How do I handle degenerate eigenvalues and eigenvectors in quantum mechanics?

[M. next]

In Quantum, I ran across the eigenvalue problem.
They gave me a matrix, and i was asked to find eigenvalues and then eigenvectors.
But the eigenvalues, were degenerate and thus i couldn't find the exact normalized eigenvector.
What to do in this case? Shoukd i choose arbitrary values?

My other question is about another problem, they gave me a matrix and i got no degenrate eigenvakues, anyway when i wanted to find eigenvector, i tried normalizing it, so i got let's say:
y^2=4 so y=±2

What do i choose? Does it make a difference?

 

[cattlecattle]

In Quantum, I ran across the eigenvalue problem.
They gave me a matrix, and i was asked to find eigenvalues and then eigenvectors.
But the eigenvalues, were degenerate and thus i couldn't find the exact normalized eigenvector.
What to do in this case? Shoukd i choose arbitrary values?

My other question is about another problem, they gave me a matrix and i got no degenrate eigenvakues, anyway when i wanted to find eigenvector, i tried normalizing it, so i got let's say:
y^2=4 so y=±2

What do i choose? Does it make a difference?

There is no unique eigenvector corresponding to degenerate eigenvalues. Instead, all the vectors in a subspace of dimension equal to the degeneracy can be its eigenvectors. Non-degenerate eigenvalue is really a special case where that dimension is 1. In case of degeneracy, you are free to choose any vectors in the eigenspace in forming a basis, it doesn't matter, but by convention, you choose orthonormal vectors with simple coordinates.

This also answers your 2nd question, if it has no degeneracy, you are choosing an orthonormal basis in 1D, but you still have freedom to choose its direction. In general, you can multiply a ket with unitary complex number without changing its physical significance.

 

[M. next]

Thanks you a lot