# Do measurements of commuting obervables really commute?

Say I have two commuting operators $A$ and $B$, with joint eigenvectors $|n\rangle$.

Say I have a state $S = \sum_n a_n | n \rangle$. If I measure $A$ first, I pick out say an eigenstate $|k\rangle$ and then measure $B$ which just gives me $|k\rangle$ again. But if I measure $B$ first, it might pick out a different eigenstate, as $|m\rangle$ and then measuring $A$ gives $|m\rangle$. So it seems measurements don't necessarily commute unless acting on a joint eigenstate. So my question is whether this statement is true and statements about measurements of commuting observable being order independent is false.

• In quantum mechanics, you can never study individual measurements like this. You must always talk about a collection of measurements and then the probability distribution of outcomes. – Prahar Mar 7 '17 at 20:06
• The probability distribution of the results will be identical in both cases. – knzhou Mar 7 '17 at 20:32
• That's fine. But there are a lot of sloppy statements in textbooks about order of measurements. – JoeB Mar 7 '17 at 20:34

If the eigenvalues are both degenerate, then this is not really a problem as there is no such "different" eigenstate $|m\rangle$.

If one or both of the eigenspaces is degenerate, though, then the measurement does not return any single eigenvalue - instead, obtaining the result $a$ in a measurement of $A$ will take your state $|\psi\rangle$ and project it to the eigenspace, i.e. the subspace spanned by all the eigenstates of $A$ with eigenvalue $a$. In Dirac notation, this reads $$|\psi\rangle\mapsto \Pi_a|\psi\rangle \quad\text{where}\quad \Pi_a = \sum_i|\psi_{a,i}\rangle\langle\psi_{a,i}|,$$ where the $|\psi_{a,i}\rangle$ are all the different eigenstates of $A$ corresponding to the given eigenvalue.

This means that the question of whether measurements of compatible observables commute to the question of whether, if $[A,B]=0$, then $[\Pi_a,\Pi_b]=0$ for all eigenprojectors $\Pi_a$ of $A$ and $\Pi_b$ of $B$; of course, the answer here is yes, though you need to be careful to include all the relevant eigenstates in the summation. For more details, see Commutator with subspaces belonging to the same eigenvalue

This same argument can be applied to show that a measurement does not commute with itself. Let's take an operator $\hat{A}$ with eigenvectors $\{|n\rangle\}$. Then if our state is $\sum_n a_n|n\rangle$, I can measure once with $\hat{A}$ and get a state $|k\rangle$, and if I measure again with $\hat{A}$, I'll still measure $|k\rangle$.

But now let's do it in the opposite order. I first measure $\hat{A}$ instead of $\hat{A}$. This time, I get a state $|m\rangle$, and if I now measure $\hat{A}$ I stay in $|m\rangle$.

Thus, $\hat{A}$ and $\hat{A}$ don't really commute!

This should give you an indication that your sense of what "commute" means is wrong, since any sensible definition of "commuting" must have that everything commutes with itself.

• But isn't this the whole point of "contextuality" that measurements in a different context don't necessarily yield the same results. Measurements aren't projection operators nor represented by the operators themselves. That is the measurement of AB and CB don't project out the same state even if A and B, or C and B commute. – JoeB Mar 8 '17 at 4:17
• Or rather, a measurement of B followed by a measurement of A is different that a measurement of A followed by a measurement of B. If A=B, then the two statements are the same, so no difference. But if A not equal to B, we have different statements. – JoeB Mar 8 '17 at 4:27
• @JoeB No, the point is that measurement of $AB$ and measurement of $BA$ are the exact same thing if $[A,B]=0$. – Jahan Claes Mar 8 '17 at 15:43
• I guess what really is happening is that a measurement of AB or BA is not at all the same as a measurement of B followed by a measurement A or vice versa. E.g., consider a spin one state and the operator Sz or (Sz)^2. If I measure (Sz)^2 and get 1, the state is left in a combination of Sz= 1 or -1 eigenstates. But if I first measure Sz, I collapse to either a +1 or -1 eigenstate, and when I do it again, I'm still in either the eignenstate for +1 or -1, and not both. So the measurement of (Sz)^2 is not the same as the measurement of Sz followed by another measurement of of Sz. – JoeB Mar 8 '17 at 16:48
• @JoeB The measurements are not the same, in that you measure different eigenvalues. But the states after measurement are the same. (At least, the probability to be in any specific eigenvector is the same if you measure A and then B, or just measure AB right away.) – Jahan Claes Mar 8 '17 at 19:37