126
4 From Orbital Models to Accurate Predictions
circumstances, the list of determinants that do not affect the energy difference of the
two states contained by the model space can be extended with the 2h-1p and 1h-2p
classes. Taking as an example Φ R =| aapb| and Φ S =| bbpa| (2h-1p determinant
generated from Φ I and Φ J , respectively), the same reasoning will be followed as
above. In the first place it is easily seen that the second-order contribution to the
off-diagonal elements is zero
Φ R :
hhab| ˆ
V |aapbaapb| ˆ
V |hhba
E J − E R
= 0
(4.36a)
Φ S :
hhab| ˆ
V |bbpabbpa| ˆ
V |hhba
E J − E S
= 0
(4.36b)
The first integral in the numerator of the Φ R contribution is non-zero because the
determinants in the bra and the ket only differ by two columns, but |aapb| differs at
three places from |hhba|, and hence, leads to a zero contribution. The same holds for
Φ S . At first sight, the contribution to the diagonal elements of the model space may
seem non-uniform:
Φ R :
||hhab| ˆ
V |aapb| 2
E I − E R
= 0
||hhba| ˆ
V |aapb| 2
E J − E R
= 0
(4.37a)
Φ S :
||hhab| ˆ
V |bbpa| 2
E I − E S
= 0
||hhba| ˆ
V |bbpa| 2
E J − E S
= 0
(4.37b)
At difference with the 2h-2p determinants discussed above, the two non-zero integrals
are not necessarily equal in this case. However, the effect of Φ ′
R =| aabp| and
Φ ′
S =|bbap| exactly compensates this disequilibrium:
Φ
′
R :
||hhab| ˆ
V |aabp| 2
E I − E ′
R
= 0
||hhba| ˆ
V |aabp| 2
E J − E ′
R
= 0
(4.38a)
Φ
′
S :
||hhab| ˆ
V |bbap| 2
E I − E ′
S
= 0
||hhba| ˆ
V |bbap| 2
E J − E ′
S
= 0
(4.38b)
The denominators in the non-zero contributions of Φ R and Φ ′
R are equal since
E R = E ′
R , and E I = E J in a degenerate model space. Furthermore, the integral
hhab| ˆ
V |aapb in Eq. 4.37a is exactly the same as the integral hhba| ˆ
V |aabp of
Eq. 4.38a
hhab| ˆ
V |aapb==hhab| ˆ
H|aapb=−−hhab| ˆ
H|paab=
−−hh|
1 − ˆ
P 12
r 12
|pa=
h(1)h(2)p(1)a(2)
r 12
dτ 1 dτ 2
(4.39a)
4 From Orbital Models to Accurate Predictions
circumstances, the list of determinants that do not affect the energy difference of the
two states contained by the model space can be extended with the 2h-1p and 1h-2p
classes. Taking as an example Φ R =| aapb| and Φ S =| bbpa| (2h-1p determinant
generated from Φ I and Φ J , respectively), the same reasoning will be followed as
above. In the first place it is easily seen that the second-order contribution to the
off-diagonal elements is zero
Φ R :
hhab| ˆ
V |aapbaapb| ˆ
V |hhba
E J − E R
= 0
(4.36a)
Φ S :
hhab| ˆ
V |bbpabbpa| ˆ
V |hhba
E J − E S
= 0
(4.36b)
The first integral in the numerator of the Φ R contribution is non-zero because the
determinants in the bra and the ket only differ by two columns, but |aapb| differs at
three places from |hhba|, and hence, leads to a zero contribution. The same holds for
Φ S . At first sight, the contribution to the diagonal elements of the model space may
seem non-uniform:
Φ R :
||hhab| ˆ
V |aapb| 2
E I − E R
= 0
||hhba| ˆ
V |aapb| 2
E J − E R
= 0
(4.37a)
Φ S :
||hhab| ˆ
V |bbpa| 2
E I − E S
= 0
||hhba| ˆ
V |bbpa| 2
E J − E S
= 0
(4.37b)
At difference with the 2h-2p determinants discussed above, the two non-zero integrals
are not necessarily equal in this case. However, the effect of Φ ′
R =| aabp| and
Φ ′
S =|bbap| exactly compensates this disequilibrium:
Φ
′
R :
||hhab| ˆ
V |aabp| 2
E I − E ′
R
= 0
||hhba| ˆ
V |aabp| 2
E J − E ′
R
= 0
(4.38a)
Φ
′
S :
||hhab| ˆ
V |bbap| 2
E I − E ′
S
= 0
||hhba| ˆ
V |bbap| 2
E J − E ′
S
= 0
(4.38b)
The denominators in the non-zero contributions of Φ R and Φ ′
R are equal since
E R = E ′
R , and E I = E J in a degenerate model space. Furthermore, the integral
hhab| ˆ
V |aapb in Eq. 4.37a is exactly the same as the integral hhba| ˆ
V |aabp of
Eq. 4.38a
hhab| ˆ
V |aapb==hhab| ˆ
H|aapb=−−hhab| ˆ
H|paab=
−−hh|
1 − ˆ
P 12
r 12
|pa=
h(1)h(2)p(1)a(2)
r 12
dτ 1 dτ 2
(4.39a)