Homework #10
1. Square lattice, free electron energies. (a) Show for a simple square lattice (two
dimensions) that the kinetic energy of a free electron at a corner of the first Brillouin
zone is higher than that of an electron at the midpoint of a side face of the zone by a
factor of 2. (b) What is the corresponding factor for a simple cubic lattice (three
dimensions)? (c) What bearing might the result of (b) have on the conductivity of
divalent metals?
Solution:
(a)
At
the
corner
of
the
first
Brillouin
zone,
r ⎛π π ⎞
k = ⎜ , ⎟,
⎝a a⎠
thus
h 2 k 2 h 2π 2 2 2
h 2π 2
(
1
1
)
.
=
+
=
ma 2
2m 2ma 2
r ⎛π ⎞
h 2π 2 2
h 2π 2
2
+
=
.
At the midpoint of a side face, k = ⎜ ,0 ⎟ , thus E(1,0 ) =
(
1
0
)
2ma 2
2ma 2
⎝a ⎠
Therefore E(1,1) / E(1,0 ) = 2 .
r ⎛π π π ⎞
(b) For a simple cubic lattice, at the corner of the first Brillouin zone, k = ⎜ , , ⎟ ,
⎝a a a⎠
E(1,1) =
h 2π 2 2 2 2
3h 2π 2
+
+
=
.
(
1
1
1
)
2ma 2
2ma 2
r ⎛π
h 2π 2 2
h 2π 2
⎞
2
(1 + 0 ) =
.
At the midpoint of a side face, k = ⎜ ,0,0 ⎟ , thus E(1,0,0 ) =
2ma 2
2ma 2
⎝a
⎠
Therefore E(1,1,1) / E(1,0,0 ) = 3 .
thus E(1,1,1) =
(c) For a divalent metal, the electron density is n = 2 / a 3 . Then the Fermi energy is
h2
h 2π 2
h2
2/3
π
=
.
EF =
(3π 2 n ) 2 / 3 =
(
6
/
)
1
.
53
2m
2ma 2
2ma 2
Thus E(1,0,0 ) < E F < E(1,1,1) . The Fermi sphere is greater than the first Brillouin zone
in the [100] direction, but smaller than the first Brillouin zone in the [111] direction.
The electrons should start to fill in the second Brillouin zone before filling up the first
Brillouin zone.
(c) The electrons start to fill in the 2nd Brillouin zone before completely filling out
the 1st Brillouin zone. Then both the first and second bands are not completely filled
so that the material will be a metal.
2. Free electron energies in reduced zone. Consider the free electron energy bands of
an fcc crystal lattice in the approximation of an empty lattice, but in the reduced zone
scheme in which all k’s are transformed to lie in the first Brillouin zone. Plot roughly
in the [111] direction the energies of all bands up to six times the lowest band energy
at the zone boundary at k=(2π/a)(1/2,1/2,1/2). Let this be the unit of energy. This
problem shows why band edges need not necessarily be at the zone center. Several of
the degeneracies (band crossings) will be removed when account is taken of the
crystal potential.
Solution:
h 2k 2 h 2 2
E=
=
( k x + k y2 + k z2 ) .
2m 2m
r 2π
( n1 , n2 , n3 ) , where n1 , n2 , n3 are all even
By adding a reciprocal lattice vector, G =
a
r
r
or all odd, we fold the k vector into k ' in the first Brillouin zone along [111]
r r r
π
π
π
direction, i.e., k = k '+G and k x ' = k y ' = k z ' ≡ β . Because − ≤ k x ' = k y ' = k z ' ≤ ,
a
a
a
there is − 1 ≤ β ≤ 1 . Then
2
2
2
h 2 ⎡⎛
2πn1 ⎞ ⎛
2πn2 ⎞ ⎛
2πn3 ⎞ ⎤
E=
⎟ + ⎜ k y '+
⎟ + ⎜ k z '+
⎟ ⎥
⎢⎜ k x ' +
a ⎠ ⎝
a ⎠ ⎝
a ⎠ ⎦⎥
2m ⎣⎢⎝
=
2πn1 ⎞ ⎛ π
2πn2 ⎞ ⎛ π
2πn3 ⎞
h 2 ⎡⎛ π
⎟ +⎜ β +
⎟ +⎜ β +
⎟
⎢⎜ β +
2m ⎢⎣⎝ a
a ⎠ ⎝a
a ⎠ ⎝a
a ⎠
=
h 2π 2
3β 2 + 4 β ( n1 + n2 + n3 ) + 4( n12 + n22 + n32 )
2ma 2
2
2
[
2
⎤
⎥
⎥⎦
]
h 2π 2
×
2ma 2
( n1 , n2 , n3 )
E=
(000)
(1 1 1 ), ( 1 1 1 ), ( 1 1 1)
(11 1 ), (1 1 1), ( 1 11)
(002 ), (02 0), ( 200)
(002), (020), ( 200)
(111)
(1 1 1)
(004)
(004 )
3β 2
3β 2 − 4 β + 12
3β 2 + 4 β + 12
3β 2 − 8β + 16
3β 2 + 8β + 16
3β 2 + 12 β + 12
3β 2 − 12 β + 12
3β 2 + 16 β + 64 > 6 × 3
3β 2 − 16 β + 64 > 6 × 3
3. Kronig-Penney model. (a) For the delta-function potential and with P<<1, find at
k=0 the energy of the lowest energy band. (b) For the same problem find the band gap
at k=π/a.
