Download PHY 140A: Solid State Physics Solution to Homework #5

PHY 140A: Solid State Physics
Solution to Homework #5
Xun Jia1
November 16, 2006
1 Email:
[email protected]
Fall 2006
c Xun Jia (November 16, 2006)
°
Physics 140A
Problem #1
Temperature dependence of the electrical resistivity of a metal. The electrical resistivity ρ of a metal is proportional to the probability that an electron is
scattered by the vibrating atoms in the lattice, and this probability is in turn proportional to the mean square amplitude of vibration of these atoms. How does the
electrical resistivity ρ of the metal depend on its absolute temperature in the range
near room temperature, or above, where classical statistical mechanics can validly be
applied to discuss the vibrations of the atoms in the metal?
Solution:
As indicated by the problem, the electrical resistivity ρ of a metal is proportional
to the probability p that an electron is scattered by the vibrating atoms in the lattice,
and this probability p is proportional to the mean square amplitude of vibration of
these atoms hA2 i. Since hA2 i is proportional to the average of potential energy hEp i,
which is proportional to T due to equipartition theorem. Thus, the the electrical
resistivity ρ is proportional to T .
Problem #2
Specific heat of anharmonic oscillators. Consider a one-dimensional oscillator
(not simple harmonic) which is described by a position coordinate x and by a mop2
mentum p and whose energy is given by ² =
+ bx4 , where the first term on the
2m
right is its kinetic energy and th second term is its potential energy. Here m denotes
the mass of the oscillator and b is some constant. Suppose that this oscillator is in
thermal equilibrium with a heat reservoir at a temperature T high enough so that
the approximation of classical mechanics is a good one. (Hint: there is no need to
evaluate explicitly any integral to answer these questions.)
(a). What is the mean kinetic energy of this oscillator?
(b). What is its mean potential energy?
(c). What is its mean total energy?
(d). Consider an assembly of weakly interacting particles, each vibrating in one
dimension so that its energy is given as above. What is the specific heart at
constant volume per mole of these particles?
Solution:
1
(a). The mean kinetic energy of this oscillator is hEk i = kT due to equipartition
2
theorem.
1
Fall 2006
c Xun Jia (November 16, 2006)
°
Physics 140A
(b). From Boltzmann statistics, the mean potential energy is:
+∞
RR
hEp i =
p2
R∞
4]
dpdx bx4 e−β[ 2m +bx
−∞
+∞
RR
=
p2
4]
dpdx e−β[ 2m +bx
dx bx4 e−βbx
−∞
R∞
4
dx e−βbx4
−∞
−∞
R∞
∂
4
 ∞

dx e−βbx
Z
∂β −∞
∂
4
=
= − ln
dx e−βbx 
R∞
∂β
dx e−βbx4
−∞
−∞
 



Z∞
Z∞
1/4
∂
∂
1
4
4
t=β
x
−−−−→ − ln β −1/4 dt e−bt  = − − ln β + ln dt e−bt 
∂β
∂β
4
−
∞
∞
1
1
=
= kT
4β
4
(1)
Another way of evaluating this is to use integral by parts to calculating the
integral directly:
R∞
hEp i =
dx bx4 e−βbx
4
−∞
R∞
dx e−βbx4
−∞


