CASINO manual - Theory of Condensed Matter

as expected from Eq. (291). Using Eqs. (370) and (376) one can show that ˜S αβ (k = 0, k ′ = 0) = 0
and ˜S αβ T (k = 0) = 0.
The rotational average of Sαβ T (r) is
S T R
αβ (r) =
The Fourier transform of Sαβ T R (r) is
=
∫
1
4πr 2 S T
Ω
αβ(r ′ ) δ(|r ′ | − r) dr ′ (377)
∫
1 |Ψ|
2 ∑ ∑
i,α j,β δ(|r iα − r jβ | − r) dR
∫
4πr 2 (378)
Ω
|Ψ|2 dR
∫
− 1
4πr 2 Ω
∫
Sαβ T R (k) = 1 |Ψ|
2 ∑ ∑
i,α j,β
∫
Ω 2 |Ψ|2 dR
− 4π ∑ 1
Ω kG ñα(G) ñ β (−G)
G
n α (r ′′ )n β (r ′′ − r ′ ) δ(|r ′ | − r) dr ′ dr ′′ . (379)
sin(k|r iα−r jβ |)
k|r iα−r jβ |
∫ ∞
0
dR
(380)
sin(kr) sin(Gr) dr. (381)
Unfortunately the integral in the second term is undefined. However, we can use the argument that,
for large enough r, Sαβ T (r) → 0 because the effects of exchange and correlation will tend to zero. The
Fourier transforms of Sαβ T R
(r) and ST
αβ
(r) are therefore well defined. The contributions from the first
and second terms in Eq. (379) must therefore cancel at large distances. In practise we will include
only pairs of particles whose separation is within the radius of the sphere inscribed in the WS cell
of the simulation cell, L WS . We can therefore include only pairs of particles whose separation is less
than L WS and set the upper limit on the integral to L WS , in which case it can be evaluated,
∫ LWS
sin(kr) sin(Gr) dr = 1 [ sin(k − G)LWS
− sin(k + G)L ]
WS
. (382)
2 k − G
k + G
0
For k − G small or k + G small the relevant sin function should be expanded, as described by Rene
Gaudoin. In this method pairs of electrons in the ‘corners’ of the WS cell whose separation is larger
than L WS are not included so that Sαβ T R (k = 0) ≠ 0. This would need to be corrected afterwards. The
structure factor Sαβ T R(k)
should satisfy Sαβ T R (k → ∞) → δ αβ . (383)
33.4.2 Homogeneous and isotropic systems
For a homogeneous and isotropic system, we have from Eq. (367),
S H αβ(r) = N α
Ω
The Fourier transform of Sαβ H (r) is
S H αβ(k) = 1 Ω
= 1 Ω
∫ [
Nα
Ω
∫
Nα
Ω
N β
Ω [g αβ(r) − 1] + δ αβ δ(r) N α
Ω . (384)
N β
Ω [g αβ(r) − 1] + δ αβ δ(r) N ]
α
e ik·r dr
Ω
(385)
N β
Ω [g αβ(r) − 1] e ik·r N α
dr + δ αβ
Ω 2 . (386)
As the system is homogeneous and isotropic Sαβ H (k) is a function of k only, and we have
S H αβ(k) = 1 Ω
∫
Nα
Ω
Using the sum rules of Eq. (344) we find that
N β
Ω [g αβ(r) − 1] 4πr 2 sin(kr) N α
dr + δ αβ
kr
Ω 2 . (387)
S H αβ(k = 0) = 0 (388)
S H αβ(k → ∞) → δ αβ
N α
Ω 2 . (389)
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