11.1: The Heisenberg picture and the Schrödinger picture

11.1.1: State does not move--- the Heisenberg picture ---

We consider that

"only one measurement" $\Longrightarrow$"state does not move"

That is because
 $\qquad \quad$Fig. 1.1: The history of world-descriptions
 $(a):$ In order to see the state movement, we have to take measurement at least more than twice. However, the "plural measurement" is prohibited. Thus, we conclude "state does not move"

We want to believe that this is associated with Parmenides' words:
There is no movement
which is related to the Heisenberg picture. This will be explained in what follows.

Theorem 11.1 [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 ${\overline{\mathcal A}_2}$.

Proof. Let $\Xi$ $(\in {\cal F} )$. And consider the countable decomposition $\{\Xi_1, \Xi_2, \ldots, \Xi_n, \ldots\}$ of $\Xi$ $\Big($ i.e., $\Xi = \bigcup\limits_{n=1}^\infty \Xi_n$, $\Xi_n \in {\cal F}, (n = 1, 2, \ldots)$, $\Xi_m \cap \Xi_n =\emptyset \;\; (m \not= n )$ $\Big)$. Then we see, for any $\rho_1 (\in ({{\mathcal A}_1})_*)$, \begin{align} & {}_{\stackrel{{}}{(\overline{\mathcal A}_1)_* }}\Big(\rho_1, \Phi_{1,2} F_2( \bigcup\limits_{n=1}^\infty \Xi_n ) \Big){}_{\stackrel{{}}{\overline{\mathcal A}_1 }} = {}_{\stackrel{{}}{(\overline{\mathcal A}_1)_* }}\Big( ({\Phi}_{1,2})_* \rho_1, F_2 ( \bigcup\limits_{n=1}^\infty \Xi_n ) \Big){}_{\stackrel{{}}{\overline{\mathcal A}_2 }} \\ = & \sum\limits_{n=1}^\infty {}_{\stackrel{{}}{(\overline{\mathcal A}_1)_* }}\Big( ({\Phi}_{1,2})_* \rho_1, F_2 ( \Xi_n ) \Big){}_{\stackrel{{}}{\overline{\mathcal A}_2 }} = \sum\limits_{n=1}^\infty {}_{\stackrel{{}}{(\overline{\mathcal A}_1)_* }}\Big( \rho_1, {\Phi}_{1,2} F_2 ( \Xi_n ) \Big){}_{\stackrel{{}}{\overline{\mathcal A}_2 }} \end{align} Thus,$\Phi_{1,2} {\mathsf O}_2$ $=$ $(X , {\cal F} , \Phi_{1,2} F_2)$ is an observable in $\overline{\mathcal A}_1$.
$\square \quad$

Let us begin from the simplest case. Consider a tree $T=\{0,1\}$. For each $t \in T$, consider the basic structure:

\begin{align} [ {\mathcal A}_t \subseteq \overline{\mathcal A}_t \subseteq {B(H_t)}] \qquad (t=0,1) \end{align}

And consider the causal operator $\Phi_{0,1}: \overline{\mathcal A}_1 \to \overline{\mathcal A}_0$. That is,

\begin{align} \overline{\mathcal A}_0 \xleftarrow[]{\Phi_{0,1}} \overline{\mathcal A}_1 \tag{11.2} \end{align}

Therefore, we have thepre-dual operator $(\Phi_{0,1})_*$ and the dual operator $\Phi_{0,1}^*$:

\begin{align} (\overline{\mathcal A}_0)_* \xrightarrow[(\Phi_{0,1})_*]{} (\overline{\mathcal A}_1)_* \qquad \qquad {\mathcal A}_0^* \xrightarrow[\Phi_{0,1}^*]{} {\mathcal A}_1^* \tag{11.3} \end{align}

If $\Phi_{0,1}: \overline{\mathcal A}_1 \to \overline{\mathcal A}_0$ is deterministic, we see that

\begin{align} {\mathcal A}_0^* \supset {\frak S}^p{({\mathcal A}_0^*)} \ni \rho \xrightarrow[\Phi_{0,1}^*]{} \Phi_{0,1}^* \rho \in {\frak S}^p{({\mathcal A}_1^*)} \subset {\mathcal A}_1^* \tag{11.4} \end{align}

Under the above preparation,we will explain the Heisenberg picture and the Schrödinger picture in what follows.

