# How does the Runge-Lenz vector explain accidental degeneracies?

1. May 15, 2012

### lugita15

Two spherically symmetric potentials, the Coulomb potential and the isotropic harmonic oscillator potential, lead to solutions of the Schrodinger equation that have more degeneracies than usual. Historically, people seem to have thought they had no deeper significance, so they were termed "accidental degeneracies". We know know however, that the 1/r potential and the r^2 potential share a special conservation law, that of the Runge-Lenz vector, which explains their special properties including stable classical orbits. So how does the Hamiltonian operator commuting with the Runge-Lenz vector operator lead to accidental degeneracies?

And while we're at it, how does the Runge-Lenz vector work quantum mechanically? Since it is a conserved quantity, by Noether's theorem it has a corresponding symmetry, so what is the Lie group corresponding to it, and what is the representation theory of that Lie group?

Any help would be greatly appreciated.

Thank You in Advance.

2. May 15, 2012

### scijeebus

Isn't this just for a hydrogen atom? It's probably because there is some other quantity model such as that it is conserved but not for other states otherwise the equivalent to a specific potential would be altered and would not hold the equality that is required by the conservation, I can't remember what it's called, but I think it was related to this Noether's theorum.

3. May 15, 2012

### lugita15

Yes, there is such a conserved quantity, the Runge-Lenz vector, and I said that in my OP.

4. May 15, 2012

### scijeebus

Well doesn't it lead to degeneracy because that quantity has to be conserved then?

5. May 15, 2012

### lugita15

Yes, I just want to know the details.

6. May 15, 2012

### scijeebus

So are you asking then "why" it must be conserved? Because otherwise it seems like you answered your own question, and I don't think classical motions can really account for things in the quantum realm. With vector states in QM, there not exactly like, physical vectors, it's sort of hard to describe, there isn't really an actual circular motion, but rather what's more likely is the direction of a field oscillation, which even at that point you can't account for with purely classical notions, and quantum operators don't really do much better, they are just pure math. What all these different things mean is more or less open to interpretation right now. Like the Hamiltonian operator doesn't "account" for degeneracy, it's just these symbols on a piece of paper that say "when I have this value, these values have this" which follows steps of logic based off of axioms. What it means, or why the operators actually cause those degeneracies to occur is in the fuzzy mess that QM is.

Last edited: May 15, 2012
7. May 15, 2012

### strangerep

lugita15,

The dynamical symmetry group for the 1/r potential (hydrogen, Kepler, etc) is SO(4,2). Wybourne discusses its representation theory in the context of the hydrogen atom. And you could perhaps try Barut & Raczka as well. It's quite an interesting subject since it shows how to deal with a system by quantizing a dynamical group, and analyzing its spectral features that way instead of solving the Schrodinger equation.

Also: http://en.wikipedia.org/wiki/Hydrogen_atom
i.e., the section "Alternatives to the SchrÃ¶dinger theory" and the references there.

8. May 16, 2012

### lugita15

Thanks strangerep. The Wikipedia article on the Runge-Lenze vector mentioned that the Noether-related symmetry corresponding to the Runge-Lenz vector has to do with transformations of a four-dimensional sphere, but what does this mean in concrete physical terms?

Wikipedia also said that the symmetry is related to the fact that for inverse-square law orbits, the hodograph (velocity diagram) is circular, meaning that the particle has equal velocity changes in equal angles, a property I'm well acquainted with having read Feynman's Lost Lecture. Are you aware of what the connection is between the mysterious (to me) symmetry group and hodographs being circular?

Finally, what is the connection between the Lie algebra associated with the Runge-Lenz vector and the corresponding Lie group? It seems like it can't be the same straightforward relationship we usually see in quantum mechanics, like momentum and the translation group, where you have the simple correspondence of a one-parameter group and the infinitesimal generators. Wikipedia says that classically, there is no variable that is made cyclic by Runge-Lenz conservation, unlike other conservation laws, so the Noether-related symmetry requires something to do with Poisson brackets. How does this all work quantum mechanically?

9. May 16, 2012

### strangerep

There, Wiki is talking about the SO(4) subgroup of the full dynamical group SO(4,2).
The "four-dimensional sphere" business should be regarded as abstract. The SO(4) subgroup simply happens to coincide with the group of rotations on a 4D sphere -- by definition. Personally, I never got anything useful out of thinking about it in this way.

I haven't studied hodographs in this context.

But note that SO(4,2) is not a symmetry group, but a dynamical group. Symmetry generators commute with the Hamiltonian, but dynamical generators map amongst themselves under commutation with the Hamiltonian. Symmetry generators are a (relatively) uninteresting subgroup of a dynamical group, whereas dynamical generators map solutions into other solutions, broadly speaking.

I'd have to go refresh my memory on the details, but I believe that the (components of) the RL vector do indeed correspond to some of the generators in the dynamical group, just as the (components of) angular momentum correspond to generators of the ordinary rotation group SO(3). Wiki gives the Lie products between the RL components and the angular momentum components.

A comprehensive answer to this would fill a couple of textbooks. But here's a very simplified summary:

Poisson brackets are how we represent a Lie algebra in classical Hamiltonian mechanics.
If you're not familiar with this, then definitely go get a textbook like Goldstein, or Jose & Saletan. The whole technique of how we start with a classical dynamics represented by functions on phase space and Poisson brackets, and then pass to the quantum case by representing the same Lie algebra as Hermitian operators on Hilbert space (possibly symmetrizing any higher order products of the basic generators), is extremely important for understanding advanced quantization.

