18.5 Angular Momentum in the Helium Atom 777
ThêL^2 and̂S^2 operators are not easy to use, and we will not operate explicitly with
them.^2 We will work with thêLzand̂Szoperators, and will deduce values ofLand
Sfrom theMLandMSvalues. We can find the values ofMLandMSsince thez
components of the one-electron operators add algebraically as in Eqs. (18.5-11) and
(18.5-13). For two electrons
MLm 1 +m 2 (18.5-14a)
and
MSms 1 +ms 2 (18.5-14b)EXAMPLE18.2
Find the values of the quantum numbersMLandMSfor each of the wave functions in
Eq. (18.4-2) for the (1s)(2s) configuration of the helium atom.
Solution̂LzΨ 1 (̂lz 1 +̂lz 2 )Ψ 1 √^1
2[̂lz 1 ψ 1 s(1)ψ 2 s(2)−ψ 2 s(1)(̂lz 2 ψ 1 s(2))]α(1)α(2) 0 + 0 0 (18.5-15)so thatML0. The other wave functions also contain onlysorbitals, so thatML0 for
all of them. We operate witĥSzto find the value ofMS.̂SzΨ 1 (̂sz 1 +̂sz 2 )Ψ 1
1
√
2(
ψ 1 s(1)ψ 1 s(2)−ψ 1 s(1)ψ 1 s(2))(
̂sz 1 α(1)α(2)+α(1)̂sz 2 α(2))
1
√
2(
ψ 1 s(1)ψ 1 s(2)−ψ 1 s(1)ψ 1 s(2))
(
h ̄
2
α(1)α(2)+α(1) ̄
h
2
α(2))(
h ̄
2+ ̄
h
2)
Ψ 1 ̄hΨ 1 (18.5-16)so thatΨ 1 corresponds toMS1. Similarly,̂SzΨ 2 (
− ̄
h
2
− ̄
h
2)
Ψ 2 −h ̄Ψ 2 (18.5-17)so thatMS−1 forΨ 2.̂SzΨ 3 ^1
2
[ψ 1 s(1)ψ 2 s(2)−ψ 2 s(1)ψ 1 s(2)]×[̂sz 1 α(1)β(2)+̂sz 1 β(1)α(2)+α(1)̂sz 2 β(2)+β(1)̂sz 2 α(2)]1
2
[ψ 1 s(1)ψ 2 s(2)−ψ 2 s(1)ψ 1 s(2)]×[(
h ̄
2)
α(1)β(2)+(
−h ̄
2)
β(1)α(2)+α(1)(
−h ̄
2)
β(2)+β(1)(
h ̄
2)
α(2)] 0 (18.5-18)(^2) I. N. Levine,op. cit., p. 318ff (note 1).