Suppose that $H$ is some (complex) Hilbert space and that $\{T_\alpha: \alpha \in I\}$ is some collection of bounded self-adjoint operators on $H$. A version of the spectral theorem states that for each $\alpha$, there exists a measure space $(X,M,\mu)$, a bounded real-valued function $\phi$ and a unitary operator $U: H \to L^2(\mu)$ such that
$$T_\alpha = U^*M_\phi U$$
where $M_\phi$ denotes the multiplication operator induced by $\phi$. I was wondering under which conditions on $\{T_\alpha\}$ we can find a measure space $(X,M,\mu)$ such that all of the $T_\alpha$ are unitarily equivalent to a multiplication on $L^2(\mu)$? Even better, under what conditions can we choose the same $U$ to work for all $T_\alpha$?
For example (although these are not bounded) all constant coefficient partial differential operators on $L^2(\mathbb{R}^n)$ are Fourier multipliers. What makes them special?
The situation that's usually considered is when you want to use the same $U$ for all $T_\alpha$. In this case, it's easy to see that it's necessary for the $T_\alpha$s to commute. As I recall, this is also sufficient: there's a spectral theorem for commuting families of self-adjoint operators. Unfortunately, I can't immediately find a reference so I could be missing a hypothesis. I've made this CW so somebody else could edit one in.
I don't know about the possibility of using a different $U_\alpha$ for each $T_\alpha$. It's not really clear to me what such a representation would be good for.