Hi
I was reading a book on quantum mechanics (thats branch of physics). There is this
theorem which I didn't understand. Now the theorem is about physical observable
quantities like energy,momentum etc.. In quantum mechanics , such physical
observable quantity is represented by an operator. In matrix mechanics , which is a
version of quantum mechanics , such operators are written mathematically as
matrices. So the theorem is basically about matrices. I am going to translate the
theorem in mathematical language , removing words which relate to physics.
Theorem-- If two matrices commute, they possess a common set of eigenvectors.
This is true for both degenerate and non-degenerate eigenvectors.
Now when we have non-degenerate eigenvectors , two such commuting matrices
happen to have common eigenvectors anyway. But I have a question about
degenerate case. In such case , how do we obtain the common eigenvectors ?
we also need the base field to be algebraically closed. so we have a finite dimensional -vector space and the operators with since is algebraically closed, for some and let
then is a non-zero subspace of and for any we have so thus is an operator on let be any eigenvector of . clearly is also an eigenvector of because hence is an eigenvector of both and .
Observables in quantum mechanics are always represented by Hermitian operators. The eigenvectors of an Hermitian operator can be chosen to form an orthonormal basis, which implies that they are not degenerate. At least, while you may have degenerate eigenvalues, the geometric multiplicity of degenerate eigenvalues of an Hermitian operator is equal to its algebraic multiplicity. So I'm not entirely sure why you're talking about degenerate eigenvectors. In the degenerate eigenvalue case, you find the eigenvectors in the usual manner. You're not even talking about generalized eigenvectors here.
Does that answer your question?
Noncomm, thanks for the proof. makes thing clear.
ackbeet , oh yes , i am actually talking about the degenerate eigenvalues ,
not the degenerate eigenvectors..
after checking with the problem i was having , i am able to get the set of
common eigenvectors for the two commuting hermitian matrices having degenerate
eigenvalues. i needed to choose the indeterminate constants properly so that
the set of eigenvectors matched.
but the theorem i quoted doesn't say anything about the hermitian nature of the
matrices. so is it more general theorem , valid for non hermitian commuting matrices as well ?