Let's consider a separable Hilbert space $(\mathcal H, \langle\cdot, \cdot\rangle_{\mathcal H})$ with Norm $||\cdot||_{\mathcal H} := \langle\cdot, \cdot\rangle^{1/2}_{\mathcal H},$ orthonomal basis $(e_j)$ of $\mathcal H$ and let $s\colon \mathcal H \rightarrow \mathcal H$ be a Hilbert Schmidt operator, denoted by $s \in \mathcal{S_H}.$ It's known that $(\mathcal{S_H}, \langle\cdot, \cdot\rangle_{\mathcal{S_H}})$ is a (separable ??) Hilbert space and for $s_1, s_2 \in \mathcal{S_H}\colon$ $\langle s_1, s_2\rangle_{\mathcal{S_H}}=\sum_{j=1}^\infty\langle s_1(e_j), s_2(e_j)\rangle_{\mathcal H},$ such that $||s_1||^2_{\mathcal{S_H}} = \sum_{j=1}^\infty||s_1(e_j)||^2_{\mathcal H}.$ Now, assuming that $\mathcal{ S_H}$ indeed is separable, let $(\phi_j)$ be an orthonormal basis of $\mathcal S_{\mathcal H}.$ It easily can be shown, since $||\cdot||_{\mathcal{S_H}}$ is sub-multiplicative, that for $s \in \mathcal{S_H}\colon s\colon \mathcal{S_H} \rightarrow \mathcal{S_H}$ is a bounded operator as well, denoted by $s \in \mathcal{L_{S_H}}.$

Is $s\colon\mathcal{S_H} \rightarrow \mathcal{S_H}$ "Hilbert-Schmidt" as well ($s \in \mathcal{S_{S_H}}$)?

Thank you in advance!