Solid State Physics II Lattice Dynamics and Heat Capacity

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1 SEOUL NATIONAL UNIVERSITY SCHOOL OF PHYSICS ssphy2/ SPRING SEMESTER 2004 Chapter 3 Solid State Physics II Lattice Dynamics and Heat Capacity Jaejun Yu jyu@snu.ac.kr jyu/ c 2001, 2002, 2004, Jaejun Yu

2 Lattice Dynamics and Heat Capacity In the previous chapter, we have introduced a model and method for the description of atomic motions about their equilibrium positions. Based on the harmonic approximation (and also the adiabatic Born-Oppenheimer approximation), the equations of motion for the displacements of atoms from their equilibrium position simply becomes those of the coupled harmonic oscillators, which can be completely decoupled by means of introducing the normal modes. In an 1-dimensional periodic solid, a plane-wave state becomes a normal mode with a given quantum number k and dispersion relation ω k. Since any one of these normal modes can gain or lose energy independently of the others, each quantized normal mode of the harmonic oscillators, i.e., phonons, can be described quantum-mechanically by the states: { n k } with ε nk = ω k (n k ) From now on, we can describe all the physical properties related to the lattice dynamics by counting the number of phonon particles.

3 Lattice dynamic model in 3D Dynamical matrix D(q) U({u(R j )}) = 1 2 u(r i ) D(R i, R j ) u(r j ) ij D(R i, R j ) = 2 U({u(R j )}) u(r i ) u(r i ) Equation of motion: Mü(R i ) = j D(R i R j ) u(r j )

5 Solutions for the normal mode: u(r i, t) = e ν exp[i(q R i ω ν (q)t)] D(q) = j D(R i R j )e iq R j D(q) e ν = λ ν (q)e ν j ω ν (q) = λ ν (q)/m

6 Here e ν is called the polarization vector, where the index ν spans all the internal degrees of freedom of atoms in the unit cell. If there are r atoms in the unit cell, ν should be in a 3-dimensional solid. ν = 1, 2,..., 3r In this purpose, we should consider the u as a 3r-dimensional vector. u = (u 11, u 12, u 13,..., u α1, u α2, u α3,..., u r1, u r2, u r3 )

7 Here e ν is called the polarization vector, where the index ν spans all the internal degrees of freedom of atoms in the unit cell. If there are r atoms in the unit cell, ν should be in a 3-dimensional solid. ν = 1, 2,..., 3r In this purpose, we should consider the u as a 3r-dimensional vector. u = (u 11, u 12, u 13,..., u α1, u α2, u α3,..., u r1, u r2, u r3 ) Question:Show that the dynamical matrix of a fcc crystal can be obtained by D(q) = B R n.n. sin 2 ( 1 q R) ˆR ˆR 2

11 Thermal energy of a harmonic oscillator Consider a system of harmonic oscillators in equilibrium with a heat bath at temperature T. The oscillator has a set of accessible states { n } with energy {ε n = ω k (n + 1/2)} (n = 0, 1, 2,..., ). The probability P n of finding the oscillator in a state n With the normalization n P n = 1, The thermal average energy ε(ω, T ) is P n e ε n/k B T P n = e nω/k BT (1 e ω/k BT ) ε(ω, T ) = n=0 ε n P n = ω( e ω/k BT 1 ) Alternatively, the average occupation number n T : n T = 1 e ω/k BT 1

12 Thermal energy of a harmonic oscillator Consider a system of harmonic oscillators in equilibrium with a heat bath at temperature T. The oscillator has a set of accessible states { n } with energy {ε n = ω k (n + 1/2)} (n = 0, 1, 2,..., ). The probability P n of finding the oscillator in a state n P n e ε n/k B T With the normalization n P n = 1, The thermal average energy ε(ω, T ) is ε(ω, T ) = P n = e nω/k BT (1 e ω/k BT ) n=0 ε n P n = ω( e ω/k BT 1 ) Alternatively, the average occupation number n T : n T = 1 e ω/k BT 1 Bose-Einstein statistics

13 How do we determine the probability P n of an oscillator in the state n when the system is in equilibrium with a heat bath at temperature T?

14 How do we determine the probability P n of an oscillator in the state n when the system is in equilibrium with a heat bath at temperature T?

17 Density of states What is the density of state? To find the total energy U of phonons with 3rN-normal modes, we need to sum the average energies over all the nodes: U = k,r ε(k, r; T ) = k,r n(k, r) ω k,r

18 Density of states What is the density of state? To find the total energy U of phonons with 3rN-normal modes, we need to sum the average energies over all the nodes: U = k,r ε(k, r; T ) = k,r n(k, r) ω k,r Here the question is how to evaluate the sum k,r. Since ε(k, r; T ) depends only on ω, it is expected to be convenient to do the evaluation over ω instead of k.

