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Chap 13 Phonons • classical theory of vibration • 1-dim, 3-dim • quantum theory of vibration • phonon specific heat • Einstein model, Debye model • thermal expansion • neutron scattering Dept of Phys M.C. Chang One dimensional vibration • consider only longitudinal motion • consider only NN coupling d 2un M = α (un +1 − un ) − α (un − un −1 ) 2 dt PBC: Assume un = Aei ( kX n −ωt ) , where X n = na , then we'll get M ( −ω 2 )eikna = −α 2eikna − eik ( n +1) a − eik ( n −1) a , which leads to ω (k )=ωM sin(ka / 2) , ωM = 2 α / M ω (k )=ωM sin(ka / 2) Dispersion curve ω λ < 2a λ > 2a λ > 2a λ < 2a (redundant) k -π/a π/a The waves with wave numbers k and k+2π/a describe the same atomic displacement Therefore, we can restrict k to within the first BZ [-π/a, π/a] Displacement of the n-th atom un (t ) = Aei ( kX n −ωt ) , X n = na Pattern of vibration: • k ~ 0, exp(ikXn) ~ 1. Every atom move in unison. Little restoring force. • k ~ π/a,exp(ikXn) ~ (-1)n. Adjacent atoms move in opposite directions. Maximum restoring force. Velocity of wave: • k ~ 0, ω = (ωMa/2)k Linear dispersion, phase velocity = group velocity • k ~ π/a, group velocity ~ 0 PBC Number of “normal modes”: u N = u0 ⇒ exp(ikNa) = 1 m 2π △k=2π/Na ∴k = N a , m∈Z Within the 1st BZ, there are N k-points. Each k describes a normal mode of vibration (i.e. a vibration with a specific frequency) Vibration of a crystal with 2 atoms in a unit cell a d 2u2 n +1 = −α (2u2 n +1 − u2 n − u2 n + 2 ), M2 2 dt d 2u2 n + 2 = −α (2u2 n + 2 − u2 n +1 − u2 n +3 ). M1 dt 2 ⎛ u2 n +1 ⎞ ⎛ A1eikX 2 n+1 ⎞ − iωt X 2 n +1 = (2n + 1)a / 2 Assume ⎜ e , ⎟=⎜ ikX 2 n+ 2 ⎟ X 2 n + 2 = (n + 1)a ⎝ u2 n + 2 ⎠ ⎝ A2 e ⎠ ⇒ ⎛ 2α − M 2ω 2 −2α cos(ka / 2) ⎞ ⎛ A1 ⎞ ⎜ ⎟ ⎜ ⎟ = 0, 2 2α − M 1ω ⎠ ⎝ A2 ⎠ ⎝ −2α cos(ka / 2) ⇒ ⎛ 2α − M 2ω 2 −2α cos(ka / 2) ⎞ det ⎜ ⎟ = 0. 2 2 α cos( ka / 2) 2 α M ω − − 1 ⎝ ⎠ 2 ⎛ 1 ⎛ 1 1 ⎞ 1 ⎞ 4sin 2 (ka / 2) 2 ⇒ ω± = α ⎜ . + + ⎟ ±α ⎜ ⎟ − M M M M M M ⎝ 1 ⎝ 1 2 ⎠ 2 ⎠ 1 2 Two branches of dispersion curves (assume M2 > M1) a d c b Patterns of vibration similar See a nice demo at http://dept.kent.edu/projects/ksuviz/leeviz/phonon/phonon.html Three dimensional vibration Along a given direction of propagation, there are 1 longitudinal wave and 2 transverse waves, each may have different velocities Sodium (bcc) 3D crystal with atom basis FCC lattice with 2-atom basis cm-1 Rules of thumb • For a 3-dim crystal, if each unit cell has p atoms, then there are 3 acoustic branches, 3(p-1) optical branches • If a crystal has N unit cells, then each branch has N normal modes。 • As a result, the total number of normal modes are 3pN (= total DOF of this system) Quantum theory of vibration Review: 1D simple harmonic oscillator (DOF=1) p2 α 2 H= + x 2m 2 • Classically, it oscillates with a single freq ω=(α/m)1/2 • The energy ε can be continuously changed. Quantization: [ x, p ] = i 1 ⎛ i ⎞ ω + m x p ⎜ ⎟ 2 ⎝ mω ⎠ define a= then ⎡⎣ a, a ⎤⎦ = 1 1⎞ ⎛ H = ⎜ a†a + ⎟ ω 2⎠ ⎝ 1⎞ ⎛ H n = ⎜n+ ⎟ ω n 2⎠ ⎝ † Note: a n = n n −1 a† n = n + 1 n + 1 a†a n = n n 1⎞ ⎛ • After quantization, the energy becomes discrete ε n = ⎜ n + ⎟ ω 2⎠ ⎝ • The number n of energy quanta depends on the amplitude of the oscillator. Quantization of a 1-dim vibrating lattice (DOF=N) ⎡ p2 α H = ∑⎢ + ( u +1 − u 2 =1 ⎣ 2m N ) 2 ⎤ ⎥ ⎦ • Classically, for a given k, it vibrates with a single frequency ω(k). The amplitude ( and hence energy ε) can be continuously changed. Quantization: [u , p ' ] = i Fourier transf. u = then δ ' 1 eik a uk ∑ N k 1 p = eik a pk ∑ N k ⎡⎣uk , pk† ' ⎤⎦ = i δ kk ' ⎛ 1 † mωk2 † ⎞ ⇒ H = ∑⎜ pk pk + uk uk ⎟ 2 2 m k ⎝ ⎠ Note: u† = u ; → uk† = u− k ; p† = p pk† = p− k k=2πm/L, L=Na. A collection of N independent oscillators ! Similarly, define 1 ak = 2 ⎛ ⎞ i pk ⎟ ⎜ mωk uk + ⎜ ⎟ mωk ⎝ ⎠ → ⎡⎣ ak , ak†' ⎤⎦ = δ kk ' 1⎞ ⎛ → H = ∑ ⎜ ak† ak + ⎟ ωk 2⎠ k ⎝ eigenstate n1 , n2 ...nk ,... 1⎞ ⎛ H n1 , n2 ...nk ,... = ∑ ⎜ nk + ⎟ ωk n1 , n2 ...nk ,... 2⎠ k ⎝ • The number of energy quanta (called phonons) for the k-mode is nk. • There are no interaction between phonons → “free” phonon gas. • If there are p-atoms in a unit cell (p branches), then the total vibrational energy of the lattice is 1⎞ ⎛ U = ∑∑ ⎜ nk , s + ⎟ ωk , s 2⎠ k s =1 ⎝ p Specific heat: experimental fact Specific heat approaches 3R (per mole) at high temperature (Dulong-Petit law) Specific heat drops to zero at low temperature After rescaling the temperature by θ (Debye temperature), which differs from material to material, a universal behavior emerges: Debye temperature In general, a harder material has a higher Debye temperature Specific heat: theoretical framework • Internal energy U of a crystal is