Solutions 369
Carrying out the summation over spin variables, the first two terms become identical
with (
5.8), while the third term (which appears because the Slater determinant is
an antisymmetrized product of N single-particle wave functions) contributes only
if ψ
α
and ψ
β
are states with the same spin. The variational principle leads for the
first two terms to the Hartree equations, which become modified by a contribution
from the third term, the exchange term (
5.11).
5.2: For free electrons with ψ
k
(r)=exp(ik · r)/
√
V the averaged exchange density
reads
¯n
HF
(r, r
′
)=−
2
N
k,k
′
|k|,|k
′
|≤k
F
1
V
e
ik
′
·(r−r
′
)
e
−ik·(r−r
′
)
and with the Fourier transform of 1/|r − r
′
| the exchange potential becomes
V
x,Slater
(r)=−
2e
2
ε
0
NV
k,k
′
|k|,|k
′
|≤k
F
q
1
q
2
1
V
V
e
i(k
′
−k+q)·(r−r
′
)
d
3
r
′
δ
k−k
′
,q
= −
2e
2
ε
0
NV
k,k
′
|k|,|k
′
|≤k
F
1
|k − k
′
|
2
= −
3
8π
2
ε
0
e
2
k
F
.
With the exchange energy ǫ
x
(n)=−3e
2
k
F
/16π
2
ε
0
from Sect.
4.4 one finds
V
LDA
x
(r)=
4
3
ǫ
LDA
x
(n(r)) =
2
3
V
x,Slater
(r).
5.3: The number of discrete k =(k
1
,k
2
,k
3
)withk
i
=2πn
i
/L
i
,i=1, 2, 3with
0 ≤ n
i
<N
i
in a Brillouin zone is N = N
1
N
2
N
3
, which is the number of unit cells
in the crystal or periodicity volume. Thus, for each electron in the unit cell with
given spin there is one state in the energy band, i.e. each band can accommodate
2N electrons.
5.4: A point at the Brillouin zone boundary is characterized by the relation k
′
=
k − G. The condition of degeneracy is k
2
= k
′2
,thus(k − G)
2
=(k
′
)
2
becomes
2k · G = G
2
, which is the condition for Bragg reflection.
5.5: The primitive reciprocal lattice vectors of the square lattice are b
1
=(1, 0)2π/a,
and b
2
=(0, 1)2π/a. Write the free electron energies
E(k)=
¯h
2
2m
(k + G)
2
=
¯h
2
2m
2π
a
!
2
κ
2
for the smallest G at the points Γ,M,andX and connect corresponding points by
parabolas defined by G.
If ν =1, 2, 3 is the number of electrons per atom then, for one atom per unit
cell, n
s
= ν/a
2
is the areal electron density. The radius of the Fermi circle is given
by k
F
=
√
2πn
s
and the Fermi energy by
E
F
=
¯h
2
2m
k
2
F
=
¯h
2
2m
2π
a
!
2
ν
2π
or κ
2
F
= ν/2π which is 0.159 for ν =1.