Chapter 4 Phonons I Crystal Vibrations Chapter 4 Phonons I Crystal Vibrations Chapter 4 Phonons Ì Crystal Vibrations Crystal Vibrations � The study of lattice vibrations is important for understanding[.]
Chapter 4: Phonons I Crystal Vibrations Crystal Vibrations • • • • The study of lattice vibrations is important for understanding how energy is absorbed in solids At low temperatures, most of the energy that is absorbed is of two types: lattice energy, and electronic energy – the heat can go to the electrons, or to vibrations of the positive charges in the lattice We will eventually show (by the end of Chapter 6) that this can be measured with the heat capacity: C = dU/dT For a metal, C = γT + βT3 Electronic term Lattice term C/T Slope = β Y-intercept = γ T2 C/T = γ + βT2 Vibrations of Crystals with Monatomic Basis • • • • • • To simplify things, we will start by looking at simple monatomic cubic crystals (one type of atom) The model we will use is that each atom is attached to the next atom by a spring force F = -C (displacement) = -kx (Hooke’s law) For monoatomic lattice, the C constants are all the same The vibrations are quantized in energy, and are called phonons For a force applied to the crystal, there will be an associated wavevector, K For each wavevector, there are three modes of vibration: of longitudinal polarization, of transverse We will call the displacement of each atom us (for atom s) Longitudinal Phonon Transverse Phonon Vibrations of Crystals with Monatomic Basis • • We will assume that each atom only feels its nearestneighbours The total force on atom s from the planes s +/- is then: Fs = -C(us - us+1) - C(us – us-1) From plane s+1 • • • From plane s-1 This looks like Hooke’s law The constant, C, will be different for longitudinal and transverse waves We can now write the equation of motion for this plane: Longitudinal Phonon d 2u s F = Ma = M = C (u s +1 + u s −1 − 2u s ) dt Vibrations of a Monatomic Lattice a us Vibrations of Crystals with Monatomic Basis • The solutions to this equation will look like harmonic oscillators: us = A cos(Ksx – ωst) Amplitude Wavevector Frequency • We will then look for solutions that have the form: d2us/dt2 = -ωs2 us • We can now rewrite our equation to look like: d 2u s M = − Mω s u s = C (u s +1 + u s −1 − 2u s ) dt C C s-1 s s+1 Solution • In keeping with the theme of this chapter, the solution to this will be us = A exp(i(Ksx-ωt)) • Remember, we can always write this as a cos or sin function • We have to modify this slightly to include the fact that our x values are now discrete instead of continuous (we can only reference our wave with respect to x = sa, s is an integer, a is a lattice constant) • So, we now have: us = A exp(i(Kssa – ωt)) Solution • We can solve this: d 2u s M = C (u s +1 + u s −1 − 2u s ) dt − Mω s Aei ( K sa −ωt ) = C ( Ae i ( K ( s +1) a −ωt ) + Aei ( K s +1 s s −1 ( s −1) a −ωt ) − Aei ( K s sa −ωt ) ) − Mω s = C (e iK s a + e −iK s a − 2) − Mω s = C (2 cos K s a − 2) − Mω s = −2C (1 − cos K s a) ω s = (2C / M )(1 − cos K s a ) ω s = ωO sin( K s a / 2) (cos 2x = – sin2x) (where ωO = (C/M)1/2) This is sometimes called the dispersion relation Dispersion relation • • • • • Why is this called a dispersion relation? In E&M, k ~ index of refraction, so ω ~ k ~ index of refraction for long wavelengths (k = 2π/(wavelength)) This shows how the energy changes, or disperses as a function of the wavelength Why has this been plotted between –π/a ≤ K ≤ π/a? These are the only physically relevant values for K = 2π/(wavelength) This is often called the “first Brillouin Zone” (remember, in reciprocal space, the units are 2π/a) ω(0) = ω(π/a) = (C/M)1/22sin(π/2) = (4C/M)1/2 Periodic boundary conditions • • • • • • • • We will now show that K has to lie between –π/a ≤ K ≤ π/a For a lattice of N ions, we will use periodic boundary conditions such that the wavefunctions will match up: uN = u0 Using this, we have: uN = Aei(KNa-ωt) = u0 = Aei0 = Aei2πm = A(1) So: eiKNa = ei2πm, or KNa = 2πm Rearranging this, we have: K = (2π/a)(m/N) Let’s look at what this condition is in more detail Let’s add 2π/a to K: uK+2π/a = ei(KNa-ωt)+iNa(2π/a) = ei(KNa-ωt)ei2πN = ei(KNa –ωt) So, using this we have shown that K must lie between and 2π/a, or that –π/a ≤ K ≤ π/a Another way of saying this is that by adding 2π/a, we are into the Brillouin Zone of the next unit cell in reciprocal space Just like adding a lattice constant to an atom has no information in real space, adding a wavevector 2π/a has no new information in reciprocal space Dispersion relation • Another way of showing this, physically, is by looking at the wavelengths of vibration: • The wave represented by the solid curve conveys no new information not given by the dashed curve Only wavelengths longer than 2a are needed to represent the motion This means that K is limited by –π/a ≤ K ≤ π/a Dispersion relation • • Yet another way to show this is by looking at how adjacent atoms are in or out of phase with one another It makes no physical sense to say that the waves are out of phase by more than π They can be completely in phase (symmetric vibrations), or out of phase (antisymmetric), or anything in between, but anything more than π does not convey new information Vibrating in phase (phase difference = 0) • • Vibrating out of phase + (phase difference =- π) So, with these two limits, we know that for two adjacent atoms, -π ≤ Ka ≤ π, or that –π/a ≤ K ≤ π/a This is the range of the 1st Brillouin Zone for the lattice Features of the Dispersion Relation • • • Most of the waves described by the wavevector K are travelling waves (meaning that the waves propagate through the lattice) However, at the zone boundary, which is K = +/- π/a the wave becomes a standing wave This means that the wave itself moves neither to the left or two the right It is also a standing wave at K = (the wavelength is infinite) + - (at Ka = +/-π these are vibrating out of phase, and so the wave does not propagate) Features of the Dispersion Relation: Sound Waves • • • • • What happens at the zone center, where K → 0? The wavelengths become very long – this is the region of sound waves As K → 0, this becomes the speed of sound The speed at which the wave travels is the group velocity, which is vG = dω/dK We can solve for this for the monatomic lattice: ω = (C/M)1/2 2sin(Ka/2) Therefore, vG = (Ca2/M)1/2cos(Ka/2) This is a maximum at K = 0, where vG = (Ca2/M)1/2 And zero at K = +/- π/a (boundary) Sound Waves • • • As K → 0, we can solve for how the dispersion relation looks: using ω2 = (2C/M) (1cos Ka), we can expand for cos Ka = – ½ (Ka)2 when Ka is small: ω2 = (2C/M)(1 – (1 – ½ (Ka)2) ω2 = (C/M)K2a2 So, the frequency is directly proportional to the wavevector in the long wavelength limit (the dispersion relation looks linear) Also, vG = ω/K, which is what is found for light and other continuous media This is linear in this region