Let $\mathcal{H}$, $\mathcal{K}$ be Hilbert spaces, let $S_2(\mathcal{H})$ denote Hilbert-Schmidt operators on $\mathcal{H}$ and let $\mathcal{C}_{\mathcal{K}}$ stand for the column Hilbert space, that is, $\mathcal{C}_{\mathcal{K}}:= B(\mathbb{C}, \mathcal{K})$ (bounded linear operators from $\mathbb{C}$ to $\mathcal{K}$). Denote the Banach space of $n$ by $n$ matrices with complex entries by $M_n$.
Let $u \in \mathcal{K}$ define $\omega_{u} \colon \mathbb{C} \rightarrow \mathcal{K}$ by $\omega_u (\alpha) = \alpha u$. Clearly, $\omega_{u} \in \mathcal{C}_{\mathcal{K}}$. It is easy to show that the map $\theta_{\mathcal{H}} \colon \mathcal{H} \rightarrow \mathcal{C}_{\mathcal{H}}$ defined via $\Theta_{\mathcal{H}}(u)= \omega_u$ is an isometry between $\mathcal{H}$ and $\mathcal{C}_{\mathcal{H}}$.
I would like to show that the map $\theta_{S_2(\mathcal{H})}^{(2)} \colon S_2(\mathcal{H} \oplus \mathcal{H}) \rightarrow B(\mathbb{C}^2, S_2(\mathcal{H}) \oplus S_2(\mathcal{H}))$ defined by $\theta_{S_2(\mathcal{H})}^{(2)}:= I_{M_2} \otimes \theta_{S_2(\mathcal{H})}$, where we identify $S_2(\mathcal{H} \oplus \mathcal{H}) $ with $M_2(S_2(\mathcal{H}))$ and $B(\mathbb{C}^2, S_2(\mathcal{H}) \oplus S_2(\mathcal{H}))$ with $M_2(\mathcal{C}_{S_2(\mathcal{H} )})$, is again an isometry or at least bounded.
I can obtain the boundedness if the following holds?
Let $x=[x_{ij}]$ in $M_2(S_2(\mathcal{H}))$. If $A$ is a $2$ by $2$ matrix such that $A=[\|x_{ij}\|_{S_2(\mathcal{H})}]_{ij}$ then its Hilbert Schmidt norm is $$\| A\|_{2} = \sqrt{\sum_{i,j} \| x_{ij}\|_{S_2(\mathcal{H})}^2}.$$ Is it true that $$\| A\|_{2}=\|x\|_{M_2(S_2(\mathcal{H}))}?$$ If the answer is positive then the boundedness would follow from Cauchy-Schwartz inequality. I think that only $\| A\|_{2}\geq\|x\|_{M_2(S_2(\mathcal{H}))}$ holds, I don't know if we have the equality.