Solid State Physics
Homework Set 5
Solutions
a. Starting from the equations of motion
M ω 2 = Ak 2 ⇒ k 2 =
M 2
ω
A
Applying the BC to the free gas in a 2-D box of length L
µ ¶2
2π
L2 k 2
M L2 ω 2
∂N
M L2 ω
N = πk 2 ⇒ N =
=
⇒ g(ω) =
=
L
4π
4πA
∂w
2πA
b. For specific heat C = ∂U/∂T we need to calculate the internal energy U
Z ωD
Z ωD
Z
M L2 ω
ω2
1
M L2 h̄ ωD
dω
dωg(ω) < n(ω) > h̄ω =
dω
h̄ω
=
U=
2πA eβh̄ω − 1
2πA 0
eβh̄ω − 1
0
0
at low T limit where ωD → ∞, we can solve the integral, by substituting x = βh̄ω the above integral
can be reduced to
Z
∞ Z
∞
Γ(3) X 1
x2
dω ∞
1 X ∞
ζ(3)Γ(3)
2.0404
dxx2 e−sx =
dx x
=
=
=
dx 0
e −1
βh̄ s=1 0
βh̄ s=1 s3
βh̄
βh̄
where Γ(3) = Γ(2 + 1) = 2! = 2 and ζ(3) == 1.202. Then the U can be written as
U=
2.0404M L2 k 3 T 3
7.212M L2 k 3 T 2
⇒
C
=
2πAh̄2
2πAh̄2
c. For Debye-Waller factor
−
f (T ) = e h
P
s
<ns |(k.u)2 |ns >i
T
Assuming that k of the gamma ray is uncorrelated with Ū (0)
Z
1
< (k.u)2 >= k 2 /2 < u2 > as
dθcos2 θ = 1/2
2π
and
< u2 (0) >=
X
s
For small ω the integrand is
d. Corrugation potential
1
βω
h̄
2M N ωs
µ
2
+1
βω
e −1
¶
∼
Z
dω
µ
2
+1
βω
e −1
¶
+ 1, thus the integral diverges logarithmically and f (T ) is then zero.
Φ=
k X
∂2Φ
uα (r) α 0β uβ (r0 )
2
∂r r
0
α,βr,r
1
where
∂2Φ
= δαβ δrr0 k
∂rα r0β
thus Dαβ (r − r 0 ) is modified as
Dαβ (r) = δr,0 δαβ 2c − crα rβ /d2 + δαβ δr,0 K

 D11 (k) = 2c − 2c cos(kx a) + k ' ckx2 + K
X
−ik.r
D22 (k) = −2c cos(ky a − 1) + k ' cky2 + K
Dαβ (k) =
e
Dαβ (r) ⇒

r
D33 (k) = K
p
Thus clearly for k = 0 ω = k/m
e. For
p T → 0 only low frequency modes h̄ω ∼ kT are excited thus the dominant mode will be
ω = k/M and we will have an Einstein specific heat
½
h̄ω
U = N eh̄ω/kT
−1
p
ω = k/M
⇒ C = N k (h̄ω/kT )
2
eh̄ω/kT
' e−h̄ω/kT /T 2 ' e−h̄ω/kT
(eh̄ω/kT − 1)2
f. Here the Debye Waller factor is finite as the oscillations are now 3 dimensional. Again considering
the Einstein limit
U=
h̄ω
eh̄ω/kT
−1
=< T > + < V >= 2 < T >= K < x2 >
where we used Viral theorem
< x2 >= √
h̄
kM (eh̄ω/kT
which is finite.
2
− 1)