- #1
y.moghadamnia
- 23
- 1
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 [tex]\Omega [/tex] and [tex]\Lambda[/tex] 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 [tex]\Omega[/tex] is an eigenvector of[tex]\Delta[/tex], the basis we have for [tex]\Omega[/tex] (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 ([tex]\varpi[/tex] ,[tex]\lambda[/tex],[tex]\gamma[/tex],...)" 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
"If [tex]\Omega [/tex] and [tex]\Lambda[/tex] 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 [tex]\Omega[/tex] is an eigenvector of[tex]\Delta[/tex], the basis we have for [tex]\Omega[/tex] (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 ([tex]\varpi[/tex] ,[tex]\lambda[/tex],[tex]\gamma[/tex],...)" 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