then diagonalize that matrix for each level of degeneracyn. Actually, it is much more convenient
to change basis and use the methods of addition of angular momentum. We shall use instead the
following maximal set of commuting observables (in addition to the Hamiltonian),J^2 ,L^2 ,Jz, where
J=L+S. The operatorS^2 need not be included since its value is fixed at 3 ̄h^2 /4.
In the basis of these commuting observables, the states are labeled again by|n,j,ℓ,m〉, andH 1
is diagonal in this basis, since we have the following well-known formula,
2 L·S=J^2 −L^2 −S^2 (10.44)so that we may evaluate the first order perturbations in a straightforward manner, and we obtain,
E^1 |n,j,ℓ,m〉=1
2
(
j(j+ 1)−ℓ(ℓ+ 1)−3
4
)
̄h^2 〈n,j,ℓ,m|φ(r)|n,j,ℓ,m〉 (10.45)The last matrix element must be evaluated using the radial wave-functions of the Coulomb problem.
10.8 General development of degenerate perturbation theory
We now develop the general perturbation theory of an energy levelEd^0 ofH 0 which admits an
N-fold degeneracy. We shall denote the unperturbed states by|Ed^0 ;i〉wherei= 1,···,N, andE^0 d
is the common unperturbed energy,
H 0 |E^0 d;i〉=E^0 d|Ed^0 ;i〉 i= 1,···,N (10.46)Under the perturbation byλH 1 , theNdegenerate levels will generally split and these energies will
be denoted byEd,i=Ed,i(λ). The equation of interest is
(Ed,i−H 0 −λH 1 )|Ed,i〉= 0 (10.47)Both the energy and the state admit an expansion in powers ofλ,
Ed,i = Ed^0 +λE^1 d,i+λ^2 Ed,i^2 +O(λ^3 )
|Ed,i> = |Ed^0 ;i〉+λ|Ed,i^1 〉+λ^2 |Ed,i^2 〉+O(λ^3 ) (10.48)The key idea is to rearrange the energy eigenvalue before starting to expand in powers ofλ.
LetH 0 denote theN-dimensional subspace of the full Hilbert spaceHgenerated by the degen-
erate states|E^0 D,i〉, and denote byP 0 the projection operator ontoH 0. The orthogonal complement
ofH 0 is denotedH 1 and consists of all the eigenstates ofH 0 with eigenvalue different fromED^0.
SinceH 0 is self-adjoint, the spacesH 0 andH 1 are guaranteed to be orthogonal and their sum to
equalH. LetP 1 denote the projection operator ontoH 1. We then have
H 0 ⊕H 1 = H P 0 P 1 =P 1 P 0 = 0
P 0 +P 1 = I P 02 =P 0 , P 12 =P 1 (10.49)The operatorsP 0 andP 1 commute withH 0 , by construction.