Simultaneous diagonalization of two hermitian operators

[y.moghadamnia]

I decided to go over the mathematical introductions of QM again.The text I use is Shankar quantum, and I came across this theorem:
"If \Omega and \Lambda are two commuting hermitain operators, there exists (at least) a basis of common eigenvectors that diagonalizes them both."
in the proof part, they first consider at least one of the operators is nondegenerate, and so they use the fact that they commute, and there's the result: "since every eigenvector of \Omega is an eigenvector of\Delta, the basis we have for \Omega (the eigenbasis) will diagonalize both operators."
and then they go over the degenerate part, means if both are degenerate, and there comes by a proof I don't completely understand, but then the conclusion is as the above,
my question is that does above mean if we have two operators and they commute (hermitians) and we suppose the degeneracy is a solved case, one operators eigen vecs and eigenvalues are the othres as well?
and there's another thing, in the end they generalize by this: " in general, one can alwaysm for infinite n, a se of operators that commute with each other and that nail down a unique common eigenbasis, the element s of which maymay be labeled unambiguously as ket (\varpi ,\lambda,\gamma,...)" that the others are to be the eigenvalues of the other operators.
what does that mean? any examples in QM?
sorry if the above is not so clear, for those who have shankar QM, look at pages 43 to 46.
and for those who dont, here's a link: www.filestube.com/q/quantum+mechanics+shankar

 

[arkajad]

The simplest way I know is this one: each Hermitian operator A has a spectral decomposition:

A=\sum_i\lambda_iE_A(i)

where \lambda_i are eigenvalues and E_A(i) are orthogonal projection on corresponding eigenspaces. Then comes a little lemma:

[A,B]=0 if and only if all E_A(i) commute with all E_B(j).

Now, suppose A and B commute. Then E_{AB}(i,j)=E_A(i)E_B(j)[/tex] are also projections and \sum_{i,j}E_{AB}(i,j)=I<br /> <br /> It is enough to choose any orthonormal basis in each subspace defined by (i,j) and you have orthonormal basis in the whole space consisting of common eigenvectors of both operators. The same method works for any number of mutually commuting operators.<br /> <br /> It may sound a little bit abstract, but I think that is the best way of looking at the problem.

 

[kerrigan]

Hello.I am reading Sharkar too, but for the first time. Very glad to talk about sth on what I am trying to understand.
I think the answer to your question is yes. The two commuting hermitian operators have the same eigenbasis. Check p29, the active and passive tranformations. What we do is just to multiply two unitary matrices to the left and rignt side of the operator. That can be seen as a unitary transformation on the basis of the space. Have a review of the proof, he just transforms the basis. In the new basis, the two are diagonalized.
So the eigenbasis is just about the basis of the space, and has nothing to do with the operators.
I am not an English speaker. sorry for grammar mistakes.