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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?

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The correct generalization of the statement for unbounded operators is the joint spectral measure theorem:

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.

See chapter 5.5 in "Unbounded Self-adjoint Operators on Hilbert Space" by Schmüdgen for more details, in particular technical details relating to the construction of the product measure.

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