# Simultaneous diagonalization of self-adjoint operators on Hilbert space

Suppose I'm given a finite set of possibly unbounded commuting self-adjoint operators $T_i : \mathfrak H \supset \mathscr D(T_i)\to \mathfrak H, i = 1 , \dots , N$ on a Hilbert space (in the sense of commuting resolvents $\forall i, j: \exists z_i \in \rho (T_i) , z_j \in \rho (T_j):[R_{T_i} (z_i ) , R_{T_j} (z_j)] = [R_{T_i} (\overline{z_i} ) , R_{T_j} (z_j)] = 0$). In linear algebra this implies that these operators can be simultaneously diagonalized.

Is there any general rigorous notion of simultaneous diagonalization in this functional analysis setting? Intuitively I would expect something like a projection-valued measure

$$E: \mathscr B(\mathbb R^N ) \to \mathbb C^N \otimes \mathfrak L(\mathfrak H)$$

supported on $\sigma (T_1) \times \dots \times \sigma (T_N)$, such that $E^{(T_i)} (A) = E(\mathbb R \times \dots \times \mathbb R \times \underbrace{A}_{i} \times \mathbb R \times \dots \times \mathbb R )$.

Or maybe a guarantee that there is a common measure space $(\Omega , \mu )$ and a common unitary transformation $U : \mathfrak H \to L^2(\Omega, \mathrm d\mu )$ such that

$$\forall i : \exists f_i \in \text{Meas}(\Omega, \mathrm d\mu): T_i = U^* M_{f_i} U$$

Is there anything like that?

Another way to do it is to consider the bounded operators $(T_i + iI)^{-1}$, check that these are normal and commute so they can be simultaneously represented as multiplication operators on an $L^2$ space, and then untransform to represent the original $T_i$ as unbounded multiplication operators.
While there is no problem constructing a unique product measure $\mu\otimes\nu$ for $\sigma$-finite measure spaces $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ the analogous question for projection-valued measures encounters a snag unless $X$ and $Y$ are sufficiently nice like $\mathbb{R}^n$. See the remark after Theorem 4.10 in the same book and the reference to the article by Birman, Vershik and Solomjak.