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Suppose we have got a triple of observables $A,B$ and $C$. Suppose furthermore, that $[A,B]=0$ and $[B,C]=0$ but $[A,C]\neq 0$ . Suppose, also now we do a measurement of $A$ then accordingly we would lose all information about $C$ because of the uncertainty in $C$. But notice that $[B,C]=0$ thus it follows that $B$ must be in the same eigenstates as $C$ after the measurement of $A$ and therefore we cannot measure $B$ . But since $[A,B]=0$ we can measure $B$ exactly subsequent to $A$ because they are in the same eigenstates. Therefore, QM tells us that we can simultaneously measure $B$ after $A$ because they are in the same eigenstates but we also cannot measure $B$ because after the measurement of $A$, $C$ is affected and $B$ is in the same eigenstates as $C$. How is this possible?

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  • $\begingroup$ "we would lose all information about $C$" Well, not all information, unless $[A,\,C]$ has no eigenvectors of eigenvalue $0$. $\endgroup$ – J.G. Jun 30 at 16:32
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I think that the misunderstanding follows from the following fact. Given that $[A,B]=0$ then we can build a simultaneous eigenbasis for $A$ and $B$, call it $\mathcal{B}_1$.

On the other hand we have that $[B,C]=0$ which means that we can construct a simultaneous eigenbasis for $B$ and $C$ but, and here is the catch, this second simultaneous eigenbasis is not the same as the first one. Call this eigenbasis $\mathcal{B}_2$. This means that a simultaneous eigenstate of $B$ and $C$ is not and eigenstate of $A$ since is build up on the $\mathcal{B}_2$ eigenbasis, the one of $B$ and $C$.

This fact relates directly to $[A,C]\neq 0$ since this implies that we cannot construct a simultaneous eigenbasis of $A$ and $C$.

Using bra-ket notation this can be easily seen: if we call $|a,b\rangle$ a state such that $$A|a,b\rangle = a|a,b\rangle\qquad B|a,b\rangle = b|a,b\rangle$$ it's clear that $C|a,b\rangle$ is going to be a linear combination of all the $|a_i,b_i\rangle$ eigenstates of $A$ and $B$. But we can construct in the same manner the basis for $B$ and $C$. Call one of the basis state $|b,c\rangle$ such that $$B|b,c\rangle = b|b,c\rangle \qquad C|b,c\rangle = c|b,c\rangle$$ then $A|b,c\rangle$ is going to be a linear combination of the basis $|b_i,c_i\rangle$. In this sense, the two basis are different even if the are basis for the same Hilbert space.

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  • $\begingroup$ my question is about the measurement of the observable B. What I am arguing is that the measurement of B is affected by the measurement of A, since C is affected because $[A,C]\neq 0$ and B commutes with C, it follows that B must also be affected. But [A,B]=0 so B should not be affected by the measurement of A. Therefore, we have a "seeming" contradiction i.e. the measurement of B is not affected by the measurement of A and it is affected by the measurement of A. $\endgroup$ – user11937 Jun 29 at 23:34
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    $\begingroup$ The user BioPhysicist explained very well were's the problem in your line of reasoning in this. $\endgroup$ – Davide Morgante Jun 30 at 7:39
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A mistake made here is this statement and all similar ones:

But since $[A,B]=0$ we can measure $B$ exactly subsequent to $A$ because they are in the same eigenstates.

The collapse to an eigenstate of $A$ after measurement does not necessarily guarantee anything about a further measurement of $B$ - it can still be in a superposition of common eigenstates of $A$ and $B$, so long as the eigenvalue in $A$ is the same. In that case, the value upon the measurement of $B$ is not necessarily guaranteed.

$[A,B]=0$ only says that there are common eigenvectors of $A$ and $B$, not that they must both be determined together. $[B,C]=0$ says a similar thing. Combining the two does not tell you $A$, $B$, $C$ are determined together upon measurement, and therefore there is no contradiction with $[A,C]≠0$ telling you that $A$ and $C$ cannot be determined together.

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Suppose, also now we do a measurement of $A$

Ok, so our system is in a state of definite $a$ after this measurement.

then accordingly we would lose all information about $C$

Yes, since $[A,C]\neq0$, we know our current state can be described as a superposition of various states with definite $c$.

But notice that $[B,C]=0$ thus it follows that $B$ must be in the same eigenstates as $C$ after the measurement of $A$ and therefore we cannot measure $B$ .

This doesn't make any sense to me. We just made a measurement of $A$ on our system, so it is in a state of definite $a$. This has nothing to do with other measurements, as you haven't said we made another measurement yet. We are totally able to make a measurement of $B$, and since we were in a state of definite $a$, and since $[A,B]=0$, if we do such a measurement we can say that now our system will be in a state with definite $a$ and definite $b$. None of this right now has anything to do with $C$ or how it relates to the other observables.

But since $[A,B]=0$ we can measure $B$ exactly subsequent to $A$ because they are in the same eigenstates.

There exist a set of simultaneous eigenstates for $A$ and $B$, yes. This is described above.

Therefore, QM tells us that we can simultaneously measure B after A because they are in the same eigenstates but we also cannot measure B because after the measurement of A, C is affected and B is in the same eigenstates as C. How is this possible?