19 Density of states What is the density of state? To find the total energy U of phonons with 3rN-normal modes, we need to sum the average energies over all the nodes: U = k,r ε(k, r; T ) = k,r n(k, r) ω k,r Here the question is how to evaluate the sum k,r. Since ε(k, r; T ) depends only on ω, it is expected to be convenient to do the evaluation over ω instead of k. θ(ω k,r ω)θ(ω + dω ω k,r ) = D(ω)dω k

21 DOS of 1d Systems Starting from the boundary condition: u(la) = u(la + L) e ikl = 1 k = 0, 2π L, 4π L,..., (N 1)2π L N = L/a, that is, by defining k max = π/a. π 2 < k π 2

22 k = ( 2π L ) n = ( 2π L ) with n = 1 Counting the total number of nodes for k < k o, Note that ω(k) = ω( k) N(k o ) = k nθ(k o k ) = L 2π kθ(k o k ) = L 2π k ko k o dk = L π k o D(ω)dω = N(k ω+dω ) N(k ω ) = L π dk ω D(ω) = L π ( ) dω 1 dk

23 DOS of 3d Systems k x = ( 2π L ) n x, k y = ( 2π L ) n y, k z = ( 2π L ) n z

24 In an isotropic system with ω(k) = ω( k ), counting the total number of nodes for k < k o, N(k o ) = k ( n)θ(k o k ) = ( ) 3 L ( 3 k)θ(k o k ) = 2π k ( ) 3 L ko 2π 0 dωk 2 dk = V 2π 2 ko 0 k 2 dk D(ω)dω = N(k ω+dω ) N(k ω ) = D(ω) = V k2 2π 2 For the acoustic branch with ω = v s k, ( ) dω 1 dk V 2π 2 k2 ωdk ω D a (ω) = V k 2 2π 2 = v s V ω 2 2π 2 vs 3 The total density-of-state D(ω) should be D(ω) = ν D ν (ω)

25 When the dispersion is not isotropic any more, i.e., ω(k) ω( k ), the density-ofstate is modified by D(ω) = V (2π) 3 ω(k)=ω,(const) ds ω k ω(k)

26 Phonon density-of-states of Pb metal

27 Debye model for the specific heat capacity The internal energy U is U(T ) = 0 ε(ω, T )D(ω)dω u(t ) = U(T )/V The Debye cutoff frequency is determined by 3rN = 3 V 2π 2 ωd 0 ω 2 v 3 s dω where the factor 3 comes from the three acoustic modes being considered.

28 The specific heat c v (T ) becomes c v (T ) = du(t ) dt = 0 D(ω) d dt ε(ω, T )dω c v (T ) = 9rN V 1 ω 3 D d dt ωd 0 ω 3 dω e ω/k BT 1

29 Introducing the Debye temperature Θ D k B Θ D = ω c v (T ) = 2rNk B V ( T 3 Θ D ) 3 ΘD /T 0 y 4 e y dy (e y 1) 2 (i) k B T > ω D cv(t ) = 3r ( N V ) k B (ii) k B T ω D c v (T ) = 3r ( N V ) k B 4π 4 5 ( T Θ D ) 3

30 Heat capacity of Si and Ge

31 Einstein model for the phonon DOS D(ω) = Nδ(ω ω o ) U(T ) = N n ω o = Nω o e ω/k BT 1 ( ) ω 2 e ω/k BT c v (T ) = Nk B k B T (e ω/kbt 1) 2 ( ) ω 2 Nk B e ω/kbt 0 k B T as T 0

32 Comparison of experimental values of the heat capacity of diamond with the Einstein model

33 Thermal Expansion effects due to anharmonicity linear expansion coefficient α: α = 1 l dl dt volume expansion coefficient α V (isotropic case) α V = 3α = 1 V dv dt Typical values for linear expansion coefficients 10 5 K 1 Thermodynamic pressure p is where F is the free energy p = ( F V ) T F v = k B T ln Z

34 For phonons, the lattice free energy F v can be written by F v = 1 2 ω + k BT ln(1 e ω/k BT ) Thus the total free energy including the potential energy of the lattice at equilibrium position Φ: F = Φ + F v Since ω/ a = 0 for the harmonic oscillator, there is no thermal expansion! When anharmonic terms are included, U(x) = cx 2 gx 3 fx 4 x = 3g 4c 2 k BT

35 Lattice constant of solid Ar as function of T

36 With the force constant d, Φ = Φ o (a o ) d(a a o) 2 the equilibrium condition at a a o can be obtained by ( ) F = 0 a T d(a a o ) + 1 ω ω a ε(ω, T ) = 0 The linear expansion coefficient α: α(t ) = 1 da α o dt = 1 ln ω a 2 od ln a T ε(ω, T ) Grüneisen parameter γ = ln ω ln V

Thermal conductivity In solids, heat is transported by phonons and by free electrons. Electrical insultors are not necessarily poor conductors of heat.

37 Thermal conductivity In solids, heat is transported by phonons and by free electrons. Electrical insultors are not necessarily poor conductors of heat. For metals, it is the electronic contribution that dominates the thermal conductivity. But, some of the insulators like Al 2 O 3 and SiO 2 is a good heat conductor. Heat conduction by phonons Thermal current dentiy j Q : j Q = κ T

38 In a simple kinetic theory, j xq = ncv x T = ncv x l x dt dx = 1 3 n v2 cτ dt dx = 1 3 c vvl dt dx where the specific heat c v = nc, the mean free path l = vτ, and the thermal gradient Thermal conductivity κ: T = dt dx l x = dt dx v xτ κ = 1 3 c vvl

39 Temperature dependence of κ (i) At low temperature (< 10 K), the scattering processes with q-conservation does not influence the thermal conductivity. Thus, the dominant scattering processes are crystal defects l const. surface scattering κ c v (T ) T 3 (ii) At higher (intermediate) temperatures, Umklapp processes intervenes in the scattering process (no q-conservation) mean free path l e Θ/bT due to the activation energy of k B Θ/b κ c v (T )e Θ/bT

40 (iii) At high temperature, mean free path drops l T 1 (random scattering with all the excited modes?) specific heat also saturates c v 3Nk B κ 1 T

41 Thermal conductivity of NaF

Phonons II. Thermal Properties

Chapter 5. Phonons II. Thermal Properties Thermal properties of phonons As mentioned before, we are now going to look at how what we know about phonons will lead us to a description of the heat capacity