Atomic Structure Theory

7.3 Atoms with One Valence Electron 209
in (7.44) by summing over the 38 × 38 possible basis functions. The resulting
contributions to the correlation energy are tabulated for l = 0 to 9 in Table 7.3.
As l →∞, the partial-wave contributions are known to fall off as
E (2)
l
→−
a
.
(l +1/2) 4
We use this known asymptotic behavior to estimate the remainder from l =10
to ∞. The resulting second-order correlation energy for helium is E (2) =
−0.03738 a.u.. Adding this value to the HF energy EHF = −2.86168 a.u., we
obtain E0 + E (1) + E (2) = −2.89906 a.u.. The theoretical ionization energy,
which is obtained by subtracting the energy of the one-electron helium-like ion,
is 0.89906 a.u. compared with the experimental value Eexp =0.90357 a.u., the
difference being less than 0.5%. Including the third-order energy -0.00368 a.u.
leads to a theoretical value of 0.90275 a.u., which is within 0.08% of the
measured value.
Table 7.3. Contributions to the second-order correlation energy for helium.
l El l El
0 -0.013498 5 -0.000168
1 -0.018980 6 -0.000088
2 -0.003194 7 -0.000050
3 -0.000933 8 -0.000031
4 -0.000362 9 -0.000020
10-∞ -0.000053
Total -0.037376
7.3 Atoms with One Valence Electron
Let us now turn to the problem of determining the second-order correlation
energy for an atom with one valence electron. For simplicity, we start with a
“frozen-core” Hartree-Fock formulation. The unperturbed state of the atom
is Ψ0 = a † v|0c〉, where|0c〉 is the HF core state, and the corresponding unper-
turbed energy is E0 =
a ɛa + ɛv.
Since V1 =
ij (∆V )ij : a †
i aj := 0 for a HF potential, it follows that
the first-order correction to the energy is E (1) = V0. Thus,E0 + E (1) =
(E0 + V0)core + ɛv. In other words, there is no first-order correction to the
valence removal energy. This result, of course, depends on the fact that the
calculation starts from a frozen-core HF potential. In any other potential,
there will be a first-order correction to the energy.