Linear response and Time dependent Hartree-Fock

Linear response and Time dependent
Hartree-Fock
November 6, 2008
1 General Linear Response Theory
This derivation follows essentially te one found in the book by Pines and
Nozieéres [1] it is a bit more general because it does not assume translational
invariance. In fact, Pines and Nozieres are to some extent misleading.
Assume that our many-body system is subjected to weak, local, timedependent
field
δH ext (t) = ∑ ∫
δh ext (r i , t) = d 3 rˆρ † (r)δh ext (r, t) (1)
i
Here, ˆρ(r) = ∑ i δ(r − r i ) is the density operator. We will look at monochromatic
external fields, this is best done in frequency space:
∫
h ext (r, ω) = dte −iωt h ext (r, t)
∫
⇒ h ∗ ext(r, ω) = dte −iωt h ext (r, t) = h ext (r, −ω) (2)
since the external field is real. Thus
and, therefore we have
δH ext (t) =
∫
h ∗ ext(r, ω) = h ext (r, −ω) (3)
d 3 rˆρ(r) [ h ext (r, ω)e −iωt + h ∗ ext(r, ω)e iωt]
= δH ext (ω)e −iωt + δH ∗ ext(ω)e iωt . (4)
To preserve causailty, we assume that the external field has been turned on
very slowly in the distant past, i.e. instead of the above we should have
δH ext (t) = [ δH ext (ω)e −iωt + δH ∗ ext(ω)e iωt] e ηt (5)
1

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)
2

with ω n0 = ω n − ω 0 .
We can now insert the perturbing Hamiltonian
∫
〈n| δH ext (t) |0〉 = d 3 r ′ 〈n| ˆρ(r ′ ) |0〉 [ h ext (r ′ )e −iωt + h ∗ ext(r ′ )e iωt] e ηt (13)
we can write
∫
i¯hȧ n (t) = d 3 r ′ 〈n| ˆρ(r ′ ) |0〉 [ h ext (r ′ )e i(ωn0−ω)t+ηt + h ∗ ext(r ′ )e ] i(ω n0+ω)t+ηt
e ηt
or
a n (t) =
−
(14)
∫
d 3 〈n| ˆρ(r) |0〉
rh ext (r, ω)
¯hω − ¯hω n0 + iη ei(ω n0−ω)t+ηt
∫
d 3 rh ∗ 〈n| ˆρ(r) |0〉
ext(r, ω)
¯hω n0 + ¯hω − iη ei(ω n0+ω)t+ηt . (15)
(We can redefine η ↔ ¯hη because η is infinitesimal.)
Inserting the expansion of |δΨ(t)〉 into the transition density gives generally
δρ(r, t) = ∑ n>0
a n (t) 〈0| ˆρ(r) |n〉 e −iω n0t + c.c.
=
−
=
∫
∫
∫
d 3 r ′ h ext (r ′ ) ∑ n>0
d 3 r ′ h ∗ ext(r ′ ) ∑ n>0
〈0| ˆρ(r) |n〉 〈n| ˆρ(r ′ ) |0〉
e −iωt+ηt
¯hω − ¯hω n0 + iη
〈0| ˆρ(r) |n〉 〈n| ˆρ(r ′ ) |0〉
e iωt+ηt + c.c.
¯hω + ¯hω n0 − iη
d 3 r ′ h ext (r ′ ) ∑ n>0
〈0| ˆρ(r) |n〉 〈n| ˆρ(r ′ ) |0〉 × (16)
[
]
1
×
¯hω − ¯hω n0 + iη − 1
e −iωt+ηt
¯hω + ¯hω n0 + iη
+c.c. (17)
We can then define the coordinate space representation of the density-density
reponse function as
χ(r, r ′ ; ω) = ∑ [ 〈0| ˆρ(r) |n〉 〈n| ˆρ(r ′ ) |0〉
− 〈0| ˆρ(r) |n〉 〈n| ]
ˆρ(r′ ) |0〉
. (18)
n>0
¯hω − ¯hω n0 + iη ¯hω + ¯hω n0 + iη
We have already seen the definition of the dynamic structure function.
We can immediately read off the relationship, noting that
lim
η→0
1
¯hω − ¯hω n0 + iη = πδ(¯hω − ¯hω n0) (19)
3

