24.10.2014 Views

Get my PhD Thesis

Get my PhD Thesis

Get my PhD Thesis

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

Part 2<br />

Atomic Orbital Based Response Theory<br />

the linear response function in Eq. (2.42) can be obtained as<br />

[1]<br />

B<br />

AB ; ω<br />

=−A N ( ω)<br />

. (2.44)<br />

2.2.5 Pairing<br />

The excitation energies are identified as the poles of the linear response function of Eq. (2.42) and<br />

are therefore solutions to the generalized eigenvalue problem<br />

[2] [2]<br />

E X = ωS X. (2.45)<br />

In the MO formulation of response theory, it has been shown that the excitation energies are<br />

paired 63 , so that if ω i is an eigenvalue for Eq. (2.45) then so is -ω i . It is important to understand how<br />

pairing appears in the AO basis, in particular since this structural feature is exploited when the<br />

equations are solved iteratively as is necessary for large problems. This is further discussed in<br />

Section 2.3. Since the proof of the pairing given in the MO formulation cannot be directly<br />

transferred to the AO formulation due to the presence of the diagonal operators D m , this section<br />

gives the proof in the AO formulation.<br />

The structure of E [2] and S [2] in the AO formulation is analyzed for the purpose of examining the<br />

pairing structure. Dividing Λ into the tree classes of Eq. (2.28), the matrix E [2] may be written as<br />

†<br />

⎛ 0 ⎡⎣Q, ⎡⎣H0, Q ⎤⎤ ⎦⎦ 0 0 [ Q, [ H0, D]<br />

] 0 0 [ Q, [ H0, Q]<br />

] 0 ⎞<br />

[2]<br />

⎜<br />

⎟<br />

†<br />

E = ⎜ 0 ⎣⎡D, ⎣⎡H0, Q ⎦⎦ ⎤⎤ 0 0 [ D, [ H0, D]<br />

] 0 0 [ D, [ H0, Q]<br />

] 0 ⎟. (2.46)<br />

⎜ † † † †<br />

⎟<br />

⎝ 0 ⎣⎡Q , ⎣⎡H0, Q ⎦⎦ ⎤⎤ 0 0 ⎣⎡Q ,[ H0, D] ⎦⎤ 0 0 ⎣⎡Q ,[ H0, Q]<br />

⎦⎤<br />

0 ⎠<br />

If we assume for simplicity that all orbitals and integrals for the unperturbed system are real, the<br />

†<br />

elements of for example the block 0 ⎡⎣Q ,[ H0<br />

, Q ] ⎤⎦<br />

0 are trivially rewritten as<br />

† †<br />

∗<br />

0 ⎡⎣Qm, [ H0, Qn ] ⎤⎦ 0 = 0 ⎡⎣Qm, [ H0, Qn<br />

] ⎤⎦<br />

0<br />

(2.47)<br />

†<br />

= 0 ⎡⎣Qm, ⎡⎣H0<br />

, Qn<br />

⎤⎤ ⎦⎦ 0 .<br />

The nine blocks in Eq. (2.46) can then all be written in terms of the following four matrices<br />

and we obtain<br />

†<br />

mn m 0 n<br />

A = 0 ⎡⎣Q , ⎡⎣H , Q ⎤⎤ ⎦⎦ 0 ,<br />

Bmn = 0 ⎡⎣Qm , ⎡⎣H0<br />

, Qn<br />

⎤⎤ ⎦⎦ 0 ,<br />

(2.48)<br />

Fmn = 0 ⎡⎣Qm , ⎡⎣H0<br />

, Dn<br />

⎤⎤ ⎦⎦ 0 ,<br />

Gmn = 0 ⎡⎣Dm , ⎡⎣H0<br />

, Dn<br />

⎤⎤ ⎦⎦ 0 ,<br />

⎛ A F B ⎞<br />

[2] ⎜ T T<br />

E = F G F<br />

⎟<br />

. (2.49)<br />

⎜<br />

⎟<br />

⎝ B F A ⎠<br />

66

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!