Question about dense property in tensor product of Hilbert space

Question

Let $H_1$ and $H_2$ be two Hilbert Spaces, $A_i$ is dense in $H_i, i=1,2$. Let $D:=\{a_1\otimes a_2:a_i\in A_i,i=1,2 \}$, can we claim that $D$ is dense in $H_1\otimes H_2$?

Answer 1

Extending on what John Doe pointed out in their comment, $D$ is dense in $H_1\otimes H_2$ if and only if $\operatorname{dim}H_1=1$ or $\operatorname{dim}H_2=1$. While $\Leftarrow$ is quite simple to verify, the converse leads to the concept of entangled states. Indeed as soon as both $\operatorname{dim}H_1,\operatorname{dim} H_2\geq 2$ there exist vectors in $H_1\otimes H_2$ which cannot be written as $\psi_1\otimes\psi_2$—the prime example being (a multiple of) the Bell state $(1,0,0,1)^\top$. That's why the statement to consider here is the following:

Proposition. Given Hilbert spaces $H_1,H_2$ as well as dense subsets $A_1\subseteq H_1,A_2\subseteq H_2$ define $D:=\{\psi_1\otimes\psi_2:\psi_1\in A_1,\psi_2\in A_2\}$. Then $\operatorname{span}D$ is dense in $H_1\otimes H_2$.

Proof. The idea is to combine density of the span of the orthonormal basis (cf. Kenny Wong's comment) and density of $A_1,A_2$ "in a clever way". As for the deails: Fixing $x\in H_1\otimes H_2$, given any $\varepsilon>0$ we have to find $x_\varepsilon\in \operatorname{span}D$ such that $\|x-x_\varepsilon\|<\varepsilon$ (w.l.o.g. $x\neq 0$ as the case $x=0$ is straightforward). For this we first pick arbitrary orthonormal bases $\{e_j\}_{j\in J}$, $\{f_k\}_{k\in K}$ of $H_1,H_2$, respectively, which is always possible. Then $\{e_j\otimes f_k\}_{j\in J,k\in K}$ is an orthonormal basis of $H_1\otimes H_2$ meaning there exist finite subsets $J_F\subseteq J$, $K_F\subseteq K$ such that $$ \Big\|x- \sum_{j\in J_F}\sum_{k\in K_F}\langle e_j\otimes f_k,x\rangle e_j\otimes f_k \Big\|<\frac\varepsilon2\,. $$ Thus, intuitively, if we can approximate $e_j\otimes f_k$ by elements in $D$ "well-enough" then we should be done. Indeed—denoting $m:=|J_F|\cdot|K_F|\in\mathbb N$—by assumption on $A_1,A_2$ for every $(j,k)\in J_F\times K_F$ there exist $v_j\in A_1$, $w_k\in A_2$ such that $$ \|e_j-v_j\|<\frac\varepsilon{8m\|x\|}\quad\text{ and }\quad\|f_k-w_k\|<\min\Big\{\frac\varepsilon{4m\|x\|},1\Big\}\,. $$ In particular applying the triangle inequality repeatedly yields \begin{align*} \|e_j\otimes f_k-v_j\otimes w_k\|&=\|e_j\otimes f_k-e_j\otimes w_k+e_j\otimes w_k-v_j\otimes w_k\|\\ &\leq\|e_j\|\|f_k-w_k\|+\|e_j-v_j\|\|w_k\|\\ &\leq\|f_k-w_k\|+\|e_j-v_j\|(\|f_k\|+\|w_k-f_k\|)\\ &<\frac\varepsilon{4m\|x\|}+\frac\varepsilon{8m\|x\|}(1+1)=\frac\varepsilon{2m\|x\|} \end{align*} for all $j\in J_F,k\in K_F$. Then $x_\varepsilon:=\sum_{j\in J_F}\sum_{k\in K_F}\langle e_j\otimes f_k,x\rangle v_j\otimes w_k$ is an element of $\operatorname{span}(D)$ and \begin{align*} \|x-x_\varepsilon\|&= \Big\|x-\sum_{j\in J_F}\sum_{k\in K_F}\langle e_j\otimes f_k,x\rangle e_j\otimes f_k+\sum_{j\in J_F}\sum_{k\in K_F}\langle e_j\otimes f_k,x\rangle e_j\otimes f_k-x_\varepsilon\Big\|\\ &\leq \Big\|x-\sum_{j\in J_F}\sum_{k\in K_F}\langle e_j\otimes f_k,x\rangle e_j\otimes f_k\Big\|+\Big\|\sum_{j\in J_F}\sum_{k\in K_F}\langle e_j\otimes f_k,x\rangle e_j\otimes f_k-x_\varepsilon\Big\|\\ &<\frac\varepsilon2+\Big\|\sum_{j\in J_F}\sum_{k\in K_F}\langle e_j\otimes f_k,x\rangle( e_j\otimes f_k-v_j\otimes w_k)\Big\|\\ &\leq\frac\varepsilon2+\sum_{j\in J_F}\sum_{k\in K_F}\underbrace{\big|\langle e_j\otimes f_k,x\rangle\big|}_{\leq\|e_k\|\|f_j\|\|x\|=\|x\|}\|e_j\otimes f_k-v_j\otimes w_k\|\\ &\leq\frac\varepsilon2+\|x\|\underbrace{\Big(\sum_{j\in J_F}\sum_{k\in K_F}1\Big)}_{=m}\frac\varepsilon{2m\|x\|}=\varepsilon\,. \tag*{$\square$} \end{align*}