Linear response and Time dependent Hartree-Fock

with an infinitesimal parameter η.
We now want to calculate the induced density fluctuation δρ(r, t) due
to this external field. A priori, the infinitesimal perturbation causes an infinitesimal
change in the ground state wave function:
The change in density is then, to first order,
|Ψ 0 〉 → |Ψ(t)〉 = |Ψ 0 〉 + |δΨ(t)〉 (6)
δρ(t) = 〈Ψ(t)| ˆρ(r) |Ψ(t)〉 − 〈Ψ 0 | ˆρ(r) |Ψ 0 〉
≈ 〈Ψ 0 | ˆρ(r) |δΨ(t)〉 + 〈δΨ(t)| ˆρ(r) |Ψ 0 〉 . (7)
This δρ(t) is called the transition density because it is the transition matrix
element of the density operator between the ground state |Ψ 0 〉 and the perturbation
|δΨ(t)〉. The time dependence of the perturbation causes the same
time dependence of the density fluctuation,
δρ(r; t) ∼ δρ(r)e −iωt e ηt . (8)
(This is to be taken with a little caution, of course the transition density also
must be real.)
The time dependent part of the wave function can be calculated by peerturbation
theory.
Ansatz for the solution
i¯h ∂ ∂t |Ψ(t)〉 = (H + δH ext(t)) |Ψ(t)〉 . (9)
|Ψ(t)〉 = ∑ n
a n (t)e −iωnt |n〉 (10)
where {E n , |n〉} ≡ {¯hω n , |n〉} is the energy/eigenfunction pair of the unperturbed
Hamiltonian. The initial condition is a n (−∞) = δ n,0 . Inserting the
above in the Schrödinger equation gives
∑
(i¯hȧ n (t) + E n a n (t)) e −iωnt |n〉 = ∑ (E n + δH ext (t))a n (t)e −iωnt |n〉
n
n
∑
i¯hȧ n (t) |n〉 = ∑ δH ext (t)a n (t)e −iωnt |n〉 (11)
n
n
Now only a 0 (t) = 1 is 0 th order in δH ext (t), all other a n (t) are first order. We
therefore need to keep only the first term on the right hand side. Projecting
the equation on a state |n〉 then gives
i¯hȧ n (t) = 〈n| δH ext (t) |0〉 e i(ωn−ω 0)t ≡ 〈n| δH ext (t) |0〉 e iω n0t
(12)
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