# Eigenvaules and quantum numbers

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?

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## Answers and Replies

tom.stoer
Science Advisor
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>

dextercioby
Science Advisor
Homework Helper
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).

kith
Science Advisor
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
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.

tom.stoer
Science Advisor
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.
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.

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?

dextercioby
Science Advisor
Homework Helper
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.

kith
Science Advisor
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. ;-)

Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
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.

Correct. This is also responsible for the fact that the Schrodinger equation is separable in two coordinate systems for those two potentials: spherical coordinates, and rectangulars for the harmonic oscillator and parabolic for the Coulomb potential.