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If I have 2 observable operators $A$ and $B$, if $A$ and $B$ commute: $[A, B] = 0$, then they must necessarily form a complete set of commuting observables (CSCO). Essentially, if 2 observables are compatible this seems to be quite significant.

I just wanted to get an intuition for what this means. Does it have something to do with precision of measurement?

For example, I know that if the Hamiltonian is time independent, it commutes with itself: $$[H, H] = 0$$ However, if the Hamiltonian is time-dependent, then this is not true at all times: $$[H(t_1), H(t_2)] \neq 0$$ Is this because the Hamiltonian is changing and thus doesn't necessarily act on itself the same way at all times anymore?

Also I know that the position and momentum operators commute if they are in different directions, but don't commute if they're in the same direction: $$[x_i, p_j] = \delta_{ij}$$ Does this imply that if 2 observables don't commute, this corresponds to the idea that we can't measure them both simultaneously to high precision?

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  • $\begingroup$ A set of CSCO per definition is such that all possible common eigenstates are uniquely determined by the eigenvalues of these observables. In terms of classical states, that the state is uniquely determined by the values of these observables. $\endgroup$
    – doetoe
    Commented Nov 9, 2018 at 13:36
  • $\begingroup$ Related to conservation law if an observable commutes with the Hamilton. $\endgroup$
    – K_inverse
    Commented Nov 9, 2018 at 13:43
  • $\begingroup$ @K_inverse Could you elaborate a little more on that? If for example $[S_x, H] = 0$, this implies some spin conservation law? $\endgroup$ Commented Nov 9, 2018 at 13:48
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    $\begingroup$ @CuriousHegemon, FYI en.wikipedia.org/wiki/Heisenberg_picture $\endgroup$
    – K_inverse
    Commented Nov 9, 2018 at 15:10
  • $\begingroup$ That's right, the Heisenberg Equation of motion! $\frac{dA}{dt} = \frac{1}{i \hbar} [A, H]$. So if $A$ commutes with $H$, then $\frac{dA}{dt} = 0$. Thanks! $\endgroup$ Commented Nov 9, 2018 at 15:35

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In quantum mechanics, a measurement (almost) always modifies the system being measured. Intuitively, if two measurements commute then the way that one measurement changes the system doesn't affect the results of the other measurement. So if you repeat those two measurements as many times as you want, in any order, then you'll always get the same results for each of the two measurements.

On the other hand, if the two measurements don't commute, then every time you perform one, it will (at least partially) "reset" the other observable to an indeterminate value (I'm glossing over some subtleties).

This explains why if an observable commutes with the Hamiltonian, then its value is conserved over time: the Hamiltonian "pushes the system into the future", so in some sense it is continuously acting on the system and changing it in a way that modifies the value of most observables - except for the special ones that commute with the Hamiltonian, and so are not affected by time evolution.

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    $\begingroup$ Great answer, thank you. I think it makes a lot more sense now. So if we have two observables $A$ and $B$, then if $AB|\psi\rangle \neq BA|\psi\rangle$, where $|\psi\rangle$ is a state ket, this implies that it matters the order in which we measure A and B, implying the measurement of either A or B affects the measurement of the other. Awesome! $\endgroup$ Commented Nov 9, 2018 at 15:37

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