PHY 140A: Solid State Physics

Solution to Homework #5

Xun Jia1

November 16, 2006

1Email: jiaxun@physics.ucla.edu

Fall 2006 Physics 140A c© Xun Jia (November 16, 2006)

Problem #1

Temperature dependence of the electrical resistivity of a metal. The elec-trical resistivity ρ of a metal is proportional to the probability that an electron isscattered by the vibrating atoms in the lattice, and this probability is in turn pro-portional to the mean square amplitude of vibration of these atoms. How does theelectrical resistivity ρ of the metal depend on its absolute temperature in the rangenear room temperature, or above, where classical statistical mechanics can validly beapplied 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 ofthese atoms 〈A2〉. Since 〈A2〉 is proportional to the average of potential energy 〈Ep〉,which is proportional to T due to equipartition theorem. Thus, the the electricalresistivity 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 mo-

mentum p and whose energy is given by ε =p2

2m+ bx4, where the first term on the

right is its kinetic energy and th second term is its potential energy. Here m denotesthe mass of the oscillator and b is some constant. Suppose that this oscillator is inthermal equilibrium with a heat reservoir at a temperature T high enough so thatthe approximation of classical mechanics is a good one. (Hint: there is no need toevaluate 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 onedimension so that its energy is given as above. What is the specific heart atconstant volume per mole of these particles?

Solution:

(a). The mean kinetic energy of this oscillator is 〈Ek〉 =1

2kT due to equipartition

theorem.

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Fall 2006 Physics 140A c© Xun Jia (November 16, 2006)

(b). From Boltzmann statistics, the mean potential energy is:

〈Ep〉 =

+∞∫∫−∞

dpdx bx4e−β[ p2

2m+bx4]

+∞∫∫−∞

dpdx e−β[ p2

2m+bx4]

=

∞∫−∞

dx bx4e−βbx4

∞∫−∞

dx e−βbx4

=

− ∂

∂β

∞∫−∞

dx e−βbx4

∞∫−∞

dx e−βbx4

= − ∂

∂β

ln

∞∫

−∞

dx e−βbx4

t=β1/4x−−−−→ − ∂

∂β

ln

β−1/4

∞∫

∞

dt e−bt4

= − ∂

∂β

−1

4ln β + ln

∞∫

∞

dt e−bt4

=1

4β=

1

4kT

(1)

Another way of evaluating this is to use integral by parts to calculating theintegral directly:

〈Ep〉 =

∞∫−∞

dx bx4e−βbx4

∞∫−∞

dx e−βbx4

=1

∞∫−∞

dx e−βbx4

x

−4βe−βbx4

∣∣∣∣∞

−∞+

1

4β

∞∫

−∞

dx e−βbx4

=

1

4β=

1

4kT

(2)

(c). Thus the mean total energy is:

〈E〉 = 〈Ek〉+ 〈Ep〉 =3

4kT (3)

(d). For the weakly interacting particles, the internal energy per mole is just thesum of each particles, i.e.:

U = NA〈E〉 =3

4kNAT =

3

4RT (4)

where NA is the Avogadro constant. Thus the specific heat at constant volumeper mole is:

cV =

(∂U

∂T

)

V

=3

4R (5)

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Fall 2006 Physics 140A c© Xun Jia (November 16, 2006)

Remark: A general equipartition theorem states that: in thermal equilibrium, the

mean value of each term in the energy proportional to ζ2n is equal to1

2nkT , where ζ

could be any independent variable, for example, momentum p or coordinate x. This isnot hard to proof. Note here the exponent is 2n, which ensures the energy is boundedfrom below, as it should be in any physical context.

Problem #3

Specific heat of a highly anisotropic solid. Consider a solid which has a highlyanisotropic crystalline layer structure. Each atom in this structure can be regardedas performing simple harmonic oscillations in three dimensions. The restoring forcesin directions parallel to a layer are very large; hence the natural frequencies of oscil-lations in the x and y direction lying within the plane of a layer are both equal to avalue ω‖ which is so large that h̄ω‖ À 300k, the thermal energy kT at room temper-ature. 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 toa layer is so small that h̄ω⊥ ¿ 300k. On the basis of this model, what is the molarspecific heat (at constant volume) of this solid at 300K?

Solution:Since h̄ω‖ 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 degreesof freedom associating with h̄ω‖ is frozen. On the other hand, 300k À h̄ω⊥, the solidatom can only move along the direction perpendicular to the plane.

