Linear response and Time dependent Hartree-Fock

where the ɛ p , ɛ h are the Hartree-Fock single particle energies.
I calculate only one of the half-off-diagonal terms, p ≠ p ′ , h = h ′ . In a
uniform system, it is clear that these matrix elements must vanish because the
left state has a different momentum as the right state. In a inhomogeneous
system, calculate
〈φ HF | a † h a pH 1 a † p ′a h |φ HF〉 = ∑ αβ
= ∑ αβ
= ∑ αβ
〈α| h |β〉 〈φ HF | a † h a pa † αa β a † p ′a h |φ HF〉
〈α| h |β〉 〈φ HF | δ αp a † h a β a† p ′a h + a† h a† αa β a p a † p ′a h |φ HF〉
〈α| h |β〉 δ αp δ βp ′ = 〈p| h |p ′ 〉 (43)
because, to get from the second to the third line, we could use that p ≠ p ′ .
The potential term is a little tedious:
〈φ HF | a † h a pV a † p ′a h |φ HF〉 = 1 2
= 1 2
= ∑ βδ
∑
αβγδ
∑
βγδ
〈αβ| v |γδ〉 〈φ HF | a † h a pa † αa † β a δ a γa † p ′a h |φ HF〉
〈pβ| v |γδ〉 〈φ HF | a † h a† β a δ a γa † p ′a h |φ HF〉
〈pβ| v |p ′ δ〉 a
〈φ HF | a † h a† β a δ a h |φ HF〉
= ∑ h ′ 〈ph ′ | v |p ′ h ′ 〉 a
. (44)
The last line (i.e. that we can restrict the sum over the δ states to occupied
states follows because of δ were a particle state it would destroy the ground
state |φ HF 〉.
Thus, the result is
〈φ HF | a † h a pHa † p ′a h |φ HF〉 = 〈p| h |p ′ 〉 + ∑ h ′ 〈ph ′ | v |p ′ h ′ 〉 a
(45)
In the above, we recover the matrix element of the Hartree-Fock operator.
Thus, if all states are generated by the Hartree-Fock equation, then these
“half-off-diagonal” term vanish. Since this is a reasonable assumption (how
else would one generate the particle wave functions ?) we can go on and
assume either p = p ′ , h = h ′ or p ≠ p ′ , h ≠ h ′ .
In the latter case, we can have only contributions from the interaction:
〈φ HF | a † h a pV a † p ′a h ′ |φ HF〉 = 1 2
∑
αβγδ
〈αβ| v |γδ〉 〈φ HF | a † h a pa † αa † β a δ a γa † p ′a h ′ |φ HF〉
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