Chemistry Reference
In-Depth Information
we require that
q
l
be equal to
hn
l
1
ihn
l
i
, which gives
q
0
¼
4
dn
. The electronic
part to be solved is then
H
el
¼T
X
ls
X
ðC
lþ
1
s
C
ls
þ
l
DÞn
ls
;
h.c
:Þþ
ððU=
2
þ
2
VÞþð
1
Þ
(8.6)
ls
where
T
and
D
are defined as
t
0
þ V m
and
4
V UÞdn
, respectively. To
diagonalize this hamiltonian, we introduce one-electron states
C
ms
ð
8
S þ
using one-
f
ms
,as
C
ms
P
l
f
ms
ðlÞC
ls
electron wave functions
. Then, the eigenvalue equation
becomes
ll
0
f
ms
ðl
0
Þ¼e
ms
f
ms
ðlÞ;
M
HF
(8.7)
where
l
DÞd
ll
0
:
M
HF
ll
0
Tðd
l;l
0
þ
1
þ d
l;l
0
1
ÞþððU=
2
þ
2
VÞþð
1
Þ
(8.8)
This equation is easily solved writing the wave functions in the form of a Bloch
state, namely, as
e
ikl
u
k
ðlÞ
and a momentum
k
.
We here drop the spin indices because the CDW state has no spin polarization.
Since the spatial period of this case is two, we can express the periodic function as
u
k
ðlÞ¼aðkÞþð
f
m
ðlÞ¼
, with a periodic function
u
k
ðlÞ
l
bðkÞ
1
Þ
, and then the eigenvalue equation is reduced to the
following 2
2 form:
a
b
¼ e
m
:
2
T
cos
ðkÞþðU=
2
þ
2
VÞ
D
a
b
(8.9)
D
2
T
cos
ðkÞþðU=
2
þ
2
VÞ
Solving this equation in an usual way, we find
p
D
2
e
m
¼ðU=
2
þ
2
VÞ
þ
4
T
2
cos
2
ðkÞ
ðU=
2
þ
2
VÞe
k
(8.10)
and the signs
correspond to valence (v) and conduction (c) bands. It turns out here
that the label
corresponds to the sets of (v,
k
) and (c,
k
). As for the values of
k
,
we impose a periodic boundary condition (PBC), i.e.,
m
f
m
ðl þ NÞ¼f
m
ðlÞ
and find
k ¼ð
p=NÞm
with
m
being an integer. We also emphasize that the total
number of the one-electron states is
N
, except for spins. We then restrict the
range of
k
to
2
2 so as to count
N
/2 states for each band, which
procedure is nothing but a band folding typical to a superlattice structure. Finally,
we display a normal orthogonal basis set for the one-electron states thus obtained:
p=
2
k<p=
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