Do the eigenvectors of different observables span the same Hilbert space?

Question

The Hilbert space is spanned by independent bases. The textbook said that the eigenvectors of observable spans the Hilbert space. Do the eigenvectors of multiple observables span the same Hilbert space?

So what I mean is that let's assume we have a state $|\Psi\rangle$ which lives in the Hilbert space. We have two operators which correspond to observables denoted as $\hat{O}_{1}$ and $\hat{O}_{2}$. Let's measure the observable $O_{1}$. Our state $|\Psi\rangle$ will collapse and we measure one of the eigenvalues. The state $|\Psi\rangle$ will be one of the eigenvectors, let's say $|\Psi\rangle = |\psi_{1}\rangle$ where $|\psi_{1}\rangle$ one of the eigenvector of observable $\hat{O}_{1}$ is. Let's measure now the observable $O_{2}$. Our state $|\Psi\rangle = |\psi_{1}\rangle$ will collapse to one of the eigenstates of observable $O_{2}$. The eigenvectors $\hat{O}_{1}$ span the Hilbert space. The eigenvectors $\hat{O}_{2}$ span the Hilbert space. Do both eigenvectors span the same Hilbert space where state $|\Psi\rangle$ lives?

Answer 1

Yes, they span the same Hilbert space. If the two observables commute, \begin{equation} [O_1 , O_2 ] = O_1 O_2 - O_2 O_1 = 0, \end{equation} they share an eigenbasis, i.e. there is a basis $\{\vert \psi_n \rangle\}$ with \begin{equation} O_1 \vert \psi_n \rangle = \lambda_n \vert \psi_n \rangle,\ O_2 \vert \psi_n \rangle = \rho_n \vert \psi_n \rangle, \end{equation} where $\lambda_n$ are the eigenvalues of $O_1$ and $\rho_n$ are the eigenvalues of $O_2$. If the observables do not commute, however, \begin{equation} [O_1 , O_2 ] \neq 0, \end{equation} they do not share and eigenbasis. In this case, there are different bases, $\{\vert \psi^i_n \rangle\},\ i = 1,2$, with \begin{equation} O_1 \vert \psi^1_n \rangle = \lambda_n \vert \psi^1_n \rangle,\ O_2 \vert \psi^2_n \rangle = \rho_n \vert \psi^2_n \rangle. \end{equation} The two bases both span the Hilbert space and you can go from one to the other by a unitary transformation \begin{equation} \vert \psi^2_n \rangle = \sum_{m} U_{mn} \vert \psi^1_m \rangle,\ U_{mn} = \langle \psi^1_m \vert \psi^2_n \rangle. \end{equation} This unitary transformation is analogous to rotating your frame of reference in a real vector space like $\mathbb{R}^2$ in the complex Hilbert space of your quantum system. Its still the same space only in a different basis.

Now, as you suggest prepare the system in an eigenstate of $O_1$, say $\vert \psi^1_1 \rangle$. Then, upon measuring $O_2$, we end up in state $\vert \psi^2_n \rangle$ with probability $\vert\langle \psi^1_m \vert \psi^2_n \rangle\vert^2$.