Question

Can an azimuthally symmetric perturbation lift the 2l+1 degeneracy of angular momentum eigenstates?

Answer 1

Yes, I can see how that is true in the particular case of k=1k=1. However, if kk is an even number I think the energy splitting of states with same nn and ll, and same absolute value of mm is the same. I would like to find a simple way to prove or generalize this idea to all values of kk.

Answer 2

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Can an azimuthally symmetric perturbation lift the 2l+1 degeneracy of angular momentum eigenstates?

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♀️Assume the initial Hamiltonian of a spinless, non relativistic particle is

H0(r,θ,ϕ)=p²/2m+V0(r)

♀️Such that the eigenstates are angular momentum eigenstates |n,l,m>, and the energy spectrum is (2l+1)-fold degenerate.

♀️The, we have a perturbation potential of the form

V(r,θ,ϕ)=f(r)Y0k(θ,ϕ)

♀️Where f(r) is a function of r, and Y0k is an spherical harmonic with m=0,

therefore the potential is azimuthally symmetric.

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