10.2: Causality---Mathematical preparation


10.2.1:The Heisenberg picture and the Schrödinger picture





First, let us review the general basic structure (cf. $\S$2.1.3 ) as follows.



(A):General basic structure and State spaces
General basic structure:$[{\mathcal A} \subseteq \overline{\mathcal A} \subseteq B(H)]$ and states


Remark 10.3 [$\overline{\mathcal A}_\ast \subseteq {\mathcal A}^*$]

Consider the basic structure $[{\mathcal A} \subseteq \overline{\mathcal A} ]_{B(H)}$. For each $\rho \in \overline{\mathcal A}_\ast$, $F \in {\mathcal A} (\subseteq \overline{\mathcal A} \subseteq B(H) )$, we see that

\begin{align} \Big| {}_{\stackrel{{}}{\overline{\mathcal A}_* }}\Big(\rho, F \Big){}_{\stackrel{{}}{\overline{\mathcal A} }} \Big| \le C \| F\|_{B(H)} = C \| F\|_{\mathcal A} \qquad \tag{10.3} \end{align}

Thus,we can consider that $\rho \in {\mathcal A}^*$. That is, in the sense of (10.3),we consider that

\begin{align} \overline{\mathcal A}_\ast \subseteq {\mathcal A}^* \tag{10.4} \end{align}

When $\rho (\in \overline{\mathcal A}_* )$ is regarded as the element of ${\mathcal A}^*$, it is sometimes denoted by $\widehat{\rho}$. Therefore,

\begin{align} {}_{\stackrel{{}}{\overline{\mathcal A}_* }}\Big(\rho, F \Big){}_{\stackrel{{}}{\overline{\mathcal A} }} = {}_{\stackrel{{}}{{\mathcal A}^* }}\Big(\widehat{\rho}, F \Big){}_{\stackrel{{}}{{\mathcal A} }} \qquad (\forall F \in {\mathcal A} (\subseteq \overline{\mathcal A})) \tag{10.5} \end{align}
Definition 10.4 [Causal operator (= Markov causal operator)] Consider two basic structures: \begin{align} [{\mathcal A}_1 \subseteq \overline{\mathcal A}_1 \subseteq {B(H_1)}] \mbox{ and } [{\mathcal A}_2 \subseteq \overline{\mathcal A}_2 \subseteq {B(H_2)}] \end{align}

A continuous linear operator $\Phi_{1,2}:\overline{\mathcal A}_2 \to \overline{\mathcal A}_1 $ is called a causal operator ( or, Markov causal operator , the Heisenberg picture of "causality"), if it satisfies the following (i)$\mbox{---}$(iv):

(i): $F_2 \in \overline{\mathcal A}_2 \;\; F_2 {\; \geqq \;}0$ $\Longrightarrow$ $\Phi_{12}F_2 {\; \geqq \;}0$
(ii):$ \Phi_{12} I_{\overline{\mathcal A}_2} =I_{\overline{\mathcal A}_1} $ $\qquad$ (where,$I_{\overline{\mathcal A}_1} (\in {\overline{\mathcal A}_1} )$ is the identity)
(iii): there exists the continuous linear operator $({\Phi}_{1,2})_*:(\overline{\mathcal A}_1)_* \to ( \overline{\mathcal A}_2)_* $ such that \begin{align} & \mbox{(a)} \quad {}_{\stackrel{{}}{(\overline{\mathcal A}_1)_* }}\Big(\rho_1, \Phi_{1,2} F_2 \Big){}_{\stackrel{{}}{\overline{\mathcal A}_1 }} = {}_{\stackrel{{}}(\overline{\mathcal A}_2)_* }\Big( ({\Phi}_{1,2})_\ast \rho_1, F_2\Big){}_{\stackrel{{}}{\overline{\mathcal A}_2 }} \qquad (\forall \rho_1 \in (\overline{\mathcal A}_1)_*, \forall F_2 \in \overline{\mathcal A}_2 ) \\ &\mbox{(b)} \quad ({\Phi}_{1,2})_\ast (\overline{\frak S}^m( ( \overline{\mathcal A}_1)_*) ) \subseteq \overline{\frak S}^m((\overline{\mathcal A}_2)_*) \tag{10.7} \end{align} This $({\Phi}_{1,2})_*$ is called the pre-dual causal operator of ${\Phi}_{1,2}$.
(iv): there exists the continuous linear operator ${\Phi}_{1,2}^*:{\mathcal A}_1^* \to {\mathcal A}_2^* $ such that \begin{align} & \mbox{(a)} \quad {}_{\stackrel{{}}{(\overline{\mathcal A}_1)_* }}\Big(\rho_1, \Phi_{1,2} F_2 \Big){}_{\stackrel{{}}{\overline{\mathcal A}_1 }} = {}_{\stackrel{{}}{\mathcal A}_2^* }\Big( {\Phi}_{1,2}^\ast \widehat{\rho}_1, F_2\Big){}_{\stackrel{{}}{{\mathcal A}_2 }} \qquad \tag{10.8} \\ & \qquad \qquad \qquad \qquad (\forall \rho_1 = \widehat{\rho}_1 \in (\overline{\mathcal A}_1)_* (\subseteq {\mathcal A}_1^*), \forall F_2 \in {\mathcal A}_2 ) \nonumber \\ & \mbox{(b)} \quad ({\Phi}_{1,2})^* ({\frak S}^p({\mathcal A}_1^*) ) \subseteq {\frak S}^m ({\mathcal A}_2^*) \tag{10.9} \end{align} This ${\Phi}_{1,2}^*$ is called the dual operator of ${\Phi}_{1,2}$.
In addition, the causal operator ${\Phi}_{1,2}$ is called a deterministic causal operator, if it satisfies that \begin{align} ({\Phi}_{1,2})^* ({\frak S}^p({\mathcal A}_1^*) ) \subseteq {\frak S}^p({\mathcal A}_2^*) \tag{10.10} \end{align}

