Phonons

Question 1

Simple 1D Model of Atoms

Description

Answer 1

-atoms with mass M separated by a distance a and connected by springs with spring constant K

Question 2

Simple 1D Model of Atoms

Equation of Motion

Answer 2

-an equation of motion can be written down in terms of the displacement of the nth atom, un:
M d²un/dt² = -2Kun + K(un+! + un-1)
-we therefore expect a time dependence proportional to e^(iωt)
-so:
un’’ = -ω²un

Question 3

Simple 1D Model of Atoms

Difference Equation for Displacements

Answer 3

-sub un’’ = -ω²un into the equation of motion:
-Mω²un = K(un+1 + un-1 - 2un)
-therefore we can expect travelling wave solutions of the form:
u±1 = ue^(inqa)*e^(±iqa)
-where a is the atomic separation distance and q is the phonon wave vector

Question 4

Simple 1D Model of Atoms

Derivation of the Dispersion Relation for Phonons

Answer 4

-substitute the travelling wave solutions for un, un+1 and un-1 into the equation of motion:
-Mω²ue^(inqa) = Ku[e^(i(n+1)qa) + e^(i(n-1)qa) - 2e^(inqa)]
-cancel u*e^(inqa) and use trigonometric double angle formula for cos to find the dispersion relation:
ω = √[4K/M] * sin(qa/2)

Question 5

Phonon Dispersion Relation

Answer 5

ω = √[4K/M] * sin(qa/2)

-where M is the mass of an atom, K is the spring constant, q is the phonon wave vector and a is the interatomic distance

Question 6

What are phonons?

Answer 6

-the normal modes of a collection of harmonic oscillators characterised by wavevectors k and angular frequencies ω, and have a dispersion relation given by;
ω = √[2K/M] * sin(qa/2)

Question 7

Energies of Phonons

Answer 7

-the same as the energies of the quantum harmonic oscillator:
En = (n + 1/2)*ℏω

Question 8

Distribution Function for Phonons

Answer 8

-phonons are bosons so are governed by the Bose-Einstein distribution function;
n(ω) = 1 / [exp(ℏω/kb*T) - 1]
-note that the chemical potential is missing from the numerator of the Boltzmann factor since bosons are not conserved

Question 9

Relationship Between Phonon Wavevector and Wavelength

Answer 9

q = 2π/λ

Question 10

Minimum and Maximum Phonon Wavelengths

Answer 10

• there is no maximum phonon wavelength since it will be as big as the size of the sample of the metal
• the minimum allowed wavelength is the distance between atoms ~3Å, this corresponds to maximum energy

Question 11

Phonon Heat Capacity

Einstein

Answer 11

• the first person to expain heat capacity due to lattic vibrations was Einstein
• he modelled each atom as a harmonic oscillator independent from the others and calculated the occupation of each energy level using a Boltzmann factor and then summed over all levels
• although this works at high temperatures predicting 3R, it predicts a low temperature heat capacity which does not agree to well with experiment
• the problem with this method is that the atoms don’t vibrate independently but collectively

Question 12

Number of States as a Function of Wavevector

Answer 12

-consider a linear chain with spacing L/2π = Na/2π
-and a sphere of radius q for the phonon wave vector
-multiply by three for the three polarisations of the phonons:
Ν(q) = q³V/2π²
-where Ν is greek letter Nu

Question 13

Debye Wavevector Equation

Answer 13

-starting with the number of states in terms of q:
Ν(q) = q³V/2π²
-but Ν=3N, where 3 is the number of degrees of translational freedom and N is the number of atoms
-taking qD to be the Debye wavevector, analogous to kf, we can write:
6π²n = qD³

Question 14

Alternative name for 3N

Answer 14

-3N, the number of translational degrees of freedom multiplied by the number of atoms, is also called the number of normal modes of the system

Question 15

Phonon Density of States Derivation

Answer 15

-start with the number of states in terms of q:
Ν(q) = q³V/2π²
-the velocity of sound is given by vs=ω/q, so:
Ν(q) = V/2π² * ω³/vs³
-to find the density of states, differentiate with respect to ω:
D(ω) = V3ω²/2π² * 1/vs³
-as for electrons the density of states is defined to be the number of states in the range ω -> ω+dω and is written:
D(ω)dω = V3ω²/2π²vs³ dω

Question 16

Phonon Derivation of States

Formula

Answer 16

D(ω)dω = V 3ω²/2π²vs³ dω

-the integral of this is obviously equal to 3N

Question 17

Phonon Total Internal Energy

Answer 17

U(T) = ∫ ℏω D(ω) fb(ω,T) dω
-where the integral is taken between 0 and ωD
-after substitution:
U(T) = 3Vℏ/2π²vs³ * ∫ω³/[e^(ℏω/kbT)-1] dω
-where the integral is taken between 0 and ωD

Question 18

Phonon Heat Capacity

Derivation

Answer 18

C = d/dT U(T)
-sub in for U(T):
C = 3Vℏ/2π²vs³kbT² * ∫[e^(ℏω/kbT)ω^4]/[e^(ℏω/kbT)-1]² dω
-changing variable x=ℏω/kbT we can write:
C = 9Nkb(T/θD)³ *
∫[x^4e^x]/[e^(x)-1]² dx

Question 19

Debye Temperature Formula

Answer 19

θD = ℏvs/kb * (6π²n)^(1/3)

Question 20

Phonon Heat Capacity at Low Temperatures

Answer 20

-at low temperatures (T<

Question 21

Phonon Heat Capacity at High Temperatures

Answer 21

-at high temperatures (T»θD), i.e. x<1 sp the integrand becomes ~x²:
C = 3R
-which is the Dulong-Petit Law

Question 22

Heat Capacity From Electrons and Phonons

Answer 22

C = γT + αT³

• the linear term is the electron contribution and the cubic term is the phonon contribution
• data is normally plotted as C/T against T² giving γ as the y intercept and α as the gradient

Question 23

What is the relationship between the velocity of sound and the Debye temperature

Answer 23

-velocity of sound is proportional to the Debye temperature