# Spectral families of commuting self-adjoint operators

I don't know if math stack exchange is more suitable for this question, but I'll try here first.

It is often stated in quantum mechanics textbooks (e.g. the first volume of Cohen-Tannoudji, Diu, Laloë, page 135) that

“If two observables $$A$$ and $$B$$ commute, it is possible to construct an orthonormal basis of the Hilbert space made of common eigenvectors of $$A$$ and $$B$$”.

However, if one is to be a little more precise (mathematically speaking), a self-adjoint operator doesn't even necessarily have an orthonormal basis of eigenvectors of its own. Instead, a self-adjoint operator $$A$$ has a “spectral family” or “resolution of the identity” $$P(\lambda)$$ with $$\lambda \in \sigma(A)$$ (the spectrum of $$A$$), and, roughly speaking, $$\mathrm d P(\lambda) = P(\lambda + \mathrm d\lambda) - P(\lambda)$$ is the “projection” on the subspace corresponding to the “eigenvalue” $$\lambda$$. Of course, $$\lambda$$ is not necessarily an eigenvalue and there isn't always a subspace associated with it.

However, by analogy, I would expect something like “the spectral families of commuting operators are the same”, meaning that their “generalized eigenvectors” and thus the “generalized subspaces” are the same. I didn't find any such result anywhere I looked: the only statement I was able to find concerning spectral families of commuting operators is that they commute, which is nothing more than the definition of commuting operators in the unbounded case.

I am wondering what am I missing: is there a theorem, which concerns spectral families of commuting operators, analogous to the statement found e.g. in Cohen-Tannoudji? Is my intuition wrong?

Let $$A_i, i = 1,\dots, n$$ be self-adjoint operators that all pairwise commute. Then there exists a joint spectral measure $$\mathrm{d}E(\vec \lambda) = \mathrm{d}E(\lambda_1,\dots,\lambda_n)$$ on $$\mathbb{R}^n$$ such that $$A_i = \int \lambda_i \mathrm{d}E(\vec \lambda)$$ and $$\mathrm{supp}(\mathrm{d}E) = \{ \vec \lambda\in\mathbb{R}^n \mid \lambda_i \in \sigma(A_i)\}$$.
In the discrete/bounded case this measure is simply the product of projectors, i.e. $$\mathrm{d}E(\vec \lambda) = \begin{cases} \prod_i P^{(i)}_{\lambda_i} & , \lambda_i \in \sigma(A_i) \forall i \\ 0 & \text{else}\end{cases}$$ where $$P^{(i)}_{\lambda_i}$$ is the projection onto the eigenspace of $$A_i$$ with eigenvalue $$\lambda_i$$. Note that this is well-defined precisely because the $$P^{(i)}_{\lambda_i}$$ all commute, and is equivalent to the existence of a simultaneous eigenbasis - just choose a basis of the images of the $$\prod_i P^{(i)}_{\lambda_i}$$.
The construction of $$\mathrm{d}E(\vec \lambda)$$ is exactly the obvious one: It is the product measure of the individual spectral measures $$\mathrm{d}E_i$$ defined by $$\mathrm{d}E(\vec \lambda) = \prod_i \mathrm{d}E_i(\lambda_i),$$ and again this makes sense because the $$A_i$$ commute and so their spectral measures commute, too.