$\fbox{Note 10.3}$ [ Causal operator in Classical systems] Consider the two basic structures: \begin{align} [C_0(\Omega_1) \subseteq L^\infty (\Omega_1, \nu_1 )]_{B(H_1)} \mbox{ and } [C_0(\Omega_2) \subseteq L^\infty (\Omega_2, \nu_2 )]_{B(H_2)} \end{align} A continuous linear operator $\Phi_{1,2}:L^\infty (\Omega_2) \to L^\infty (\Omega_1) $ called a causal operator, if it satisfies the following (i)$\mbox{---}$(iv):
$(i):$ $f_2 \in L^\infty (\Omega_2), \;\; f_2 {\; \geqq \;}0$ $\Longrightarrow$ $\Phi_{12}f_2 {\; \geqq \;}0$
(ii): $\Phi_{12} 1_2 = 1_1$ where, $1_k (\omega_k ) = 1$ $( \forall \omega_k \in \Omega_k, k=1,2)$
(iii): There exists a continuous linear operator $({\Phi}_{1,2})_\ast :L^1 (\Omega_1) \to L^1_{}(\Omega_2) $ (and $({\Phi}_{1,2})_\ast :L^1_{+1} (\Omega_1) \to L^1_{+1}(\Omega_2) $ ) such that \begin{align} & \int_{\Omega_1} [\Phi_{1,2} f_2](\omega_1) \;\;\rho_1 ( \omega_1 ) \nu_1( d \omega_1 ) = \int_{\Omega_2} f_2(\omega_2) \;\; [({\Phi}_{1,2})_\ast \rho_1](\omega_2 ) \nu_2 ( d \omega_2 ) \\ & \hspace{4cm} (\forall \rho_1 \in L^1 (\Omega_1), \forall f_2 \in L^\infty (\Omega_2)) \end{align} This $({\Phi}_{1,2})_\ast$ is called a pre-dual causal operator of ${\Phi}_{1,2}$.
(iv): There exists a continuous linear operator ${\Phi}_{1,2}^\ast :{\mathcal M} (\Omega_1) \to {\mathcal M}(\Omega_2) $ (and ${\Phi}_{1,2}^\ast :{\mathcal M}_{+1} (\Omega_1) \to {\mathcal M}_{+1}(\Omega_2) $ ) such that

\begin{align} & {}_{\stackrel{{}}{L^1(\Omega_1) }}\Big(\rho_1, \Phi_{1,2} F_2 \Big){}_{\stackrel{{}}{L^\infty(\Omega_1) }} = {}_{\stackrel{{}}{{\mathcal M}(\Omega_2) }}\Big( {\Phi}_{1,2}^\ast \widehat{\rho}_1, F_2\Big){}_{\stackrel{{}}{{C_0}(\Omega_2) }} \\ & \qquad (\forall \rho_1=\widehat{\rho}_1 \in {\mathcal M}(\Omega_1), \forall F_2 \in {C_0}(\Omega_2) ) \end{align} where, $\widehat{\rho}_1(D)= \int_D \rho_1 ( \omega_1) \nu_1(d \omega_1)$ $(\forall D \in {\mathcal B}_{\Omega_1} )$. This $({\Phi}_{1,2})^\ast$ is called a dual causal operator of ${\Phi}_{1,2}$.
In addition, a causal operator ${\Phi}_{1,2}$ is called a deterministic causal operator, if there exists a continuous map $\phi_{1,2}:\Omega_1 \to \Omega_2$ such that \begin{align} [\Phi_{1,2}f_2](\omega_1) = f_2(\phi_{1,2}(\omega_1)) \quad (\forall f_2 \in C(\Omega_2 ), \forall \omega_1 \in \Omega_1 ) \tag{10.11} \end{align} This $\phi_{1,2}:\Omega_1 \to \Omega_2$ is called a deterministic causal map. Here, it is clear that \begin{align} \Omega_1 \approx {\frak S}^p(C_0(\Omega_1)^*) \ni \delta_{\omega_1} \xrightarrow[\Phi_{12}^*]{} \delta_{\phi_{12}(\omega_1)} \in {\frak S}^p(C_0(\Omega_2)^*) \approx \Omega_2 \end{align}




