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PHYSICS 5520 (SPRING 2011); HOMEWORK #1 SOLUTIONS

Problem 1; structure of diamond (10 points)

The primitive basis of the diamond structure has identical atoms at x1 = y1 = z1 = 0 and x2 = y2 = z2 = 1/4(Kittel, page 16). The structure factor becomes

S =∑

j

fj exp[−2πi(v1xj + v2yj + v3zj)] = f[

1 − e−i π

2(v1+v2+v3)

]

. (1)

The value of S is zero when v1 + v2 + v3 = 2(2n + 1), and is 2f when v1 + v2 + v3 = 4n, with n integer. The lastrelation yields allowed reflections of the diamond structure.

Problem 2; fcc tight-binding expression (15 points total)

a) The tight-binding of an s-band in a fcc structure is (Ashcroft and Mermin Eq. 10.22)

ε(k) = Es − β − 4γ

[

cos

(

1

2kxa

)

cos

(

1

2kya

)

+ cos

(

1

2kya

)

cos

(

1

2kza

)

+ cos

(

1

2kza

)

cos

(

1

2kxa

)]

. (2)

i) Along ΓX; inserting ky = kz = 0, kx = 2πa

µ, 0 ≤ µ ≤ 1 into Eq. (2) yields

ε = Es − β − 4γ [1 + 2 cos πµ] .

ii) Along ΓL; inserting (kx = ky = kz = 2πa

µ, 0 ≤ µ ≤ 1/2 into Eq. (2) yields

ε = Es − β − 12γ cos2 πµ.

iii) Along ΓK, plugging kz = 0, kx = ky = 2πa

µ, 0 ≤ µ ≤ 3/4 into Eq. (2) yields

ε = Es − β − 4γ[

cos2 πµ + 2 cos πµ]

.

iv) Along ΓW ; plugging kz = 0, kx = 2πa

µ, ky = πaµ, 0 ≤ µ ≤ 1 into Eq. (2) yields

ε = Es − β − 4γ [cos πµ cos (πµ/2) + cos πµ + cos (πµ/2)] .

b) The gradient of the energy reads

∇ε(k) = 2γa sin

(

1

2kxa

)(

cos

(

1

2kya

)

+ cos

(

1

2kza

))

x

+2γa sin

(

1

2kya

)(

cos

(

1

2kxa

)

+ cos

(

1

2kza

))

y

+2γa sin

(

1

2kza

)(

cos

(

1

2kxa

)

+ cos

(

1

2kya

))

z. (3)

The square face which contains the point X is specified by relation kx = 2πa

. The derivative along the normal to thisfaces, which is along ΓX and thus parallel to x, is thus

2γa sin (π)

(

cos

(

1

2kya

)

+ cos

(

1

2kza

))

= 0

c) The derivative along the normal to the hexagonal face containing the point L is along ΓL and thus along (1, 1, 1).It is thus given by

2γa√3

[

sin

(

1

2(kx + ky)a

)

+ sin

(

1

2(kx + kz)a

)

+ sin

(

1

2(ky + kz)a

)]

.

It is easy to see that its value is zero along the line WL, which is given by πa(1, 1, 1)+β(π/a, 0,−π/a) with 0 ≤ β ≤ 1.

The fact that the above derivative vanishes along all six diagonals of the hexagons follows from the symmetry.

2

Problem 3; one dimensional crystal (10 points total)

Let the electron dispersion, E(k), is given by

E(k) =~

2

mea2

[

7

8− cos ka +

1

8cos 2ka

]

. (4)

(i) In tight binding approximation one has

E(k) = E0 + γnn

∑

R,nn

coskR + γnnn

∑

R,nnn

coskR + · · ·

Then from Eq. (4) it follows that two nearest neighbor shells are involved.(ii)(iii) By definition, m∗(k) = ~

2(d2E/dk2)−1. At the bottom of the band, k = 0, one finds m∗ = 2me.At the top of the band, k = ±π/a, one gets m∗ = −2me/3.(iv)

dv(k)

dt=

F

m∗(k);

1

~

d

dt

(

dE(k)

dk

)

=1

~2

(

d2E

dk2

)

d(~k)

dt=

1

~2

(

d2E

dk2

)

F =F

m∗(k)⇒ m∗(k) =

~2

d2E/dk2

Problem 4; gap as a function of temperature (5 points)

The temperature dependence of the gap is given by

Eg = Eg(0) −αT 2

β + T.

with the given numbers one finds

Eg = 1.17 − 4.73 × 10−4T 2

636 + T(eV).

At T = 300K one finds

Eg = 1.125 eV.