174
11 Intermediate-State Representation (ISR)
It remains to deal with the three second-order contributions to the first term on
the right-hand side:
(1) ≡≡
(1)
0 |c
†
k
ˆ
Dc l |
(1)
0
(11.75)
(2, 3) ≡≡
(2)
0 |c
†
k
ˆ
Dc l | 0 + h.c.
(11.76)
While the evaluation of (2) and (3) is rather simple, that of (1) is more demanding.
In both cases, it is advisable to make use of the commutators,
[c
†
k , ˆ
D] = −
r
d rk c
†
r , [ ˆ
D, c l ] = −
s
d ls c s
(11.77)
We skip the somewhat lengthy derivation and jump directly to the final expressions [6]:
˜
D kl = D 0 (2)δ kl − d lk −
a
(d ak ρ
(2)
la + d la ρ
(2)
ak ) + ˜
D
(2,1)
kl
+ ˜
D
(2,2)
kl
+ ˜
D
(2,3)
kl
where
˜
D
(2,1)
kl
= −
b,c,d
j
v
∗
bdl j v cdk j d bc
˜
D
(2,2)
kl
=
1
2
c,d
i, j
v
∗
dcl j v dcki d i j
˜
D
(2,3)
kl
= −
1
4
c,d
i, j
v
∗
dci j v dcki d l j + h.c.
(11.78)
Here, D 0 (2) is given by Eq. (11.66), and ρ
(2)
rs denotes the second-order contributions
to the one-particle density matrix (Eq. 11.63).
The order relations and separability structure applying to ˜
D will be addressed in
Sects. 12.1 and 12.2.
Exercises
11.1 Use ground-state PT to expand the overlap matrix for the 1h CE states (Eq. 11.7)
through second order.
11.2 Derive the ISR(2) expressions for the transition amplitudes (11.49).
11.3 Use the direct ADC(2) secular equation to derive the ionization energy for
a single-hole main state through second order, and compare the result to
Eq. (8.68).