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I'm new to QM and trying to understand the basics of commutators. If you have two operators that commute, then it is said you can measure two observables simultaneously. However, I'm a bit confused:

Assume $\hat{A}$ and $\hat{B}$ commute. If we express $|\psi\rangle$ in terms of the common eigenbasis, we get $|\psi\rangle = c_1 \psi_1 + ... + c_n \psi_n$.

Now, if I apply $\hat{B}$ only, then my state is forced into some eigenfunction $|\psi_n\rangle$ with probability $|c_n|^2$, which we will say has corresponding eigenvalue $b_n$ .

Say I apply $\hat{A}$, then $\hat{B}$. Then, $\hat{A}$ forces my state into $|\psi_n\rangle$ with probability $|c_n|^2$, and $\hat{B}$ does not further collapse the wavefunction since it is already collapsed into a particular common eigenfunction. So I observe eigenvalues $b_n$ and $a_n$ simultaneously, with probability $|c_n|^2$.

What confuses me is that this would imply I only observe $a_n$ and $b_n$ together for any $n$ (e.g. only $a_1$ and $b_1$, $a_2$ and $b_2$, etc., but never $a_1$ and $b_2$). After applying the first operator, performing an observation with $\hat{B}$ becomes a deterministic process determined by the eigenfunction "chosen" by $\hat{A}$. But surely if the position is in a superposition state then we should be able to observe other pairings such as $a_1$ and $b_2$ together—the fact that we would only measure $a_1$ with $b_1$ seems to be a result of a flawed measurement process rather than a true constraint of the system.

(Note on terminology: Let $a_n$ and $b_n$ be the eigenvalues associated with eigenfunction $\psi_n$).

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    $\begingroup$ If you denote the common orthonormal eigenbasis as something like $\psi_{ijk}$, where $i$ enumartes the eigenvalues of $A$, $j$ the eigenvalues of $B$ and $k$ some possible degeneracy, then your doubt in the last paragraph should be answered... Of course, you can re-label everything with a single index $n$, as you did, but then $c_n$ is not the probability amplitude to measure $a_n$ and $b_n$ or so (in general). $\endgroup$ Commented Nov 11, 2024 at 6:39
  • $\begingroup$ For a basis independent notion: If $P_a$, $Q_b$ project on the eigenspace of $A$ with eigenvalue $a$ and on the eigenspace of $B$ with eigenvalue $b$, respectively, then $P_aQ_b=Q_bP_a$ (holds because hermitian $A$ and $B$ commute) projects on the common eigenspace corresponding to the respective eigenvalues. The probability to measure $a$ and $b$ in some normalized state $\psi$ is then $\langle \psi, P_aQ_b \psi\rangle$. Everything is nicely explained in Isham's QM book, for example. $\endgroup$ Commented Nov 11, 2024 at 6:42
  • $\begingroup$ The first comment resolved my confusion—thank you! $\endgroup$ Commented Nov 11, 2024 at 20:44

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I think the good comments of @TobiasFünke and the good answer of @hft already showed the missing point, but I want to give you two examples.

Ex. 1: No degeneracy

Consider the matrices (I'll consider finite dimension Hilbert spaces for simplicity)

$$ A = \begin{pmatrix} a_1 & 0 & 0\newline 0 & a_2 & 0\newline 0 & 0 & a_3 \end{pmatrix} \quad B = \begin{pmatrix} b_1 & 0 & 0\newline 0 & b_2 & 0\newline 0 & 0 & b_3 \end{pmatrix} $$

Both observables are already diagonal in our initial basis, let's say $\{|0\rangle,|1\rangle,|2\rangle\}$. The probabilities of the results are given by the expressions

$$ p(a_j) = \operatorname{tr}(|\psi\rangle\!\langle \psi| \, |j\rangle \!\langle j |)\quad p(b_k) = \operatorname{tr}(|\psi\rangle\!\langle \psi| \, |k\rangle \!\langle k |) $$

where I'm considering the linear projectors related to each measurement result. Since there are no degeneracy and they are simultaneously diagonalized, the same protectors describe the probabilities of $A$ results and $B$ results. Now, if we want to calculate joint probability distributions

$$ p(a_j,b_k) = \operatorname{tr}(|\psi\rangle\!\langle\psi| \, |j\rangle\!\langle j|k\rangle\!\langle k| ) = 0, \quad j\neq k. $$

So in this case, the probability of results of $A$ and $B$ are zero for results with different labels. But labelling is just an artificial property of nature, and we could in principle give any other label for $A$ results. In this case, the list of one-dimensional projectors is enough to describe the physics, and it's quite nonsense to talk about $A$ and $B$ as being different physical quantities.

This example is very artificial and it is rare to find commuting observables without degeneracy. Usually, degeneracies are a consequence of the symmetry of the problem, which is connected with commutation relations between the system's Hamiltonian and some unitary operator representing the symmetry.

