theorem 12.8 [The realized causal observable of deterministic sequential causal observable in classical systems]

Let $(T(t_0), {{\; \leqq \;}})$ be a finite tree. For each $t \in T(t_0)$, consider the classical basic structure

\begin{align} [C_0(\Omega_t) \subseteq L^\infty (\Omega_t, \nu_t ) \subseteq B(L^2 (\Omega_t, \nu_t ))] \end{align}

Let $[{}{\mathbb O}_T{}]$ $=$ $[{}\{ {\mathsf O}_t \}_{ t \in T} , \{ \Phi_{t_1,t_2}{}:$ ${L^\infty (\Omega_{t_2})} \to {L^\infty (\Omega_{t_1})} \}_{(t_1,t_2) \in T^2_{\leqq}}$ $]$ be deterministic causal observable. Then, the realization $\widehat{\mathsf O}_{{t_0}{}}$ $\equiv ({\times}_{t \in T} X_t,{{\boxtimes}_{t \in T} {\cal F}_t}, {\widehat F}_{t_0})$ is represented by

\begin{align} \widehat{\mathsf O}_{{t_0}{}} = \times_{t\in T} \Phi_{{t_0},t} {\mathsf O}_t \end{align}

That is, it holds that

\begin{align} & [\widehat{F}_{t_0} ( \times_{t\in T} \Xi_t \ )] (\omega_{t_0} ) = \times_{t\in T} [\Phi_{{t_0},t} {F}_t (\Xi_t )](\omega_{t_0} ) = \times_{t\in T} [{F}_t (\Xi_t )](\phi_{{t_0},t} \omega_{t_0} ) \\ & \quad \qquad \quad \qquad \quad \qquad \quad \qquad (\forall \omega_{t_0} \in \Omega_{t_0}, \forall \Xi_t \in {\cal F}_t ) \end{align}

Proof

It suffices to prove the simple classical case of Example 12.5. Using Theorem 10.6 repeatedly, we see that

\begin{align} & {\widehat F}_0 = F_0 \times (\times_{t=1,6,7} \Phi_{0,t} {\widehat F}_t) \\ = & F_0 \times ( \Phi_{0,1} {\widehat F}_1{} \times \Phi_{0,6} {\widehat F}_6{} \times \Phi_{0,7} {\widehat F}_7{} ) = F_0 \times ( \Phi_{0,1} {\widehat F}_1{} \times \Phi_{0,6} {F}_6{} \times \Phi_{0,7} { F}_7{} ) \\ = & \Big( \times_{t=0,6,7} \Phi_{0,t} F_t \Big) \times ( \Phi_{0,1} {\widehat F}_1{} ) = \Big(\times_{t=0,6,7} \Phi_{0,t} F_t \Big) \times \Phi_{0,1} ( F_1 \times (\times_{t=2,5} \Phi_{1,t} {\widehat F}_t) {} ) \\ = & \Big( \times_{t=0,1,6,7}\Phi_{0,t} F_t \Big) \times \Phi_{0,1}(\times_{t=2,5} \Phi_{1,t} {\widehat F}_t) = \Big( \times_{t=0,1,6,7}\Phi_{0,t} F_t \Big) \times \Phi_{0,1}( \Phi_{1,2} {\widehat F}_2{} \times \Phi_{1,5} {\widehat F}_5) \\ = & \Big( \times_{t=0,1,5,6,7}\Phi_{0,t} F_t \Big) \times \Phi_{0,1}( \Phi_{1,2} {\widehat F}_2{} ) = \Big( \times_{t=0,1,5,6,7}\Phi_{0,t} F_t \Big) \times \Phi_{0,1}( \Phi_{1,2} (F_2 \times (\times_{t=3,4} \Phi_{2,t} {\widehat F}_t)) {} ) \\ = & \times_{t=0}^7 \Phi_{0,t} F_t \end{align} This completes the proof.
$\square \quad$