《固体物理学》课程教学课件（PPT讲稿）Chapter 4 lattice dynamics and lattice capacity 4.1 Normal Modes of Vibration 4.2 Density of States 4.3 Harmonic approximation and normal mode coordinates 4.4 phonon

Chap 4:lattice dynamics and lattice capacity Objectives At the end of this Chapter,you should: 1.Know the dispersion relation of 1D monoatom chain and diatom chain. 2.Know the difference between acoustic branch and optic branch. 3.Understand the concept of density of states. 4.Know the capacity difference between Eistein modes and Debye modes. 5.Establish the concept of phonons. 6.Understand the anharmonic effects

Chap 4: lattice dynamics and lattice capacity Objectives At the end of this Chapter, you should: 1. Know the dispersion relation of 1D monoatom chain and diatom chain. 2. Know the difference between acoustic branch and optic branch . 3. Understand the concept of density of states. 4. Know the capacity difference between Eistein modes and Debye modes. 5. Establish the concept of phonons. 6. Understand the anharmonic effects

4.1 Normal Modes of Vibration 4.1.1 One dimensional model 1:The Monatomic Chain Consider a Monatomic Chain of Identical Atoms with nearest-neighbor,"Hooke's Law"type forces(F=-kx) between the atoms.Aforce-spring model,with masses m spring constants K. (+1)a(+2a 1(m2)a m-1)a This is the simplest possible solid.Assume that the chain contains a very large number (N->oo)of atoms with identical masses m. Let the atomic separation be a distance a.we must Assume that the atoms move only in a direction parallel to the chain,and that only nearest-neighbours interact with each other

4.1 Normal Modes of Vibration 4.1.1 One dimensional model 1: The Monatomic Chain • Consider a Monatomic Chain of Identical Atoms with nearest-neighbor,“Hooke’s Law” type forces (F = - kx) between the atoms. Aforce-spring model, with masses m & spring constants K. • This is the simplest possible solid.Assume that the chain contains a very large number (N → ) of atoms with identical masses m. Let the atomic separation be a distance a. we must Assume that the atoms move only in a direction parallel to the chain, and that only nearest-neighbours interact with each other

Consider just two neighboring atoms.Assume that they interact with a known potential V(r).Expand V(r)in a Taylor's series in displacements about the equilibrium separation, V(R苏 ☒无法显示该图片 Repulsive Attractive Un+1 13T V(r)=V(a+δ)=V(a)+ dr 63+. （d2 2 31dr3 The potential energy is the same as that associated with a spring with spring constant: d" B= 6=-B6 dr dr

• Consider just two neighboring atoms. Assume that they interact with a known potential V(r). Expand V(r) in a Taylor’s series in displacements about the equilibrium separation, The potential energy is the same as that associated with a spring with spring constant: r R V(R) 0 a Repulsive Attractive 2 3 2 3 2 3 d 1 d 1 d ( ) ( ) ( ) d 2 d 3! d a a a V V V V r V a V a r r r           = + = + + + +              a a Un-1 Un Un+1 2 2 d d d d a V V f r r     = − = − = −     2 2 d d a V r    = −   

n=f+f方=-B(4n-+)-B(4,-n1)=B(41+4n1-24n) dw=B(4+4-2n.） m The Equation of Motion ofeach atom is of this form.Only the value of 'n'changes. 。Assume Lng Aeflo-nag) 2π q A,amplitude;o,frequency;q,wavenumber;A,wavelength -mo2=B(e-iag +eiag-2)=2B(cosaq-1) 0=21 m 2 max m The only allowed vibration frequencies o must be related to the wavenumber in this way."Phonon Dispersion Relation"for the chain,even though these are classical lattice vibrations in the classical theory

The Equation of Motion of each atom is of this form. Only the value of ‘n’ changes. ( ) ( ) ( ) f f f n n n n n n n n = + = − − − − = + − 1 2 1 1 1 1           + − + − 2 ( ) 2 2 1 1 d 2 d n n n n u m t = + −     + − • Assume 1 2 sin 2 aq m   = ( ) ( ) 2 2 2 cos 1 iaq iaq m e e aq    − − = + − = − ( ) nq i t naq Ae   − = 2 q   = A, amplitude; ω, frequency; q, wavenumber；λ，wavelength The only allowed vibration frequencies ω must be related to the wavenumber in this way. “Phonon Dispersion Relation” for the chain, even though these are classical lattice vibrations in the classical theory

Dispersion curve (k)=sin(ka/2) ↑0 @max =2 m λ2aλ>2a<2a V,=olk (redundant) -n/a n/a The waves with wave numbers k and k+2n/a describe the same atomic displacement Therefore,we can restrict k to within the first BZ [-/a,/a]

max 2 / s K m V k   = =

If one atom starts vibrating,it does not continue with constant amplitude,but transfers energy to the others in a complicated way That is,the vibrations of individual atoms are not simple harmonic because of this exchange of energy among them. They are a collective property of the system as a whole not a property of any of the individual atoms.Each mode represented by o(k)oscillates independently of the other modes.Also,it can be shown that the number of modes is the same as the original number of equations N. Because of periodicity,the nth atom is equivalent to the(N+n)th atom.This means that the assumed solution for the displacements: u,Aexpi(k-ot) must satisfy the periodic boundary condition: Un =uN+n

