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October 19th, 2010 Class 12 Crystal Vibrations: Monoatomic basis We found that the dispersion relation for monoatomic basis was 4C 4C Ka Ka sin 2 sin 2 what leads to the dispersion relation K M M 2 2 That for the long wavelength limit (small K) leads to, cosKa 1 2C C 1 Ka2 and 2 1 cos Ka Ka 2 2 M M Ca 2 or K M and for K=±/a leads to stationary waves (the derivative is zero at those points). These are the limits of the first Brillouin zone. for Kmax=/a 2/min=/a or min=2a and s u s u exp is a u exp is u 1 a Thus, consecutive planes move in opposite phase. All the physics of the problem is already in the first Brillouin zone, second and others are redundant and do not produce new information. To understand this, notice that us are actually the amplitude of the oscillation at each atom, thus any wave for which the amplitude at each atomic position is the same and NOT a new solution since there are no atoms to see intermediate amplitudes. Notice in Figure 5 (page 93) that the two waves predict the same amplitude at each point. Notice that the ratio of the amplitude of the atomic displacement between consecutive planes is u s 1 u expi s 1Ka expiKa us u expi sKa All independent values of this ratio are obtained for Ka between - and . Given any value of K outside the mentioned [- ,] interval, there exist another wavevector K’=K-2n/a (with n and integer) that it is in the interval. For this wavevector K’ in the first Brillouin zone us 1 2n 2n exp iKa exp i K ' a exp iKa exp i a exp iK ' a us a a Thus the relative amplitude is exactly the same for K as it is for a K’ in the first Brillouin zone. 1 Group Velocity When a packet of waves with different wave lengths (and frequency) travel together, it is not the speed of each of the components what matters, but the velocity of the group (imagine a pulse sent through space). In non-dispersive mediums, frequency and wavevector are proportional and, as we will see, phase velocity (the one of individual components) is equal to group velocity (like light in vacuum), all the components travel at the same speed. But in general that is not the case, each component travels at a different speed, and the actual package travels at a speed giving by: d dK To see this, consider two waves with a slightly different frequency and wavevector (and just for illustration purpose, the same amplitude too). The composed waves will be: vg ue i t Kx ue i ' t K ' x This equation can be re-arranged to give K 2ue i t Kx cos t x Where and K are the average frequency and wave vector. 2 2 That is a high frequency wave with a low frequency envelope. The velocity at which the package moves is the velocity of the envelope, thus vg thus the vg indeed give the speed of the package. K If for a particular medium, the frequency and the wavevector are proportional, then group and phase velocity are the same, and equal to the speed of each component. In this case the medium is non-dispersive and all the components move at the same speed, thus that package does not change shape. If instead different components travel at different speeds, the package will spread, or disperse as it moves on (thus the relationship between and K is known as the dispersion relationship) d For a continuum of frequencies, v g dK In the case of phonons, using the dispersion relation obtained above. Ca 2 Ka cos 0 Ka d M 2 vg dK Ca 2 Ka cos Ka 0 M 2 vg is discontinuous for K=0 where it goes from -1 to 1 in going from negative values of K to positive values of K. Which is not surprising, it only means that the wave now moves in the opposite direction. In the limits of the Brillouin zone, Ka=± and vg=0, consistent with standing waves. 2 As we saw before, for the long wavelength limit (small K), Ca 2 K M Thus the group velocity is equal to /K. This can be understood by realizing that in the long wave length limit K~0, >>a and the interaction with the lattice is minimal leading to a nondispersive behavior. Force constant from the interaction with non-consecutive planes If we consider now the effect of a plane p places from the reference plane s then us p a u exp i s p Ka from where the dispersion condition can be easily shown to be 2 2 C p 1 cos pKa and considering all the planes M 2 2 M C 1 cos pKa p 0 p Where Cp, the force constant associated with the interaction between plane s and p is given by: (See derivations provided) /a Ma Cp 2 cos pKa dK 2 / a More than one type of atoms per primitive basis When the basis consist on one atom, there are three independent vibration modes, 1 longitudinal (the atoms in the plane oscillate in a direction perpendicular to the plane) and 2 transversal where the atoms oscillates parallel to the plane (the two modes come from the two dimensions of the plane). When the basis have more than one type of atom, then each type of atom give rise to three independent vibration modes, thus in a crystal with a p-atoms basis, there are 3p independent modes. Out of the 3p modes, 3 and called acoustical modes and 3p-3 are called optical modes for a reason that will be evident soon. Also both optical and acoustical modes have transversal and longitudinal in the rate of 2-to-1. In what follows we will consider a cubic crystal with a two atoms base with alternate planes of atoms 1 (with mass M1) and atoms 2 (with mass M2). d 2u s C vs vs1 2us dt 2 d 2v Atom 2 : M 2 2s C us us 1 2vs dt Atom1 : M 1 3 Solutions to the above equations can be proposed to be of the form u s u exp it sKa and v s v exp i t sKa Where a is the distance between plane of atoms of the same kind Using these solutions in