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 px and py 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. .
Examples worked out for all first and second period atoms:
H: Configuration (1s)1.
Only one determinant: |1sα(1)|
He: Configuration (1s)2.
Li: Configuration (2s)1.
Only one determinant: |2sα(1)|
Be: Configuration (2s)2.
B: Configuration (2p)1.
Ground multiplet 2P.
D1 = |p1α(1)| = |2P;ML=1;MS=1/2〉
ρα = |p1|2 = (-px/√2 - ipy/√2)* (-px/√2 - ipy/√2)
= 1/2(px - ipy)(px + ipy) = 1/2(px2 + py2)
The occupation numbers for D1 are
pxα = pyα = 1/2; pzα = 0; pxβ = pyβ = pzβ = 0
Another determinant belonging to 2P is
D2 = |..p0α(1)|
with occupations pzα = 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 MS and ML
(1+ ≡ p1α, ...)
|D1 = |1+ 1-|||0||2|
|D2 = |1+ 0+|||1||1|
|D3 = |1+ 0-|||0||1|
|D4 = |1+ -1+|||1||0|
|D5 = |1+ -1-|||0||0|
|D6 = |1- 0+|||0||1|
|D7 = |1- 0-|||-1||1|
|D8 = |1- -1+|||0||0|
|D9 = |1- -1-|||-1||0|
|D10 = |0+ 0-|||0||0|
|D11 = |0+ -1+|||1||-1|
|D12 = |0+ -1-|||0||-1|
|D13 = |0- -1+|||0||-1|
|D14 = |0- -1-|||-1||-1|
|D15 = |-1+ -1-|||0||-2|
The presence of a determinant with ML = 2, MS = 0 and no ML = 2, MS > 0
determinant indicates the presence of a 1D multiplet, and E(1D) = E(D1).
There is also a 3P, the determinant with MS = 1, ML = 1. We should have E(3P) = E(D2) = E(D4). The two determinants D3 and D6 in the MS = 0, ML = 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 MS = 0, ML = 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 ρα(D6) = ρβ(D3) and ρβ(D6) = ρα(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):
|3P → 1D:||1.066 to 1.389||1.26||1.55|
|3P → 1S:||2.741 to 3.376||2.684||3.78|
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;ML=2;MS=1/2〉 = |p1α p1β p0α| = D2
ρα = 1/2 px2 + 1/2 py2 + pz2
ρβ = 1/2 px2 + 1/2 py2
E(D2) = - 0.745 according to the table in the Appendix (VWN + Becke)
The energy of D1, with ρα = px2 + py2, ρβ = pz2, is E(|1A 2B 3A|) = -1.9702.
The energy of D3, with ρα = 1/2 px2 + 1/2 py2 + pz2, ρβ = pz2 corresponding to |1A 2A 2B| or |2A 2B 3A|, is E(D3) = - 0.158.
Finally, D4 has ρα = px2 + py2, ρβ = 1/2 px2 + 1/2 py2, corresponding to |1A 1B 3A| and |1A 3A 3B|, and E(D4) = - 0.109.
The ML=1, MS=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 ML=0, MS=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 ρα = px2 + py2 + pz2, ρβ = pz2 corresponds to |1A 2A 2B 3A| , the ML=1, MS=1 determinant of 3P: E(3P) = -1.836
D2 with ρα = px2 + py2 + pz2, ρβ = 1/2 px2 + 1/2 py2, corresponds to |1A 1B 2A 3A|, the determinant of 3P: E(3P) = -1.568
D3, with ρα = 1/2 px2 + 1/2 py2 + pz2, ρβ = px2 + py2, corresponds to |1A 1B 2A 3B|, and ML=1, MS=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 = |...(p0α)1 (p0β)0| → E(D1) = - 0.715.
D2 = |(p-1)2 (p0)2 (p1α)1 (p1β)0| → E(D2) = - 0.467.
Ne: Configuration p6.