Theorem 10.5 [Continuous map and deterministic causal map] Let $(\Omega_1, {\cal B}_{\Omega_1}, \nu_1)$ and $(\Omega_2, {\cal B}_{\Omega_2}, \nu_2)$ be measure spaces. Assume that a continuous map $\phi_{1,2}:\Omega_1 \to \Omega_2$ satisfies:

\begin{align} D_2 \in {\cal B}_{\Omega_2}, \; \; \nu_2 ( D_2 ) = 0 \quad \Longrightarrow \quad \nu_1 ( \phi_{1,2}^{-1} ( D_2 )) = 0. \end{align}

Then, the continuous map $\phi_{1,2}:\Omega_1 \to \Omega_2$ is deterministic, that is, the operator ${\Phi}_{1,2} :L^{\infty} (\Omega_2, \nu_2) \to L^{\infty} (\Omega_1, \nu_1)$ defined by (10.11) is a deterministic causal operator.



Proof For each ${\overline \rho}_1$ $\in L^1(\Omega_1 , \nu_1 )$, define a measure $\mu_2$ on $(\Omega_2 , {\cal B}_{\Omega_2})$ such that

\begin{align} \mu_2 ( D_2 ) = \int_{\phi_{1,2}^{-1} ( D_2 ) } {\overline \rho}_1 (\omega_1) \; \nu_1 ( d \omega_1 ) \qquad (\forall D_2 \in {\cal B}_{\Omega_2} ) \end{align}

Then, it suffices to consider the Radon-Nikodym derivative $[\Phi_{1,2}]_* ({\overline \rho}_1) = {d \mu_2}/{d \nu_2 }$. That is because

\begin{align} D_2 \in {\cal B}_{\Omega_2}, \; \; \nu_2 ( D_2 ) = 0 \quad \Longrightarrow \quad \nu_1 ( \phi_{1,2}^{-1} ( D_2 )) = 0 \quad \Longrightarrow \quad \mu_2 ( D_2 ) = 0 \tag{10.12} \end{align}

Thus, by the Radon-Nikodym theorem, we get a continuous linear operator $[{\Phi}_{1,2}]_\ast :L^{1} (\Omega_1, \nu_1) \to L^{1} (\Omega_2, \nu_2)$.

$\square \quad$

Theorem 10.6

Let ${\Phi}_{1,2}:L^\infty (\Omega_2) \to L^\infty (\Omega_1) $ be a deterministic causal operator. Then, it holds that

\begin{align} {\Phi}_{1,2} (f_2 \cdot g_2 ) = {\Phi}_{1,2} (f_2 ) \cdot {\Phi}_{1,2} (g_2 ) \qquad (\forall f_2, \forall g_2 \in L^\infty (\Omega_2 )) \end{align}

Proof Let $f_2$, $g_2$ be in $L^\infty (\Omega_2)$. Let $\phi_{1,2}:\Omega_1 \to \Omega_2$ be the deterministic causal map of the deterministic causal operator ${\Phi}_{1,2}$. Then, we see

\begin{align} & [{\Phi}_{1,2} (f_2 \cdot g_2 )] (\omega_1) = (f_2 \cdot g_2 )(\phi_{1,2}(\omega_1)) = f_2(\phi_{1,2}(\omega_1)) \cdot g_2(\phi_{1,2}(\omega_1)) \\ = & [{\Phi}_{1,2} (f_2 )] (\omega_1) \cdot [{\Phi}_{1,2} (g_2 )] (\omega_1) = [{\Phi}_{1,2} (f_2 ) \cdot {\Phi}_{1,2} (g_2 )] (\omega_1) \qquad (\forall \omega_1 \in \Omega_1) \end{align} This completes the theorem.
$\square \quad$