nullnullLecture 11Chapter 2
The lattice dynamics:
Thermal and optical properties of crystals Lattice dynamics - a 1d homogeneous solidLattice dynamics - a 1d homogeneous solidConsider the one dimensional solid above, in an arbitrary state of strainu is the displacement of the indicated element from its equilibrium positionthe associated strain is e = du/dx, (ie the fractional change of length)At the left hand end of the element we have a strain e(x)where c is an elastic modulusLecture 11Dynamics in a 1d homogeneous solidDynamics in a 1d homogeneous solidThis is the well-known wave equation with solutions in the form of a travelling waveThese waves are sound waves, and vo is the velocity of sound in an homogeneous medium must be an integral multiple of /L where L is the
length of the body, so the dispersion relation is not
strictly continuous but a series of closely spaced points or modes of vibrationLecture 11A line of atomsA line of atomsWe know that matter is discontinuous, so we have to take into account the atomic structureThis is best done by considering the 1d solid as a linear chain of atoms of mass M spaced at a distance a connected by springs, of spring constant C, which obey an ideal Hooke’s law equilibriumstrainedLecture 11Dynamics of a chain of atomsDynamics of a chain of atomsAs before we can set up a force equationTaking the nth atom we have forces of C(un - un-1) to the left C(un+1 -un) to the right This equation of motion is the same for all atoms, only the subscripts varyReplacing the x coordinate by na, where a is the lattice parameter we haveLecture 11Dynamics of a chain of atomsDynamics of a chain of atomsLecture 11Dynamics of a chain of atomsDynamics of a chain of atomsLecture 11Dynamics in the Brillouin zonesDynamics in the Brillouin zones1st Brillouin zoneFrom what we know about Brillouin zones the points A and B (related by a reciprocal lattice vector) must be identicalThis implies that the wave form of the vibrating atoms must also be identical.Lecture 11Equivalence of points in 1st and 2nd zonesEquivalence of points in 1st and 2nd zonesBut: note that point B represents a wave travelling right, and point A one travelling leftLecture 11nullLecture 11A zone boundary modeA zone boundary modeConsider point C at the zone boundaryWhen =/a, =2a, and motion becomes that of a standing wave (the atoms are bouncing backward and forward against each otherLecture 11nullLecture 11Zone boundary modeZone boundary modeAt higher frequencies (eg near the boundary of the 1st Brillouin zone) the group velocity and phase velocity are no longer equalIf 0, vg vs, but if /a, vg 0Similarlyand if 0, vp vs, but if /a, vp 2vs/Vg is the quantity that represents the transfer of signal or energy, so at =/a no signal or energy is propagated: the wave is a standing waveThis situation is equivalent to a Bragg reflection in a diffraction experimentRemember that the zone boundary is the locus of points for which a Bragg reflection is possibleLecture 11Simple dispersion relations:Simple dispersion relations:Lecture 12The diatomic linear latticeThe diatomic linear latticeAt s, M is displaced a small amount usAt s+1/2, m is displaced a small amount vs+1/2The equations of motion for the two atom types are: andAs before we assume only near neighbour interactions with a spring constant CLecture 12The diatomic latticeThe diatomic latticeThere are solutions to these equations of motion of the type:where s is an integerSubstituting these solutions back into the equations of motion we havesowhich we must now solveLecture 12The diatomic latticeThe diatomic latticeSolving these equations for u and v we have Solving this quadratic equation for 2 givesTwo solutions!Lecture 12The diatomic latticeThe diatomic latticeWe have therefore found two possible solutions giving two possible dispersion relations for the diatomic latticeWhen =0, ie at the zone centreWhen =/a, ie at the zone boundary+sign-signLecture 12Dispersion relation for a diatomic latticeDispersion relation for a diatomic latticeLecture 12Dispersion relation for a diatomic latticeDispersion relation for a diatomic latticeWhat do the two branches represent?Evidently the amplitudes of the masses M and m are in phase in branch A of the dispersion relation and out of phase in branch BLecture 12At the zone boundaryAt the zone boundaryand only the light atoms vibrateand only the heavy atoms vibrateLecture 12The low energy branchThe low energy branch….and the two types of atom vibrate with the same amplitude, direction and phase This branch is called the Acoustic BranchLecture 12Acoustic mode at =0.2/a Acoustic mode at =0.2/a Lecture 12The acoustic mode at the zone boundaryThe acoustic mode at the zone boundaryLecture 12The high energy branchThe high energy branch….and the two types of atom vibrate against one another.The centre of mass remains stationary, and the amplitude of the heavy atoms is m/M that of the light atoms
This branch is called the Optic BranchLecture 12The optic mode at the zone centreThe optic mode at the zone centreLecture 12The optic mode at =0.2/aThe optic mode at =0.2/aLecture 12The optic mode at the zone boundaryThe optic mode at the zone boundaryLecture 12In the extended zone schemeIn the extended zone schemeWe have drawn the dispersion relation in the reduced zone scheme - everything mapped into the first BZWe could also draw it in the extended zone schemeWhen we do it is apparent what happens as mM(a) the gap starts to close and when m=M it looks like the monatomic lattice(b) also, when m=M the first BZ extends to =2/a - this is to be expected as we know have an effective lattice parameter of only a/2Lecture 12Discrete excitations Discrete excitations We have already suggested that the dispersion relation is not strictly continuous but a series of closely spaced points or modes of vibration - a consequence of the finite length of the sample. The same is also true hereS+N-1SS+1S+2and each value is a “mode” or ”state”Note that is independent