# What does it mean for 2 observables to be compatible?

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?

• 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. Commented Nov 9, 2018 at 13:36
• Related to conservation law if an observable commutes with the Hamilton. Commented Nov 9, 2018 at 13:43
• @K_inverse Could you elaborate a little more on that? If for example $[S_x, H] = 0$, this implies some spin conservation law? Commented Nov 9, 2018 at 13:48
• @CuriousHegemon, FYI en.wikipedia.org/wiki/Heisenberg_picture Commented Nov 9, 2018 at 15:10
• 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! Commented Nov 9, 2018 at 15:35

• 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! Commented Nov 9, 2018 at 15:37