and therefore, for ω >=
Imχ(r, r ′ ; ω) = π ∑ n>0
〈0| ˆρ(r) |n〉 〈n| ˆρ(r ′ ) |0〉 δ(¯hω − ¯hω n0 ) (20)
(The definition of common factors and signs differs in different textbooks !)
In scattering experiments, one observes the momentum transfer, so the
∫
χ(q, ω) = d 3 rd 3 r ′ e iq·(r−r′) χ(r, r ′ ; ω)
= ∑ [ |〈0| ˆρq |n〉| 2
n>0
¯hω − ¯hω n0 + iη − |〈0| ˆρ q |n〉| 2 ]
. (21)
¯hω + ¯hω n0 + iη
2 Excited state wave function and the action
integral
It comes now to the point where we must some definite model for the excited
states of a physical system. This can not be done in general; we must make
some approximations. With all the reservations one can have about the
Hartree-Fock approximation, we start with that one.
We approximate the wave function of excited states by
|ψ(t)〉 = 1
∑
N (t) e−iE HF t/¯h e 1 2 ph c ph(t)a † pa h |φHF 〉 (22)
where |φ HF 〉 is a Hartree-Fock ground state. E HF the Hartree-Fock energy
expectation value, and the c ph (t) “particle–hole amplitudes”.
N 2 (t) = 〈φ HF | e 1 2
∑
ph c∗ ph (t)a† h a p
e 1 2
∑
ph c ph(t)a † pa h |φHF 〉 (23)
is the normalization integral. Note that states that are labeled with
h, h i , h ′ , . . . are understood to be occupied (“hole”) states (i.e. states that are
present in the Slater determinant φ HF and states labeled with p, p i , p ′ , . . . are
unoccupied “particle” states. As above, assume that the system is described
by a Hamiltonian that is written in coordinate space as
H = H 1 + V (24)
H 1 = ∑ h(r i ) = ∑ ]
[− ¯h2
i
i
2m ∇2 i + U ext (r i )
(25)
V = ∑ v(r i , r j ) (26)
i

or, in second quantized form
H 1 = ∑ αβ
〈α| h ext |β〉 a † αa β (27)
V = 1 ∑
〈αβ| v |γδ〉 a †
2
αa † β a δ a γ . (28)
αβγδ
We furthermore subject the system to a weak, scalar external field
δH ext (t) = ∑ i
δh ext (r i ; t) (29)
or, in second quantized form,
δH 1 (t) = ∑ αβ
〈α| δh(t) |β〉 a † αa β . (30)
Similar to the determination of the single-particle wave functions of the
Hartree-Fock theory by a variational principle, the “particle-hole amplitudes”
are determined by the “Kerman-Koonin” stationarity principle
∫ t1
δS = δ dtL(t) (31)
t 0
L(t) = 〈ψ(t)| H + δH ext − i¯h ∂ |ψ(t)〉 . (32)
∂t
We want to derive linear equations of motion, hence we must keep all
second order terms in the Lagrangian. The driving quantity is the external
perturbing field δH 1 (t) which is, by assumption, of first order. Linear equations
of motion will lead to c ph (t) that are of the same order. Thus, to get
linear equations of motion, we must keep the linear terms in c ph (t) with the
perturbing field, and the quadratic terms in all others.
Thus:
〈ψ(t)| δH ext |ψ(t)〉
≈ 1 ∑ [
c
∗
2 ph 〈φ HF | a † h a pδH ext (t) |φ HF 〉 + c ph 〈φ HF | δH ext (t)a † pa h |φ HF 〉 ]
= 1 2
= 1 2
= 1 2
ph
∑
phαβ
∑
phαβ
∑
ph
〈α| δh ext (t) |β〉 [ c ∗ ph 〈φ HF | a † h a pa † αa β |φ HF 〉 + c ph 〈φ HF | a † αa β a † pa h |φ HF 〉 ]
〈α| δh ext (t) |β〉 [ c ∗ phδ α,p δ β,h + c ph δ β,p δ α,h
]
[
c
∗
ph 〈p| δh ext (t) |h〉 + c ph 〈h| δh ext (t) |p〉 ] . (33)
5

