Recent content by yeahright

yeah right. If you start with a three dimensional square hermitian matrix, you'll get a 3D eigenspace. period. If you have degenerate eigenvalues - like you do, then you get one unambiguous eigenvector (for the eigenvalue 0) and a 2D eigenspace (corresponding to the degenerate eigenvalue 1)...

yeah right. only if your " box " happens to be an atom - in which case - what are you putting in it again? if not, any basic QM text will tell you that for the same potential you could choose position or momentum eigenstates (or eigenstates of any other operator) which would have...