14.3: The Schrödinger picture of the sequential deterministic causal operator

14.3.2: The preparation of the next section ($\S$14.4: Zeno's paradox)

The linguistic interpretation ($\S$3.1) says that

\begin{align} \mbox{ a state does no move, } \end{align}

which is called the Heisenberg picture (i.e., a state does not move, and, an observable moves). This is formal. On the other hand, we sometimes use the Schrödinger picture (i.e., a state moves, and, an observable does not move), which is handy and makeshift.

In this section, we explain something about the Schrödinger picture in classical deterministic systems. This section is the preparation of the next section (Zeno's paradoxes).

Let $(T{(t_0)}, {{\; \leqq \;}})$ be an infinite tree with the root $t_0$. For each $t \in T$, 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}

Definition 14.4 [State changes --- the Schrödinger picture]

Let $\{ \Phi_{t_1,t_2}{}:$ ${L^\infty (\Omega_{t_2}, \nu_{t_2})}$ $\to {L^\infty (\Omega_{t_1})},$ ${\nu_{t_1})} \}_{(t_1,t_2) \in T^2_{\leqq}}$ be a deterministic causal relation with the deterministic causal maps $\phi_{t_1,t_2 }:\Omega_{t_1} \to \Omega_{t_2}$ $(\forall {(t_1,t_2) \in T^2_{\leqq}} )$. Let $\omega_{t_0} \in \Omega_{t_0}$ be an initial state. Then, the $\{ \phi_{t_0,t} (\omega_{t_0} ) \}_{t \in T }$ (or, $\{ \delta_{\phi_{t_0,t} (\omega_{t_0} )} \}_{t \in T }$ is called the Schrödinger picture representation.

The following is the infinite version of Theorem 12.8.

Theorem 14.5 [Deterministic sequential causal operator and realized causal observable]

Let $(T(t_0), {{\; \leqq \;}})$ be an infinite tree with the root $t_0$. Let $[{}{\mathbb O}_T{}]$ $=$ $[{}\{ {\mathsf O}_t \}_{ t \in T} , \{ \Phi_{t_1,t_2}{}:$ ${L^\infty (\Omega_{t_2}, \nu_{t_2})} \to {L^\infty (\Omega_{t_1}, \nu_{t_1})} \}_{(t_1,t_2) \in T^2_{\leqq}}$ $]$ be a deterministic sequential 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 The proof is similar to that of Theorem 12.8

$\square \quad$

Theorem 14.6 Let $[{}{\mathbb O}_{T(t_0)}]$ ${{=}}$ $[{}\{ {\mathsf O}^{\scriptsize{\mbox{exa}}}_t \}_{ t \in T} , \{ \Phi_{t_1,t_2}{}:$ ${L^\infty (\Omega_{t_2}, \nu_{t_2})} \to {L^\infty (\Omega_{t_1}, \nu_{t_1})} \}_{(t_1,t_2) \in T^2_{\leqq}}$ $]$ be a deterministic sequential causal exact observable, which has the deterministic causal maps $\phi_{t_1,t_2 }:\Omega_{t_1} \to \Omega_{t_2}$ $(\forall {(t_1,t_2) \in T^2_{\leqq}} )$. And let ${\widehat{\mathsf O}}_{{{t_0}}}$ ${{=}}$ $(\mathop{\Large{\mbox{$\times$}}}_{{t} \in {T}}X_{{t}},$ $\mathop{\Large{\mbox{$\times$}}}_{{t}\in{{T}}} {\cal F}_{{t}},$ ${\widehat F}_{{T}} )$ be its realized causal observable in $L^\infty(\Omega_{t_0}, \nu_{t_0} )$. Assume that the measured value $(x_t )_{t\in T }$ is obtained by ${{\mathsf M}}_{L^\infty(\Omega_{t_0})} (\widehat{\mathbb O}_{T{}}$ ${{=}}$ $(\times_{t \in T} X_t, \mathop{\Large{\mbox{$\times$}}}_{{t}\in{{T}}} {\cal F}_{{t}} ,$ ${\widehat F}_0), S_{[\omega_{t_0}]})$. Then, we surely believe that

\begin{align} x_t = \phi_{t_0,t } (\omega_{t_0} ) \qquad (\forall t \in T ) \end{align}

Thus, we say that, as far as a deterministic sequential causal observable,

 $(a):$ exact measured value $(x_t)_{t\in T}$ = the Schrödinger picture representation $(\phi_{t_0,t } (\omega_{t_0} ))_{t\in T}$

Proof Let $D=\{t_1,t_2,\ldots,t_n\} (\subseteq T)$ be any finite subset of $T$. Put ${\widehat\Xi}=\Large{\mbox{$\times$}}_{t\in {T} }^D\Xi_t$ $=$ $(\times_{t\in D} \Xi_t ) \times (\times_{t\in T\setminus D} X_t )$, where $\Xi_t$ $\subseteq$ $X_t ( = \Omega_t)$ is an open set such that $\phi_{t_0, t} (\omega_{t_0}) \in \Xi_t$ $(\forall t\in D)$. Then, we see that

 $(b):$ the probability that the measured value $(x_t )_{t\in T }$ belongs to ${\widehat\Xi}=\Large{\mbox{$\times$}}_{t\in {T} }^D\Xi_t$ is equal to $1$.

