Results for first and second period atoms

In the appendix tables are given for the energy lowering in going from the converged spherically symmetric spin-restricted atom (the 'master' fragment) to specific one-determinant wavefunctions with the orbital occupations as specified. Note that the p_x and p_y populations are always equal; only their sum is given. In many cases the determinant corresponds to a specific state, which is then given in the last column. For each atom, the first calculation is for the spherically symmetric spin-unrestricted atom. These tables are now obsolete, all information needed to obtain the atomic reference energies, i.e. the groundstate multiplet energy with respect to the AOC, can be found in ref. [3].

Examples worked out for all first and second period atoms:

H: Configuration (1s)¹.
Only one determinant: |1sα(1)|

He: Configuration (1s)².
Closed shell.

Li: Configuration (2s)¹.
Only one determinant: |2sα(1)|

Be: Configuration (2s)².
Closed shell.

B: Configuration (2p)¹.
Ground multiplet ²P.
D₁ = |p₁α(1)| = |²P;M_L=1;M_S=1/2⟩
ρ^α = |p₁|² = (-p_x/√2 - ip_y/√2)* (-p_x/√2 - ip_y/√2)
      = 1/2(p_x - ip_y)(p_x + ip_y) = 1/2(p_x² + p_y²)
The occupation numbers for D₁ are
p_x^α = p_y^α = 1/2; p_z^α = 0; p_x^β = p_y^β = p_z^β = 0
Another determinant belonging to ²P is
D₂ = |..p₀α(1)|
with occupations p_z^α = 1 and all other p-occupations zero. This determinant is ~ 0.04 eV lower in energy than D₁ for VWN, but ~ 0.15 eV for VWN + Becke's GGA exchange correction.

C: Configuration p².
Multiplet states are ³P, ¹S and ¹D.
We use this atom as an example of the general procedure. First write down all determinants belonging to p² and group them according to M_S and M_L
(1⁺ ≡ p₁α, ...)

The presence of a determinant with M_L = 2, M_S = 0 and no M_L = 2, M_S > 0 determinant indicates the presence of a ¹D multiplet, and E(¹D) = E(D₁).
There is also a ³P, the determinant with M_S = 1, M_L = 1. We should have E(³P) = E(D₂) = E(D₄). The two determinants D₃ and D₆ in the M_S = 0, M_L = 1 box will mix, and the solutions of the 2 by 2 secular problem will be E(¹D) and E(³P). Since the sum of the eigenvalues is equal to the sum of the initial diagonal elements, we have E(¹D) + E(³P) = E(D₃) + E(D₆).
We should also have E(D₃) + E(D₆) = E(D₁) + E(D₂). Such a relation provides a test on the consistency of the results.
Finally we have the M_S = 0, M_L = 0 block. The sum of the energies of D₅, D₈ and D₁₀ should be the sum of the energies of ¹S, ³P and ¹D. Since E(³P) and E(¹D) are already known, E(¹S) can be calculated.
In the appendix we first locate for C the spherical unrestricted atom. Next we have E(D₄), yielding E(³P) = -1.345 eV (VWN + Becke). Next E(D₂) = E(³P) = -1.189 (always VWN + Becke). The difference is substantial: ~ 0.15
Next we have E(D₃) = - 0.812. Since E(D₆) = E(D₃), because ρ^α(D₆) = ρ^β(D₃) and ρ^β(D₆) = ρ^α(D₃), we should have 2E(D₃) = -1.624 = E(¹D) + E(³P). Therefore E(¹D) = -1.624 - (-1.345) = - 0.279 or E(¹D) = - 1.624 - (-1.189) = - 0.435.
These numbers can be checked against E(D₁) which also should be E(¹D):
E(D₁) =+ 0.044. The discrepancy is large!
Finally, ¹S can be obtained:
E(D₁₀) = + 0.319
(D₈) = E(D₁) = + 0.044
E(D₅) = E(D₁) = + 0.044
So 0.407 = E(¹S) + E(³P) +E(¹D).
Different results for E(¹S) are obtained depending on the E(³P) and E(¹D) we choose:
e.g. E(¹S) = 0.407 -(-1.345) - (- 0.279) = 2.031
or E(¹S) = 0.407 - (-1.189) - (0.044) = 1.552.
Comparing to experiment we might calculate the excitation energies w.r.t. E(³P):

N: Configuration p³.
Ground multiplet ⁴S, corresponds to the spherical unrestricted atom, energy -2.943 eV. Other multiplets: ²P, ²D. According to the printed output for configuration p³ we have
|²D;M_L=2;M_S=1/2⟩ = |p₁^α p₁^β p₀^α| = D₂
ρ^α = 1/2 p_x² + 1/2 p_y² + p_z²
ρ^β = 1/2 p_x² + 1/2 p_y²
E(D₂) = - 0.745 according to the table in the Appendix (VWN + Becke)
The energy of D₁, with ρ^α = p_x² + p_y², ρ^β = p_z², is E(|1A 2B 3A|) = -1.9702.
The energy of D₃, with ρ^α = 1/2 p_x² + 1/2 p_y² + p_z², ρ^β = p_z² corresponding to |1A 2A 2B| or |2A 2B 3A|, is E(D₃) = - 0.158.
Finally, D₄ has ρ^α = p_x² + p_y², ρ^β = 1/2 p_x² + 1/2 p_y², corresponding to |1A 1B 3A| and |1A 3A 3B|, and E(D₄) = - 0.109.
The M_L=1, M_S=1/2 determinants are |1A 1B 3A| and |1A 2A 2B|. Therefore E(²D) + E(²P) = E(D₄) + E(D₃), so E(²P) = - 0.109 - 0.158 - (- 0.745) = + 0.478.
We can use D₁ in the M_L=0, M_S=1/2 block, from which we find E(⁴S) + E(²D) + E(²P) = 2E(D₂) =+ E(D₁).
Hence E(²P) = -1.490 - 1.9702 - (- 0.745) - (- 2.943) = + 0.2278.

O: Configuration p⁴.
Multiplet states ³P, ¹S, ¹D.
D₁, with ρ^α = p_x² + p_y² + p_z², ρ^β = p_z² corresponds to |1A 2A 2B 3A| , the M_L=1, M_S=1 determinant of ³P: E(³P) = -1.836
D₂ with ρ^α = p_x² + p_y² + p_z², ρ^β = 1/2 p_x² + 1/2 p_y², corresponds to |1A 1B 2A 3A|, the determinant of ³P: E(³P) = -1.568
D₃, with ρ^α = 1/2 p_x² + 1/2 p_y² + p_z², ρ^β = p_x² + p_y², corresponds to |1A 1B 2A 3B|, and M_L=1, M_S=0 belonging to ³P as well as ¹D.

F: Configuration p⁵.
Ground multiplet ²P.
As in B, we have two determinants with different energies belonging to this state:
D₁ = |...(p₀^α)¹ (p₀^β)⁰| → E(D₁) = - 0.715.
D₂ = |(p_-1)² (p₀)² (p₁^α)¹ (p₁^β)⁰| → E(D₂) = - 0.467.

Ne: Configuration p⁶.
Closed shell.