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### 3.2.1 Tensor Hilbert space (i.e., the tensor product of Hilbert spaces)

The linguistic interpretation ($\S$3.1) says \begin{align*} \mbox{ "Only one measurement is permitted"} \end{align*} which implies "only one measuring object" and "only one state". Thus, if there are several states, these should be regarded as "only one state". In order to do it, we have to prepare "tensor operator algebra". That is,

 $(A):$ "several states" $\xrightarrow[\mbox{ by tensor operator algebra}]{\mbox{ combine several into one}}$ "one state"

In what follows, we shall introduce the tensor operator algebra.

Let $H,K$ be Hilbert spaces. We shall define the tensor Hilbert space $H \otimes K$ as follows. Let $\{ e_m \; | \; m \in {\mathbb N}\equiv \{1,2, \ldots \} \}$ be the CONS (i.e, complete orthonormal system ) in $H$. And, let $\{ f_n \; | \; n \in {\mathbb N}\equiv \{1,2, \ldots \} \}$ be the CONS in $K$. For each $(m,n) \in {\mathbb N}^2$, consider the symbol "$e_m \otimes f_n$". Here, consider the following "space":

\begin{align} & H \otimes K = \Big\{ g=\sum_{(m,n) \in {\mathbb N}^2} \alpha_{m,n}e_m \otimes f_n \; \Big| \; ||g||_{H \otimes K} \equiv [\sum_{(m,n) \in {\mathbb N}^2} |\alpha_{m,m}|^2]^{1/2} < \infty \Big\} & {} \\ \tag{3.4} \end{align} Also, the inner product $\langle \cdot , \cdot \rangle_{H \otimes K}$ is represented by \begin{align} \langle e_{m_1} \otimes f_{n_1} , e_{m_2} \otimes f_{n_2} \rangle_{H \otimes K} & \equiv \langle e_{m_1} , e_{m_2} \rangle_{H } \cdot \langle f_{n_1} , f_{n_2} \rangle_{ K} \nonumber \\ & = \left\{\begin{array}{ll} 1 \quad & ({m_1} , {n_1}) =({m_2} , {n_2}) \\ 0 \quad & ({m_1} , {n_1}) \not=({m_2} , {n_2}) \end{array}\right. \tag{3.5} \end{align} Thus, summing up, we say
 $(B):$ the tensor Hilbert space ${H \otimes K}$ is defined by the Hilbert spac with the CONS $\{ e_m \otimes f_n \; | \; (m,n) \in {\mathbb N}^2 \}$.

For example, for any $e=\sum_{m=1}^\infty \alpha_m e_m \in H$ and any $f=\sum_{n=1}^\infty \beta_n f_m \in H$, the tensor $e \otimes f$ is defined by

\begin{align*} e \otimes f = \sum_{(m,n) \in {\mathbb N}^2} \alpha_{m} \beta_n (e_m \otimes f_n) \end{align*}

Also, the tensor norm $|| \widehat{u}||_{H \otimes K}$ $(\widehat{u} \in H \otimes K)$ is defined by

\begin{align*} || \widehat{u}||_{H \otimes K} = | \langle \widehat{u} , \widehat{u} \rangle_{H \otimes K}|^{1/2} \end{align*}

Example 3.2[Simple example: tensor Hilbert space ${\mathbb C}^2 \otimes {\mathbb C}^3$] Consider the 2-dimensional Hilbert space $H={\mathbb C}^2$ and the 3-dimensional Hilbert space $K={\mathbb C}^3$. Now we shall define the tensor Hilbert space $H \otimes K = {\mathbb C}^2 \otimes {\mathbb C}^3$ as follows. Consider the CONS $\{e_1, e_2 \}$ in $H$ such as

\begin{align*} e_1 = \left[\begin{array}{l} 1 \\ 0 \end{array}\right], \quad e_2 = \left[\begin{array}{l} 0 \\ 1 \end{array}\right] \end{align*}

And, consider the CONS $\{f_1. f_2, f_3 \}$@in@$K$@such as

\begin{align*} f_1 = \left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right], \quad f_2 = \left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right], \quad f_2 = \left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] \end{align*}

Therefore, @the tensor Hilbert space@$H \otimes K = {\mathbb C}^2 \otimes {\mathbb C}^3$@has the CONS@such as

\begin{align*} & e_1 \otimes f_1 = \left[\begin{array}{l} 1 \\ 0 \end{array}\right] \otimes \left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right], \;\; e_1 \otimes f_2 = \left[\begin{array}{l} 1 \\ 0 \end{array}\right] \otimes \left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right], e_1 \otimes f_3 = \left[\begin{array}{l} 1 \\ 0 \end{array}\right] \otimes \left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right], \\ & e_2 \otimes f_1 = \left[\begin{array} 0 \\ 1 \end{array}\right] \otimes \left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right], \;\; e_2 \otimes f_2 = \left[\begin{array}{l} 0 \\ 1 \end{array}\right] \otimes \left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right], e_2 \otimes f_3 = \left[\begin{array}{l} 0 \\ 1 \end{array}\right] \otimes \left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] \end{align*} Thus, we see that \begin{align*} H \otimes K = {\mathbb C}^2 \otimes {\mathbb C}^3= {\mathbb C}^6 \end{align*}

