Solid State Physics. Lecturer: Dr. Lafy Faraj

Transcription

1 Solid State Physics Lecturer: Dr. Lafy Faraj

2 CHAPTER 1 Phonons and Lattice vibration

3 Crystal Dynamics Atoms vibrate about their equilibrium position at absolute zero. The amplitude of the motion increases as the atoms gain more thermal energy at higher temperatures. In this chapter we discuss the nature of atomic motions, sometimes referred to as lattice vibrations. Atomic motions are governed by the forces exerted on atoms when they are displaced from their equilibrium positions. 3

4 Hooke's Law One of the properties of elasticity is that it takes about twice as much force to stretch a spring twice as far. That linear dependence of displacement upon stretching force is called Hooke's law. F spring = K. x Spring constant K F It takes twice as much force to stretch a spring twice as far. F 4

5 Monoatomic Chain The simplest crystal is the one dimensional chain of identical atoms. Chain consists of a very large number of identical atoms with identical masses. Atoms are separated by a distance of a. Atoms move only in a direction parallel to the chain. Only nearest neighbours interact (short-range forces). a a a a a a U n- U n-1 U n U n+1 U n+ 5

6 Monoatomic Chain Start with the simplest case of monoatomic linear chain with only nearest neighbour interaction The force on the n th atom; a a The force to the right; K( u ) n 1 un The force to the left; K( u ) n u 1 n U n-1 U n U n+1 Hooks law F=-Kx The total force = Force to the right Force to the left 6

7 Monoatomic Chain since.. mu n K( un un 1 un 1) Eqn s of motion of all atoms are of this form, only the value of n varies All atoms oscillate with a same amplitude A and frequency ω. Then we can offer a solution; Where (k) is wavevector F = m d u dt = mu n 0 un Aexp i kxn t.. u n u n Where (m) is mass of atom H.W 0 0 x n na Undisplaced position x na n u n Displaced position 7

8 Equation of motion for n th atom mu n = K(u n+1 u n + u n 1 ) By substituting the solution in eq. of motion mω Ae i kna ωt = K Ae i k(n+1)a ωt Ae i kna ωt + Aei k(n 1)a ωt By using cosx = eix + e ix We get mω = K(1 coska) By using 1 cosx = sin x Maximum value of it is 1 We get ω = 4K m sin ka When ka = π k = ± π a The maximum allowed frequency is: ω max = 4K m 8

9 Phonon Dispersion Relations or Normal Mode Frequencies or ω versus k relation for the monatomic chain. C B A V max s / k K m л / a 0 0 л / a л / a Because of Brillouin zone periodicity with a period of π/a, only the first BZ is needed. Points A, B & C correspond to the same frequency, so they all have the same instantaneous atomic displacements. k 9

10 Briefly look in more detail at the group velocity, v g. The dispersion relation is: So, the group velocity is: v g (dω/dk) = a(k/m) ½ cos(½ka) 4K v g = 0 at the BZ edge [k = (π/a)] (H.W) This tells us that a wave with λ corresponding to a zone edge wavenumber k = (π/a) will not propagate. m That is, it must be a standing wave! ka sin v g Ka /m 1/ k 1 st BZ Edge 10

11 First Brillouin Zone What range of k is physically significant for elastic waves? The ratio of the displacements of two successive planes is given by u n+1 u n = uei n+1 ka ue inka = e ika The range -π +π for (ka) covers all independent values of e ika π a < k π a First Brillouin Zone If k outside these limits, we use formula: k = k πn/a Then displacement ratio u n+1 u n = e ik a where (e iπn = 1) H.W πn/a is reciprocal lattice vector 11

12 Monoatomic Chain Since there is only one possible propagation direction and one polarization direction, the 1D crystal has only one sound velocity. In this calculation we only take nearest neighbor interaction although this is a good approximation for the inert-gas solids, its not a good assumption for many solids. If we use a model in which each atom is attached by springs of different spring constant to neighbors at different distances many of the features in above calculation are preserved. Wave equation solution still satisfies. The detailed form of the dispersion relation is changed but ω is still periodic function of k with period π/a Group velocity vanishes at k=(±)π/a There are still N distinct normal modes Furthermore the motion at long wavelengths corresponds to sound waves with a velocity given by (velocity formulü) 1

13 Chain of two types of atom Two different types of atoms of masses M and m are connected by identical springs of spring constant K; (n-) (n-1) (n) (n+1) (n+) M K K K K m M m M a) a b) U n- U n-1 U n U n+1 U n+ This is the simplest possible model of an ionic crystal. Since a is the repeat distance, the nearest neighbors separations is a/ 13

14 Chain of two types of atom We will consider only the first neighbour interaction although it is a poor approximation in ionic crystals because there is a long range interaction between the ions. The model is complicated due to the presence of two different types of atoms which move in opposite directions. Our aim is to obtain ω-k relation for diatomic lattice Two equations of motion must be written; One for mass M, and One for mass m. 14

