Let

- $H$ be a separable $\mathbb R$-Hilbert space
- $H\:\hat\otimes_\pi\:H$ denote the projective tensor product of $H$ and $H$
- $(\Omega,\mathcal A)$ be a measurable space
- $\mu$ be a $H\:\hat\otimes_\pi\:H$-valued vector measure on $(\Omega,\mathcal A)$ of bounded variation
- $f:\Omega\to\mathfrak L(H,\mathfrak L(H,\mathbb R))$ be bounded and strongly $\mathcal A$-measurable

Note that $L(H,\mathfrak L(H,\mathbb R))\cong\mathfrak L(H\:\hat\otimes_\pi\:H,\mathbb R)$ and hence $f(\omega)$ can be identified with $$H\:\hat\otimes_\pi\:H\ni\sum_{n=1}^\infty x_n\otimes y_n\mapsto\sum_{n=1}^\infty (f(\omega)x_n)y_n\tag1$$ for all $\omega\in\Omega$. With respect to $(1)$, we can consider an integral $$\int f\:{\rm d}\mu.\tag2$$

Now, if $E,F,X,Y$ are $\mathbb R$-Banach spaces, $S\in\mathfrak L(X,E)$ and $T\in\mathfrak L(Y,F)$, there is a unique bounded linear operator $S\otimes_\pi T$ from $X\:\hat\otimes_\pi\:Y$ to $E\:\hat\otimes_\pi\:F$ with $$(S\otimes_\pi T)\left(\sum_{n=1}^\infty x_i\otimes y_i\right)=\sum_{n=1}^\infty (Sx_i)\otimes (Ty_i)\tag3.$$

Since $L(H,\mathfrak L(H,\mathbb R))\cong\mathfrak L(H)$, $f(\omega)$ can be identified with $$H\ni x\mapsto\sum_{n=1}^\infty((f(\omega)x)f_n)f_n\tag4,$$ where $(f_n)_{n\in\mathbb N}$ is an orthonormal basis of $H$, for all $\omega\in\Omega$. Thus, with respect to $(3)$, we can consider an integral $$\int f\otimes_\pi\operatorname{id}_H{\rm d}\mu,\tag5$$ where $\operatorname{id}_H$ denotes the identity on $H$.

My question is: How are $(2)$ and $(5)$ related? To be precise: I want to express $(5)$ in terms of $(2)$. Is that possible? (I'm pretty sure that there is a bounded linear operator which transforms $(5)$ into $(2)$.)