Solution:
h 2κ 2
p sin(κa )
ε
=
+ cos(κa ) = cos(ka ) and
κa
2m
p sin(κa )
(a) At k = 0 ,
+ cos(κa ) = 1 .
κa
For p << 1 , κ ≈ 0 .
From the text book,
(κa )3
(κa ) 2
Then sin(κa ) ≈ κa −
and cos(κa ) ≈ 1 −
.
6
2
The equation becomes
⎡ (κa ) 2 ⎤
(κa ) 2
+
1
−
=1
p ⎢1 −
6 ⎥⎦
2
⎣
(κa ) 2
p−
(1 + p / 3) = 0
2
2p
2p
≈ 2 .
κ2 = 2
a (1 + p / 3) a
Then ε =
h 2κ 2 h 2 p
≈
2m
ma 2
p sin(κa )
+ cos(κa ) = −1 .
κa
a
For p << 1 , κ ≈ π / a .
(b) At k =
π
,
Let κa = π + δ , then ε =
h 2κ 2 h 2 (π + δ ) 2
=
with δ << 1 .
2m
2ma 2
The equation becomes
− p sin δ
− cos δ = −1 .
π +δ
With sin δ ≈ δ and cos δ ≈ 1 − δ 2 / 2 .
δ2
− pδ
−1+
= −1
2
π +δ
2p
Then δ 1 = 0 and δ 2 =
.
π
The energy gap is the energy difference at δ 1 and δ 2 .
εg =
h 2 (π + δ 2 ) 2 h 2 (π + δ 1 ) 2 h 2πδ 2 2h 2 p
−
≈
=
.
2ma 2
2ma 2
ma 2
ma 2
4. Square lattice. Consider a square lattice in two dimensions with the crystal potential
U ( x, y ) = −4U cos(2πx / a ) cos(2πy / a ) .
Apply the central equation to find approximately the energy gap at the corner point
(π/a,π/a) of the Brillouin zone. It will suffice to solve a 2 × 2 determinantal equation.
Solution:
U ( x, y ) = −4U cos(2πx / a ) cos(2πy / a )
= −U [exp(i 2πx / a ) + exp( −i 2πx / a )][exp(i 2πy / a ) + exp( −i 2πy / a )]
⎧ ⎡ 2π ( x + y ) ⎤
⎡ 2π ( x − y ) ⎤
⎡ 2π ( − x + y ) ⎤
⎡ 2π ( x + y ) ⎤ ⎫
i
i
exp
exp
exp
= −U ⎨exp ⎢i
+
+
+
⎥⎦
⎢⎣
⎥⎦
⎢⎣
⎥⎦
⎢⎣ − i
⎥⎦ ⎬
a
a
a
a
⎭
⎩ ⎣
r r
= ∑ U gr exp(ig ⋅ r )
r
g
Therefore the Fourie component of U(x,y) are: U gr = −U
r 2π
r 2π
r 2π
g=
(1,−1) , g =
( −1,1) , and g =
( −1,−1) .
a
a
a
r π
r r r 2π
For k = (1,1) , only k − g = k −
(1,1) is important.
a
a
Then the equation is
⎧ ( λk − ε )ck + U g ck − g = 0
⎨
⎩( λk − g − ε )ck − g + U − g ck = 0
r 2π
for g =
(1,1) ,
a
h2k 2
λk =
, U g = U − g = −U
2m
r π
r r π
r
2π
π
At k = (1,1) , k − g = (1,1) −
(1,1) = − (1,1) = − k . The equation becomes
a
a
a
a
2 2
⎧ ⎛h k
⎞
− ε ⎟⎟ck − Uc− k = 0
⎪ ⎜⎜
⎪ ⎝ 2m
⎠
⎨
2 2
⎪− Uc + ⎛⎜ h k − ε ⎞⎟c = 0
k
⎜ 2m
⎟ −k
⎪⎩
⎝
⎠
⎞
⎛ h2k 2
⎜
−ε
−U ⎟
⎟ = 0.
Let det ⎜ 2m
h2k 2
⎟
⎜
−ε ⎟
⎜ −U
2m
⎠
⎝
2 2
hk
We have ε =
± U . Therefore, the energy gap is
2m
⎛ h2k 2
⎞ ⎛ h2k 2
⎞
ε g = ⎜⎜
+ U ⎟⎟ − ⎜⎜
− U ⎟⎟ = 2U .
⎝ 2m
⎠ ⎝ 2m
⎠