# Degeneracy important

1. Jun 4, 2014

### Jano L.

Sometimes it happens that hamiltonian has degenerate eigenvalue. This degeneracy can be removed by modifying the Hamiltonian. If the modification is additional term proportional to an adjustable parameter, the resulting difference between the eigenvalues can be made arbitrarily small.

Is there some phenomenon or experiment that requires the eigenvalues to be exactly degenerate, or is it possible to always use eigenvalues that are close enough but different with virtually the same result ? In other words, is degeneracy necessary for explanation of some experiments, or is it just an artefact maintained by the use of simple hamiltonians?

2. Jun 4, 2014

### Simon Bridge

The degeneracy is important, for eg, chemistry.
It is what determines the shell occupations and thus how covalent bonds form.

Remember, your hamiltonian is not supposed to be arbitrary. The aditions to the hamiltonian that remove the degeneracy must reflect something that happens in Nature.

What you seem to be asking is if the degeneracy we get is just because we have only got an approximation - do degenerate eigenvalues appear in Nature? The short answer is "yes".

3. Jun 5, 2014

### atyy

Last edited: Jun 5, 2014
4. Jun 5, 2014

### Jano L.

But shells refer to "mean central field approximation" theory. Degeneracy may be just a result of this simplification. Shells are important in practice but not necessary in theory. In principle one can work directly with the exact Schroedinger equation for $N$-particle system.
Can you give a concrete example? And why it is different from two eigenvalues being very close to each other?

5. Jun 5, 2014

### HeavyMetal

I think I see what you are saying...are you talking about modifying the Hamiltonian as a function of something else, such as changing Coulombic forces along an individual axis?

There's something called Jahn-Teller distortion that is exhibited in certain molecules, which in turn affects the degeneracy. Maybe something like this happens in those situations. As a vibrational stretch occurs along one of the axes in an octahedral molecule, the orbitals that lie along that direction go down in energy. A regular octahedral d-orbital splitting pattern becomes distorted as this stretching occurs. The orbitals that lie along the stretching axis go down in energy, and the orbitals that lie along the other two axes go up in energy.

I could see what you're talking about serving as a practical way to computationally model this phenomenon. I could be totally wrong though...will someone with more knowledge on this subject please chime in?

6. Jun 5, 2014

### Simon Bridge

OK - what you are doing here is a kind of philosophical hair splitting.

There is, indeed, no practical difference between saying that an energy level is degenerate and saying that it has discrete energy levels so very close together that none of our equipment is able to measure them individually.

In QM, these statements are functionally identical so we use the one with the fewest additional associated baggage.

i.e. An atom by itself would have degenerate states in energy ... this is just an effect of the symmetry: there is more than one configuration that can produce that particular energy eigenvalue (even if we include the various internal coupling perturbations). However, IRL, an atom is never completely by itself. There is always some external field that could, in principle, result in a small difference in energy for each configuration, by some amount, however miniscule.

I could try an example where two different configurations of atoms in a molecule may have the same binding energy. Would that lead to a two-fold degeneracy for energy states? Like the previous example we can argue that the molecule is never alone - there is always some external field somewhere that could ... etc etc etc. The same applies for anything I could name.

So - when we say that degenerate states exist in Nature, we have a silent rider: "to the extent that we can currently measure". This rider exists on every scientific statement anyone can make.

The rest is philosophy, and we don't discuss that here.
It is standard science to talk about degenerate states in Nature and makes sense to do so.

7. Jun 6, 2014

### Jano L.

Of course, it makes no difference as to the numerical value of the eigenvalue. But degeneracy may have other effects. For example, in perturbation theory one has to calculate differently with eigenfunctions belonging to degenerate eigenvalues. This is just an approximate calculation, but it makes me feel that there is a possibility that using Hamiltonian with close eigenvalues may not lead to the same conclusions as using the Hamiltonian where the close values are collapsed to one degenerate eigenvalue. It seems quite probable to me that it makes no difference and I would like that result, but I am not sure if there isn't some phenomenon that requires degenerate eigenvalues, otherwise the theory leads to wrong result or does not work.

8. Jun 6, 2014

### Simon Bridge

There will always be some situation where it makes a difference ... but where the perturbing fields are vanishingly small - the eigenvalue is degenerate to a very high degree of approximation. Whether it is actually composed of individual barely separated states or not is impossible to tell.

Circumstances where the calculation shows a difference but experiment does not are circumstances where the calculation was not confirmed. There has to be a reason for the states to separate out.

9. Jun 7, 2014

Thanks.