of , and is determined by N, hence size of sampleThis is an indication of the quantisation of lattice vibrationsHowever a linear chain 1cm long with a=1Å, has =1cm-1. But the BZ is 108cm-1 wide and will thus have 108 modes distributed evenly through the BZLecture 12PhononsPhononsAlthough we have treated lattice vibrations classically we can see that they are in fact quantisedThe normal modes are harmonic and independent and therefore lend themselves to quantum mechanical treatmentA lattice vibration mode of frequency w behaves like a simple harmonic oscillator with an energy En can be considered as the sum of n excitation quanta added to the ground stateThe quantum of thermally excited lattice vibration energy is called a PHONON in analogy to the thermally excited photons of black body electromagnetic radiationThe two systems are both equivalent to a quantum harmonic oscillatorBoth photons and phonons are bosons - they can be created and destroyed in collisionsLecture 12Spatial localisation of a phononSpatial localisation of a phononNormal modes are plane waves extending throughout the crystal hence phonons are not localised particlesTheir position cannot be determined because ħ is known exactlyHowever a reasonably localised wave packet can be constructed within the limits of the uncertainty principle by combining modes of differing w and lThus if we take waves with a spread of of say p/10a we will construct a wave packet localised to within 10 unit cells moving with a group velocity of dw/dkLecture 12Phonon momentumPhonon momentumThe phonon momentum is given by ħpNeutrons, photons, electrons, etc can interact with the phonon as if it were a particle with this momentum The phonon therefore does not carry physical momentum…….…..a phonon coordinate involves the relative coordinates of the participating atoms….…..as the atoms are only vibrating about their centre of mass no linear momentum is transferred.ħ is sometimes called the crystal momentumLecture 13Phonon interactionsPhonon interactionswhere k is the wavevector of the incident radiation, k’ that of the scattered radiation and Ghkl is a reciprocal lattice vectorIn the scattering process the momentum of the system as a whole is rigorously conservedLecture 13Studying phononsStudying phononsLight scatteringLight scattering, ie Raman scattering, is produced by the polarisation of atoms and subsequent emission of dipole radiationThe maximum wave vector transfer is 2k=2x2p/l=2x10-3Å-1, which is only 1/1000 of a reciprocal lattice vector.Raman scattering can therefore only be used to study modes close to the zone centreX-ray inelastic scatteringThe wavelength of x-rays (~1Å) overcomes this problem, but the associated energy is ~10keV.Lattice vibration energies are typically ~1meV, so we would need to monochromate with Dl/l~10-7, which, assuming Bragg’s Law, implies a crystal monochromator for which Dd/d<10-7 both in accuracy and stabilityDifficult to the point of impossible….But neutrons are just right…1Å neutrons have 80meV energy, implying Dl/l~10-3Lecture 13Neutrons and phononsNeutrons and phononswhere ħ is the energy of the phonon (created or absorbed in the scattering)which can be solved by measuring k’ , the final neutron momentumLecture 13..and on the Ewald diagram..and on the Ewald diagramThis is phonon would be difficult to measure as the total scattering angle, f, is very small, and the scattered neutron beam is close to the incident neutron beam Lecture 13..and on the Ewald diagram..and on the Ewald diagramExactly the same phonon can be measured close to the reciprocal lattice point, where the geometry is much more favourableLecture 13The triple axis spectrometerThe triple axis spectrometerLecture 13The triple axis spectrometerThe triple axis spectrometerLecture 13The triple axis spectrometerThe triple axis spectrometerLecture 13The triple axis spectrometerThe triple axis spectrometerLecture 13Dispersion relationshipsDispersion relationshipsMonatomic FCC NeonLecture 13Dispersion relationshipsDispersion relationshipsMonatomic BCC NaLecture 13Dispersion relationshipsDispersion relationships“diatomic” Si
(diamond structure)Lecture 13Phonons in MgB2 Phonons in MgB2 Phonon modes in the anomalous 40K superconductor MgB2Lecture 13AnharmonicityAnharmonicityThe measured phonon dispersion relations suggest that this is not unreasonable, however the consequences for other properties are severe:So far we have considered lattice vibrations only within the “harmonic approximation”, ie the interatomic potential is treated as being parabolic(ii) Thermal conductivity is infinite(i) There can be no thermal expansion in the harmonic approximationIn the harmonic approximation phonons do not interact with one anotherso, in the absence of any defects there will be no hindrance to the transfer of heat energy by phononsLecture 13Phonon-Phonon CollisionsPhonon-Phonon CollisionsIn the harmonic approximation two or more phonons may pass through each other without interaction, ie the phonon modes are uncoupled.When an atom vibrates it excites a phonon.This is because the elastic constants are everywhere the same and are constant with timeThe heat in the system is carried by phonons - one might expect at the speed of soundHowever experiments show that conductivity of heat is more like a slow diffusion processThis suggests that heat conduction might be limited by phonon-phonon interactionsLecture 13Anharmonicity and CollisionsAnharmonicity and CollisionsIn the anharmonic limit a phonon causes a periodic elastic strain which in turn modulates - in space and time - the elastic constantTwo crossing phonons therefore “see” one another and hence interactw1,k1 and w2,k2 coalesce to produce a phonon with w3,k3Conservation of energy and momentum givew3= w2 + w1k3 = k2 + k1However, the second equation must be modified to take into account that we normally represent phonon momenta within the 1st Brillouin zone ie -p/a
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