In the time derivative term we must be a bit more careful and keep the
normalization N (t).
〈ψ(t)| i¯h ∂ ∂t |ψ(t)〉 = E HF + i¯h 2 〈ψ 0(t)| ∑ ph
= E HF + i¯h 4
∑
ph
ċ ph (t)a † pa h |ψ(t)〉 − i¯h 2
˙ N (t)
N (t)
〈ψ(t)| ċ ph (t)a † pa h − ċ ∗ ph(t)a † h a p |ψ(t)〉 (34) .
This formula is still exact, we now expand to second order. We can immediately
leave out all terms that can be written as a time-derivative. Also, first
order terms are zero because 〈φ HF | a † pa h |φ HF 〉 = 0. Hence
〈ψ(t)| i¯h ∂ ∂t |ψ(t)〉
= E HF + i¯h 4
= E HF + i¯h 4
∑
〈φ HF | c ∗ p ′ h ′(t)ċ ph(t)a † h ′a p ′a† pa h − ċ ∗ ph(t)c p ′ h ′(t)a† h a pa † p ′a h |φ HF〉
′
php ′ h ′
∑ [
c
∗
ph (t)ċ ph (t) − ċ ∗ ph(t)c ph (t) ] . (35)
ph
It should also be noted that this term occurs under the time integral in the
action (31), hence we can integrate by parts. In other words, the two term
appearing above are the same.
Finally we must expand 〈ψ(t)| H − E HF |ψ(t)〉 to second order in the
c ph (t). The E HF comes from the time-derivative term, it evidently cancels
the zeroth order term in our expansion. The normalization integral (23) is
to second order
N 2 (t) ≈ 〈φ HF | 1 + 1 ∑
c ph (t)a †
2
pa h + 1 ∑
c ∗
ph
2
ph(t)a † h a p + O(c 2 ph) |φ HF 〉
ph
= 1 + O(c 2 ph) (36)
and has therefore no influence.
Now
〈ψ(t)| H − E HF |ψ(t)〉
= 1 ∑ [
c
∗
2 ph 〈φ HF | a † h a p(H − E HF ) |φ HF 〉 + c ph 〈φ HF | (H − E HF )a † pa h |φ HF 〉 ]
ph
∑
+ 1 c ∗
8
phc ∗ p ′ h 〈φ HF| a † ′ h a pa † h ′a p ′(H − E HF ) |φ HF 〉
pp ′ hh ′
+ 1 ∑
c ph c p
8
′ h ′ 〈φ HF| (H − E HF )a † pa h a † p ′a h |φ HF〉
′
pp ′ hh ′
6

+ 1 ∑
c ∗
4
phc p ′ h ′ 〈φ HF| a † h a p(H − E HF )a † p ′a h |φ HF〉 . (37)
′
pp ′ hh ′
The situation is the same as the usual one in the theory of small oscillations:
The first order terms must be zero in order to obtain a sensible second order
equation. We must therefore require that
〈φ HF | a † h a p(H − E HF ) |φ HF 〉 = 〈φ HF | (H − E HF )a † pa h |φ HF 〉 = 0 . (38)
These conditions are called the “Brillouin conditions”. In the case of a
Hartree–Fock ground state the condition is identical to the requirement that
the single particle orbitals have been obtained by the Hartree-Fock equation.
We can now turn to the evaluation of the second-order terms: In the terms
that contain c ∗ phc ∗ p ′ h or c phc ′ p ′ h ′, the matrix element of E HF is taken between
a 2p − 2h state and the ground state and is zero because of orthogonality
〈φ HF | a † h a pa † h ′a p |φ HF〉 = 0. Moreover, the matrix element of the one-body
′
part of the Hamiltonian is zero:
〈φ HF | a † h a pa † h ′a p ′H 1 |φ HF 〉 = ∑ αβ
〈α| h |β〉 〈φ HF | a † h a pa † h ′a p ′a† αa β |φ HF 〉 = 0 (39)
because, for example α = p and β = h and then we would have the overlap
between a 1p-1h state and the ground state. Therefore, this term has
contributions form the pair potential only:
〈φ HF | a † h a pa † h ′a p ′V |φ HF〉 = 1 2
Likewise, we get
= 1 2
= 1 2
∑
αβγδ
∑
αβγδ
∑
αβγδ
〈αβ| v |γδ〉 〈φ HF | a † h a pa † h ′a p ′a† αa † β a δ a γ |φ HF 〉
〈αβ| v |γδ〉 〈φ HF | a † h a pa † h ′a p ′a† αa † β a δ a γ |φ HF 〉
〈αβ| v |γδ〉 [δ αp δ βp ′ − δ αp ′δ βp ] [δ δh ′δ γh − δ δh δ γh ′]
= 〈pp ′ | v |hh ′ 〉 a
(40)
〈φ HF | V a † pa h a † p ′a h ′ |φ HF〉 = 〈hh ′ | v |pp ′ 〉 a
. (41)
Finally we need to work out the last term in Eq. (37). We have to
distinguish four cases, namely p = p ′ , h = h ′ , p = p ′ , h ≠ h ′ , p ≠ p ′ , h = h ′
and p ≠ p ′ , h ≠ h ′ . The case p = p ′ , h = h ′ has already been worked out, we
have obtained
〈φ HF | a † pa h (H − E HF )a † pa h |φ HF 〉 = ɛ p − ɛ h , (42)
7

