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

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:

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.

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

That is, it holds that

Proof The proof is similar to that of Theorem 12.8

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

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

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

That is because Theorem 14.5 says that

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

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:

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

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:

in Einstein and Tagore's conversation. This should be compared with Berkley's words (i.e., To be is to be perceived ).

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