the summation of vibrational energies (consider an insulator so there’s no electronic energies) U (T ) = ∑ (nk , s + 1/ 2) ωk , s , s =L/T, A/O... k ,s • For a crystal in thermal equilibrium, the average phonon number is nk , s = • Therefore, • Specific heat 1 e ωk ,s / kT ⎛ U (T )= ∑ ⎜ k ,s ⎝ e −1 1 ωk ,s / kT Bose-Einstein distribution 1⎞ + ⎟ ωk ,s −1 2 ⎠ CV = ( ∂U / ∂T )V • Density of states (similar to electron energy band) ∑f k k =V ∫ d 3k ( 2π ) ⎛ L ⎞ D (ω ) = ⎜ ⎟ ⎝ 2π ⎠ 3 ∫ 3 f k = ∫ dω D(ω ) fω dSω ∇k ω Ex: In 3D if ω = vk , then D(ω ) = V ω 2 / 2π 2 v 3 Einstein model (1907) Assume that each atom vibrates independently of each other, and every atom has the same vibration frequency ω0 DOS D(ω ) = 3N δ (ω − ω0 ) 3 dim × number of atoms ω0 ω0 1⎞ ⎛ + 3N U = 3N ⎜ n + ⎟ ω0 = 3 N 2⎠ exp( ω0 / kT ) − 1 2 ⎝ e ω0 / kT ⎛ ω0 ⎞ CV = (∂U / ∂T )V = 3Nk ⎜ ⎟ ⎝ kT ⎠ e ω0 / kT − 1 2 ( ≈ e− ω0 / kT as T → 0 K (Activation behavior) ) 2 Debye model (1912) Atoms vibrate collectively in a wave-like fashion. U (T ) = ∑ nk , s ωk , s ( ωk ,s /2 neglected) k ,s = ∑ ∫ d ω Ds (ω ) s ω e ω / kT −1 • Debye assumed a simple dispersion relation: ω = vsk. Therefore, Ds (ω ) = V ω 2 / 2π 2 vs3 Also, the 1st BZ is approximated by a sphere with the same volume 3 ∑ ∫ dω D (ω ) = 3N s =1 s V ωD3 → ∑ 2 3 = 3N s =1 6π vs 3 3 3 1 ≡ ∑ v3 s =1 vs3 → ωD = v(6π n) , n = N / V 2 1/3 If vg = ∇ k ω = 0, then there is "van Hove singularity". 3 U (T ) = ∑ s =1 = V 2π 2 vs3 ωD ∫ d ωω 2 0 3V ⎛ kT ⎞ ⎜ ⎟ 2π 2 v3 ⎝ ⎠ ⎛T ⎞ =9 NkT ⎜ ⎟ ⎝θ ⎠ 4 xD ω e ∫ dx 0 ω / kT −1 Debye temperature ωD θ x , x = = , kθ ≡ ωD D x e −1 kT T 3 3 xD x3 ∫0 dx e x − 1 = π4/15 as T→ 0 12π 4 ⎛T ⎞ Nk ⎜ ⎟ ∝ T 3 as T → 0 ∴ CV = 5 ⎝θ ⎠ 3 solid Argon (θ=92 K) At low T, Debye’s curve drops slowly because long wavelength vibration can still be excited. A simple explanation of the T3 behavior: Suppose that 1. All the phonons with wave vector k<kT are excited with thermal energy kT. 2. All the modes between kT and kD are not excited. kD kT Then the fraction of excited modes = (kT/kD)3 = (T/θ)3. energy U ~ kT⋅3N(T/θ)3, and the heat capacity C ~ 12Nk(T/θ)3 Thermal expansion Coeff. Of volume expansion: Bulk modulus: 1 ⎛ ∂V ⎞ ⎜ ⎟ , V ⎝ ∂T ⎠ P ⎛ ∂P ⎞ B = −V ⎜ ⎟ ⎝ ∂V ⎠T β= 1 ⎛ ∂P ⎞ ⎜ ⎟ B ⎝ ∂T ⎠V ⎛ ∂F ⎞ P = −⎜ ⎟ ⎝ ∂V ⎠T →β = F = Eelastic + Fphonon , Eelastic Free energy: Next page we’ll