Phonons (Classical theory)

Transcription

1 Phonons (Classical theory) (Read Kittel ch. 4) Classical theory. Consider propagation of elastic waves in cubic crystal, along [00], [0], or [] directions. Entire plane vibrates in phase in these directions we will treat this as a D problem. atom per primitive cell equilibrium positions U s- U s+ U s+ U s+3 longitudinal wave U i : displacements from equilibrium s- s s+ s+ s+3 U s- U s+ U s+ U s+3 transverse wave modes - why? ( polarizations) s- s s+ s+ Assume nearest neighbor interations only, atomic mass, M F s CU ( s + U s ) + CU ( s U s ) - 5 -

2 look for harmonic solutions U s Aexp[ i( ωt ksa) ] then ω C ( coska) M and so, ω 4C M sin ka ω π a st Brillouin zone π a only k values inside st BZ are physically significant. Why? long wavelength limit (also continuum limit): ka << cos ka - (ka) / - 6 -

3 so, ω v sound k where the velocity of sound, v sound now consider atoms per primitive basis a u s v s u s + v s + let atom displacements {v } M atom M displacements {u } M v s Cu ( s + u s + v s ) M u s Cv ( s + v s u s ) again, take travelling wave solutions u s v s Plug into equations and obtain dispersion relation u v exp[ i( ωt ska) ] ω C --- ± μ C --- μ 4C -- sin M M ka where μ is the reduced mass, - 7 -

4 One way to see the origin of the optical and acoustic modes: Check M M case (a' a -- ω C M ---- cos ka' sin ka' for + for - C M ---- C M π a' π a' π a' --- π a' k Same as before. The cos curve duplicates the sin curve. Each pt. on the cos curve represents a mode on the sin curve (extended zone representation). Now let M differ from M. Primitive cell now larger ( a' ). Brillouin zone boundary at - 8 -

5 ± π a'. Gaps open at zone boundaries. C μ ω optical branch C M Acoustic branch C M π π a' a' π a' --- π a' New BZ for optic modes, the two atoms vibrate opposite to each other, for acoustic-vibrate together acoustic mode optical mode - 9 -

6 if binding is partly ionic, then optic mode has a dipole moment interacting with radiation. Optic mode can be excited by light. For p atoms in primitive cell, there are 3p branches 3 acoustic, 3p-3 optical Phonons in real crystal long range forces. In metals, the coduction electrons tend to modulate forces at long distances. In ionic crystals, coulomb forces from the charged ions also act at long distance. Even in semiconductors, forces turn out to be long range due to higher order multipolar interactions. bond bending forces θ in addition to springs shown, there is stiffness to the bond angles. Leads to additional complications. Our model is qualitative only, but shows the salient features. Tables of phonon dispersions are given in: Landoldt-Börnstein, New Series, III,, O. Modelung & M. Schulz, eds. (Springer Verlag, Berlin, 987)

7 Quantum Mechanics of Phonons (read Kittel, Appendix C) Review of quantum mechanics of the simple harmonic oscillator (SHO) Hamiltonian for D SHO, mass m, resonant frequency ω: p: momentum operator, x: displacement operator p ih - d dx A quantum mechanical commutator is defined as: out to be a useful object. Take linear combinations of the p and x operators: [ x, p] xp px ih -. This will turn a a m h - ω ωx m h - ω ωx i + ---p m i ---p m Clearly, x and p can be expressed in terms of a and a : x p h ( a+ a ) mω h - - m ω ( a a) Also, it is easy to show that [a, a ]. Using these relations, we can re-express the Hamiltonian in terms of the canonical operators a and a : h - ω a a + -- (using the commutator) - -

8 We can thus express the Hamiltonian quite compactly: with N a a. N is called the number operator. Both H and N can be shown to be Hermitian operators, which implies they have real eigenvalues. Dirac bra-ket notation We write: n to represent the eigenstate of the N operator with eigenvalue n. e.g. These are also the eigenstates of H, with: Nn nn We also write: n which represents the complex conjugate of n. Finally, n' n represents the overlap integral of n with n. Using these relations, it is easy to show that: relation tells us that: Na n ( n )an. Examination of this an is an eigenstate of N, with eigenvalue n -, so that with c n some constant to be determined. Since it reduces n by, a is called the annihilation operator. Similarly, a n is an eigenstate of N, with eigenvalue n + Since it increases n by, a is called the creation operator.,. - -

9 We therefore find that the energy levels of the harmonic oscillator form a ladder: ( a ) n a n The ladder must have a lower bound. To see this take:.... n Nn n a an. Using the property of the Hermitian conjugate of an operator, we can write this as: n an ( a) n an an c n, which must be 0. But we also have n Nn n, so therefore, n 0. Also, c n n (we choose the positive root, by convention). Similarly, Finally, we can prove that n must be an integer. If we apply the a operator to the state on the ladder with 0 < n <, the next lower state would have a negative n, which we have just shown is not allowed. But if n must be an integer, then we would reach a 0 0, which terminates the ladder. Application of harmonic oscillator theory to lattice vibrations The wavefunction for the crystal lattice, φ(r, R,... R N ) is a function of the positions of all of the ions, R, R,... R N, where N is the total number of ions in the crystal. The Hamiltonian for φ(r, R,... R N ) is: H N k P k N VR M k ( i R j ) ij, i j - 3 -