¯∞
Z∞
¯
1
1
1
1
4
 x e−βbx4 ¯
+
= kT
= R∞
dx e−βbx  =
¯
−4β
4β
4β
4
−∞
dx e−βbx4
−∞
(2)
−∞
(c). Thus the mean total energy is:
3
hEi = hEk i + hEp i = kT
4
(3)
(d). For the weakly interacting particles, the internal energy per mole is just the
sum of each particles, i.e.:
3
3
U = NA hEi = kNA T = RT
4
4
(4)
where NA is the Avogadro constant. Thus the specific heat at constant volume
per mole is:
µ
¶
∂U
3
cV =
= R
(5)
∂T V
4
2
Fall 2006
c Xun Jia (November 16, 2006)
°
Physics 140A
Remark: A general equipartition theorem states that: in thermal equilibrium, the
1
mean value of each term in the energy proportional to ζ 2n is equal to kT , where ζ
2n
could be any independent variable, for example, momentum p or coordinate x. This is
not hard to proof. Note here the exponent is 2n, which ensures the energy is bounded
from below, as it should be in any physical context.
Problem #3
Specific heat of a highly anisotropic solid. Consider a solid which has a highly
anisotropic crystalline layer structure. Each atom in this structure can be regarded
as performing simple harmonic oscillations in three dimensions. The restoring forces
in directions parallel to a layer are very large; hence the natural frequencies of oscillations in the x and y direction lying within the plane of a layer are both equal to a
value ωk which is so large that h̄ωk À 300k, the thermal energy kT at room temperature. On the other hand, the restoring force perpendicular to a layer is quite small;
hence the frequency of oscillation ω⊥ of an atom in the z direction perpendicular to
a layer is so small that h̄ω⊥ ¿ 300k. On the basis of this model, what is the molar
specific heat (at constant volume) of this solid at 300K?
Solution:
Since h̄ωk À 300k, at room temperature T = 300K, the thermal fluctuation is
not high enough to excite the degrees of freedom within the plane, i.e. the degrees
of freedom associating with h̄ωk is frozen. On the other hand, 300k À h̄ω⊥ , the solid
atom can only move along the direction perpendicular to the plane.
Hence, from equipartition theorem, the average kinetic and potential energy per
1
atom are both kT , and then the average total energy is kT , which therefore gives a
2
specific heat per atom of cV = k, or equivalently, cV = kNA = R is the molar specific
heat.
Problem #4
Kittel 5.1 Singularity in the density of states.(Ashcroft and Mermin 23.3, essentially the same problem, explains that this is called a ”van Hove singularity”.) (a)
From the dispersion relation derived in Chapter 4 for a monatomic linear lattice of N
atoms with nearest-neighbor interactions, show that the density of modes is:
D(ω) =
1
2N
· 2
π (ωm − ω 2 )1/2
where ωm is the maximum frequency. (b) Suppose that an optical phonon branch
has the form ω(K) = ω0 − AK 2 , near K = 0 in three dimensions, Show that
D(ω) = (L/2π)3 (2π/A3/2 )(ω0 − ω)1/2 for ω < ω0 and D(ω) = 0 for ω > ω0 . Here the
3
Fall 2006
c Xun Jia (November 16, 2006)
°
Physics 140A
density of modes is discontinuous.
Solution:
(a). A general expression of D(ω) in d dimension lattice is (a generation of Eqn. (35)
in Kittel):
µ ¶d Z
L
D(ω)dω =
dd K
(6)
2π
shell
where the ”shell” refers to a region in K space bounded by two surfaces on
which the phonon frequency are ω and ω + dω.
R
In 1-D, the ”shell” is just composed by two points, and hence shell dd K =
|2dK|. Therefore, from above equation:
¯
µ ¶ ¯
¯ dK ¯
L
¯
¯
D(ω) =
·¯
(7)
π
dω ¯
From the dispersion relation of a monatomic linear chain, we have:
¯
¯
¯ dK ¯
1
2
1
¯
¯
¯=
¯= ¯
¯ dω ¯ = ¯¯
2
¯
¯
¯
¡1
¢¯ a(ωm − ω 2 )1/2
¯ dω ¯ ¯ a
¯
¯ ¯ ωm cos 2 Ka ¯
¯ dK ¯ ¯ 2
¯
(8)
thus, from above equations, note that N = L/a, it follows that:
D(ω) =
1
2N
· 2
π (ωm − ω 2 )1/2
(9)
(b). For the optical branch in 3-D case, ω(K) = ω0 − AK 2 . Now the ”shell” is
2
Rindeed da spherical 2shell with thickness of dK and a surface area of 4πK , thus
shell d K = |4πK dK|. Therefore from (6), we have:
¯
¯
µ ¶3
¯
¯
L
2 ¯ dK ¯
D(ω) =
(10)
· 4πK ¯
2π
dω ¯
From the dispersion relation:
¯
¯
¯ dK ¯
1
1
1
¯
¯
¯=
=
¯ dω ¯ = ¯¯
1/2
1/2
¯
¯ dω ¯ | − 2AK| 2A (ω0 − ω)
¯
¯
¯ dK ¯
thus, we obtain the density of states:
µ ¶3 µ
¶
L
2π
D(ω) =
(ω0 − ω)1/2
3/2
2π
A
(11)
(12)
However, this expression is only correct for ω < ω0 . From the dispersion relation
ω(K) = ω0 − AK 2 , there is no K states such that ω > ω0 , then D(ω) = 0 when
ω > ω0 . Therefore, there is a discontinuity at ω = ω0 .
4
Fall 2006
c Xun Jia (November 16, 2006)
°
Physics 140A
Problem #5
Kittel 5.4 Heat capacity of a layer lattice in the Debye approximation (a)
Consider a dielectric crystal made up of layers of atoms, with rigid coupling between
layers so that the motion of the atoms is restricted to the plane of the layer. Show
that the phonon heat capacity in the Debye approximation in the low temperature
limit is proportional to T 2 . (b) Suppose instead, as in many layer structures, that
adjacent layers are very weakly bound to each other. What form would you expect
the phonon heat capacity to approach at extremely low temperature?
Solution:
(a). For Debye model in this 2-D system, the density of states can be obtained from
the dispersion relation ω = cK:
µ ¶2
dK
L2 ω
L
2πK
=
(13)
D(ω) =
2π
dω
2πc2
thus the internal energy is given by:
µ 2 ¶µ
¶
Z
Z ωD
Lω
h̄ω
dω
U = dω D(ω)hn(ω)ih̄ω =
h̄ω
2πc2
e kT − 1
0
¶
µ
Z
xD
x2
k3T 3
L2
dx
= 2
2πc2
ex − 1
h̄
0
(14)
where x = h̄ω/kT , and xD = h̄ωD /kT . In the low temperature limit, we can
set xD → ∞, therefore:
µ 2 ¶Z ∞
L
x2
k3T 3
dx x
(15)
U= 2
2πc2
e −1
h̄
0
since U ∝ T 3 , the heat capacity in the low temperature will be:
µ
¶
∂U
CV =
∝ T2
∂T V
(16)
(b). In this case, since the coupling in between layers are weaker than the coupling
inside the layers, at extremely low temperatures, the degrees of freedom for the
motion inside the layers will be frozen, and the motion of the atoms is restricted
in the direction perpendicular to the layers. Therefore the system becomes 1-D.
Follow a similar calculation as in part (a), we can find the heat capacity for 1-D
system in low temperature will be proportional to T .
Remark: A general statement is: for a d dimensional system with a dispersion
relation in the form ω = AK α , the heat capacity of the system in low temperature
limit will be proportional to T d/α . You can easily prove this statement in the way
similar as in above.
5