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Consider the observables $A$ and $B$ with $[A, B]=0$. Suppose that the state of the system satisfies the equation $A|a_1\rangle=a_1 |a_1\rangle$. After a measurement of the observable $A$, in what state will the system be? If we then measure the observable $B$, what will the final state of the system be?

I am a bit confused by this question. By the measurement postulate, I know that after a measurement of $A$ the system will jump into one of the eigenstates of $A$. But the question doesn't say anything about all the eigenvectors of $A$, I only know that it has an eigenvector $|a_1\rangle$. Is it safe to assume that $A$ has only this eigenvector? If this is the case, the question is trivial: after a measurement of $A$, the system will still be in the state $|a_1\rangle$ and if we then measure $B$ the system I think will still be in the state $|a_1\rangle$ since $A$ and $B$ commute. But what if $A$ has more than one eigenvector? What conclusions could be drawn?

Qmechanic
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QuantumBoy
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  • See https://physics.stackexchange.com/a/542961/284199 – Cameron Gibson May 22 '23 at 02:12
  • Are there any more assumptions about the relationship between the operators $A$ and $B$? From just the given information, I wouldn't know what the state is post-measurement of observable $B$ because $\lvert a_1 \rangle$ is not even necessarily an eigenstate of $B$. – Silly Goose May 22 '23 at 04:00
  • To see why what I claimed above is true (particularly in the case of degenerate spectrums) look at ACM's answer here: https://physics.stackexchange.com/questions/457392/commuting-observables-share-common-eigenstates – Silly Goose May 22 '23 at 04:01

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Your initial state is an eigenstate $\lvert a_1 \rangle$ of the observable $A$. In the time leading up to your measurement of observable $A$, there is no interaction occurring which would change the initial state to another state. In other words, for all time $t$ leading up to measurement of observable $A$, the state remains in the state $\lvert a_1 \rangle$. Thus, the state remains in an eigenstate of the observable you are measuring up to the point in time at which you measure it. Recall that if a state is already is an eigenstate $\lvert a_1 \rangle$ of an observable $A$ and you perform a measurement of observable $A$ that you will read measurement value $a_1$ 100% of the time, and the state post-measurement is still $\lvert a_1 \rangle$.

If $A$ and $B$ have nondegenerate spectrums, $[A, B] = 0$ implies that they share a unique set of eigenvectors which form a basis for the Hilbert space said operators are defined over. This means that $$A\lvert a_1 \rangle = a_1 \lvert a_1 \rangle$$ and $$B\lvert a_1 \rangle = b_1 \lvert a_1 \rangle$$ and the same reasoning we applied in the first paragraph would then be applied again. Note that if $\lvert \psi \rangle$ is a shared eigenstate of operators $A$ and $B$ with eigenvalues $a_1$ and $b_1$, respectively, it is customary to write the eigenstate as $\lvert \psi \rangle = \lvert a_1, b_1 \rangle$ which makes it clear that the ket is an eigenstate of both operators with the corresponding eigenvalues.

If the spectrums are degenerate for either operator or both, then the above possibly does not hold true. And I would say what happens when you measure $B$ is unknowable given the information provided in the quotation. At the beginning of Quantum texts, it is usually implicit that spectrums are nondegenerate; however, I am not sure that is the case here.

Silly Goose
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