10 where P k represents the momentum of the kth ion, and V(R i - R j ) represents the pairwise potential between the ith and the jth ion. Near the equilibrium positions of the ions, the potential has the form of a constant plus a quadratic variation for small motion about equilibrium. Define the magnitude of the displacement from equilibrium for the ith ion as δr i. Dropping the constant, and making a Taylor series expansion of V(R i - R j ) in the small displacements, we have: H N k As in the classical analysis, we make a Fourier analysis of the atomic displacements: The potential term becomes (neglecting the polarization aspects): P k N V + δr M k i R i R δr j i, j i j - u q e iq R j e ˆ q N q j δr j pairwise interatomic force constant unit polarization vector e iq R i N i j qq', V R i R j e iq' R j u q u q' u N q u q' e iq R i R j qq', i j ( ) V R i R j e iq ( + q' ) R j only a function of R i - R j Hold R i - R j fixed, and sum over R j. For each R i - R j, we get: The Fourier Transform of the momentum is simply expressed as obtain a Fourier domain representation of the Hamiltonian: P q d M uq. Finally, we dt H qq', Mdu ---- q dt du q' dt c q u q u q' δ( q + q' ) - 4 -

11 with the force constant: ( ) c q e iq R i R j i j V R i R j now complete the sum over all R i - R j H q Mdu ---- q dt du q c q u dt q u q H q -P M q P q c q u q u q P q is the conjugate momentum to u q in the Fourier representation. Each Fourier component of the lattice vibration (phonon wave) has a Hamiltonian which has the exact form of a D SHO. [This is identical to the treatment of the quantization of the electromagnetic field, which leads to photon modes that are also harmonic oscillators.] We identify c q Mω q. Now go back and use the analysis we did on the SHO. We define phonon creation and annihilation operators: a q M h _ ω q -- ω q u q ----P M q i a q M h _ ω q -- i ω q u q P M q The phonon eigenstates are. The energy per phonon mode is n q E n h _ ω q n q + -- Since the total Hamiltonian is a sum of single phonon oscillator terms, the total phonon wavefunction can be written as a product: Φ n n n N We now have a very powerful and compact representation of the phonon physics. The creation and annihilation operators will be extremely useful when we begin to study electronphonon scattering, where electrons gain or lose energy via the annihilation or creation of phonons

12 Phonons and Heat Capacity of the Lattice (read Kittel ch.5) This subject serves to illustrate a number of the concepts we have developed thusfar, and is important in it s own right. Beginning with a classical approach: For N harmonic oscillators, total energy is: Equipartion theorem of statistical mechanics: mean value of each quadratic term in energy --kt (kt/degree of freedom). So the average total thermal energy is: 3Nk B T Heat Capacity cons tant volume U C V T V 3Nk Dulong-Petit Law Why? -- The classical expres- Experimentally valid for high temperatures. Fails as sion for the energy is wrong! T 0 Quantum statistical mechanics: n q : quantum number of phonons of wavevector q Quantum statistical mechanics avg. thermal excitation given by Bose-Einstein distribution - 6 -

13 If we know phonon density - of - states, D λ ( ω), (λ: polarization index) then U ωd λ ( ω) n q + -- d h ω λ correct quantum thermal energy zero - point energy : --hω per mode a constant q Problem is to find D( ω). Follows from phonon dispersion relation ω(q). In 3 dimensions: dζ( ω) D( ω) -- dω ζ # of modes with frequency ω Do mode counting :. Phonon spatial dependence is sinusoidal. Assume crystal is a cube with side L. Recall that the ion displacements are given by the Fourier series representation: δr j - u q e iq R j e ˆ q N q This form of the series satisfies periodic boundary conditions : L L L δr j ( x + L, y, z) δr j ( x, y, z) and similarly for y, z, if q satisfies: In calculating the density of states, it will be useful to calculate the total number of phonon modes, ζ(q) that have a wavevector amplitude less than or equal to a particular value, q. This is easily calculated by considering the space of allowed q vectors given by the above condition. The number we want is given by taking the volume of a sphere of radius q 4 --πq 3, and dividing by the volume taken up by each allowed point in q-space 3-7 -

14 π L. We want dζ -- ( ω), which, using the chain rule, can be written: dω. So, D( ω) dζ( ω) Vq dq dω π dω Once we know the phonon dispersion relation q(ω), we can calculate the density of phonon states, D(ω). One example would be the simple, classical dispersion: ω 4C M sin qa -----, This one leads to somewhat messy algebra. There are some simplified dispersions that are often used. Einstein model Einstein assumed only one oscillator frequency ω E : Einstein frequency Decent representation of the optic phonons. (Also of some historical interest.) then U ωd λ ( ω) n q + -- _ d h ω λ 3Nh _ ---- ω E _ + exp( h ωe k B T) Z.P.E. zero point energy The constant zero point energy is unimportant since we will be taking a derivative of the energy

