Let $A_1$ and $A_2$ be two $C^*$-algebras considered as closed $*$-subspaces of some $\mathcal{B}(H_1)$ and $\mathcal{B}(H_2)$. More preciesly there exist faithfull $*$-representations $\pi_1:A_1\to\mathcal{B}(H_1)$, $\pi_2:A_2\to\mathcal{B}(H_2)$. By definition, the uncompleted spatial tensor product $A_1\otimes_{sp}A_2$ of $C^*$-algebras $A_1$ and $A_2$ is $ A_1\otimes_{sp}A_2={\rm span}\{\pi_1(a_1)\otimes_2 \pi_2(a_2) : a_1\in\mathcal{B}(H_1),a_2\in\mathcal{B}(H_2)\}\subset\mathcal{B}(H_1\otimes_2 H_2) $
This definition doesn't depend on the choice of representations $\pi_1$ and $\pi_2$. This proved in Murphy's book.
Now consider arbitrary normed space $X$. Denote by $B_{X^*}$ closed unit ball of $X^*$ at $0$. It is easy to check that operators $ i_1:X\to l_\infty(B_{X^*}):x\mapsto(f\mapsto f(x))\qquad x\in X,\quad f\in B_{X^*} $ $ i_2:l_\infty(B_{X^*})\to\mathcal{B}(l_2(B_{X^*})):g\mapsto(f\mapsto g\cdot f)\qquad g\in l_\infty(B_{X^*}),\quad f\in l_2(B_{X^*}) $ are isometric. Thus each normed space $X$ we can embed into some space of bounded operators acting on Hilbert space by isometric operator $i :X\to\mathcal{B}(l_2(B_{X^*})):x\mapsto i_2(i_1(x))$.
Now given some isometric inclusions $i_X:X\to\mathcal{B}(H_X)$ and $i_Y:Y\to\mathcal{B}(H_Y)$ we can define spatial tensor product of normed spaces: $ X\otimes_{sp} Y={\rm span}\{i_X(x)\otimes_2 i_Y(y): x\in X, y\in Y\}\subset\mathcal{B}(H_X\otimes_2 H_Y) $
My question: Will this definition of spatial tensor product depend on particular choice of inclusions. For what kind of spaces this definition doesn't depend on the choice? May be it is necessary to put some restrictions on embeddings to get independence withing this class of embeddings?