Chapter 4 Phonons Chapter 4 Phonons Two Atoms/Primitive Basis � Last time, we derived the dispersion relation for a monatomic cubic solid � Now, we are going to do the same analysis for a solid with 2[.]
Chapter 4: Phonons Two Atoms/Primitive Basis • • • • Last time, we derived the dispersion relation for a monatomic cubic solid Now, we are going to the same analysis for a solid with atoms/primitive basis (eg NaCl, diamond cell, ZnS, CsCl, etc.) We will find that there are two branches on this dispersion curve (the acoustic, or low energy branch, and the optical, or high energy branch) How will this curve change for something like diamond? ω A sneak peak at inelastic neutron scattering • • How we measure dispersion curves? Inelastic neutron scattering If you can measure the energy lost by the neutron (by causing a vibration in the solid), and you can measure which direction you created the wave (the wavevector), then you can construct a dispersion curve Inelastic neutron scattering Triple-axis neutron spectrometer Defines incoming wavelength Defines outgoing wavelength, which may be different if the neutron lost a bit of energy to make a phonon A nerdy neutron scatterer Inelastic neutron scattering Bert Brockhouse, a Canadian scientist, pioneered this technique in Canada (He invented the triple axis spectrometer) Acoustic/Optical Branches • • The acoustic branch has this name because it gives rise to long wavelength vibrations – speed of sound The optical branch is a higher energy vibration (the frequency is higher, and you need a certain amount of energy to excite this mode) The term “optical” comes from how these were discovered – notice that if atom is +ve and atom is –ve, that the charges are moving in opposite directions You can excite these modes with electromagnetic radiation (ie The oscillating electric fields generated by EM radiation) Transverse optical and acoustical modes Two Atoms/Primitive Basis • For each one of these branches, there is also a transverse mode, and a longitudinal mode Two Atoms/Primitive Basis • • If there are p atoms in the unit cell, there has to be 3p branches in the dispersion relation: acoustical branches, and 3p-3 optical branches (there are more ways for the atoms to move for the optical modes) So, for Germanium, which has atoms/unit cell (diamond structure: 000 and ¼ ¼ ¼ ), there has to be branches: one longitudinal acoustic (LA), transverse acoustic (TA), one longitudinal optical (LO), transverse optical (TO) (phonon dispersion curve in the [111] direction for Germanium at 80 K, as determined by inelastic neutron scattering) Two Atoms/Primitive Basis • • • Inelastic neutron scattering is the experimental method of choice to look at lattice vibrations We will look at this later in the chapter The dispersion curves of KBr (2 atoms/unit cell) look similar as for diamond How can we calculate what these curves look like? (Phonon dispersion curve for KBr at 90 K, as determined by neutron scattering) Two Atoms/Primitive Basis • • • • • Let’s look at a cubic crystal where we have one set of atoms of mass M1 and another M2 which alternate Actually, it’s not essential that the masses have to be different –we just need atoms in the basis For simplicity sake (to capture the shape of these curves), we are going to look only for vibrations that excite a row of the same type of atom Examples of this are the [111] direction of the NaCl structure, or the [100] direction of the CsCl structure How can we write the equations of motion? M1 M2 [100] direction Two atoms/primitive basis Assuming that the force constants are identical, we have: d us M = C (vs + vs −1 − 2u s ) dt d vs M 2 = C (u s +1 + u s − 2vs ) dt us-1 vs-1 us vs us+1 vs+1 M1 M2 M1 M2 M1 M2 C C C C a C Two atoms/primitive cell • • • We now assume that our solutions will look like harmonic oscillators: us=uexp(isKa)exp(-iωt), vs=vexp(isKa)exp(-iωt) The amplitudes, u and v, can now be different for each plane Substituting this into the diff equation above, we have: d 2u s M = C (v exp(isKa ) exp(−iωt ) + v exp(i ( s − 1) Ka ) exp(−iωt ) − 2u exp(isKa ) exp(−iωt )) dt − M 1ω 2u exp(isKa ) exp(−iωt ) = exp(isKa ) exp(−iωt )C[v + v exp(−iKa ) − 2u ] − M 1ω 2u = Cv[1 + exp(−iKa )] − 2Cu − M 2ω v = Cu[1 + exp(iKa )] − 2Cv We can solve this by taking the determinant: (2 eq, unknowns) (and similarly, solving for the M2 term) 2C − M 1ω − C[1 + exp(−iKa )] =0 − C[1 + exp(iKa )] 2C − M 2ω M 1M 2ω − 2C ( M + M )ω + 2C (1 − cos Ka ) = Two atoms/primitive cell • • We can solve this exactly for ω2, but its easier for now to just look at the limiting case (ie When Ka is small, and when Ka is at the zone boundary) For small Ka, cos Ka ~ – ½ (Ka)2, so we have: M 1M 2ω − 2C ( M + M )ω + 2C (1 − (1 − ( Ka ) ) = M 1M 2ω − 2C ( M + M )ω + C ( Ka ) = (2C ( M + M )) − M 1M 2C ( Ka ) 2C ( M + M ) ± ω = M 1M 2 M 1M 2 1 + ω ≅ 2C M M 2 1 C K 2a ω2 ≅ M1 + M 2 Simplifying this, we have: (assignment #4) (optical branch) (acoustic branch) Acoustic/optical branch • Compare this to the experimental curves 1 + ω ≅ 2C M M 2 1 C K 2a ω2 ≅ M1 + M 2 (optical) (acoustic) Acoustic mode is linear in this region, just like the one atom basis lattice Zone boundary • • At the zone boundary, Ka = +/- π, so cos(Ka) = -1 Substitute this back into our original equation: 2 ( C ( M + M )) − M M ( − C ) 2C ( M + M ) 2 ± ω = M 1M 2 M 1M (assignment #4) Roots: 2C ω ≅ M1 2C ω ≅ M2