Angular momentum effect on quantum energy I'm doing some computational research into quarkonium states and I've written a code that determines energy levels by finding a solution to the Schrodinger equation for a given angular momentum. I.e. if you tell it L=2, the ground state it will return will be for n=3,L=2. The next solution it will find will be the energy of n=4,L=2.
The thing is, the energy of n=2,L=0 is not only not equal to n=2,L=1, the energy of the state will decrease with increasing angular momentum. E(n=3,L=2)

It seems counter-intuitive that higher energy is achieved by lower angular momentum. All my chemistry education tells me this is wrong, but the results of the code are undeniable.
Can anyone explain?
 A: If anyone's interested, I found out what was wrong. The limit of L less than N is only a happenstance of the 1/r potential. In the case of a non-geometric potential like that for quarkonium, this limit is removed. What I thought was the energy for N=2,L=1 was actually N=1,L=1 which shifted the results in the table I produced to return to a more expected range of results.
A: ...This is old, so I'm not sure it is still viewed....  I do not have an answer really, but I'd just refer you to R. Feynman's work on orbital jumps (up) under conditions of energy/potential reduction...  I point this out to young people all the time, e.g., take a quarter (25 cents I mean...) and drop it on a table top in such a way that it wobbles...  Well then, why does it speed up as it loses energy?
(Feynman wrote the QM equation of motion for this increase in orbital velocity, by mythical anecdote, after his wife dropped a plate on the kitchen table.). His equation explained a not previously understood phenomena whereby elections jumped to higher orbits under less then intuitively clear circumstances. I'm not sure this is related, but it sounds like a similar "anomaly"...
