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Lecture Note on Solid State Physics Lattice waves Masatsugu Suzuki and Itsuko S. Suzuki Physics, Department, State University of New York at Binghamton (December 6, 2007) Abstract A lecture note on the lattice waves in the solid is presented. In a crystal each atom are coupled with the neighboring atoms by spring constants. The collective motion of atoms leads to a well-defined traveling wave over the whole system, leading to the collective motion, so called phonon. Here the equation of motion of atoms around thermal equilibrium position will be discussed in terms of several methods, which include numerical calculation on the eigenvalue problem (based on the Mathematica) and the translation operators in the quantum mechanics. We show that all these methods lead to the same conclusion, the existence of lattice wave, phonon in the quantum mechanics. For students who just start to study the solid state physics, it may be difficult to understand the validity of the assumption that the deviation of the displacement of the atoms from the thermal equilibrium is well described by a traveling wave. We numerically solve the eigenvalue-problem of the motion of atoms in the linear chain (typically 50 - 100 atoms) using Mathematica. We will give a bit of evidence for the existence of the normal modes propagating along the linear chain. Content 1. Introduction 2. Lattice waves 2.1 Overview 2.2 One dimensional case: longitudinal mode 2.3 Born-von Karman boundary condition 2.4 First Brillouin zone 2.5 Normal modes 2.6 Continuum wave equation 3. Eigenvalue-problem; solution using Mathematica 4. First Brillouin zone and group velocity 4.1 Definition of the group velocity 4.2 The physical meaning of the first Brillouin zone 4.3 Standing wave 4.4 General property of the group velocity 5. Determination of force constants 5.1 The system with the nearest neighbor interaction 5.2 System with long-ranged interactions 6. Vibration of square lattice 7. Two atoms per primitive basis 8. The number of modes; degree of freedom 8.1 One-dimensional case 8.2 Three-dimensional case 9. Classical Model 1 9.1 Theory of the transverse wave in a string 9.2 Energy density of the elastic wave 10. Quantum mechanical approach: phonon 10.1 Annihilation and creation operators 10.2 Symmetry of lattice and translation operator 11. Crystal momentum 12. Semi-classical approach 12.1 Simple case 12.2 General case 13. Inelastic neutron scattering by crystal with lattice vibration 13.1 Scattering cross section 13.2 Energy and momentum conservation Conclusion References Appendix A. Simple harmonics (1D) in quantum mechanics B. Translation operator in quantum mechanics 1. Introduction Phonons are a quantum mechanical version of a special type of vibrational motion, known as normal modes in classical mechanics, in which each part of a lattice oscillates with the same frequency. These normal modes are important because, according to a well-known result in classical mechanics, any arbitrary vibrational motion of a lattice can be considered as a superposition of normal modes with various frequencies; in this sense, the normal modes are the elementary vibrations of the lattice. Although normal modes are wave-like phenomena in classical mechanics, they acquire certain particle-like properties when the lattice is analyzed using quantum mechanics (see wave-particle duality.) They are then known as phonons. There have been many excellent textbooks on the physics of lattice waves. Typical books1-14 which we read during the preparation of writing this lecture note, are presented in References. These books are very useful for our understanding physics. 2 2.1 Lattice waves Overview Consider the elastic vibrations of a crystal with one atom in the primitive cell. We want to find the frequency of an elastic wave in terms of the wavevector k and the elastic constants. When a wave propagates along the x-direction, entire planes of atoms move in phase with displacements either parallel or perpendicular to the direction of k. We can describe with a single co-ordinate us the displacement of the plane s from its equilibrium position. 2 Fig,.1 (Left figure) (Dashed lines) Planes of atoms when in e quilibrium. (Solid lines) Planes of atoms when displaced as for a longitudinal wave. The coordinate u measures the displacement of the planes. (Right figure) Plane of atoms as displaced during passage of transverse wave. For each wavevector there are three modes; one of longitudinal polarization, two of transverse polarization. We assume that the elastic response of the crystal is a linear function of the forces. Or the elastic energy is a quadratic function of the relative displacement of any two points in the crystal. The forces on the plane s caused by the displacement of the plane s+p is proportional to the difference us+p - us of their displacements. For brevity, we consider only nearest-neighbor interactions, so that p = ±1. The total force on s comes from planes s ± 1. Fig.2 The displacements of atoms with mass M are denoted by us-1, us, and us+1. The repeat distance is a in the direction of the wavevector k. The direction of us is parallel to the direction of the wavevector k for the longitudinal wave and is perpendicular to the direction of the wavevector k for the transverse wave 3 2.2 One dimensional case: longitudinal mode We start our discussion with the Lagrangian for the displacement of the s-th plane, given by Ls = Ts − Vs 1 1 mu s2 − C[(u s +1 + a − u s − a ) 2 + (u s + a − u s −1 − a) 2 ] . 