Ex.2: Degenerate case

Consider the case

$$ A = \begin{pmatrix} a_1 & 0 & 0\newline 0 & a_2 & 0\newline 0 & 0 & a_3 \end{pmatrix} \quad B = \begin{pmatrix} b_1 & 0 & 0\newline 0 & b_1 & 0\newline 0 & 0 & b_2 \end{pmatrix} $$

Now we have degeneracy in $B$ observable in the initial basis. It means that the projector associated to the result $b_2$ is a two-dimensional projector

$$ |b_1\rangle\!\langle b_1| = |0\rangle\!\langle 0| + |1\rangle \!\langle 1| $$

In this case, $p(a_1,b_1) = p(a_2,b_1)$ and these probabilities are not necessarily zero (depending on the state, of course). But more than that, the system could be in the state

$$ |\psi\rangle = \frac{1}{\sqrt 2} \left ( |0\rangle + |1\rangle\right ) $$

which is an eigenstate of $B$ but not an eigenstate of $A$, for example. This example shows that not all eigenstates of $B$ are eigenstates of $A$ even if they commute.

There are other situations where a commutative property is obtained, for example due to tensor product structure in composite systems. I recommend you to work out specific examples, preferable with clear physical meaning like angular momentum and other stuff. To get intuition through mathematical examples is quite hard, but if you want, try to work out finite dimensional example.

You tried to assume a mathematical condition, get mathematically consistent results and asked the physical plausibility of the results, but I recommend you to assume physically plausible condition, get mathematically consistent results and then ask the physical plausibility of the results.

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If you have two operators that commute, then it is said you can measure two observables simultaneously.

Yes, that is fine to say.

Assume $\hat{A}$ and $\hat{B}$ commute. If we express $|\psi\rangle$ in terms of the common eigenbasis, we get $|\psi\rangle = c_1 \psi_1 + ... + c_n \psi_n$.

I think this may be just a typo on your part, but the above expression should read: $$ |\psi\rangle = c_1 |\psi_1\rangle +\ldots + c_n |\psi_n\rangle\;. $$ The eigenbasis is a basis of kets. The LHS is a ket and so the RHS should be a weighted sum of kets.

Now, if I apply $\hat{B}$ only, then my state is forced into some eigenfunction $|\psi_n\rangle$ with probability $|c_n|^2$, which we will say has corresponding eigenvalue $b_n$ .

No. This is false. Applying the operator is not the same as measuring an eigenvalue associated with the operator.

You stated that the basis $|\psi_i\rangle$ is a common eigenbasis for both $\hat A$ and $\hat B$. Applying $\hat B$ does not "force" your state into a single one of eigenbasis states. Rather, what happens is: $$ \hat B |\psi\rangle = c_1 b_1|\psi_1\rangle + \ldots + c_n b_n|\psi_n\rangle\;, $$ where $b_i$ is the eigenvalue of $\hat B$ associated with the eigenbasis state $|\psi_i\rangle$.

Say I apply $\hat{A}$, then $\hat{B}$.

OK, no problem.

Then, $\hat{A}$ forces my state into $|\psi_n\rangle$ with probability $|c_n|^2$,

No. Nothing collapses, if you are just applying the operator $\hat A$ to the state. Applying the operator is not the same as measuring an eigenvalue associated with the operator.

If you apply $\hat A$ and then $\hat B$ you get: $$ \hat B \hat A |\psi\rangle = c_1 a_1\hat B|\psi_1\rangle + \ldots + c_n a_n\hat B|\psi_n\rangle $$ $$ = c_1 a_1 b_1|\psi_1\rangle + \ldots + c_n a_n b_n|\psi_n\rangle $$ $$ = c_1 b_1 a_1|\psi_1\rangle + \ldots + c_n b_n a_n|\psi_n\rangle $$ $$ =\hat A \hat B |\psi\rangle\;, $$ where $a_i$ is the eigenvalue of $\hat A$ associated with $|\psi_i\rangle$.

and $\hat{B}$ does not further collapse the wavefunction since it is already...

Nothing is "collapsing" when you apply either A or B to the state. Acting with an operator on a state does not "collapse" anything. Measurement is said to "collapse," but that is different than acting on the state with the operator associated with the measurement. If you actually perform a measurement, you measure one of the eigenvalues and the state is said to "collapse," but this is not performed by acting with the observable operator.

After applying the first operator...

No. Applying the operator and measurement/collapse are not the same.

If you want to talk about what happens when you measure, say, an eigenvalue of $\hat A$, then you just have to assert that your measuring apparatus measured, say, $a_4$. All you can say beforehand is that the probability to measure $a_4$ is $|\langle \psi_4|\psi\rangle|^2 = |c_4|^2$. You cannot know what will be measured, but of course any actual measurement will result in some value.