That is because Theorem 14.5 says that

\begin{align} & \bigl({\widehat F}_{T} ({\widehat\Xi}) \bigr) ({\omega_{t_0}}) = \Big(\times_{k=1}^n \bigl(\Phi_{t_0 , t_k} F^{\scriptsize{\mbox{exa}}}(\Xi_{t_k} ) \bigl) \Big)({\omega_{t_0}}) \\ = & \Big(\times_{k=1}^n F^{\scriptsize{\mbox{exa}}} (\phi_{t_0, t_k}^{-1}(\Xi_{t_k} ) \bigl) \Big)({\omega_{t_0}}) = \times_{k=1}^n \chi_{{}_{\Xi_{t_k}}}({\phi_{t_0, t_k} (\omega_{t_0})}) =1 \end{align}

Thus, from the arbitrariness of $\Xi_t$, we surely believe that

 $(c):$ $(x_t )_{t\in T } = \phi_{t_0 , t} (\omega_{t_0}) \qquad (\forall t \in T)$
$\square \quad$

 $\fbox{Note 14.2}$ Note that "(b) $\Leftrightarrow$(c)" in the above. That is, (b) is the definition of (c).

Thus, we have the following corollary, which is the generalization of Theorem 3.15.

Corollary 14.7 [System quantity and exact observable] For each $t \in T(t_0)$, consider the exact observable ${\mathsf O}^{\scriptsize{\mbox{exa}}}_t=(X, {\cal F}_t, F^{\scriptsize{\mbox{exa}}}) (=(\Omega_t , {\cal B}_t , \chi ))$ in $L^\infty(\Omega_t, \nu_t)$ and a system quantity $g_t:\Omega_t \to {\mathbb R}$ on $\Omega_t$. Let ${\mathsf O}'_t = ({\mathbb R}, {\cal B}_{\mathbb R}, G_t)$ be the observable representation of the quantity $g_t$ in $L^\infty (\Omega_t)$. Assuming the simultaneous observable ${\mathsf O}^{\scriptsize{\mbox{exa}}}_t \times{\mathsf O}'_t$, define the sequential deterministic causal observable:

\begin{align} \mbox{ $[{}{\mathbb O}_{T(t_0)}]$ ${{=}}$ $[{}\{ {\mathsf O}^{\scriptsize{\mbox{exa}}}_t \times{\mathsf O}'_t \}_{ t \in T} , \{ \Phi_{t_1,t_2}{}:$ ${L^\infty (\Omega_{t_2}, \nu_{t_2})} \to {L^\infty (\Omega_{t_1}, \nu_{t_1})} \}_{(t_1,t_2) \in T^2_{\leqq}}$ $]$ } \end{align}

Let $\phi_{t_1,t_2 }:\Omega_{t_1} \to \Omega_{t_2}$ $(\forall {(t_1,t_2) \in T^2_{\leqq}} )$ be the deterministic causal map. Let ${\widehat{\mathsf O}}_{{{t_0}}}$ ${{=}}$ $\bigl(\mathop{\Large{\mbox{$\times$}}}_{{t} \in {T}}(X_{{t}}\times {\mathbb R}),$ $\mathop{\Large{\mbox{$\boxtimes$}}}_{{t}\in{{T}}} ({\cal F}_{{t}}\boxtimes {\cal B}_{\mathbb R}),$ ${\widehat F}_{{t_0}} \bigr)$ be the realized causal observable. Thus, we have the measurement ${{\mathsf M}}_{L^\infty(\Omega_{t_0})} (\widehat{\mathsf O}_{t_0{}},$ $S_{[\omega_{t_0}]})$. Let $(x_t , y_t)_{t\in T }$ be the measured value obtained by the measurement ${{\mathsf M}}_{L^\infty(\Omega_{t_0})} (\widehat{\mathsf O}_{t_0{}},$ $S_{[\omega_{t_0}]})$. Then, we can surely believe that

\begin{align} x_t = \phi_{{t_0},t } (\omega_{t_0} ) \;\; \mbox{ and } \;\; y_t = g_t(\phi_{{t_0},t } (\omega_{t_0} )) \qquad (\forall t \in T ) \end{align}

Remark 14.8 [Why doesn't Newtonian mechanics have measurement?]. Newtonian mechanics and quantum mechanics are formulated as follows:
 $(\sharp_1):$ $\left\{\begin{array}{ll} \fbox{Newtoinan mechanics} &=\qquad \color{magenta}{ \underset{\mbox{}}{\fbox{Nothing}} } \quad + \quad \underset{\scriptsize{\mbox{ (Newtonian equation)}}}{\fbox{Causality}} \\ \\ \fbox{quantum mechanics} &= \underset{\scriptsize{\mbox{ (Born's quantum measurement)}}}{\fbox{Measurement}} + \underset{\scriptsize{\mbox{ (Heisenberg (and Schrödinger) equation)}}}{\fbox{Causality}} \end{array}\right.$
Thus, the following question is natural:
 $(\sharp_2):$ Why doesn't Newtonian mechanics have measurement?

Some may think that the reason is due to Theorem 14.6 (or, Corollary 14.7 ), which says that we need only $\phi_{t_0, t }(\omega_{t_0} )$ and not $x_t$. However, this answer is shallow and surerfacial. The question $(\sharp_2)$ is significant in the light of Einstein's words: $(\sharp_3):$ The moon is there whether one looks at it or not.

in Einstein and Tagore's conversation. This should be compared with Berkley's words (i.e., To be is to be perceived ). That is,
 $(\sharp_4):$ $\qquad \overset{\mbox{ no measurement}}{ \underset{\mbox{ Einstein}}{ \fbox{monistic realism (physics)} } } \quad \mbox{v.s.} \quad \overset{\mbox{measurement}}{ \underset{\mbox{Berkley}}{ \fbox{dualistic idealism ( language )} } }$

which means the following $(\sharp_5)+(\sharp_6)$:

 $(\sharp_5):$ Physics should exist without measurement $(\sharp_6):$ The concept of "measurement" is metaphysical and not physical