Inner product of linear bounded operators between Hilbert spaces Let $X$ and $Y$ be Hilbert spaces, and let $L(X,Y)$ be the set of bounded linear operators between Hilbert spaces.
Can we equip $L(X,Y)$ with a natural inner product? I think it should look like
$\langle S, T \rangle = \sup_{x \in X} \dfrac{ \langle S x, T x \rangle_Y }{ \|x\|^2_X }$
where $S$ and $T$ and are from $L(X,Y)$. I have not found such a construct in standard text books on Hilbert spaces, therefore I would like to learn whether this is the way to do it.
 A: If $X$ and $Y$ are both non separable, then there is no continuous one-to-one linear mapping of $L(X,Y)$ into a Hilbert space. This is because in this case $L(X,Y)$ contains a subspace isomorphic to $c_0(\omega_1)$, and an argument of Olagunju (A Banach space that cannot be made into a BIP space, Math. Proc. Cambridge Philos. Soc., 63 (1967) pp 949-950) shows that there is no continuous one-to-one linear mapping of $c_0(\omega_1)$ into a Hilbert space.
Whether or not this rules out any "natural" inner product for you presumably depends on what you are looking for; in any case, what I've written above rules out various possibilities in the non separable setting (in particular, the identity mapping on $L(X,Y)$ being operator-norm-to-inner-product continuous).
As for the separable setting (i.e., $X$ and $Y$ both separable), $L(X,Y)$ is isomorphic to a subspace of $L_\infty([0,1])$ via the Hahn-Banach theorem, which admits a continuous one-to-one linear mapping into the Hilbert space $L_2([0,1])$ (i.e., the formal inclusion operator), hence there is a continuous one-to-one linear mapping of $L(X,Y)$ into the separable Hilbert space $L_2([0,1])$. But whether or not a "natural" such mapping exists, I don't know.
I haven't necessarily answered your question, because I restricted attention to having a continuity requirement, but I would guess (perhaps wrongly) that any "natural" inner product would satisfy at least some kind of continuity condition.
