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

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".

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.

• Just a follow-up on this: If two observables commute, does that mean that all their eigenstates are the same or that only a subset of them are? Nov 2, 2020 at 8:47
• If two observables commute, then you can always find a common set of eigenstates, which covers all the spectrum. I recently described this in some detail here, going into the subtleties of the degenerate case: youtu.be/IhJvX4H7xkA Nov 2, 2020 at 8:50
• I watched your video, but the thing that bothers me is this: Let's say $[\hat{A}, \hat{B}] = 0 = [\hat{A}, \hat{C}]$ but $[\hat{B}, \hat{C}] \neq 0$. Then, if one follows your example above, measuring $\hat{A}\hat{B}\hat{A}$ on $|\psi\rangle$ will yield the measurement sequence $a_i, b_i, a_i$ whereas $\hat{A}\hat{C}\hat{A}$ could yield a different sequence $a_j, c_j, a_i$ where the last measurement of $\hat{A}$ is $a_j \neq a_i$. It's as if the two sequences, which both supposed to be quantum non-demolition, lead to different contexts (cf. contextuality). Nov 15, 2020 at 12:46
• @Tfovid The example in my answer specifies that it is valid for non-degenerate eigenvalues. The situation you describe is trickier because it arises when you have degenerate eigenvalues. In that case, the eigenvalue of $\hat{A}$ will be $n$-fold degenerate, and you can build valid eigenstates of $\hat{A}$ making different linear combinations over the $n$-fold degenerate subspace. Then, the eigenstates common to $\hat{A}$ and $\hat{B}$ will be different to those common to $\hat{A}$ and $\hat{C}$. If you then work the maths through taking degeneracy into account, all falls into place. Nov 15, 2020 at 12:55
• @Tfovid A concrete example of the situation you are describing is where $\hat{A}$ is the squared angular momentum operator $\hat{L}^2$ and $\hat{B}$ and $\hat{C}$ are two angular momentum components, say $\hat{L}_z$ and $\hat{L}_x$. Nov 15, 2020 at 12:56