15 then C V U T 3N( h _ ω E ) e hω E k B T k B T [ exp( h _ ω E k B T) ] high temp limit: as T, C V 3Nk B and we recover Dulong - Petit AND as T 0, C V 0. This result was an early triumph of quantum mechanics. Unfortunately, experimentally, the form of the heat capacity at low temperatures was found to go as C V T 3 at low T. The Einstein result is:. The temperature dependence of C V in the Einstein theory is too steep because when kt «hω E, the occupation of all modes drops exponentially. The acoustic phonon modes have been neglected. At low temperatures, these dominate. Debye model good model for the acoustic phonon behavior. take simple linear phonon dispersion, ω v sound k. Then we get: Need a cutoff to get correct total # modes. Choose a cutoff frequency (called the Debye frequency), such that: ω D 0 D( ω) dω N (number of modes for one polarization) V ω D π N 3 v 3 sound

16 ω D now U ω Vω hω d - π 3 hω kt v s e + Z.P.E. λ 0 Assume the 3 polarizations all have the same sound velocity, U 3Vh _ - π 3 v s which now gives the Debye heat capacity: ω D 0 ω 3 dω hω kt e C V let du dt 3Vh _ π 3 v s kt ω D 0 hω kt ω ω 4 - e d ( e hω kt ) Debye Temperature hv - s 6π N 3 k V then dω kt -----dx h ω 4 kt h 4 x 4 3Vh _ C kt xe x x 4 V d π 3 v s kt h -- ( e x ) - 3V k4 T π 3 v s h 3 x D

17 Using the definition of θ D, and after a bit of algebra, we obtain: θ D T C V 9Nk T B xe x x 4 d θ D -- ( e x ) 0 Debye Heat capacity High temperature limit: as T, x 0 x D 0 dxx 4 e x -- ( e x ) x D 0 dx( x)x x x D So, C V 3Nk B recovering the Dulong - Petit high temperature heat capacity. Now consider the low temperature limit. As T 0, do the U integral x D U 9NkT T 3 θ D dx- x3 e x 0 The T 0 limit corresponds to x D. The definite integral can be done: 0 x3 dx- e x π Thus, 3π 4 U --NK----- T4 5 θ D 3 Debye T 3 law This is a very good match to what is observed experimentally for the lattice heat capacity of crystals. Later we will also consider the electronic heat capacity, which is important for metals

18 Real crystals exhibit a somewhat more complicated D(ω): D( ω) ω D Representative real crystal Debye Model Discontinuities at zone boundaries ω Higher Order Effects: Phonon Decay and Thermal Expansion Phonon decay ω TO LO LA TA q Optic modes non-dispersive: dω dq vs small. Optic phonons tend to be localized - decay to acoustic modes

19 Anharmonicity causes this coupling. potential of atom at displacement x from equilibrium: Ux ( ) Cx gx 3 fx 4 + Ux ( ) harmonic potential asymmetry of repulsion softening for large amplitude x In 3D - g a 3 rd rank tensor. Treating this quantum mechanically, Fourier transform domain: H 3 ( r ) ---- δr 3! j ( δr i g ijk δr k ) ijk H 3 ( q) u q' ( u q C qq'q'' u q'' )δ( q + q' + q'' ) 3! N q'q'' This perturbation will lead to mode q decaying into q', q''. Leaving out the polarization (tensor) effects, and expressing the Fourier coefficients by raising and lowering operators: H 3 ( q) _ - h - N M 3 -- q'q'' Ĉ qq'q'' -- ( a q + a q) ( a q' + a q' )( a q'' + a q'' )δ( q + q' + q'' ) ω q ω q' ω q'' There are 8 possible triple products of the raising/lowering operators, leading to various 3- phonon coupling effects. Here we are interested in terms where an optic phonon is destroyed, and two different acoustic phonons are created: H 3 ( q) _ - - h N M 3 -- q'q'' Ĉ qq'q'' --a q a q'a q''δ( q + q' + q'' ) ω q ω q' ω q''

20 We can calculate the phonon decay rate using Fermi s Golden Rule: Γ LO π _ n h q, n q' +, n q'' + H 3 n q, n q', n q q' δ ( E ) q' πh _ -- 4NM 3 q' Ĉ qq' ( q q' ) ----n ω q ω q' ω q ( n q' + ) ( n q q' + )δ( h _ ω q h _ ω q' h _ ω q + q' ) q + q' Only one coupling constant needed (C). The dependencies on phonon frequencies and occupations (temperature) falls out very simply. Thermal expansion. as T increases, <x> clearly increases Basic effect can be understood most easily classically: assume anharmonic terms numerator: «kt dx xe Ux ( ) kt xe cx kt d 0 0 x gx 4 - fx kt kt denominator: 3 -- g π ---- ( kt)3 4 c 5 dxe Ux ( ) kt xe cx kt d πkt --- c linear thermal expansion