• If one atom starts vibrating, it does not continue with constant amplitude, but transfers energy to the others in a complicated way. That is, the vibrations of individual atoms are not simple harmonic because of this exchange of energy among them. • They are a collective property of the system as a whole & not a property of any of the individual atoms. Each mode represented by ω(k) oscillates independently of the other modes. Also, it can be shown that the number of modes is the same as the original number of equations N. Because of periodicity, the n th atom is equivalent to the (N+n)th atom. This means that the assumed solution for the displacements: must satisfy the periodic boundary condition: ( ) 0 exp n n u A i kx t = −      un = uN+n

This,in turn requires that there are an integer number of wavelengths in the chain.So,in the first BZ,there are only N allowed values of k. Na=p入 Na=p2→=0-2→M=2p a 2=4a 4 2三 -a 2ππ 91= 2a 2π_ 5π 92 2a q is supposed to be limited in the first BZ,different from that of continuous elastic medium wave

q is supposed to be limited in the first BZ，different from that of continuous elastic medium wave。 1 2 4 4 5 a a   = = 1 1 2 2 2 2 2 5 2 q a q a       = = = = This, in turn requires that there are an integer number of wavelengths in the chain. So, in the first BZ, there are only N allowed values of k. p a Nk p k Na Na p     2 2 Na = p =  = =  =

What is the physical significance of wave numbers k outside of the First Brillouin Zone[-(π/a)≤k≤（π/a)]? ·At the Brillouin Zone edge:-2ak-27→k=a a This k value corresponds to the maximum frequency.A detailed analysis of the displacements shows that,in that mode,every atom is oscillating n radians out of phase with it's 2 nearest neighbors.That is,a wave at this value ofk is A STANDING WAVE. Black k=π/a or λ=2a Green: k=(0.85)π/a or 1=2.35a

• What is the physical significance of wave numbers k outside of the First Brillouin Zone [-(π/a)  k  (π/a)]? • At the Brillouin Zone edge: • This k value corresponds to the maximum frequency. A detailed analysis of the displacements shows that, in that mode, every atom is oscillating π radians out of phase with it’s 2 nearest neighbors. That is, a wave at this value of k is A STANDING WAVE. x Black k = π/a or  = 2a Green: k = (0.85)π/a or  = 2.35 a

Points A and C have the same frequency the same atomic displacements.group velocity v=(do/dk)there is negative,so =/K that a wave at that o and k moves to the left. The green curve(below)corresponds to -π/a0 π/a2m/ak point B in the o(k)diagram.It has the (k)(dispersion relation) same frequency displacement as points A and C,but v=(d/dk)there is positive,PointsA&Care symmetrically so that a wave at that o&that k moves to equivalent;adding a multiple of2/a the right. to k does not change either c or va so point A contains no physical information that is different from point B. The points k=±π/a have special @-m=4Ksin2 ka 2 significance n=n2r=n2r→ VA k In a 2a=2dsin90→d=a Bragg reflection occurs at k=±nπ/a

Points A and C have the same frequency & the same atomic displacements. group velocity vg = (dω/dk) there is negative, so that a wave at that ω and k moves to the left. The green curve (below) corresponds to point B in the ω(k) diagram. It has the same frequency & displacement as points A and C, but vg = (dω/dk) there is positive, so that a wave at that ω & that k moves to the right. Points A & C are symmetrically equivalent; adding a multiple of 2π/a to k does not change either ω or vg , so point A contains no physical information that is different from point B. T The points k = ± π/a have special significance x n2 2 n n k n a     = =  Bragg reflection occurs at k= ± nπ/a 2 2 4 sin 2 ka  m K = u n x u n a -π/a k V K m K s / 2   = = k  C B A 0 ω(k) (dispersion relation) 0 π/a 2π/a

So,the group velocity is: Vg=(do/dk)=a(K/m)%cos(ka) vg=0 at the BZ edge[k=±（π/a)] This tells us that a wave with A corresponding to a zone edge wavenumber k=±（π/a)will not propagate. That is,it must be a standing wave! At the BZ edge,the displacements have the form (for site n): U.=U。einka=U。ei(n/a)=U(-1)r 1.0 (CaM西 Vg=(do/dk) =a(K/m)cos(ka) 05

So, the group velocity is: vg  (dω/dk) = a(K/m)½cos(½ka) vg = 0 at the BZ edge [k =  (π/a)] – This tells us that a wave with λ corresponding to a zone edge wavenumber k =  (π/a) will not propagate. That is, it must be a standing wave! – At the BZ edge, the displacements have the form (for site n): Un= Uoe inka = Uo e i(nπ/a) = Uo (-1)n vg  (dω/dk) = a(K/m)½cos(½ka)