# Eigenvaules and quantum numbers

1. Nov 7, 2011

### khemist

Eigenvalues and quantum numbers

My professor was discussing the Schrodinger equation in 3-d today and mentioned that there are degeneracies within quantum numbers (for the hydrogen atom, the x, y, and z components are symmetric?). Does that imply that the eigenvalues of some operator are the quantum numbers of a system? If so, which operator is it?

Last edited: Nov 7, 2011
2. Nov 8, 2011

### tom.stoer

Studying the Hamiltonian H you see that it is rotationally invariant, i.e. it commutes with all components of orbital angular momentum Li acting as generators of rotations:

[H,Li] = 0

H contains an angular momentum term, namely ~L²/r² which is the familiar additional term in the effective potential (already present in the Kepler problem).

Studying L² and Lz you get eigenfunctions, the so called spherical harmonics Ylm; they are labels by two quantum numbers l and m:

L² Ylm = l(l+1) Ylm

Lz Ylm = m Ylm

The radial part of H is responsible for the quantum number n, so in total the states are labelled as |nlm> with n=1,2,..., l=0,1,...,n-1, m=-l, ...,l-1, l

It's interesting (and I have to admit that I do not understand if there is some deeper reason!) that
a) the radial equation results in a radial wave funtion Rnl(r) where l is due to the additional term in the effective potential ~L²/r² ~ l(l+1)/r², but
b) the eigenvalues Enl do in the very end not depend on l, i.e. are labelled as En ~ 1/n²

So in total you have

H |nlm> = En |nlm>
L² |nlm> = l(l+1) |nlm>
Lz |nlm> = m |nlm>

3. Nov 8, 2011

### dextercioby

What are the so-called <quantum numbers> ? Eigenvalues of operators describing observables. So what are these observables which could have a countable number of spectral values ? Energy for once, then spin, then orbital angular momentum, then total angular momentum, then electric charge, then barionic charge, parity, etc.

So for a conserved observable, (its operator, assumed time-independent in the Schrödinger picture) there's always a degeneracy of the discrete energy levels, since each closed subspace associated with an energy value is reduced by the symmetry generator, for example L2 for the spherically-symmetric potentials (free particle in 3D, isotropic oscillator in 3D, H-atom/hydrogenoid ions).

4. Nov 8, 2011

### kith

There's an additional conserved quantity in the Kepler problem -the Laplace–Runge–Lenz vector- which is due to a subtle SO(4) symmetry of the potential. This leads to the l-degeneracy of the hydrogen atom energies.

5. Nov 8, 2011

### tom.stoer

This was my conjecture, but I wasn't sure about it.

That means that for V(r)~1/r with its SO(4) symmetry in dim=3 and for V(r)~r² with its SU(N) symmetry in dim=N we have l-degeneracy, but for no other V(r)~rz (for different z-values). For the harmonic oscillator in dim=N it's trivial to prove SU(N) symmetry and degeneracy, for the hydrogen atom it's more involved.

6. Nov 9, 2011

### khemist

This was much more in depth than I intended. Thanks for the info. When does this sort of material get introduced in a physics major?

7. Nov 9, 2011

### dextercioby

Depends on the curriculum the teacher has prepared for his QM class. You may also get a separate group theory course where applications of group theory to quantum mechanics are presented.

8. Nov 9, 2011

### kith

I didn't hear about this in lectures but stumbled upon it by studying symmetry operators for an oral exam. I think that most lecturers don't want to spend half the QM course on the hydrogen atom. ;-)

9. Nov 9, 2011