BTW, the thing about cyclic coordinates only matters when you're dealing with Action-Angle variables -- see J&S. These are a special case of the Hamilton-Jacobi method
in classical mechanics wherein we canonically transform from the basic phase space variables (positions and conjugate momenta) to a set of conserved quantities and their conjugate "position" variables -- which are really initial conditions. But this is too involved to explain properly here. I can only re-urge you to go get one or both of those textbooks.

Last edited: May 16, 2012
10. May 16, 2012

### DrDu

Wikipedia cites an interesting article by Fock who showed explicitly how the Schroedinger Equation in momentum space can be mapped onto a four dimensional sphere:
http://dx.doi.org/10.1007/BF01336904

11. May 17, 2012

### strangerep

Unfortunately, I'm not sufficiently proficient in German to read that without spending many hours with a German-English dictionary. :-(

But the maximal invariance group for the (free) Schroedinger Equation is the Niederer group, which (I think) contains something like SO(4) as a subgroup.

12. May 17, 2012

### lugita15

Here's an English translation:
http://books.google.com/books?id=KzQqQx88LEoC&pg=PA286&dq=fock+hydrogen+atom&hl=en&sa=X&ei=5gS1T8f3Cc_nggehs5j5Dw&ved=0CDYQ6AEwAA

He seems to be saying, among other things, that momentum space is non-Euclidean.

13. May 17, 2012

### strangerep

OK, thanks. This is just one of the very early papers (1935) grappling with the concept of a dynamical group. Subsequent papers showed that the DG for hydrogen is in fact SO(4,2), which has SO(4) as a subgroup. Barut and collaborators did lots of work on this sort of thing. IIRC, Gilmore's textbook contains a step-by-step account of what emerges as one considers successively larger dynamical groups in this context.

The "non-Euclidean" business refers (I think) to the construction of a Hilbert space suitable in this context. I seem to recall one needs a quantum metric of 1/r in the Hilbert space inner product to model this class of system.

14. Jun 8, 2012

### lugita15

strangerep, where would I be able to find more about the Hilbert space inner product involves a "quantum metric"? Does changing the geometry of states change the geometry of the actual position space?

15. Jun 9, 2012

### strangerep

Try looking up dynamical groups and spectrum-generating algebras. It's also related to harmonic analysis on Lie groups. Try Barut & Raczka from chapter 12 onwards.

(Arnold Neumaier made me aware of this stuff. Maybe he'll stop by in this thread and give you a better answer.)

I don't think so - that's kinda backwards. One tries to construct a Hilbert space carrying a representation of a particular dynamical group. IIUC, the group properties determine the details required of the Hilbert space. E.g., one can typically construct (generalized) coherent states given a dynamical group, or even just the product between two such states and from it construct a so-called "reproducing-kernel Hilbert space". (I'm not sure of the best reference for this, but maybe J. Faraut & A. Koranyi "Analysis on symmetric cones", Clarendon, Oxford 1994.)

Last edited: Jun 9, 2012
16. Jun 9, 2012

### lugita15

I may be missing something, but if you change the Hilbert space metric, and thus the inner product, aren't you also changing the outer products (since they are operators defined in terms of inner products), so that it is no longer true that the integral of the outer product |x><x| over all space is 1, unless you put in a Jacobian factor into the integrand? And isn't the introduction of such a Jacobian an indication that the positional space is non-Euclidean? Or can we do a coordinate transformation to a set of coordinates |x'> with respect to which this Jacobian is 1?

17. Jun 10, 2012

### A. Neumaier

The conserved quantities are just those whose Poission bracket with the Hamiltonian vanishes. A sufficiently real quantity is also the infinitesimal generator of a 1-parameter group of symmetries, i.e., canonical transformations. The vanishing Poisson bracket implies that for every solution of Hamilton's equations there is a continuum of others, namely those obtained by applying one of these symmetries, and the conserved energy of all these is the same. This is the essence of degeneracy.
The quantum version has the same interpretation as the classical version, except that quantities are now operators, sufficiently regular means self-adjoint, the Poisson bracket is replaced by -i/hbar times the commutator, canonical transformations are now unitary mappings, and Hamilton's equations are replaced by the Schroedinger equation.

Thus we get from any solution of the time-independent Schroedinger equation an infinite family of solutions with the same energy, by applying a symmetry. These span an invariant subspace of the symmetry group, and its multiplicity is the eigenvalue multiplicity, or a lower bound for it. Which is the quantum version of degeneracy.

Nothing is accidental here - accidental degeneracy means degeneracy in the absence of a symmetry.

A particle in a central potential has the rotation group as a symmetry group, hence the spectrum is already degenerate, resolved by looking at solutions with fixed spin. In the case of hydrogen, the rotation group together with the Lenz-Runge vector generate a symmetry group O(4), which induces even more degeneracy (namely degeneracy at fixed spin). There is a book ''The Kepler problem'' by Gordani that gives as many details as you want about both the classical and the quantum case, and even more.

18. Jun 10, 2012

### A. Neumaier

Changing the metric just means representing the states in a nonorthogonal basis, which is better adapted to the dynamical group. This is described in Chapter 12 of Barut & Raczka,
where a complete spectral analysis of the hydrogen atom is given (including eigenvectors and continuous spectrum).

But if the sole interest is in seeing the higher degeneracy of the hydrogen spectrum, one doesn't need the dynamical group SO(2,4) but only the symmetry group O(4), and there is no need to change the inner product.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook