Chapter 4 Phonons I Chapter 4 Phonons I Two atoms/primitive basis � Last lecture, we showed that there are two branches to the phonon dispersion curve if you have 2 atoms/primitive basis � We also sol[.]
Chapter 4: Phonons I Two atoms/primitive basis • Last lecture, we showed that there are two branches to the phonon dispersion curve if you have atoms/primitive basis • We also solved for the case of Ka = (zone center) and Ka = π (zone boundary) Two atoms/primitive basis • • Let’s take another look at the phases For K = 0, we have the two solutions: 1 + M M 2 1 C ω ≅ K 2a M1 + M ω ≅ 2C • (optical branch) (acoustic branch) We can substitute these into our original equations to solve for u and v, the displacments of atom M1 and M2 − M 1ω 2u = Cv[1 + exp(−iKa )] − 2Cu − M 2ω v = Cu[1 + exp(iKa )] − 2Cv Phases at K = • So, for Ka = 0, exp(-iKa) = Our equations then become: − M 1ω 2u = Cv[2] − 2Cu − M 2ω v = Cu[2] − 2Cv • For the optical branch, we find that: 1 u = Cv[2] − 2Cu − M 2C + M M M − 1 + u = v − u M2 What does this mean? M1 u M2 − u=v→ =− M2 v M1 Phases of optical mode at K=0 • • • This means that the displacement of atom M1 is opposite to M2 Also, the magnitudes are inversely related to the mass (so the smaller mass oscillates with a larger magnitude than the larger mass) M1 M2 M1 u v us+1 M2 vs+1 The center of mass is fixed though (they vibrate out of phase, but the wave does not propagate along the chain – this is a standing wave) Phases of the acoustic mode at K=0 • At K=0, we can find the solution for u and v: C ω ≅ K 2a → ω = M1 + M • Substituting this into either one of these equations: − M 1ω 2u = Cv[2] − 2Cu − M 2ω v = Cu[2] − 2Cv • You get: u = v This means that the atoms oscillate in phase: M1 M2 u v Phases of modes at K = π/a • The solutions at K = π/a are: 2C 2C ω ≅ ,ω ≅ M1 M2 • So, using the same trick, we can solve for the phases by substitution: -1 − M 1ω 2u = Cv[1 + exp(−iKa )] − 2Cu -1 − M 2ω v = Cu[1 + exp(iKa )] − 2Cv → − M 1ω 2u = Cv[1 − 1] − 2Cu = −2Cu → − M 2ω v = Cu[1 − 1] − 2Cv = −2Cv (2C − M 1ω )u = (2C − M 2ω )v Phases of modes at K = π/a • With this solution: • We get: 2C ω ≅ M1 2C 2C (2C − ( ) M )u = (2C − ( ) M )v M1 M1 → 0u = 2C (1 − • M2 )v M1 On the right hand side, v is the only term that can be zero for M1 ≠ M2 Therefore, v=0, and u is the only term which can change Phases at K = π/a • In adjacent cells, the phase of u changes (this is because us ~ ueiKsa, and if Ka = π, then we have us ~ ueisπ So us ~ ueiπ = -u, and us+1 ~ uei2π = u) M1 u • M2 v M1 us+1 M2 vs+1 The other solution, ω2 ≈ √(2C/M2), gives oscillating M2 masses, with the M1 not moving Note that this motion is another standing wave, so the center of mass does not move for these vibrations Forbidden frequencies • • • Note that for polyatomic lattices (M1 ≠ M2), there is a gap for certain frequencies at the zone boundary (K = +/- π/a) This is a characteristic feature for elastic waves If you look for solutions for ω in this region, you will find complex solutions of K (which means that the wave is damped out and it cannot exist in the lattice) Forbidden zone Quantization of Elastic Waves • • • • From quantum mechanics, we learn that all energy is quantized (it comes only in discrete values) So, on this dispersion curve, there will only be discrete values of ω (because energy is released in packets of E = ħω) These quanta of lattice vibrations (waves) are called phonons in analogy with the photons of electromagnetic waves Thermal vibrations in crystals are thermally excited phonons, like the thermally excited photons of black-body electromagnetic radiation Discrete values Black body radiation waves Quantization of Elastic Waves • These waves are quantized just like harmonic oscillator waves (as you might expect) • For a wave of frequency ω: ε = (n + ½ ) ħω • In this case, the mode is occupied by n phonons, each has an energy of ħω (see Appendix C) • The zero point energy of the mode is ½ ħω, where n = According to quantum mechanics, there is a zero-point energy associated with every system Quantization of Elastic Waves • The mean square amplitude of the wave vibration is quantized as well We can show this to be true: • Let’s look at a standing wave mode of amplitude u = uo exp(iKa)exp(-iωt) • In a harmonic oscillator, the total energy is: E = ½ mv2 + ½ kx2 • So, averaged over time, the wave is ½ KE and ½ PE • The kinetic energy density (KE/unit volume) of a wave is ½ ρ(du/dt)2 = ½mv2/V (where ρ is the mass density) • So, taking the second derivative of u, we get: KE/(unit volume) = ½ ρω2 uO2 (exp(iKa) exp(-iωt))2 Quantization of Elastic Waves • So, the energy in this mode is ½ of the avg kinetic energy: Energy = (1/4) (Volume) ρω2 uO2 (exp(iKa) exp(-iωt))2 = ẵ (n + ẵ) ẵ ã The average of the exp terms is ½ (take the average of sin2 or cos2), so we have: (1/8) (Volume) ρω2 uO2 = ½ (n + ½) ħω and uO2 = 4(n + ẵ) /(V) ã Therefore, the mean square displacement of the motion of the atoms is quantized (since n can only be an integer) Phonon Momentum • You might think that these phonons have momentum which is p = ħK, just like photons have momentum p = E/c • Be careful! Phonons don’t carry momentum like photons due They can interact with particles like they have a momentum (for example, a neutron can hit a crystal, and start a wave by transferring momentum to the lattice) • However, you have to think of this momentum as being transferred to the whole lattice The atoms themselves are not being translated permanently from their equilibrium positions • The only exception to this rule is the K = mode, where the whole lattice translates This, of course, carries momentum Phonon Momentum • For all practical purposes, a phonon acts as if it carries a momentum p = ħK, which is sometimes called the crystal momentum • We have already seen examples of where this is used For example, if an x-ray interacts with a lattice, then we know that the scattered ray(k’) and incident ray (k) have to be related by: k’ = k + G • Where G is a reciprocal lattice vector In this process, the whole crystal recoils with momentum -ħG G k k’ Crystal Recoil -G Phonon Momentum • If this photon interacts inelastically with the lattice, then we have: (phonon is created) k’ + K = k + G • Or: (phonon is absorbed) k’ = k + K + G K phonon absorbed G k k’ Crystal K phonon created G k k’ Crystal Crystal Momentum • Note: in the above pictures, we left out the crystal momentum (the recoil of the crystal, -ħG) • We can always add or subtract a crystal momentum wavevector from these totals This is a general rule of solid state physics • Why is this true? Think about this in terms of the reciprocal lattice Just like adding a lattice constant, a, to a real lattice doesn’t change the properties of the real space lattice, adding a wavevector G does not change the properties in momentum space (you are just in the next reciprocal lattice cell) • This is only true because the lattice is periodic