# If there are two conserved quantities that do not commute, why is there typically another?

I have come to the following conjecture:

Consider two observables $A,B$ that do not commute and represent conserved quantities, there is typically a third conserved quantity.

E.g. for some rotationally invariant system (let's say a rigid rotator on a sphere), $[L_z,H]=[L_x]=0$ but $[L_x,L_z]\ne0$ so there exists some orther conserved quantity, which is $L^2$ is this case.

The argument I have managed to come up with is the following: Since A and B are both conserved, the both commute with the Hamiltonian: $$[A,H]=[B,H]=0,$$ but since $$[A,B]\ne 0$$ simultaneous eigenstates of $A,B$ cannot be found. This means that simultaneous eigenstates of $H$ and $A$ or $B$ and $H$ can be found, but these sets will not be the same. I have come to the point where I want to say that there has to be some other operator that commutes with $A$ and/or $B$ because eigenstates of $H,A$, nor $H,B$ cannot possible constitute a complete set, and because QM postulates that there exists a complete set, there has to be another commuting observable. I think my argument is OK, but I'm not sure about the last part...

• Hint: The commutator $[A,B]$ will then commute with $H$. May 28, 2016 at 11:58
• @Qmechanic Ok, so $AB-BA$ is a conserved quantity, but 'why'? May 28, 2016 at 12:14
• @Qmechanic I find this quite unsatisfactory, because then you could also just say $A+B$ or some other linear combination. May 28, 2016 at 17:16
• @fawningflagellum $A+B$ is not an independent conserved quantity. $[A,B]$ sometimes is. May 28, 2016 at 21:00

We want two things from quantised physical conserved charges. The quantisation construes the charge as a linear operator on the Hilbert space of physical states. We require this operator to be a Hermitian operator that commutes with the Hamiltonian $H$. Exercise: prove that, if $A,\,B$ are Hermitian operators that commute with $H$, then $i\left[A,\,B\right]$ is another such operator. (Note: this result has an analogue in classical mechanics; the Poisson bracket of two conserved charges is a conserved charge.)
• What in general? H(AB-BA) = AHB-BHA=ABH-BAH=(AB-BA)H because you have assumed H commutes with A as well as B. H=I, A=P,B=Q is a counterexample for anything nontrivial (different from I) existing which commutes with above A and B. Operators which commute only with A are always a lot, starting with $A^2, A^3, \ldots$. May 28, 2016 at 17:21
• The choice $H=I$ is surely atypical, isn't it? So whatever the stuff you wrote is supposed to mean, it just can't be an answer to the original question which talks about the typical case. Jun 25, 2016 at 16:06