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.


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".


1 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.

  • $\begingroup$ 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? $\endgroup$
    – Tfovid
    Nov 2, 2020 at 8:47
  • 1
    $\begingroup$ 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 $\endgroup$
    – ProfM
    Nov 2, 2020 at 8:50
  • $\begingroup$ 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). $\endgroup$
    – Tfovid
    Nov 15, 2020 at 12:46
  • $\begingroup$ @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. $\endgroup$
    – ProfM
    Nov 15, 2020 at 12:55
  • $\begingroup$ @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 $\hat{L}_x$. $\endgroup$
    – ProfM
    Nov 15, 2020 at 12:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy