CASINO manual - Theory of Condensed Matter

where n αβ (r, r ′ ) is the charge density of the N β −δ αβ electrons of spin β calculated with an electron of
spin α held fixed at r. We can now average g αβ (r, r ′ ) over the surface of a sphere of radius ρ centred
on r. The vector r ′ − r can be written in spherical polar coordinates as (ρ, θ, φ). The required pair
correlation function is then
∫
gαβ(r, F 1
ρ) =
4πρ 2 g αβ (r, r ′ ) δ(|r ′ − r| − ρ) dΩ, (355)
where dΩ = sin θ dθ dφ. This gives
∫
gαβ(r, F 1
ρ) =
4πρ 2 g αβ (r, r ′ ) δ(|r ′ − r| − ρ) dΩ (356)
=
∫
1 |Ψ F | 2 ∑ N β −δ αβ
j=1 δ(|r j,β − r ′ | − ρ) dr 2 . . . dr N
∫
4πρ 2 dΩ. (357)
|ΨF | 2 dr 2 . . . dr N n β (r ′ )
This type of averaging was used by Maezono et al. [83]. The pair correlation functions can be evaluated
in ρ bins for each r of interest. Equations (347), (349) and (357) contain one or two charge densities
in the denominator. When one or both of these charge densities is very small there will be a large
amount of noise in the quantity evaluated. We expect g αβ (r, r ′ ) to tend to 1 − δ αβ /N β as r − r ′
becomes large, while for small values of r − r ′ we expect it be less than one. One strategy for coping
with the noise would be to set a lower limit on the charge density n β (r ′ ) below which the contribution
to Eq. (357) is not accumulated.
33.3.5 Collecting in bins
Assume the system is homogeneous and isotropic. The pair correlation function g αβ (r) can be collected
in bins of width ∆, giving
g αβ (r n ) = Ω 〈 N
n 〉
αβ
, (358)
Ω n N α N β
where Nαβ n is the number of α, β-spin pairs whose separation falls within the nth bin (the pairs iα, jα
and jα, iα should be counted separately, and similarly for the parallel β-spins), and Ω n is the volume
of the nth bin. Note that if we consider a single bin, so that Ω = Ω n , then Nαβ n = N α(N β − δ αβ ).
This confirms that the normalization of Eq. (358) is correct, but note that our formulae disagree with
Eq. (35) of Ortiz and Ballone [84].
The volume of the nth bin, Ω n , is given by
Ω n = 4 3 π ( n 3 − (n − 1) 3) ∆ 3 = 4π ( n 2 − n + 1/3 ) ∆ 3 . (359)
One could define r n to be the centre of the nth bin, but a better choice is [85]
r n = 3∆ 4
[n 4 − (n − 1) 4 ]
[n 3 − (n − 1) 3 ] = 3∆ 4
[4n 3 − 6n 2 + 4n − 1]
[3n 2 . (360)
− 3n + 1]
If g were to be linear across each bin and if g were sampled exactly without statistical error then Eqs.
(358) and (359) would reproduce the exact value at each point r n .
In two dimensions one requires the area of the nth circular strip,
A n = π((n∆) 2 − ((n − 1)∆) 2 ) = π∆ 2 (2n − 1). (361)
We want to average f(r) = ar + b over a strip, and find the corresponding radius r n . Therefore
Hence
f(r n ) =
r n = 2∆ 3
∫ n∆
(n−1)∆
f(r) 2πr dr
∫ n∆
2πr dr
(n−1)∆
= [ 1 3 ar3 + 1 2 br2 ] n∆
(n−1)∆
[ 1 2 r2 ] n∆
(n−1)∆
= a 2[(n∆)3 − ((n − 1)∆) 3 ]
3[(n∆) 2 − ((n − 1)∆) 2 ] + b
= ar n + b.
[n 3 − (n − 1) 3 ]
[n 2 − (n − 1) 2 ] = 2∆ 3
[3n 2 − 3n + 1]
. (362)
[2n − 1]
190