There is no contradiction. $[A,B]=0$ does not mean that all eigenstates of $A$ are eigenstates of $B$. It just means that we can find a common eigenbasis of both operators.

To go through the entire process, let's first measure $A$, then we are in some state of definite $a$. Now let's measure $B$, then we are in a state with definite $a$ and definite $b$. Now let's measure $C$, then we are in a state with definite $b$ and definite $c$. No contradiction.

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I would like to answer this from the perspective of angular momentum algebra in Quantum Mechanics, and also using ideas from vectors in 2 dimensional space.

Let's consider a two dimensional space with vectors all starting with their tip at the origin. Let M be a matrix. The purpose of M is to rotate and scale(squeeze or stretch) any vector in this 2D space. However, there are a few vectors that it only scales and does not rotate. Such vectors are the eigenvectors of M. Similarly for another matrix N. In 2D, there can be at most 2 linearly independent (l.i.) eigenvectors. Let's assume M and N both has only 2 l.i. eigenvectors, but each of them are different, that is, no eigenvector of M overlaps with eigenvector of N. So far you probably know all these stuff.

Now, consider the Identity Matrix of order 2. The role of identity matrix is to scale "all" of the vectors in 2D by the same amount (scale by 1). Thus, the entire space of vectors are eigenvectors of Identity.

Now, we know all matrices commute with Identity. Thus in our case, [M,I] = 0 and [N,I] = 0. However, does that imply that the set of eigenvectors of M or N (or any other matrix in fact) is the entire 2D space of vectors? No, right? So commuting matrices does not necessarily mean that the complete set of eigenvectors of both has to be the same in number. It can just mean that one of the matrices has a bigger set of eigenvectors - some of which happens to exactly match with the set of eigenvectors of the other matrix.

So, why is this relevant in this discussion?

Consider now the case of angular momentum algebra. In line with your question, let A be $S_x$ (the spin operator to measure spin along x-axis), B be $S^2$ (the total spin measuring operator) and C be $S_z$ (the spin operator to measure spin along z-axis).

You see that the commutation relations you gave are satisfied by these 3 matrices, namely [$S_x$,$S^2$] = 0, [$S^2$,$S_z$] = 0 but [$S_x$,$S_z$] = $i\hbar\ S_y\ \ne 0$

Now, lets visualize these spin matrices to act on a space of vectors in 2D (although inappropriately).

$S_x$ has the eigenvectors $\frac{1}{\sqrt{2}}{1\choose 1}$ and $\frac{1}{\sqrt{2}}{1\choose -1}$, which are like $\frac{\hat{\imath}+ \hat{\jmath}}{\sqrt{2}}$ and $\frac{\hat{\imath} - \hat{\jmath}}{\sqrt{2}}$

$S_z$ has the eigenvectors ${1\choose 0}$ and ${0\choose 1}$, which are like $\hat{\imath}$ and $\hat{\jmath}$

Note that $S_x$ and $S_z$ has non-overlapping set of eigenvectors.

While, $S^2$ is literally the Identity matrix of order 2, hence its set of eigenvectors is the entire 2D space of vectors, including $\frac{1}{\sqrt{2}}{1\choose 1}$, $\frac{1}{\sqrt{2}}{1\choose -1}$, ${1\choose 0}$ and ${0\choose 1}$

Now, when you measure $S_x$ (which is A in your question), the state collapses to one of the eigenvectors let say to $\frac{1}{\sqrt{2}}{1\choose 1}$

Now, $\frac{1}{\sqrt{2}}{1\choose 1}$ = $\frac{1}{\sqrt{2}}{1\choose 0} + \frac{1}{\sqrt{2}}{0\choose 1}$.

Thus, we find that there is equal probability to get ${1\choose 0}$ as is the probability to get ${0\choose 1}$. In other words, the z-component of the spin of the particle is completely uncertain. This is exactly what you say when you say that C becomes uncertain after measuring A.

But, notice one thing, even though we do not know absolutely anything about the z-component, we still find that the both the "uncertain eigenvectors" of $S_z$ are still the eigenvectors of $S^2$.

So, even though measurement of $S_x$ completely made $S_z$ uncertain, yet not only does the collapsed eigenstate of $S_x$ overlaps with one of the eigenstate of $S^2$ (which means $S_x$ and $S^2$ can be measured simultaneously) but also the uncertain eigenvectors of $S_z$ overlaps with $S^2$.

Thus, you probably now understand that the only assumption that was at fault was that if [B,C] = 0, then C becoming uncertain implies B becoming uncertain. It is not necessary. When B has a bigger set of eigenvectors than C, then even if eigenvectors of C becomes uncertain, it might be possible that the range of uncertainty is maintained within the set of eigenvectors of B, so that [B,C] = 0 is still obeyed.

P.S. If A and B has the same number of overlapping eigenvectors, and B and C also has the same number of overlapping eigenvectors, then it necessarily implies that A and C must also have the same number of overlapping eigenvectors, and hence [A,C] must be 0. It is only when B has a bigger set of eigenvectors than at least A or C, only then can [A,C] may not be 0.

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