multiplication of Hilbert Schmidt operators and bounded operators

Question

Suppose $T_1,T_2 \in B(H)$,where $H$ is an infinite dimensional Hilbert space. $S\in \mathcal{HS}(H)$,$\mathcal{HS}(H)$ is the set of Hilbert Schmidt operators on $H$.

Does there exist nonzero operators $T_3,T_4 \in B(H)$ such that $T_1T_3ST_4T_2=\alpha S$,where $\alpha$ is a nonzero complex number.

Answer 1

Yes. $T_3$ or $T_4$ being the zero operator, we have $T_1T_3ST_4T_2=0=0S$.

If you want $\alpha\neq0$, then the answer is no. If either $T_1$ or $T_2$ is rank one, then $T_1T_3ST_4T_2$ will be rank one (or zero). But $\alpha S$ will have the same rank as $S$, which can be any natural number, or infinite.