Why do q's form a complete commuting set of observables?

[bikashkanungo]

I have a doubt from Dirac’s book which states that , Pr =-iħ∂/∂qr
This equation was concluded after stating that the linear operator iħ∂/∂qr satisfy the same commutation relations with the q’s and with each other as p’s do.
Next it is stated that(Dirac, p-92) “ This possibility enables us to see that the q’s must form a complete commuting set of observables since it means that any function of the q’s and p’s could be taken to be a function of q’s and iħ∂/∂q ‘ s and then could not commute with all the q’s unless it is a function of the q’s only.
I could not understand the above statement. Can anyone help me in the explanation of it ??

Also there is stated that ∂fr/∂qs =∂fs /∂qr showing that the functions fr are all of the form :
fr = ∂F/∂qr, where F is independent of r . How is this equation concluded ??

 

[dextercioby]

That q's commute is assumed by hypothesis. That they form a CSCO, well, Dirac's heuristic argument is quite self explanatory. Q's commute with an arbitrary function of P and Q, iff the function is dependent of Q's only.

Please, see another good heuristic argument in Ballentine's book, i.e. his chapter on dynamics derived from the Galileo group and especially the appendix on the irreducibility of Q's and P's based on Schur's lemma.