Download SOLID STATE PHYSICS HW#7 Question 1. Square Lattice, free

SOLID STATE PHYSICS HW#7
Question 1. Square Lattice, free electron energies(Kittel 7.1).
(a) The free electron kinetic energy is given by
E=
~2 k 2
.
2m
(1)
At the corner of the first zone ( πa , πa ), |~k|2 for the two dimensional wave vector ~k = kx i + ky j is
equal to
to
π2
a2
2π 2
,
a2
which leads to E =
leading to E =
~2 π 2
.
2ma2
~2 π 2
.
ma2
At midpoint of a side face of the zone ( πa , 0), |~k|2 is equal
The ratio of former to latter is 2.
(b) In 3 dimensions, |~k|2 for a wave vector ~k = kx i + ky j + kz k is equal to
of the first zone ( πa , πa , πa ). Then the free electron energy is E =
corresponding to an energy E =
~2 π 2
.
2ma2
3~2 π 2
2ma2
3π 2
a2
at the corner
. At ( πa , 0, 0), |~k|2 is
π2
,
a2
The ratio is 3. In d dimensions, the ratio is equal to d.
(c) For divalent metals (metals with even number of valence electrons per primitive cell) a band
overlap occurs: Instead of one filled band giving an insulator, we can have two partially filled bands
giving a metal. The kinetic energy difference between the boundary and the middle of the 1st BZ
causes such an overlap which leads to partially filled orbitals, leaving room for electron hopping,
namely resulting in a finite conductivity.
Question 2. Kronig-Penney Model (Kittel 7.3).
(a) For a square well potential (U = 0 for 0 < x < a, and U = U0 for −b < x < 0), consider
two solutions of the Schrödinger equation:
ψI = AeiKx + Be−iKx
for U = 0 ,
ψII = CeQx + De−Qx
for U = U0 ,
(2)
For a square well potential, there are 4 boundary conditions. 2 of them are from continuity of
wave functions and their derivatives:
ψI (a) = ψII (a) ,
dψI
dψII
|x=a =
|x=a ,
dx
dx
(3)
The other 2 boundary conditions are derived from Bloch theorem, that is, from periodicity of the
wavefunctions:
~
ψk (~r) = uk (~r)eik·~r
where uk (~r) = uk (~r + T~ ) . T~ : lattice translation vector .
(4)
These 4 boundary conditions result in 4 equations, namely:
A+B = C +D,
(5)
ik(A − B) = Q(C − D) ,
(6)
AeiKa + Be−iKa = (Ce−Qb + DeQb )eik(a+b) ,
(7)
iK(AeiKa − Be−iKa ) = Q(Ce−Qb − DeQb )eik(a+b) .
(8)
The determinant of matrix of the coefficients A, B, C, D is then equal to
· 2
¸
Q − K2
sinh Qb sin Ka + cosh Qb cos Ka = cos k(a + b) ,
2QK
(9)
which should vanish in order to obtain finite coefficients.
Consider the dirac delta potential limit of the determinant equation above. The limit in order
to obtain dirac delta function is b → 0, U0 → ∞, Q >> K, and Qb << 1. Under these conditions,
the equation reduces to
µ
P
Ka
¶
sin Ka + cos Ka = cos ka ,
where P ≡
Q2 ab
.
2
(10)
Setting k = 0 as in the question and using Ka << 1 for finding the energy of the lowest energy
band, one should obtain
P
K 2 a2
Ka + 1 −
' 1.
Ka
2
Then the energy for the lowest energy band is E =
~2 K 2
2m
=
(11)
~2 P
.
ma2
(b) For k = πa , eq. (10) reduces to
µ
¶
P
sin Ka + cos Ka = −1 .
Ka
(12)
In the vicinity of the zone boundary, in order to find the band gap, taking Ka = π + δ, where
δ << 1 and looking for the solutions of δ is convenient. Eq. (12) is rewritten as
µ ¶
P
δ2
(−δ) − 1 +
= −1 .
π
2
There are two solutions for δ: δ1 = 0 and δ2 =
2P
π .
(13)
Here, δ2 causes a finite energy gap. The total
energy is
E=
~2 π 2
~2 πδ
~2 (π + δ)2
=
+
.
2ma2
2ma2
ma2
(14)
The second term corresponds to the energy gap at k = πa .
Question 3. Square Lattice (Kittel 7.6).
The potential energy is given by
U (x, y) = −4U cos
2πx
2πy
cos
.
a
a
(15)
In terms of Fourier components, the potential energy is equal to
U (x, y) =
XX
G
0
UG,G0 eiGx eiG y = 4
G0
X
cos Gx
0
1
UG,G0 cos G0 y .
(16)
G0 >0
G>0
The coefficient UG,G0 is found as follows:
Z 1 Z
UG,G0 =
dx
X
dyU (x, y) cos Gx cos G0 y .
(17)
0
Here the reciprocal lattice vectors G = G0 are equal to
2π
a ,
since the energy gap we are looking for is
equivalent to the energy difference of two ends of the Brillouin zone, namely ( πa , πa ) and (− πa , − πa ).
Actually, the coefficient UG,G0 gives us the potential energy difference between two lattice layers
corresponding to ( πa , πa ) and (− πa , − πa ) in k-space, respectively.
UG,G0 is found using the following orthogonality condition,
Z L
2nπx
2mπx
L
dx cos
cos
= δn,m ,
a
a
2
0
(18)
and is equal to UG,G0 = −U .
In 2 dimensions, the central equation is
(λk − ²)C(kx , ky ) +
X
C(kx − G, ky − G0 ) = 0 .
(19)
G,G0
Here, C(kx , ky ) are the coefficients of the Fourier components of the wave function (see Kittel),
and λk ≡
~2 k2
2m .
Remember, the central equation is the characteristic equation (the determinant
equation which is used for finding eigenvalues) of the Hamiltonian matrix in k-space. ² corresponds
to energy eigenvalues.
At the boundary of 1st Brillouin zone, k = ± πa = ± 12 G. Then the two equations derived from
the central equation are as follows:
1
1
(λ − ²)C( G) − U C(− G) = 0 ,
2
2
1
1
(λ − ²)C(− G) − U C( G) = 0 .
2
2
(20)
The determinant of the 2 × 2 coefficient matrix above gives the two energies, λ1 = ² + U and
λ1 = ² − U . Ultimately, the energy gap is 2U at the corner point of the Brillouin zone.