MTPA 1 : Condensed Matter Theory I
Exercise sheet 8, 10.12.2014
LPTM, Université de Cergy-Pontoise
Prof. Andreas Honecker
Exercise 8.2
Tight-binding model on a face-centered cubic lattice
Consider a face-centered cubic lattice with lattice constant a and only one s-band that contributes
~ = 0, γ(R)
~ = t if R
~ 6= ~0 is one of the lattice
to the tight-binding dispersion. Assume that α(R)
vectors closest to zero.
Solution: It is convenient to first construct the three-dimensional tight-binding dispersion. For
this purpose, we note that the origin ~0 of the face-centered cubic lattice has 12 nearest neighbors
at a/2 (±1, ±1, 0), a/2 (±1, 0, ±1), and a/2 (0, ±1, ±), compare the following sketch:
In the present case, the tight-binding dispersion reads
E(~k)
=
α=0
=
~ cos(~k · R)
~
γ(R)
P
~ cos(~k · R)
~
1 + R∈Λ
α(R)
~
X
~ cos(~k · R)
~
γ(R)
Es − β −
Es −
β+
P
~
R∈Λ
~
R∈Λ
=
Es − β − t
X
~ .
cos(~k · R)
~ nearest neighbors
R
~ and −R
~ give identical contributions. Thus, for the faceDue to symmetry of the cosine, R
centered cubic lattice we get 6 contributions, multiplied by a factor two:
a (kx + ky )
a (kx − ky )
a (kx + kz )
a (kx − kz )
+ cos
+ cos
+ cos
E(~k) = Es − β − 2 t cos
2
2
2
2
a (ky + kz )
a (ky − kz )
+ cos
+ cos
.
2
2
This can be simplified with the help of the trigonometric identity cos(α ± β) = cos α cos β ∓
sin α sin β. In the present case, the sines cancel in pairs and one is left with
a
k
a
k
a
k
a
k
a
k
a
k
x
y
x
z
y
z
E(~k) = Es − β − 4 t cos
cos
+ cos
cos
+ cos
cos
.
(1)
2
2
2
2
2
2
1
(a) Construct the tight-binding dispersion E(~k) along the following lines (for a sketch of the
first Brillouin zone of the face-centered cubic lattice compare below):
i) ΓX:
ky = kz = 0, kx = κ 2π/a.
Solution: We insert the given parametrization for the line ΓX into (1) and get:
E(~k) = Es − β − 4 t (2 cos(π κ) + 1) .
ii) ΓL:
kx = ky = kz = κ 2π/a.
Solution: Again, we insert the given parametrization for the line ΓL into (1) and get
this time:
E(~k) = Es − β − 12 t cos(π κ)2 .
iii) ΓK:
kz = 0, kx = ky = κ 2π/a.
Solution: As before, we insert the given parametrization for the line ΓK into (1) and
get now:
E(~k) = Es − β − 4 t cos(π κ)2 + 2 cos(π κ) .
iv) ΓW:
kz = 0, kx = κ 2π/a, ky = κ π/a.
Solution: Once more, we insert the given parametrization for the line ΓW into (1)
and get for a final time:
π κ
πκ
~
+ cos(π κ) + cos
E(k) = Es − β − 4 t cos(π κ) cos
2
2
(b) Show that on the square faces of the zone the normal derivative of E vanishes.
Solution: We start by computing the gradient of (1):
 a kx

sin 2 cos a 2ky + sin a 2kx cos a 2kz
~ ~ E(~k) = 2 a t cos a kx sin a ky + sin a ky cos a kz  .
∇
k
2
2
2
2
cos a 2kx sin a 2kz + cos a 2ky sin a 2kz
(2)
Now we consider the six square faces by the first Brillouin zone. Since they are all equivalent,
we may pick one of them, for example the one defined by1 kx = 2π/a. If one inserts
1
Recall that the reciprocal lattice of a face-centered cubic lattice with lattice constant a when viewed as a
conventional cubic lattice with basis is a body-centered cubic lattice with lattice constant 4π/a when regarded
as a conventional cubic lattice with basis.
2
kx = 2π/a into (2) and remembers that sin π = 0, cos π = 1, one finds


0
~ ~ E(~k)
∇
= 2 a t sin a 2ky + sin a 2ky cos a 2kz  .
k
kx =2π/a
sin a 2kz + cos a 2ky sin a 2kz
The normal direction of the chosen plane is the x-direction and indeed the x-component
of the result vanishes.
(c) Show that on the hexagonal faces of the zone the normal derivative of E does not vanish
in general.
Solution: The first Brillouin zone of the face-centered cubic lattice has eight hexagonal
faces and by symmetry we may pick one of them, for example the one defined by the normal
vector (1, 1, 1), i.e.,
 
1
3π   ~
= 1 · k = kx + ky + kz .
(3)
a
1
In particular, the (un-normalized) normal derivative on this face is
~
~
∇~k E(k)
n
 
 
 a kx

sin 2 cos a 2ky + sin a 2kx cos a 2kz
1
1
~ ~ E(~k) (2)
= 1 · ∇
= 1 · 2 a t cos a 2kx sin a 2ky + sin a 2ky cos a 2kz 
k
1
1
cos a 2kx sin a 2kz + cos a 2ky sin a 2kz
a kx
a ky
a kx
a kz
a kx
a ky
= 2 a t sin
cos
+ sin
cos
+ cos
sin
2
2
2
2
2
2
a ky
a kz
a kx
a kz
a ky
a kz
+ sin
cos
+ cos
sin
+ cos
sin
. (4)
2
2
2
2
2
2
To show that this does not vanish in general, it suffices to look at some selected points.
Let us start with the high-symmetry point L = (π/a, π/a, π/a). In this case, we always
have one factor cos π2 = 0 in (4) and thus
~
~
∇~k E(k) = 0 .
n,L
In fact, it may be expected that the normal derivative at such high symmetry points
vanishes due to symmetry. So, let us look at the less symmetric point K = ( 23πa , 23πa , 0).
Upon insertion into (4) we have
3π
3π
3π
3π
3π
3π
~
~
∇~k E(k)
= 2 a t sin
cos
+ sin
+ cos
sin
+ sin
4
4
4
4
4
4
n,K
3π
3π
3π
1
1
= 4 a t sin
cos
+ sin
= 4at − + √
.
4
4
4
2
2
This result is manifestly non-zero!
For completeness, let us look at W = ( πa , 2aπ , 0). Upon insertion of this point into (4) we
now have
π
π
π
~
~
∇~k E(k)
= 2 a t sin cos π + sin + cos sin π + sin π = 0 .
2
2
2
n,W
3
Here, the normal gradient happens to vanish again2 . Still, the normal gradient (4) is
a continuous function on the plane defined by (3) and thus is necessarily non-zero in a
neighborhood of the point K.
Remark: For a weak periodic potential, the normal derivative would have to vanish on any
Bragg plane, in particular also on the boundaries of the first Brillouin zone. The example of
part (c) shows that this is not true in general if the potential is not weak. Remember that the
tight-binding approach is based on the assumption of electrons strongly bound to the nuclei and
thus outside the range of validity of a weak potential.
We also observed that the normal derivative vanishes at highly symmetric points. For example,
the square face of part (b) belongs to a plane that is a reflection plane for the reciprocal lattice
and indeed on such a plane the normal derivative vanishes also for a tight-binding model.
2
In fact, W is another high-symmetry point where in the extended zone scheme three hexagonal faces meet.
4