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 =~2k2

2m. (1)

At the corner of the first zone (πa , π

a ), |~k|2 for the two dimensional wave vector ~k = kxi + kyj is

equal to 2π2

a2 , which leads to E = ~2π2

ma2 . At midpoint of a side face of the zone (πa , 0), |~k|2 is equal

to π2

a2 leading to E = ~2π2

2ma2 . The ratio of former to latter is 2.

(b) In 3 dimensions, |~k|2 for a wave vector ~k = kxi + kyj + kzk is equal to 3π2

a2 at the corner

of the first zone (πa , π

a , πa ). Then the free electron energy is E = 3~2π2

2ma2 . At (πa , 0, 0), |~k|2 is π2

a2 ,

corresponding to an energy E = ~2π2

2ma2 . 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 Schrodinger 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

dx|x=a =

dψII

dx|x=a ,

(3)

The other 2 boundary conditions are derived from Bloch theorem, that is, from periodicity of the

wavefunctions:

ψk(~r) = uk(~r)ei~k·~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[Q2 −K2

2QK

]sinhQb sinKa + coshQb cosKa = cos k(a + b) , (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(

P

Ka

)sinKa + cosKa = cos ka , where P ≡ Q2ab

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

KaKa + 1− K2a2

2' 1 . (11)

Then the energy for the lowest energy band is E = ~2K2

2m = ~2Pma2 .

(b) For k = πa , eq. (10) reduces to

(P

Ka

)sinKa + cosKa = −1 . (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(

P

π

)(−δ)− 1 +

δ2

2= −1 . (13)

There are two solutions for δ: δ1 = 0 and δ2 = 2Pπ . Here, δ2 causes a finite energy gap. The total

energy is

E =~2(π + δ)2

2ma2=~2π2

2ma2+~2πδ

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 cos2πx

acos

2πy

a. (15)

In terms of Fourier components, the potential energy is equal to

U(x, y) =∑

G

∑

G′UG,G′e

iGxeiG′y = 4∑

G>0

cosGx∑

G′>0

UG,G′ cosG′y . (16)

The coefficient UG,G′ is found as follows:

UG,G′ =∫ 1

0dx

∫ 1

0dyU(x, y) cos Gx cosG′y . (17)

Here the reciprocal lattice vectors G = G′ are equal to 2πa , since the energy gap we are looking for is

equivalent to the energy difference of two ends of the Brillouin zone, namely (πa , π

a ) and (−πa ,−π

a ).

Actually, the coefficient UG,G′ gives us the potential energy difference between two lattice layers

corresponding to (πa , π

a ) and (−πa ,−π

a ) in k-space, respectively.

UG,G′ is found using the following orthogonality condition,∫ L

0dx cos

2nπx

acos

2mπx

a=

L

2δn,m , (18)

and is equal to UG,G′ = −U .

In 2 dimensions, the central equation is

(λk − ε)C(kx, ky) +∑

G,G′C(kx −G, ky −G′) = 0 . (19)

Here, C(kx, ky) are the coefficients of the Fourier components of the wave function (see Kittel),

and λk ≡ ~2k2

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. ε corresponds

to energy eigenvalues.

At the boundary of 1st Brillouin zone, k = ±πa = ±1

2G. Then the two equations derived from

the central equation are as follows:

(λ− ε)C(12G)− UC(−1

2G) = 0 ,

(λ− ε)C(−12G)− UC(

12G) = 0 .

(20)

The determinant of the 2 × 2 coefficient matrix above gives the two energies, λ1 = ε + U and

λ1 = ε− U . Ultimately, the energy gap is 2U at the corner point of the Brillouin zone.