show that, U (T ) ⎛ ∂E ⎞ P = − ⎜ els ⎟ + γ V ⎝ ∂V ⎠T Indep of T ⇒ 1 ⎛ ∂P ⎞ γ = = β cV ⎜ ⎟ B ⎝ ∂T ⎠V B (cV = CV / V ) ∝ T 3 at low T use ⎛ ∂x ⎞ 1 and ⎜ ⎟ = ∂y y ∂ ⎝ ⎠z ⎛ ⎞ ⎜ ⎟ ⎝ ∂x ⎠ z ⎛ ∂x ⎞ ⎛ ∂y ⎞ ⎛ ∂z ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂y ⎠ z ⎝ ∂z ⎠ x ⎝ ∂x ⎠ y B ⎛ ΔV ⎞ = ⎜ ⎟ 2⎝ V ⎠ 2 Partition function: Z ph = ∑ e − Ei / kT (Ei are the macroscopic eigen-energies) i ⎡ 1 = ∑ exp ⎢ − ⎣ kT {nks } ⎤ 1⎞ ⎛ ω + n ∑ ⎜ ks ⎟ ks ⎥ 2⎠ ks ⎝ ⎦ ⎡ 1 =∏ ∑ exp ⎢ − ks nks = 0 ⎣ kT ∞ ⎤ 1⎞ ⎛ + n ω ⎜ ks ⎟ ks ⎥ 2⎠ ⎝ ⎦ Grüneisen parameter d ωks ωks Fph = − kT ln Z ph ( = kT ∑ ln e ωks /2 kT − e− ωks /2 kT ⇒β = ) ks ⎛ ∂Fph ⎞ 1 ⎛ ∂ ωks ⎞ e P' ≡ −⎜ ⎟ = − ∑⎜ ⎟ 2 ks ⎝ ∂V ⎠T e ⎝ ∂V ⎠T ∴P' = 1 γ ksU ks (T ), ∑ V ks ⎛ where U ks (T ) ≡ ωks ⎜ ⎝e ωks / kT ωks / kT +1 −1 ωk s d ωk s V ωk s ≡ −γ k s V ωk s dV V = = −γ ks dV V 1 ⎛ ∂P ' ⎞ ⎜ ⎟ B ⎝ ∂T ⎠V γ CV 1 1 γ C = ∑ ks V ,ks B V B V ks ∑γ C where γ ≡ ∑C ks V , ks ks V , ks ks CV = ∑ CV ,ks 1 ωks / kT 1⎞ + ⎟ −1 2 ⎠ ks (γ~1-2 for most materials) Neutron scattering En ( k) = Why neutron? 2 2mn = 2.07 k 2 meVA 2 • Neutron has no charge (can probe bulk properties) • Neutron wavelength comparable to interatomic spacings (1-5 Å) • Neutron energy comparable to phonon’s (5-100 meV) • Neutron has spin (can probe magnetic structure and magnetic excitations) Measure phonon dispersions by neutron scattering More than one phonon mode may be excited Δnq.s = n 'q.s − nq.s out E’,p’ in E, p Conservation of energy En '+ ∑ ωq , s Δnq , s = En q,s Conservation of crystal momentum (for a proof. see App. M of A+M) p '+ ∑ q Δnq , s = p + G q,s (momentum of phonon with λ< a must be shifted by G) One phonon scattering En ' = En ± ωs (q ) p' = p ± q + G Neutron energy: p '2 p2 = ± ωs (q ), 2mn 2mn En=p2/2mn Consider phonon absorption ( p '2 p2 − = ωs k '− k 2mn 2mn ⇒ f k (q ) ≡ 2 ( p' ≡ k '; p ≡ k q = ± (k '− k ) k=0 ) 2 2 | q + k |2 k − = ωs ( q ) 2mn 2mn q and ωs can be determined from the intersections in the figure. ) 2 q2 2M n ω (q ) q k≠0 several solutions at a given direction. → a series of peaks in the data Phonon dispersion curve for Si, comparing experimental data and ab initio calculation (Wei and Chou 1994) • Width of one-phonon peaks due to anharmonic effect • Multi-phonon scatterings give a continuous background Neutron scattering: formal theory 2π ∑ Ψ f V Ψ i δ ( E f − Ei ) 