Why does commutativity mean that two observables can be measured together?

Question

BACKGROUND

As far as the Heisenberg uncertainty principle is concerned, my understanding of commuting observables $\hat{A}$ and $\hat{B}$ is that the measurement outcome $a_i$ does not perturb (or correlate with) the measurement outcome $b_j$ because they $a_i$ and $b_j$ arise from projections onto orthogonal eigenvectors of $\hat{A}$ and $\hat{B}$, respectively.

QUESTION

What I don't understand is this: What does it actually mean that $\hat{A}$ does not influence (i.e., is independent of) $\hat{B}$? If I visualize some measured quantum state $\mid \psi\rangle = \alpha~\hat{a}_i + \beta~\hat{b}_j$ as, say, a vector in a Bloch-sphere, then measuring $\hat{A}$ will collapse $\mid \psi\rangle$ onto the eigenvector $\hat{a}_i$ (with probability $\alpha$). However, won't any subsequent measurement on $\hat{B}$ become completely randomized? No information about $\beta$ could then possibly be retrieved. I therefore don't understand how one can say that $\hat{A}$ and $\hat{B}$ can be measured "simultaneously".

Answer 1

If two observables commute, $[\hat{A},\hat{B}]=0$, then this means that you can always find a common set of eigenstates. In the simplest case of the eigenvalue spectra of $\hat{A}$ and $\hat{B}$ being non-degenerate, then this implies that the eigenstates $\{|u_n\rangle\}$ are the same for both: $$ \hat{A}|u_{n}\rangle=a_n|u_{n}\rangle, \\ \hat{B}|u_{n}\rangle=b_n|u_{n}\rangle. $$

If you start with your initial state written in the basis of eigenstates of $\hat{A}$, $|\psi\rangle=\alpha|u_i\rangle+\beta|u_j\rangle$, then if measuring $\hat{A}$ you get $a_i$, your state immediately after the measurement is $|\psi^{\prime}\rangle=|u_i\rangle$.

If you then want to measure $\hat{B}$, you have to write your new state $|\psi^{\prime}\rangle$ in the basis of eigenstates of $\hat{B}$. Crucially, this is $|\psi^{\prime}\rangle=|u_i\rangle$ because as $\hat{A}$ and $\hat{B}$ commute so they share the same set of eigenstates. So $|\psi^{\prime}\rangle$ is already in an eigenstate of $\hat{B}$, and when you measure $\hat{B}$ you will get $b_i$ with probability 1. If you did measure $\hat{A}$ again you would get $a_i$ again, and so on.

This discussion becomes more subtle when $\hat{A}$ and/or $\hat{B}$ have a degenerate eigenvalues spectrum, but I think the above is a good starting point to answer your question.