Any quantum analog for Betrand's theorem?

[kof9595995]

Betrand theorem in classical mechanics states that only harmonic & -1/r potential will give closed and stable orbit.
Is there any quantum analog? Just curious, I don't even know what "closed orbit" means in QM.

 

[DrDu]

I suppose in both cases the existence of closed orbits is due to the two problems having some dynamical symmetry, SU(3) in case of the harmonic oscillator and SO(4) in case of the Coulomb (or Kepler) problem. This leads to an extra degeneracy in the quantum case. E.g. in the Coulomb problem, all orbitals with the same main quantum number n are degenerate and not only those with the same n and l, as would be the case for other central potentials. This allows you to form some hybrid orbitals (e.g. sp)which in the classical limit approximate closed elliptical orbits.

 

[arkajad]

I have a vague memory that you may find some relevant references in these books:

The stability of matter: from atoms to stars : selecta of Elliott H. Lieb

http://www.cambridge.org/aus/catalogue/catalogue.asp?isbn=9780521191180"

http://www.amazon.com/dp/3540615652/?tag=pfamazon01-20

 

[kof9595995]

Emm...first thing first, what does it mean by "closed orbit" in QM?

 

[DrDu]

a closed curve, e.g. a circle, an elliptic orbit, an eight...

 

[kof9595995]

a closed curve, e.g. a circle, an elliptic orbit, an eight...

Don't those only make sense in classical mechanics?

 

[elduderino]

I googled your problem and this is a paper which exactly confronts your problem. (link at end of post)

Bertrand's theorem: the only central forces that result in closed orbits for all bound particles are the inverse square law and Hooke’s law

When looked at from a QM perspective, there's some link obviously as the Harmonic oscillator and the coulomb potential are the only exactly solvable models with infinite bound states. Well there are the step and barrier potentials, but these have a finite number of bound states.

How is an infinite number of bound states somehow an analog to a classically stable orbit I do not understand. Anyone?

Also, the authors write:

These are the only central potentials for which the corresponding Schrödinger equations can be factorized to yield both the energy and angular momentum raising and lowering operators.

I understand that once this has been done, it is possible to get an infinite number of bound states by using these to raise or lower levels. But.. how does one factorize the coulomb potential Hamiltonian to get raising and lowering operators?

http://arxiv.org/abs/quant-ph/9905011

 

[kof9595995]

Quite interesting paper, though I really don't want to go through all the math to verify the proof...

How is an infinite number of bound states somehow an analog to a classically stable orbit I do not understand. Anyone?

I have the same question .

I understand that once this has been done, it is possible to get an infinite number of bound states by using these to raise or lower levels. But.. how does one factorize the coulomb potential Hamiltonian to get raising and lowering operators?

As cited as [10] in your paper, http://www.sciencedirect.com.ezlibproxy1.ntu.edu.sg/science?_ob=ArticleURL&_udi=B6TVM-3SPKY3F-2B&_user=892051&_coverDate=06%2F30%2F1997&_alid=1583441234&_rdoc=1&_fmt=high&_orig=search&_origin=search&_zone=rslt_list_item&_cdi=5538&_sort=r&_st=13&_docanchor=&view=c&_ct=1&_acct=C000047479&_version=1&_urlVersion=0&_userid=892051&md5=3dd24ec768c5b39a06eb2a07c2649a33&searchtype=a

 

[DrDu]

Don't those only make sense in classical mechanics?

Of course it makes sense only in classical mechanics. I think the important point which makes possible an extension to qm is that the classical orbits have a definite direction (i.e. the large axis of the ellipsis, or the position of the perihelion). In general relativity, the orbits are no longer closed and the perihelion wanders, which constituted one of the first tests of general relativity. In qm the operator corresponding to the direction of the large axis still exists and is conserved only in the Kepler and harmonic oscillator problem as a consequence of the symmetries I mentioned below.

 

[kof9595995]

Emm... then what is this “the operator corresponding to the direction of the large axis” ？

 

[DrDu]

In the Coulomb problem, it is the Runge Lenz vector.

 

[kof9595995]

Hmm, it makes sense, thanks