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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...

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

2 Answers 2

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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.)

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The argument is wrong, because the set of eigenstates of H and A may be complete. For H and B too.

As a trivial (even if artificial) example use H=I and some arbitrary pair of operators which define a complete base even if taken alone, say, A=Q, B=P.

Of course, AB-BA will commute with H anyway, because H commutes with A and B. But this may not give anything new, for example for P and Q it gives only I.

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  • $\begingroup$ But will it in general? Do you know an example of this, I cannot think of any? $\endgroup$ May 28, 2016 at 13:47
  • $\begingroup$ 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 $A^2, A^3, \ldots$. $\endgroup$
    – Schmelzer
    May 28, 2016 at 17:21
  • $\begingroup$ 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. $\endgroup$ Jun 25, 2016 at 16:06

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