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?



Simultaneous diagonalization of self-adjoint operators on Hilbert space
Apply the SNAG (Stone-Naimark-Godement-Ambrose) theorem to the unitary group generated by these operators.


