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Part 2
Atomic Orbital Based Response Theory
The variational condition on the ground state, Eq. (2.75), and the orthonormality constraint
condition on the eigenvectors, Eq. (2.80), are included, and they are multiplied by the Lagrange
multipliers ω and X , respectively.
We then require the Lagrangian to be variational in all parameters
∂L f
= SDF − FDS = 0
(2.82)
∂X
f
∂L
f † [2] f
= b S b − 1=
0
(2.83)
∂ω
f
∂L
[2] f [2] f
= E b − ωS b = 0
(2.84)
f †
∂b
f
∂L
f † [2] f † [2]
= b E − ωb S = 0
(2.85)
f
∂b
f 0 f † [2] f f † [2] f
∂L
∂E
∂b E b ∂b S b ∂( FDS −SDF
)
n
= + −ω
− X n
= 0
∂X ∂X ∂X ∂X ∑
, (2.86)
∂X
m m m m n
m
where X m are the orbital rotation parameters. Due to the 2n + 1 rule, and since the gradient is a firstorder
property, we only need to solve the above equations through zero order. Eqs. (2.82)-(2.85) are
thus already taken care of, and it is seen that the multiplier ω is determined as the eigenvalue of the
linear response equations, i.e. it corresponds to the excitation energy. It is then only necessary to
determine the Lagrange multipliers X such that Eq. (2.86) is also fulfilled.
2.4.2 The Lagrange Multipliers
To evaluate the terms in Eq. (2.86), the asymmetric Baker-Campbell-Hausdorff (BCH) expansion 46
of the exponentially parameterized density is applied
DX ( ) = exp( − XSD ) exp( SX) = D+ [ DX , ] S
+ , (2.87)
where
[ AB , ] S
= ASB−BSA. (2.88)
Since the derivatives are evaluated at the expansion point, only terms of first order in X are nonzero.
The last term in Eq. (2.86) is found to be equal to 61
[2]
[ , ] [ , ] ([ , ] ) ([ , ] )
E X = F X D S− S X D F+ G X D DS−SDG X D . (2.89)
S S S S
We can thus find X by solving the set of linear equations
E
[2]
0 f † [2] f f † [2] f
∂E
∂b E b ∂b S b
X = + −ω
∂X ∂X ∂X
From the matrix expressions for b f† E [2] b f and b f† S [2] b f 61
. (2.90)
72