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AO Based Response Equations in Second Quantization
The matrix S [2] may in a similar way be written as
⎛ Σ Ω ∆ ⎞
[2] T T
S =
⎜
Ω 0 -Ω
⎟
⎜
- - -
⎟
⎝ ∆ Ω Σ ⎠
, (2.50)
where
†
mn ⎡Qm Qn
Σ = 0 ⎣ , ⎤⎦
0 ,
∆ mn = 0 ⎡⎣Qm , Qn
⎤⎦
0 ,
Ω = 0 [ Q , D ] 0 .
mn m n
(2.51)
Note that the block containing two diagonal operators vanishes as
† † † †
[ Dm
Dn
] = ⎡⎣aµ aµ aνaν ⎤⎦ = Sµν aµ aν − Sνµ aν aµ
= . (2.52)
0 , 0 0 , 0 0 0 0 0 0
To illustrate how the pairing is obtained in the AO formulation, we assume that the vector
⎛ Z ⎞
X =
⎜
U
⎟
⎜ ⎟
⎝Y
⎠
(2.53)
is an eigenvector for Eq. (2.45) with eigenvalue ω
⎛ A F B ⎞⎛ Z⎞ ⎛ Σ Ω ∆ ⎞⎛ Z ⎞
⎜ T T ⎟⎜ ⎟ T T
F G F U = ω
⎜
Ω 0 -Ω ⎟⎜
U
⎟
. (2.54)
⎟⎜ ⎟⎜ ⎜ ⎟⎜ ⎟ ⎜
- - -
⎟⎜ ⎟
⎝ B F A ⎠⎝Y⎠ ⎝ ∆ Ω Σ ⎠⎝Y
⎠
Multiplying the blocks of Eq. (2.54) gives three sets of equations
AZ + FU + BY = ω ( ΣZ + ΩU + ∆Y )
( )
T T T T
F Z+ GU+ F Y = ω Ω Z −Ω Y
BZ + FU + AY = ω ( −∆Z −ΩU −ΣY
).
(2.55)
We will now prove that the paired vector
X
P
⎛Y
⎞
=
⎜
U
⎟
⎜ ⎟
⎝ Z ⎠
(2.56)
is an eigenvector for Eq. (2.45) with eigenvalue –ω
⎛ A F B ⎞⎛Y⎞ ⎛ Σ Ω ∆ ⎞⎛Y
⎞
⎜ T T ⎟⎜ ⎟ T T
F G F U =−ω
⎜
Ω 0 -Ω ⎟⎜
U
⎟
. (2.57)
⎟⎜ ⎟⎜ ⎜ ⎟⎜ ⎟ ⎜
- - -
⎟⎜ ⎟
⎝ B F A ⎠⎝ Z⎠ ⎝ ∆ Ω Σ ⎠⎝ Z ⎠
Multiplying the blocks of Eq. (2.57) leads to the three sets of equations
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