2 • Fermi golden rule (transition rate) Ri →[ f ] = • Neutron states scattered to d3k’ (per unit time) mn k ' d ε d Ω d 3k ' Rk →k ' ⋅ V R V = ⋅ k →k ' (2π )3 (2π )3 f j = ρv • Differential cross-section Rk →[d Ω ,d ε ] ⎛ d 2σ ⎞ ⎜ ⎟ d Ωd ε ≡ ε d d incident particle flux Ω ⎝ ⎠ ⎛ d 2σ ⎞ k ' (Vmn ) 2 Rk →k ' ⇒⎜ ⎟= 3 ε π (2 ) Ω d d k ⎝ ⎠ • Quantum state of “neutron + crystal” Ψ = ψ k (r )Φ crystal , ψ k (r ) = initial: final: Ψ i = ψ k ( r )Φ i , Ψ f = ψ k ' ( r )Φ f , 1 ik ⋅r e V 2 k2 Ei = + Ei 2mn Ef = 2 2 k' + Ef 2mn Neutron Crystal energy energy = ψk 2 k 1 k = mn V mn • The energy gained by a neutron due to a phonon in crystal 2 2 2 k '2 k − = Ei − E f = ω 2mn 2mn Momentum transfer q = k '− k Neutron-ion interaction potential ( ) V ( r ) = ∑ v r − r ( R) , where r ( R ) = R + u ( R) v(r) is the atomic potential R Matrix element ( ) let v r − r ( R ) = 1 vq eiq ⋅r e − iq⋅r ( R ) ∑ V q Ψ f V Ψ i = ∑ ψ k 'Φ f v ψ k Φ i R 1 vq ψ k ' eiq⋅r ψ k Φ f e − iq ⋅r ( R ) Φ i ∑ V R ,q δ k + q ,k ' vk ' − k ∼ v0 1 since Δk for phonon ∼ 108 cm −1 = ∑ vk '− k Φ f e − i ( k '− k )⋅r ( R ) Φ i V R the range of vq ∼ 1013 cm −1 = ⎛ dσ ∴⎜ ⎝ d Ωd ε ⎞ k ' ⎛ v0 mn ⎞ ⎟= ⎜ 2 ⎟ ⎠ k ⎝ 2π ⎠ 2 ∑δ (E f f − E i ) ∑ Φ f eiq ⋅r ( R ) Φ i Sum over a complete set of phonon states R 2 ⎛ dσ ⎞ One can re-write ⎜ d Ωd ε ⎟ using dynamical structure factor ⎝ ⎠ 1 ⎛ E f − Ei δ ( E f − Ei ) = δ ⎜ ⎝ use e i ( E f − Ei ) t / ⎛ dσ then ⎜ ⎝ d Ωd ε ⎞ 1 +ω⎟= ⎠ 2π iω t dt e e ∫ i ( E f − Ei ) t / for phonon absorption (always from neutron’s viewpoint) Φ f A Φ i = Φ f A(t ) Φ i ⎞ k ' 1 ⎛ mn v0 ⎞ ⎟= ⎜ 2 ⎟ ⎠ k h ⎝ 2π ⎠ 2 ω ∫ dt e ∑ i t Φ i eiq ⋅r ( R ') e − iq ⋅r ( R ,t ) Φ i R,R ' 2 k ' 1 ⎛ mn v0 ⎞ = 2π NSi (q , ω ) ⎜ 2 ⎟ k h ⎝ 2π ⎠ Dynamical structure factor Si ( q , ω ) ≡ Density operator (for ions) 1 dt iωt iq ⋅r ( R ') − iq ⋅r ( R ,t ) e e e Φ Φi ∑ i ∫ N 2π R,R ' ρ (r ) = ∑ δ (r − r ( R)) R ρ q = ∫ d 3r eiq⋅r ρ (r ) = ∑ eiq⋅r ( R ) dt iωt 1 =∫ e Φ i ρ q (0) ρ − q (t ) Φ i 2π N R Density correlation function For a crystal at finite temperature Φ i ρ q (0) ρ − q (t ) Φ i → ρ q (0) ρ − q (t ) T = ∑e i − Ei / kT Φ i ρ q ρ − q (t ) Φ i ∑e i − Ei / kT Evaluation of the dynamical structure factor eiq ⋅r ( R ') e − iq⋅r ( R ,t ) T = eiq⋅( R '− R ) eiq ⋅u ( R ') e − iq ⋅u ( R ,t ) It can be shown that, for A, B linear in a , a e Ae B T =e 1 2 A + 2 AB + B 2 