Assume that
 $(A_1):$ Consider a deterministic causal operator $\Phi_{0,1}: \overline{\mathcal A}_1 \to \overline{\mathcal A}_0$. $(A_2):$ a state $\rho_{{}0}$ $\in {\frak S}^p({\mathcal A}_{{}0}^*):\;\; \mbox{pure state}$ $(A_3):$ Let ${\mathsf O}_1=(X_1, {\mathcal F}_1, F_1)$ be an observable in $\overline{\mathcal A}_1$.
Then, we see:
Explanation 11.2 [the Heisenberg picture] The Heisenberg picture is just the following (a):
 $(a1):$ To identify an observable ${\mathsf O}_1$ in $\overline{\mathcal A}_1$ with an $\Phi_{0,1}{\mathsf O}_1$ in $\overline{\mathcal A}_0$ }. That is, \begin{align} \underset{\mbox{ ( in $\overline{\mathcal A}_0$)}}{\Phi_{0,1}\overline{\mathsf O}_1} \qquad \xleftarrow[\scriptsize{\mbox{ identification}}]{\Phi_{0,1}} \qquad \underset{\mbox{ ( in $\overline{\mathcal A}_1$)}}{{\mathsf O}_1} \end{align}
Therefore,
 $(a2):$ a measurement of an observable ${\mathsf O}_1$ (at time $t=1$) for a pure state $\rho_{{}0}$ (at time $t=0$) $\in {\frak S}^p({\mathcal A}_{{}0}^*)$ is represented by \begin{align} {\mathsf M_{\overline{\mathcal A}_0}(\Phi_{0,1}{\mathsf O}_1}, S_{[\rho_0]}) \end{align}
Thus, {Axiom 1 ( measurement: $\S$2.7)} says that
 $(a3):$ the probability that a measured value belongs to $\Xi( \in {\mathcal F})$ is given by \begin{align} {}_{{\mathcal A}_0^* } \Big(\rho_0, \Phi_{0,1}(F_1( \Xi )) \Big) {}_{\overline{\mathcal A}_0} \tag{11.5} \end{align}
Explanation 11.3 [the Schrödinger picture] The Schrödinger picture is just the following (b):
 $(b1):$ To identify a pure state $\Phi_{0,1}^* \rho_0 (\in {\frak S}^p({\mathcal A}_1^*))$ with $\rho_0 (\in {\frak S}^p({\mathcal A}_0^*))$, That is, \begin{align} {\mathcal A}_0^* \supset {\frak S}^p{({\mathcal A}_0^*)} \ni \rho_0 \xrightarrow[\scriptsize{\mbox{ identification}}]{\Phi_{0,1}^*} \Phi_{0,1}^* \rho_0 \in {\frak S}^p{({\mathcal A}_1^*)} \subset {\mathcal A}_1^* \nonumber \end{align}
Therefore, Axiom 1 ( measurement: $\S$2.7) says that

 $(b2):$ a measurement of an observable ${\mathsf O}_1$ (at time $t=1$) for a pure state $\rho_{{}0}$ (at time $t=0$) $\in {\frak S}^p({\mathcal A}_{{}1}^*)$ is represented by \begin{align} {\mathsf M_{\overline{\mathcal A}_1}({\mathsf O}_1}, S_{[\Phi_{0,1}^* \rho_0]}) \end{align}

Thus,
 $(a3):$ the probability that a measured value belongs to $\Xi( \in {\mathcal F})$ is given by \begin{align} {}_{{\mathcal A}_1^* } \Big(\Phi_{0,1}^*\rho_0, F_1( \Xi ) \Big) {}_{\overline{\mathcal A}_1} \tag{11.6} \end{align} which is equal to \begin{align} {}_{{\mathcal A}_0^* } \Big(\rho_0, \Phi_{0,1}(F_1( \Xi )) \Big) {}_{\overline{\mathcal A}_0} \tag{11.7} \end{align}

In the above sense (i.e., (11.6) and (11.7) ), we conclude that, under the condition (A$_1$), \begin{align} \mbox{ the Heisenberg picture and the Schrödinger picture are equivalent} \end{align} That is, \begin{align} \underset{\mbox{ (Heisenberg picture)}}{ \fbox{${\mathsf M_{\overline{\mathcal A}_0}(\Phi_{0,1}{\mathsf O}_1}, S_{[\rho_0]})$} } \quad \underset{\scriptsize{\mbox{ (identification)}}}{\longleftrightarrow} \quad \underset{\mbox{ (Schrödinger picture)}}{ \fbox{ ${\mathsf M_{\overline{\mathcal A}_1}({\mathsf O}_1}, S_{[\Phi_{0,1}^* \rho_0]})$ } } \tag{11.8} \end{align}
Remark 11.4 In the above, the conditions (A$_1$) is indispensable, that is,
 $(A_1):$ Consider a deterministic causal operator $\Phi_{0,1}: \overline{\mathcal A}_1 \to \overline{\mathcal A}_0$.

Without the deterministic conditions (A$_1$), the Schrödinger picture can not be formulated completely. That is because $\Phi_{0,1}^* \rho_0$ is not necessarily a pure state. On the other hand, the Heisenberg picture is always formulated. Hence we consider that we consider that

 $\bullet$ $\left\{\begin{array}{ll} \mbox{ the Heisenberg picture is$\color{red}{formal}$} \\ \\ \mbox{ the Schrödinger picture is$\color{blue}{makeshift}$} \end{array}\right.$
$\square \quad$