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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?

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  • $\begingroup$ Could you explain your last sentence a bit more? (it will help others help you). I'm assuming you mean that in things like the Dirac hydrogen atom, increasing $\ell$ for a given principal quantum number always corresponds to an increase in the energy eigenvalue. Don't forget that angular momentum here refers to a rotational symmetry rather than something spinning, so there doesn't seem to be any universal reason why your observed behaviour cannot be, although I agree, to a nonspecialist like me, this is a bit weird. $\endgroup$
    – Selene Routley
    Commented Jan 11, 2015 at 23:20
  • $\begingroup$ Make an analogy to deuterium, where the spin-1 state is bound and the spin-0 state isn't, due to the isospin interaction. $\endgroup$
    – rob ♦
    Commented Jan 12, 2015 at 0:33
  • $\begingroup$ @buzsh You should probably write down the Hamiltonian to make the question clearer. $\endgroup$
    – Mark Mitchison
    Commented Jan 12, 2015 at 0:41
  • $\begingroup$ @buzsh To my understanding you speak of the shell model for the nucleus. Then, in the Schrodinger equation, for non-zero $ℓ$ appears an additional term of positive sign, $ℏ^2ℓ(ℓ+1)/2mr^2$. $\endgroup$
    – Sofia
    Commented Jan 12, 2015 at 0:53

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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.

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  • $\begingroup$ You might want to consider editing this into the question, rather than posting it as an answer. Either one is okay, though. $\endgroup$
    – HDE 226868
    Commented Sep 30, 2015 at 1:43
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...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"...

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  • $\begingroup$ Opps... I didn't see that you found a bug in your equations........ Well, I'll leave it in any event - Neat way to keep people curious........... $\endgroup$
    – humancl
    Commented Jan 31, 2017 at 7:20

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