Degeneracy removed when commuting observables are specified?

[Happiness]

The following is a proof that two commuting operators $A$ and $B$ possesses a complete set of common eigenfunctions. The issue I have with the proof is the claim that the eigenvalue $a_n$ together with the eigenvalue $b_n$ completely specify a particular simultaneous eigenfunction $\phi_n^{(k)}$ of $A$ and $B$, so that when both operators are considered together the degeneracy is completely removed.

(5.99) may not have $\alpha$ distinct roots. If so, $a_n$ together with $b_n$ do not completely specify a particular simultaneous eigenfunction, since an eigenvalue $b_n^{(k)}$ may have more than one solution $d_r^{(k)}$ corresponding to it.

Furthermore, the last paragraph mentions that the degeneracy is completely removed when the largest set of commuting observables are considered. This implies that if $A$ and $B$ do not form the largest set of commuting observables, the degeneracy may not be not completely removed. So I believe the claim is not proved.

 

[DrClaude]

I agree with you. To quote Cohen-Tannoudji, Diu, and Laloë, when they do a similar proof:

What we have just shown is that it is always possible to choose, in every eigenspace of A, a basis of eigenvectors common to A and B.

That's it. It does not necessarily lift the degeneracy, and often does not, hence the need for additional operators to build a C.S.C.O.

 

[blue_leaf77]

The following is a proof that two commuting operators $A$ and $B$ possesses a complete set of common eigenfunctions. The issue I have with the proof is the claim that the eigenvalue $a_n$ together with the eigenvalue $b_n$ completely specify a particular simultaneous eigenfunction $\phi_n^{(k)}$ of $A$ and $B$, so that when both operators are considered together the degeneracy is completely removed.

(5.99) may not have $\alpha$ distinct roots. If so, $a_n$ together with $b_n$ do not completely specify a particular simultaneous eigenfunction, since an eigenvalue $b_n^{(k)}$ may have more than one solution $d_r^{(k)}$ corresponding to it.

Furthermore, the last paragraph mentions that the degeneracy is completely removed when the largest set of commuting observables are considered. This implies that if $A$ and $B$ do not form the largest set of commuting observables, the degeneracy may not be not completely removed. So I believe the claim is not proved.

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I guess the authors forgot to specify that $A$ and $B$ must be the only operators that commute within the system under consideration for the degeneracy that present in either of the two operators to be removed. Instead, they introduced the concept of CSCO when the discussion is almost over.

 

[Happiness]

I guess the authors forgot to specify that $A$ and $B$ must be the only operators that commute within the system under consideration for the degeneracy that present in either of the two operators to be removed. Instead, they introduced the concept of CSCO when the discussion is almost over.

How can we show that when $A$ and $B$ are the only operators that commute, the $\alpha$ roots in (5.99) are distinct in order for the degeneracy to be removed?

 

[blue_leaf77]

Ah I see your point. I didn't read the excerpt till the end so I thought that the proof that no further operator exist without destroying the commutativity between them was already given but not stated literally.