# 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?

• 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. Jan 11, 2015 at 23:20
• Make an analogy to deuterium, where the spin-1 state is bound and the spin-0 state isn't, due to the isospin interaction.
– rob
Jan 12, 2015 at 0:33
• @buzsh You should probably write down the Hamiltonian to make the question clearer. Jan 12, 2015 at 0:41
• @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, $ℏ^2ℓ(ℓ+1)/2mr^2$. Jan 12, 2015 at 0:53