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 ?
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 .
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 ?