15 Chain of two types of atom M m M m M U n- U n-1 U n U n+1 U n+ Equation of motion for mass M (n th ): mass x acceleration = restoring force Mu n = K u n+1 u n K u n u n 1 Mu n = K u n+1 u n + u n 1 Equation of motion for mass m (n-1) th : mu n 1 = K u n u n 1 K u n 1 u n mu n 1 = K u n u n 1 + u n 15

16 Chain of two types of atom M m M m M U n- U n-1 U n U n+1 U n+ Offer a solution for the mass M 0 0 uu n Aexp i kxn t n = Aexp i kx n ωt xn x n na = na/ For the mass m; 0 uu n n 1 = Aexp αaexp i kxi kx n n t ωt α : complex number which determines the relative amplitude and phase of the vibrational wave. u n = ω Aexp i kx n ωt u n 1 = ω αaexp i kx n ωt 16

17 Equation of motion for n th atom (M): By substituting the solution: Mu n = K u n+1 u n + u n 1 ω kna i MAe ωt = K αae k(n+1)a i ωt Ae i kna ωt + αae k(n 1)a i ωt Cancel the common terms ω M = K 1 αcos ka H.W where cosx = eix + e ix 17

18 Equation of motion for the (n 1) th atom (m) By substituting the solution: mu n 1 = K u n u n 1 + u n αaω me k n 1 a i ωt kna i = K Ae ωt αae k n 1 a i ωt + Ae i k n a ωt Cancel the common terms αω m = K α cos ka H.W 18

19 Chain of two types of atom Now we have a pair of algebraic equations for α and ω as a function of k. ω M = K 1 αcos ka for M αω m = K α cos ka for m α can be found as: α = Kcos(ka ) K ω m = K ω M Kcos(ka ) H.W A quadratic equation for ω can be obtained by cross-multiplication ω 4 K m + M ω + 4K sin (ka ) = 0 The two roots are: We get: ω m + M = K x 1, = b ± ± K m + M b 4ac a 4 sin (ka ) 1/ Dispersion relation 19

20 Chain of two types of atom As there are two values of ω for each value of k, the dispersion relation is said to have two branches; A B C Optical Branch Upper branch is due to the +ve sign of the root. л/ a 0 л / a л / a k Acoustical Branch Lower branch is due to the -ve sign of the root. The dispersion relation is periodic in k with a period π /a = π /(unit cell length). This result remains valid for a chain of containing an arbitrary number of atoms per unit cell. 0

21 Let s examine the limiting solutions at 0, A, B and C. 1- In long wavelength region In this case (ka<<1); sin(ka/) ka/ ω = K m + M ± K m + M 4 sin (ka ) 1/ ω K m + M ± K m + M (ka) 1/ ω m + M K 1 ± 1 ka m + M 1/ Use Taylor expansion: (1 x) 1/ 1 1 x for x<<1 ω K m + M 1 ± 1 1 ka m + M 1

22 Taking +ve root; in case ka<<1 ω op K m + M (max value of optical branch) H.W Taking -ve root; in case ka<<1 ω ac 1 (ka) K m + M (min value of acoustical brach) By substituting these values of ω in α (relative amplitude) equation and using cos(ka/) 1 for ka<<1 we find the corresponding values of α for acoustical branch as; H.W α = Kcos(ka ) K ω m α = 1 acoustical branch H.W

23 Chain of two types of atom The relative amplitude α=1 of acoustical branch represents long-wavelength sound waves in the neighborhood of point 0 in the graph; the two types of atoms oscillate with same amplitude and phase. and the velocity of sound is v s = ω k v s = a K (M + m) Optical A B C Acoustical π/ a 0 π / a π / a k 3

24 Chain of two types of atom We substitute ω m + M op K For optical branch in equation α = Kcos(ka ) K ω m We get α = M m Optical branch H.W The relative amplitude α=-m/m of optical branch, This solution corresponds to point A in dispersion graph. This value of α shows that the two atoms oscillate in antiphase (the two atoms are oscillating 180º out of phase) with their center of mass at rest. Optical A B C Acoustical π/ a 0 π / a π / a k 4

25 Chain of two types of atom - At Brillouin Zone boundary In this case ka= π, i.e sin(ka/)=1. ω m + M = K ω = K ω = m + M K Taking -ve root; ± K m + M ± K m + M (M + m) ± (M m) ω ac = K M 4 sin (ka ) 4 1/ at point (C) 1/ Optical Acoustical π/a A B C ω op = 0 π/a π/a (max value of acoustical brach) K m ω ac = k K M Taking +ve root; ω op = K m at point (B) (min value of optical brach) At max.acoustical point C, M oscillates and m is at rest. At min.optical point B, m oscillates and M is at rest. 5

26 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 1 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) 6