After the measurement of $a_4$ the state collapses to: $$ |\psi\rangle \to |\psi_4\rangle\;. $$

Obviously any subsequent measurements of $A$ or $B$ return $a_4$ and $b_4$, respectively, with 100% certainty.

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    $\begingroup$ Thanks for the correction on the terminology. The core of my conceptual confusion comes from the final point that you made. I was trying to model the process of measuring two observables simultaneously. The idea that measuring $a_4$ can only lead to measuring $b_4$ with 100% certainty, I pointed out, "seems to be a result of a flawed measurement process rather than a true constraint of the system." I.e., if you were to measure the eigenvalues associated with $A$ and $B$ simultaneously, you would measure only a subset of possibilities $(a_i, b_i)$ rather than any possibility $(a_i, b_j)$. $\endgroup$ Commented Nov 11, 2024 at 5:10
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Before I answer your question I'm going to write some material to clarify some aspects of quantum theory.

In classical physics the evolution of a physical quantity such as the $x$ position is described in terms of a variable $x(t)$ and if you measure the $x$ position you get the value $x(t)$ and likewise for other physical quantities.

In quantum physics the situation is different. The equations of motion of quantum theory describe the evolution of a measurable quantity $A$ in terms of a Hermitian operator $\hat{A}$. The possible results of measuring $A$ are the eigenvalues of $\hat{A}$. There is another Hermitian operator called the state $\rho$ with unit trace and quantum physics predicts the expectation value of $A$ denoted $\langle\hat{A}\rangle$ given by $$\langle\hat{A}\rangle=tr(\rho\hat{A})$$

In general predicting the expectation value involves tracking what happens to all of the possible results because of quantum interference. For an example see Section 2 of

https://arxiv.org/abs/math/9911150

A measurement is a process that produces a record of whatever you measured. A record is a piece of information that can be used to test a theory and so it must be possible to copy the record so other people can look at it and discuss it.

One way to think about measurement in quantum theory is to look at what constraints the equations of motion of quantum theory place on records. A measurement of an observable would involve applying a unitary operation $U$ to the measured system $S$ and the measurement device $M$ that couples them to produce a record. If $S$ is in a state $$|\psi\rangle_S = \sum_a \alpha_a |a\rangle_S,$$ the measurement device is in a blank state $|0\rangle_M$ so the state of the overall system before the measurement is $$|\Psi(t_1)\rangle =|\psi\rangle_S|0\rangle_M$$ and the unitary operator produces the mapping $$|a\rangle_S|0\rangle_M\to|a\rangle_S|a\rangle_M$$ then the final state after the measurement is $$|\Psi(t_2)\rangle =\sum_a \alpha_a |a\rangle_S|a\rangle_M$$ This is called a perfect measurement. There are also various kinds of imperfect measurements such as indirect measurements and that sort of thing that can be described by the equations of motion of quantum theory.

Copying information out of a system in general suppresses quantum interference between the measured results - this is called decoherence:

https://arxiv.org/abs/1911.06282

So then the equations of motion of quantum theory predict that a measurement will result in the different possible measurement results evolving autonomously from one another:

https://arxiv.org/abs/1111.2189

https://arxiv.org/abs/quant-ph/0104033

As such you can set the state to reflect the measurement result you got and this is called the relative state.

Many physicists don't like this implication of quantum theory, which is typically called the many worlds interpretation, and say that all but one of the possible results are eliminated by a process called collapse. It is common for this process to be mentioned in textbooks and casual discussion without actually explaining the process by which it happens or giving its equations of motion. As such it is difficult to discuss what collapse implies in the way physicists usually talk about it. Some physicists have proposed actual models for collapse

https://arxiv.org/abs/2310.14969

These models don't currently explain the predictions of relativistic quantum theories which are the vast bulk of predictions of quantum theory:

https://arxiv.org/abs/2205.00568

In addition, quantum measurement theory involves repeated, continuous and unsharp measurement and none of these are described well by collapse:

https://arxiv.org/abs/1604.05973

I don't see much reason to discuss collapse alternatives to quantum theory instead of discussing quantum theory itself which has passed all the experimental tests described above and can be used to discuss any kind of measurement you like.

Now, let's return to your question about commuting observables. If you're doing a perfect measurement of two commuting observables $\hat{A}$ and $\hat{B}$ and the system doesn't evolve between the two measurement then the measurements will have the corresponding results. You can check this by just applying two perfect measurement operators. Indeed, since these observables have the same set of possible states you can just do one measurement and relabel the results. If the measurement of $\hat{A}$ and $\hat{B}$ is complicated and might have imperfections like indirection or loss of some results or something like that then you can describe it with the relevant evolution operator and work out the consequences of that measurement.

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