Let $H$ be a separable Hilbert space on which von Neumann algebra $R$ acts; let $\eta \in H$ be a separating vector for $R$ (i.e. the zero operator is the only $T \in R$ such that $T \eta = 0$); let $R'$ denote $R$'s commutant as usual.
Is it true that for all $x \in H$, $x = U T \eta$ for some $T \in R$ and unitary $U \in R'$? [Note -- I should stress I'm requiring $T$ to be in $R$ itself, and $U$ to be in $R$'s commutant.]
I know the answer is yes [even if $T$ is further required to be the identity operator] if $U$ is allowed to be an arbitrary member of $R'$, since $\eta$'s being separating for $R$ entails that it is cyclic for $R'$. But I'm wondering if this statement about unitary $U \in R'$ holds. Thanks!