25.2 The Helium Ground State
Calculating the first order correction to the ground state is simple inprinciple.
∆Egs=〈u 0 |V|u 0 〉=∫
d^3 r 1 d^3 r 2 |φ 100 (~r 1 )|^2 |φ 100 (~r 2 )|^2
e^2
|~r 1 −~r 2 |=
5
8
Ze^2
a 0=
5
4
Z(
1
2
α^2 mc^2 ) =5
4
(2)(13.6) = 34 eVThecalculation(see section 25.7.1)of the energy shift in first order involves an integral over the
coordinates of both electrons.
So the ground state energy to first order is
Egs=− 108 .8 + 34 =− 74 .8 eVcompared to -78.975 eV from experiment. A 10% error is not bad considering the size of the
perturbation. First order perturbation theory neglects the change in the electron’s wavefunction
due to screening of the nuclear charge by the other electron. Higher order perturbation theory
would correct this, however, it is hard work doing that infinite sum. We will find a better way to
improve the calculation a bit.
25.3 The First Excited State(s)
Now we will look at the energies of the excited states. The Pauli principle will cause big energy
differences between the different spin states, even though we neglect all spin contribution inH 1 This
effect is called the exchange interaction. In the equation below, thesstands for singlet corresponding
to the plus sign.
E( 1 s,tst) =e^2
2〈
φ 100 φ 2 ℓm±φ 2 ℓmφ 100∣
∣
∣
∣
1
|~r 1 −~r 2 |∣
∣
∣
∣φ^100 φ^2 ℓm±φ^2 ℓmφ^100〉
=
e^2
2{
2
〈
φ 100 φ 2 ℓm∣
∣
∣
∣
1
|~r 1 −~r 2 |∣
∣
∣
∣φ^100 φ^2 ℓm〉
± 2
〈
φ 100 φ 2 ℓm∣
∣
∣
∣
1
|~r 1 −~r 2 |∣
∣
∣
∣φ^2 ℓmφ^100〉}
≡ J 2 ℓ±K 2 ℓ
It’s easy to show thatK 2 ℓ>0. Therefore, the spin triplet energy is lower. We can write the energy
in terms of the Pauli matrices:
S~ 1 ·S~ 2 =^1
2
(S^2 −S^21 −S 22 ) =
1
2
[
s(s+ 1)−3
2
]
̄h^2~σ 1 ·~σ 2 = 4S~ 1 ·S~ 2 / ̄h^2 = 2[
s(s+ 1)−3
2
]
=
{
1 triplet
−3 singlet}
1
2
(1 +~σ 1 ·~σ 2 ) ={
1 triplet
−1 singlet}
E 1 (s,tst) = Jnℓ−1
2
(1 +~σ 1 ·~σ 2 )Knℓ