That is because@the CONS $\{ e_i \otimes f_j \;|\; i=1,2,3, \;\; j=1,2 \}$@in $H \otimes K$@can be regarded as@$\{ g_k \;\;|\; k=1,2,...,6 \}$@such that

\begin{align*} & g_1=e_1 \otimes f_1 = \left[\begin{array}{l} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right], \;\; g_2=e_1 \otimes f_2 = \left[\begin{array}{l} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right], g_3=e_1 \otimes f_3 = \left[\begin{array}{l} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right], \\ & g_4=e_2 \otimes f_1 = \left[\begin{array}{l} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}\right], \;\; g_5=e_2 \otimes f_2 = \left[\begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{array}\right], g_6=e_2 \otimes f_3 = \left[\begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}\right] \end{align*} This Example 3.2 can be easily generalized as follows.
Theorem@3.3[Finite tensor Hilbert space] \begin{align} {\mathbb C}^{m_1} \otimes {\mathbb C}^{m_2} \otimes \cdots \otimes \otimes {\mathbb C}^{m_n}= {\mathbb C}^{\sum_{k=1}^n m_k} \tag{3.6} \end{align} Theorem 3.4[Concrete tensor Hilbert space] \begin{align} L^2(\Omega_1, \nu_1 ) \otimes L^2(\Omega_2, \nu_2 ) = L^2(\Omega_1 \times \Omega_2 , \nu_1 \otimes \nu_2 ) \tag{3.7} \end{align} where,@$\nu_1 \otimes \nu_2$@is the product measure.

Definition@3.5@[Infinite tensor Hilbert space]@Let@$H_1, H_2,...,H_ k,...$@be Hilbert spaces.@Then,@the infinite tensor Hilbert space@$\bigotimes_{k=1}^\infty H_k$@can be defined as follows.@For each $k( \in {\mathbb N})$,@consider the CONS@$\{{e_k^j}\}_{j=1}^\infty$ in a@Hilbert space $H_k$.@For any map $b: {\mathbb N}\to {\mathbb N}$,@define the symbol@$\bigotimes_{k=1}^\infty e_k^{b(k)}$@such that

\begin{align*} \bigotimes_{k=1}^\infty e_k^{b(k)} =e_1^{b(1)} \otimes e_2^{b(2)} \otimes e_3^{b(3)} \otimes \cdots \end{align*} Then, we have: \begin{align} \Big\{ \bigotimes_{k=1}^\infty e_k^{b(k)} \;\Big|\; b: {\mathbb N}\to {\mathbb N} \mbox{ is a map} \Big\} \tag{3.8} \end{align}

Hence we can@define the infinite@Hilbert space@$\bigotimes_{k=1}^\infty H_k$@such that@it has the CONS (cf. (3.8)).

### 3.2.2: Tensor basic structure

For each continuous linear operators $F \in B(H), G \in B(K)$, the tensor operator $F \otimes G$ $\in B( H \otimes K )$ is defined by

\begin{align*} (F \otimes G) (e \otimes f ) =F e \otimes G f \quad (\forall e \in H, f \in K) \end{align*}

Definition 3.6 [Tensor $C^*$-algebra and Tensor $W^*$-algebra] Consider basic structures

\begin{align*} \mbox{ $[{\mathcal A}_1\subseteq \overline{{\mathcal A}_1}\subseteq B(H_1)]$ and $[{\mathcal A}_2\subseteq \overline{{\mathcal A}_2}\subseteq B(H_2)]$ } \end{align*}

[I]: The tensor $C^*$-algebra ${\mathcal A}_1 \otimes {\mathcal A}_2$ is defined by the smallest $C^*$-algebra ${\widehat{\mathcal A}}$ such that

\begin{align*} \{F \otimes G\; (\in B( H_1 \otimes H_2 )) \; |\; F \in {\mathcal A}_1, \; G \in {\mathcal A}_2\} \subseteq {\widehat{\mathcal A}} \subseteq B( H_1 \otimes H_2 ) \end{align*}

[II]: The tensor $W^*$-algebra $\overline{\mathcal A}_1 \otimes \overline{\mathcal A}_2$ is defined by the smallest $W^*$-algebra ${\widetilde{\mathcal A}}$ such that

\begin{align*} \{F \otimes G\; (\in B( H_1 \otimes H_2 )) \; |\; F \in \overline{\mathcal A}_1, \; G \in \overline{\mathcal A}_2\} \subseteq {\widetilde{\mathcal A}} \subseteq B( H_1 \otimes H_2 ) \end{align*} Here, note that $\overline{\mathcal A}_1 \otimes \overline{\mathcal A}_2 = \overline{ {\mathcal A}_1 \otimes {\mathcal A}_2}$.