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〉
8

= 1 ∑
〈pβ| v |γδ〉 〈φ HF | a † h
2
a† β a δ a γa † p ′a h |φ HF〉
′
βγδ
= ∑ 〈pβ| v |p ′ δ〉 a
〈φ HF | a † h a† β a δ a h |φ HF〉
′
βδ
= − 〈ph ′ | v |p ′ h〉 a
= 〈ph ′ | v |hp ′ 〉 a
. (46)
This derivation deviates from the previous one only in the last line: Now we
can not have β = δ because that would imply h = h ′ which contradicts our
assumption.
With these manipulations, we have derived our Lagrangian (32) to second
order:
L(t) = ∑ [
c
∗
ph 〈p| δh(t) |h〉 + c ph 〈h| δh(t) |p〉 ] + i¯h ∑ [
c
∗
ph
2 ph (t)ċ ph (t) − ċ ∗ ph(t)c ph (t) ]
ph
+ 1 ∑ [ ]
c
∗
2 ph c ∗ p ′ h ′ 〈pp′ | v |hh ′ 〉 a
+ c ph c p ′ h ′ 〈hh′ | v |pp ′ 〉 a
pp ′ hh ′
+ ∑
c ∗ phc p ′ h ′ [(ɛ p − ɛ h )δ pp ′δ hh ′ + 〈ph ′ | v |hp ′ 〉 a
] (47)
pp ′ hh ′
3 Equations of Motion
We are now ready to derive the equations of motion. The action integral
(47) has been expanded to second order in the particle–hole amplitudes c ph ,
c ∗ ph. The linear equations of motion are now obtained from the stationarity
principle
δS [ c ph (t), c ∗ ph(t) ] = 0 . (48)
Recall that the Lagrangian (47) appears under the time integral, we can
therefore integrate the time coordinate by parts. Doing the variation with
respect to c ph and c ∗ ph gives two equations:
0 = i¯hċ ph + 〈p| δh ext (t) |h〉
+ ∑ p ′ h ′ [
〈pp ′ | v |hh ′ 〉 c ∗ p ′ h ′ + [(ɛ p − ɛ h )δ pp ′δ hh ′ + 〈ph ′ | v |hp ′ 〉 a
] c p ′ h ′ ]
0 = −i¯hċ ∗ ph + 〈p| δh ext (t) |h〉
+ ∑ p ′ h ′ [
〈pp ′ | v |hh ′ 〉 c p ′ h ′ + [(ɛ p − ɛ h )δ pp ′δ hh ′ + 〈ph ′ | v |hp ′ 〉 a
] c ∗ p ′ h ′ ]
(49)
9