the equation above, M1 2u C v v exp iKa 2u M 2 2v C u u expiKa 2v What leads to a set of two homogeneous equations with two unknowns, 2C M u C1 e 0 C 1 e u 2C M 0 iKa 2 1 iKa 2 1 This set of equations has non-trivial solution only if the following determinant is zero 2C M1 2 C1 exp iKa C1 exp iKa 2C M 2 2 what leads to M 1 M 2 4 2C M 1 M 2 2 2C 2 1 cos Ka 0 with solutions 2C M 1 M 2 4C 2 M 1 M 2 8M 1 M 2C 2 1 cos Ka 2M 1M 2 2 2 For the limit Ka << 1 cos(Ka)~1- ½ (Ka)2 or 1- cos(Ka)~ ½ (Ka)2 2C M 1 M 2 4C 2 M 1 M 2 4 M 1 M 2 C 2 Ka 2M 1M 2 2 2 2 and for small values of Ka we can expand the square root in series values of x 2 4M1M 2C 2 Ka 2C M1 M 2 2C M1 M 2 4 C M M 1 2 2 2 M 1M 2 what provide the two solutions for small Ka 4 a 2 bx ~ a bx for small 2a 4M 1M 2C 2 Ka 1 4C M 1 M 2 1 4C M 1 M 2 2 2C 2 M 1M 2 2 M 1M 2 M M 2 1 To the first order in Ka and 2 4C M 1 M 2 4 M 1 M 2C 2 Ka 4C M 1 M 2 C Ka 2 2 2M 1M 2 2M 1 M 2 2 By using these solutions in the set of equations for u and v and solving for them, it can be seen M u that for the first of the solutions in the K~0 limit, 2 , thus the consecutive planes vibrate v M1 in opposite direction keeping the center of mass fixed. If the two atoms are ions of opposite sign, this mode can be excited by an electromagnetic field, thus this mode is called optical mode. The other solution for the K~0 leads to u=v thus the planes move in the same direction as it is for long wavelengths sound waves, thus this mode is known as acoustical mode. For the other limit, Ka=± cos(Ka)=-1 and thus the solutions are : 2C M 1 M 2 4C 2 M 1 M 2 16 M 1 M 2 C 2 2C M 1 M 2 2C M 1 M 2 2M 1M 2 2M 1M 2 2 2 Or 2 2C M2 and 2 2C M1 This leads to a solution where each type of atom vibrates independently from the other. One solution correspond to one type of atom vibrating while the other is at rest and viceversa There is a gap between optical and acoustical mode at the boundary of the first Brillouin zone where frequencies between (2C/M2)2 and (2C/M1)2 are not allowed. Figure 8 shows both modes. Also in Figure 8, it can be seen that a 3-D crystal, longitudinal and transversal vibrations split in frequency. In a crystal with a p atom base there are 3 acoustical modes (1 longitudinal and 2 transversal) and 3p-3 optical modes. 5 Quantization of Elastic Waves By describing vibration in the lattice by a quantum harmonic oscillator model, it is observed that the lattice energy is quantized, in other words, the energy of a lattice vibrating with a single frequency is: 1 2 n That means that no just any value of energy is allowed but the ones in that equation. For n=0 the lowest energy state is obtained and it is known as zero point energy. Other values of n represent exited states of the crystal vibration where n photons populate the state. Phonons are not subjected to the Pauli Exclusion Principle (as photons are not) thus any number of phonons can occupy a given state, however you add phonons one by one, a fraction of a phonon would not do. The quantization of the energy has the consequence to produce quantization on the amplitude of the oscillation. To see this consider a vibration described by u uo cos Kx cos t As for a harmonic oscillator, the temporal average of the energy is half kinetic and half potential. 1 u Lets consider the kinetic energy density K , where is the mass density 2 t 2 The volume integral of the kinetic energy density gives (notice that the integral of the cos2 is Lx/2 while the integral in y and z gives Ly and Lz), thus the volume integral is V/2 ¼ V2uo2sin2t with a time average of 1 V2uo2 thus (since <sin2t>=1/2). Making this average kinetic 8 energy equal to ½ of the total energy 1 V2uo2= ½(n+½) ħ 8 we obtain that uo2= 4(n+½) ħ/V= 4(n+½) ħ/M Thus the vibration amplitude is related to the occupancy n of the mode. The solution of the motion’s equation for atoms in the crystal gives 2 so, what is the sign of ? Conventionally is assumed positive. For unstable structures, 2 is negative and is imaginary. 6 Phonon Momentum A phonon of wavevector K can interact with photons, neutrons, electrons, and other particles as if they had a momentum ħK. Looking at this in a different way, if a photon (or other quantum particle) collides with the crystal, without losing energy, the selection rule for the incoming particle wave vector imposes that k’=k+G No energy is transferred to the crystal but the crystal as a whole recoils with a momentum -ħG. If inelastic scattering occurs, some energy is indeed lost by the incoming particle to the crystal that will excite phonon modes, besides the fact that the transferred energy should be at least ħ, the energy of a phonon of frequency, there is a generalized selection rule that can be written as k’+K=k+G where K is the phonon’s wave vector, thus the wavevector of the scattered particle is conditioned to the wavevector of the created phonon such that the above equation is satisfied. In the same way, if a phonon is annihilated in the interaction, the equivalent selection rule is: k’=K+k+G Neutrons are often used to determine the dispersion relations for the phonon. After scattering, the energy and direction of scattered neutrons is recorded thus k and k’ are determined. The kinetic energy of the neutrons is p2/2Mn and the momentum is p= ħk, so conservation of energy imposes 2 k 2 2 k' 2 Where ħ is the energy of the phonon created (+) or absorbed (-) from 2M n 2M n where is obtained. By knowing the crystal lattice (G) and from the formula k’±K=k+G, K is determined, where G is chosen such that K is in the first Brillouin zone. The ± sign applies to phonon created or absorbed. 7