Hence, from equipartition theorem, the average kinetic and potential energy per

atom are both1

2kT , and then the average total energy is kT , which therefore gives a

specific heat per atom of cV = k, or equivalently, cV = kNA = R is the molar specificheat.

Problem #4

Kittel 5.1 Singularity in the density of states.(Ashcroft and Mermin 23.3, essen-tially 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 Natoms with nearest-neighbor interactions, show that the density of modes is:

D(ω) =2N

π· 1

(ω2m − ω2)1/2

where ωm is the maximum frequency. (b) Suppose that an optical phonon branchhas the form ω(K) = ω0 − AK2, near K = 0 in three dimensions, Show thatD(ω) = (L/2π)3(2π/A3/2)(ω0 − ω)1/2 for ω < ω0 and D(ω) = 0 for ω > ω0. Here the

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Fall 2006 Physics 140A c© Xun Jia (November 16, 2006)

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):

D(ω)dω =

(L

2π

)d ∫

shellddK (6)

where the ”shell” refers to a region in K space bounded by two surfaces onwhich the phonon frequency are ω and ω + dω.

In 1-D, the ”shell” is just composed by two points, and hence∫shell d

dK =|2dK|. Therefore, from above equation:

D(ω) =

(L

π

)·∣∣∣∣dK

dω

∣∣∣∣ (7)

From the dispersion relation of a monatomic linear chain, we have:∣∣∣∣dK

dω

∣∣∣∣ =1∣∣∣∣∣dω

dK

∣∣∣∣∣

=1∣∣∣∣∣

a

2ωm cos

(12Ka

)∣∣∣∣∣

=2

a(ω2m − ω2)1/2

(8)

thus, from above equations, note that N = L/a, it follows that:

D(ω) =2N

π· 1

(ω2m − ω2)1/2

(9)

(b). For the optical branch in 3-D case, ω(K) = ω0 − AK2. Now the ”shell” isindeed a spherical shell with thickness of dK and a surface area of 4πK2, thus∫shell d

dK = |4πK2dK|. Therefore from (6), we have:

D(ω) =

(L

2π

)3

· 4πK2

∣∣∣∣dK

dω

∣∣∣∣ (10)

From the dispersion relation:∣∣∣∣dK

dω

∣∣∣∣ =1∣∣∣∣∣dω

dK

∣∣∣∣∣

=1

| − 2AK| =1

2A1/2(ω0 − ω)1/2(11)

thus, we obtain the density of states:

D(ω) =

(L

2π

)3 (2π

A3/2

)(ω0 − ω)1/2 (12)

However, this expression is only correct for ω < ω0. From the dispersion relationω(K) = ω0−AK2, there is no K states such that ω > ω0, then D(ω) = 0 whenω > ω0. Therefore, there is a discontinuity at ω = ω0.

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Fall 2006 Physics 140A c© Xun Jia (November 16, 2006)

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 betweenlayers so that the motion of the atoms is restricted to the plane of the layer. Showthat the phonon heat capacity in the Debye approximation in the low temperaturelimit is proportional to T 2. (b) Suppose instead, as in many layer structures, thatadjacent layers are very weakly bound to each other. What form would you expectthe 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 fromthe dispersion relation ω = cK:

D(ω) =

(L

2π

)2

2πKdK

dω=

L2ω

2πc2(13)

thus the internal energy is given by:

U =

∫dω D(ω)〈n(ω)〉h̄ω =

∫ ωD

0

dω

(L2ω

2πc2

)(h̄ω

eh̄ωkT − 1

)

=k3T 3

h̄2

(L2

2πc2

) ∫ xD

0

dxx2

ex − 1

(14)

where x = h̄ω/kT , and xD = h̄ωD/kT . In the low temperature limit, we canset xD →∞, therefore:

U =k3T 3

h̄2

(L2

2πc2

) ∫ ∞

0

dxx2

ex − 1(15)

since U ∝ T 3, the heat capacity in the low temperature will be:

CV =

(∂U

∂T

)

V

∝ T 2 (16)

(b). In this case, since the coupling in between layers are weaker than the couplinginside the layers, at extremely low temperatures, the degrees of freedom for themotion inside the layers will be frozen, and the motion of the atoms is restrictedin 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-Dsystem in low temperature will be proportional to T .

Remark: A general statement is: for a d dimensional system with a dispersionrelation in the form ω = AKα, the heat capacity of the system in low temperaturelimit will be proportional to T d/α. You can easily prove this statement in the waysimilar as in above.

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