A convenient representation of the resulting equations is a “supermatrix”
form, considering sets of states ph as row/column indices of a “supermatrix”.
H ≡
c(t) ≡
[ ] A B
B † A †
M ≡
[ ] 1 0
0 −1
[ ]
[ ]
c(t)
h(t)
c ∗ , h(t) ≡
(t)
h ∗ (t)
where A and B are particle–hole matrices of the form
and
(50)
(51)
A = (A ph,p ′ h ′) ≡ ((ɛ p − ɛ h ) δ pp ′δ hh ′ + 〈ph ′ | v |hp ′ 〉) , (52)
B = (B ph,p ′ h ′) ≡ (〈pp′ | v |hh ′ 〉) , (53)
c(t) ≡ (c ph (t)) h(t) ≡ (〈p| δh ext (t) |h〉) . (54)
In terms of these quantities, the linearized equations of motion assume the
form
Hc(t) + h(t) = i¯h ∂ Mc(t) . (55)
∂t
It is always convenient to rewrite the equations in terms of frequencies. To
that end, write
δh ext (r; t) = δh ext (r; ω) [ e iωt + e −iωt] e ηt (56)
and also split the c ph (t) into components with positive and negative frequency:
c(t) = [ xe −iωt + y ∗ e iωt] e ηt (57)
Insert these in the time-dependent equation and collect the terms with positive
and negative frequency gives the final form
[ ] [ ] [ ]
[ ] [ ]
A B x h 1 0 x
B † A † +
y h ∗ = ¯h(ω + iη)
. (58)
0 −1 y
These are the general “time-dependent Hartree-Fock” equations. Solving
these equations is quite common in nuclear and atomic physics. A derivation
may be found, for example, in the book by Thouless [2].
4 TDHF for the density-density response
function
The above equation gives, for the case of zero external field, the eigenfrequencies
of the system. To get a density-density response, one must calculate, to
10

first order in the particle-hole amplitudes, the density fluctuation
δρ(r; t) = 〈ψ(t)| ˆρ(r) |ψ(t)〉 − 〈φ HF | ˆρ(r) |φ HF 〉
[
c
∗
ph 〈φ HF | a † h a p ˆρ(r) |φ HF 〉 + c ph 〈φ HF | ˆρ(r)a † h a p |φ HF 〉 ]
= ∑ ph
= ∑ ph
[
c
∗
ph ϕ p (r)ϕ ∗ h(r) + c ph ϕ ∗ p(r)ϕ h (r) ]
= ∑ ph
[
xph ϕ p (r)ϕ ∗ h(r) + y ph ϕ ∗ p(r)ϕ h (r) ] e iωt+ηt + c.c.
where
= [x(r) + y(r)] e −iωt+ηt + [x ∗ (r) + y ∗ (r)] e iωt+ηt (59)
x(r) ≡ ∑ ph
ϕ ∗ h(r)ϕ p (r)x ph and y(r) ≡ ∑ ph
ϕ h (r)ϕ ∗ p(r)y ph . (60)
In principle, one can solve the TDHF equation for any external potential,
obtain the x ph and y ph , and insert these in the density fluctuation. But this
does not lead to the beloved form of the density-density response function
that is fancied by the condensed matter community. To get such a form,
one must first omit all exchange terms. Using this simplification, the explicit
forms of A and B are
∫
A ph,p ′ h ′ = [e(p) − e(h)]δ pp ′δ hh ′ + d 3 rd 3 r ′ ϕ ∗ p(r)ϕ ∗ h ′(r′ )V (r, r ′ )ϕ h (r)ϕ p ′(r ′ )
∫
B ph,p ′ h ′ = d 3 rd 3 r ′ ϕ ∗ p(r)ϕ ∗ p ′(r′ )V (r, r ′ )ϕ h (r)ϕ h ′(r ′ ) (61)
Then we can write Eq. (58)
x ph +
y ph +
∫
〈p| δh ext |h〉
ɛ p − ɛ h − ¯hω − iη + ∫
〈h| δh ext |p〉
ɛ p − ɛ h + ¯hω + iη +
d 3 rd 3 r ′ ϕ ∗ p(r)ϕ h (r)
ɛ p − ɛ h − ¯hω − iη V (r, r′ ) [x(r ′ ) + y(r ′ )] = 0
d 3 rd 3 r ′ ϕ p (r)ϕ ∗ h(r)
ɛ p − ɛ h + ¯hω + iη V (r, r′ ) [x(r ′ ) + y(r ′ )] = 0 ,
where the small imaginary parts have been introduced to maintain causality.
Now we multiply the first equation ϕ ∗ h(r ′′ )ϕ p (r ′′ ) and the second equation
with ϕ h (r ′′ )ϕ ∗ p(r ′′ ) and sum over the indices p, h to obtain after the renaming
r ↔ r ′′
x(r) =
y(r) =
∫
∫
d 3 r ′ χ (+)
{
0 (r, r ′ , ω)
d 3 r ′ χ (−)
0 (r, r ′ , ω)
∫
δh ext (r ′ ) +
{
∫
δh ext (r ′ ) +
11
}
d 3 r ′′ V (r ′ , r ′′ ) [x(r ′′ ) + y(r ′′ )] (63)
}
d 3 r ′′ V (r ′ , r ′′ ) [x(r ′′ ) + y(r ′′ )]
(62)