2 use ⎡⎣ q ⋅ u ( R ') ⎤⎦ 2 T T † D. Mermin, J Math.Phys. 7,1038 (1966) = ⎡⎣ q ⋅ u ( R, t ) ⎤⎦ and ⎡⎣ q ⋅ u ( R ') ⎤⎦ ⎡⎣ q ⋅ u ( R, t ) ⎤⎦ r ( R) ≡ R + u ( R) T T = [ q ⋅ u (0)] 2 ≡ 2W 2 T T = [ q ⋅ u (0) ] ⎡⎣ q ⋅ u ( R − R ', t ) ⎤⎦ T Translation symmetry of the system Debye-Waller factor ⇒ S (q , ω ) ≡ e e T −2W =1 + dt iωt − iq ⋅ R [ q ⋅u (0)]⎡⎣ q ⋅u ( R ,t ) ⎤⎦ T ∫ 2π e ∑R e e 1 2 + + ... T T 2 0-phonon 1-phonon process … Exact so far (for a harmonic crystal) • Rough estimate (Kittel. App.A) Zero-phonon process 2W = S0 (q , ω ) = e −2W δ (ω ) ∑ e R DebyeWaller factor Elastic scattering − iq ⋅ R N ∑ δ q ,G G Laue’s diffraction condition T T cos 2 θ T 1 = G2 u2 T 3 m 3 use ion ω 2 u 2 = kT T 2 2 −2W =e − G2 mionω 2 kT 2 ⎞ k ' 1 ⎛ mn v0 ⎞ 2π NS (q , ω ) ⎟= ⎜ 2 ⎟ ⎠ k h ⎝ 2π ⎠ k' N or = a 2 S (q , ω ) k ⎛ dσ ⎜ ⎝ d Ωd ε 2 Intensity I=I0 e-2W (I0 for a rigid lattice) Differential cross-section Cf: ) = G2 u2 → e ⎛ dσ ⎜ ⎝ d Ωd ε ( G ⋅u ⎞ N ⎟ = S (q , ω ) ⎠ 2π 2 a v0 ≡ mn For X-ray scattering (the same S) Scattering length For more discussion, see A+M, App. N use aks† aks • A more accurate evaluation 1 aks = 2 a−† ks = 1 2 uks = pks = −i ⎛ ⎞ i pks ⎟ ⋅ eˆks ⎜ mωks uks + ⎜ ⎟ mωks ⎝ ⎠ ⎛ ⎞ i ˆ m u p ω − ⎜ ks ks ks ⎟ ⋅ eks ⎜ ⎟ mωks ⎝ ⎠ (eˆ*− ks =eˆks ) a ( ω (k ) 2mion ks +a † − ks ) eˆ ks s 2mionωs (k ) (a ks − a−† ks ) eˆks ⇒ ( G ⋅u ) = nks T 2 = T 1 N G ⋅ eˆks ∑ 2m ks 2 ωs ( k ) (2nks + 1) ion In 3D Debye model (at T =0!) G2 3 2W = 4 mion v k D (Prob. 7) In the calculation, one has kD 1 kD D − 2 D −1 k k d = k dk ∫0 k ∫0 • In 3D, W weakens the diffraction peaks. • In 2D, W is finite at T=0 but infinite at finite-T. 1 uks eik ⋅R ∑ N k ,s 1 p( R) = pks eik ⋅R ∑ N k ,s u ( R) = • In 1D, no long-range order even at T=0 Mermin-Wagner theo (Mermin PR1968) There is no long-range crystalline order at finite-T in 2D. One-phonon process S1 (q , ω ) = e −2W ∫ dt iωt e ∑ e − iq⋅R [ q ⋅ u (0) ] ⎡⎣ q ⋅ u ( R, t ) ⎤⎦ 2π R T 1/2 ⎛ ⎞ ik ⋅R 1 † q ⋅ u ( R, t ) = ⎜ ⎟ e ( aks (t ) + a− ks (t ) ) q ⋅ eˆks ∑ N ks ⎝ 2mionωks ⎠ ⇒ S1 (q , ω ) = e−2W ∑ s 2mionωqs ( q ⋅ eˆks ) 2 aks (t ) = e − iωks t aks ⎣⎡(1 + nqs )δ (ω + ωqs ) + nqsδ (ω − ωqs ) ⎦⎤ Phonon emission absorption Delta peaks are broadened only if anharmonic effect (phononphonon interaction) is included. One-phonon cross-section ⎛ dσ ⎜ ⎝ d Ωd ε ⎞ k' 2 N S1 (q , ω ) ⎟= a k ⎠