Theorem 3.7 [Tensor basic structure] [I]: Consider basic structures

\begin{align*} \mbox{ $[{\mathcal A}_1\subseteq \overline{{\mathcal A}_1}\subseteq B(H_1)]$ and $[{\mathcal A}_2\subseteq \overline{{\mathcal A}_2}\subseteq B(H_2)]$ } \end{align*} Then, we have the tensor basic structure: \begin{align*} [{\mathcal A}_1 \otimes {\mathcal A}_2 \subseteq \overline{\mathcal A}_1 \otimes \overline{\mathcal A}_2 \subseteq B(H_1 \otimes H_2)] \end{align*}

[II]: Consider quantum basic structures $[{\mathcal C}(H_1)\subseteq B(H_1)\subseteq B(H_1)]$ and $[{\mathcal C} (H_2)\subseteq B(H_2)\subseteq B(H_2)]$. Then, we have tensor quantum basic structure:

\begin{align*} & [{\mathcal C}(H_1)\subseteq B(H_1)\subseteq B(H_1)] \otimes [{\mathcal C}(H_2)\subseteq B(H_2)\subseteq B(H_2)] \\ = & [{\mathcal C}(H_1\otimes H_2 )\subseteq B(H_1 \otimes H_2)\subseteq B(H_1 \otimes H_2)] \end{align*}

[III]: Consider classical basic structures $[C_0(\Omega_1 ) \subseteq L^\infty (\Omega_1, \nu_1 ) \subseteq B(L^2 (\Omega_1, \nu_1 ))]$ and $[C_0(\Omega_2 ) \subseteq L^\infty (\Omega_2, \nu_2 )\subseteq B(L^2 (\Omega_2\, \nu_2 ))]$. Then, we have tensor classical basic structure:

\begin{align*} & [C_0( \Omega_1 ) \subseteq L^\infty (\Omega_1 \subseteq \nu_1 ) \subseteq B(L^2 (\Omega_1 , \nu_1 ))] \otimes [C_0(\Omega_2 ) \subseteq L^\infty (\Omega_2 \subseteq \nu_2 ) \subseteq B(L^2 (\Omega_2 , \nu_2 ))] \\ \\ = & [C_0(\Omega_1 \times \Omega_2 ) \subseteq L^\infty (\Omega_1 \times \Omega_2 , \nu_1 \otimes \nu_2 ) \subseteq B(L^2 (\Omega_1 \times \Omega_2, \nu_1 \otimes \nu_2 ))] \end{align*}

Theorem 3.8 The $\bigotimes_{k=1}^\infty B(H_k)$ $( \subseteq B( \bigotimes_{k=1}^\infty H_k) )$ is defined by the smallest $C^*$-algebra that contains

\begin{align*} F_1 \otimes F_2 \otimes \cdots \otimes F_n \otimes I \otimes I \otimes \cdots \Big( \in B( \bigotimes_{k=1}^\infty H_k) \Big) \\ \;\; (\forall F_k \in B(H_k) ,\;k=1,2,...,n, n=1,2,...) \end{align*} Then,it holds that \begin{align} \bigotimes_{k=1}^\infty B(H_k) = B(\bigotimes_{k=1}^\infty H_k) \tag{3.9} \end{align} Theorem 3.9 The followings hold: \begin{align*} \mbox{(i)}:\quad & \rho_k \in {\mathcal A}_k^* \Longrightarrow \bigotimes_{k=1}^n \rho_k \in ( \bigotimes_{k=1}^n {\mathcal A}_k)^* \\ \mbox{(ii)}:\quad & \rho_k \in {\frak S}^m({\mathcal A}_k^*) \Longrightarrow \bigotimes_{k=1}^n \rho_k \in {\frak S}^m(( \bigotimes_{k=1}^n {\mathcal A}_k)^*) \\ \mbox{(iii)}:\quad & \rho_k \in {\frak S}^p({\mathcal A}_k^*) \Longrightarrow \bigotimes_{k=1}^n \rho_k \in {\frak S}^p(( \bigotimes_{k=1}^n {\mathcal A}_k)^*) \end{align*}
 $\fbox{Note 3.1}$ @The theory of operator algebra@is@a deep mathematical theory.@However, in this note,@we do not use more than the above preparation.