with
Noting that
χ (+)
0 (r, r ′ , ω) = ∑ ph
χ (−)
0 (r, r ′ , ω) = ∑ ph
ϕ p (r)ϕ ∗ h(r)ϕ ∗ p(r ′ )ϕ h (r ′ )
,
¯hω − ɛ p + ɛ h + iη
ϕ ∗ p(r)ϕ h (r)ϕ p (r ′ )ϕ ∗ h(r ′ )
¯hω + ɛ p − ɛ h + iη
. (64)
δρ(r; ω) = [x(r) + y(r)] (65)
and defining the Lindhard function for an inhomogeneous system as
χ 0 (r, r ′ ; ω) = χ (+)
0 (r, r ′ , ω) − χ (−)
0 (r, r ′ , ω) (66)
we get, by adding the two equations (63)
∫
∫
δρ(r; ω)− d 3 r ′ d 3 r ′′ χ 0 (r, r ′′ ; ω)V (r ′′ , r ′ )δρ(r ′ ; ω) = d 3 r ′ χ 0 (r, r ′ ; ω)δh ext (r ′ ; ω) .
(67)
This is the desired result. It is known as the “Random Phase approximation”.
Since there are about 25 derivations of this result, I refrain from explaining
what that has to do with “random phase”.
5 Homogeneous system
In the limit of a homogeneous system, the potential is a function of the
distance
V (r, r ′ ) = V (|r − r ′ |) (68)
We define the Fourier transform with a density factor
∫
Ṽ (q) ≡ ρ
d 3 re iq·r V (r) . (69)
The single-particle wave functions are just plane waves, i.e.
χ (+)
0 (r, r ′ , ω) = 1 ∑ e i(p−h)·(r−r′ )
Ω 2 ¯hω − ɛ p + ɛ h + iη
ph
∑
= ρ Ω q
∫
ρ
≡
(2π) 3
e iq·(r−r′ ) 1 N
∑
h
n(h)¯n(h + q)
¯hω − ɛ h+q + ɛ h + iη
d 3 qe iq·(r−r′) χ (+)
0 (q, ω) , (70)
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χ (−)
0 (r, r ′ , ω) = 1 ∑ e i(p−h)·(r′ −r)
Ω 2 ¯hω + ɛ p − ɛ h + iη
ph
∑
= ρ Ω q
∫
ρ
≡
(2π) 3
e iq·(r′ −r) 1 N
∑
h
n(h)¯n(h + q)
¯hω + ɛ h+q − ɛ h + iη
d 3 qe iq·(r′ −r) χ (−)
0 (q, ω) . (71)
Since the functions χ (+)
0 (q, ω) and χ (−)
0 (q, ω) are invariant under q ↔ −q, we
can add the two functions to the coordinate form of the full response function
of the non-interacting system and write
χ 0 (r, r ′ ; ω) =
ρ ∫
d 3 qe iq·(r′ −r) χ
(2π) 3 0 (q, ω) (72)
If, furthermore, we approximate the Hartree-Fock spectrum by the free singleparticle
spectrum, we obtain the familiar Lindhard function. This is actually
consistent with the approximation to leave out the exchange matrix elements
in the matrices A and B. We finally transform the full equation (67) in
momentum space and solve
or
δρ(q, ω) =
χ(q, ω) =
χ 0 (q, ω)
1 − Ṽ (q)χ 0(q, ω) δ˜h(q, ω), . (73)
χ 0 (q, ω)
1 − Ṽ (q)χ 0(q, ω) . (74)
This is the beloved “RPA” equation for the density-density response function.
The present derivation is perhaps a bit complicated, but it shows very well
what kinds of approximations have been made. These are
• Assume a local, tramslationally invariant (effective) interaction
• Omit exchange
• Assume a free single particle spectrum
References
[1] D. Pines and P. Nozieres. The Theory of Quantum Liquids, volume I.
Benjamin, New York, 1966.
[2] D. J. Thouless. The quantum mechanics of many-body systems. Academic
Press, New York, 2 edition, 1972.
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