2 2 1 1 = mu s2 − C[(u s +1 − u s ) 2 + (u s − u s −1 ) 2 ] 2 2 The Lagrange’s equation for this system is derived as d ⎛ ∂L ⎞ ∂L ⎟= ⎜ , dt ⎜⎝ ∂u s ⎟⎠ ∂u s or 1 mus = − C[2(us +1 − us )(−1) + 2(u s − us −1 )] 2 , = −C (−us +1 + us + us − u s −1 ) = (2.1) (2.2) (2.3) = C (u s +1 − 2us + us −1 ) = Fs where Fs = C (us +1 − us ) − C (us − us −1 ) is the effective force on the s-th plane (Hooke’s law), C is the force constant between nearest-neighbor planes, C L ≠ CT (CL: force constant for longitudinal wave, CT: force constant for transverse wave). It is convenient hereafter to regard C as defined for one atom of the plane, so that Fs is the force on one atom in the plane s. The equation of motion of the plane s is d 2u s M (2.4) = C (u s +1 − 2u s + u s −1 ) , dt 2 where M is the mass of an atom in the s-th plane. Suppose that this equation has the traveling wave solutions of the form u s = ue i ( ksa −ωt ) . (2.5) Note that the validity of this assumption will be verified by solving directly the eigenvalue problem (see Sec. 2.5). The boundary condition is illustrated below. Fig.3 One-dimensional array of equal masses and springs. This is the simplest model of a vibrational band.. Alternative representation of the Born-von Karman boundary condition. The object connecting the ion on the extreme left with the spring on the extreme right is a massless regid rod of length L = Na. ⇓ 4 Fig.4 The Born-von Karman periodic boundary condition for the linear chain. 2.3 Born-von Karman boundary condition The Born-von Karman or periodic boundary condition u s = u s+ N , u s = e i ( ksa −ωt ) . The periodic boundary condition requires that 2π or , k= e ikNa = 1 , a N (2.6) (2.7) (2.8) Fig.5 N modes for − π / a ≤ k ≤ π / a where N N N N ,− + 1, , − 1, . 2 2 2 2 N For each k, there are one longitudinal mode and two transverse modes. ∴(1+2)N modes = 3N modes ((very important)) N atoms: each atom has 3 freedoms 3N u s = e i ( ksa −ωt ) , 2π k′ = k + n (n: integer). a The displacement of the n-th atom for the wave k′ =− i[( k _ 2π (2.8) (2.9) n ) sa −ωt ] a (2.10) u s = e i ( k ′sa −ωt ) = e = e i ( ksa −ωt ) . So the displacement us is the same as for the wavevector k, for any atom whatsoever. Thus the wave k′ cannot be physically differentiated from the wave k. 2.4 First Brilouin zone What range of k is physically significant for elastic waves? ⇒ Only those in the first Brilloiun. u s +1 (2.11) = e ika . us The range −π to π for the phase ka covers all independent values of the exponential. 5 − π ≤ ka ≤ π − or π a ≤k≤ π a . (2.12) In the continuum limit (a = 0), k max = π a →∞. (2.13) Fig.6 The relation between k’ and k = k’+2π/a. Suppose that k′ = k + 2π n a (n: integer) (2.14) Then 2π i ( k ′+ n ) a u s +1 a = e ika = e = e ik ′a . (2.15) us Thus the displacement can always be described by a wavevector within the first Brillouin 2π zone. Note that G = n is a reciprocal lattice vector. Thus by subtraction of an a appropriate reciprocal lattice vector from k, we always obtain an equivalent wavevector in the 1st zone: k = k′ + G, where k′ is the wavevector in the first Brillouin zone. 2.5 Normal modes We solve the equations of motion d 2u s M = C (u s +1 − 2u s + u s −1 ) , dt 2 by assuming that u s = e i ( ksa −ωt ) . Then we have d 2u s = −ω 2 u s , 2 dt u s ±1 = e ± ika u s , [ (2.16) (2.17) (2.18) (2.19) ] − ω 2 Mu s = C e ika + e − ika − 2 u s , leading to the dispersion relation 2C 4C 2 ka C ω 2 = (2 − 2 cos ka) = (1 − cos ka) = sin ( ) , 2 M M M or 6 (2.20) (2.21) ω= 4C ka | sin( ) | . M 2 (2.22) The boundary of the first Brillouin zone lies at k = ± π a . dω 2 dω 2Ca = 2ω = sin ka , dk dk M dω π ⇒ = 0 at k = ± . dk a (2.23) (2.24) 2 C M Fig.7 The dispersion curve (ω vs k) for a monatomic linear lattice with nearest neighbor interactions only. The first Brillouin zone is the segment between –π/a and π/a. w is linear for small k and that dω/dk vanishes at the zone boundaries (k = ±π/a. ω2 = 4C ⎛ ka ⎞ sin 2 ⎜ ⎟ M ⎝ 2⎠ or ω= 4C ⎛ ka ⎞ sin ⎜ ⎟ M ⎝ 2⎠ 2.6 dispersion relation (2.25) Continuum wave equation We consider an original wave equation given by d 2u s M = C (u s +1 − 2u s + u s −1 ) . (2.26) dt 2 We assume that us(x) = u(x, t). In the small limit of a = Δx, sa = x and s is continuous variable. Under this assumption, u(x, t) can be expanded using a Taylor expansion, Δx ∂u ( x, t ) (Δx) 2 ∂ 2u ( x, t ) us +1 (t ) = u ( x + Δx, t ) = u ( x, t ) + + + O(Δx)3 , (2.27) ∂x 2 1! ∂x 2! Then the original wave equation can be rewritten as 7 ∂ 2u ( x, t ) M ∂t 2 = C[u ( x, t ) + Δx ∂u ( x, t ) (Δx) 2 ∂ 2u ( x, t ) Δx ∂u ( x, t ) (Δx) 2 ∂ 2u ( x, t ) + − u x t + u x t − + 2 ( , ) ( , ) ] 1! ∂x 2! 1! ∂x 2! ∂x 2 ∂x 2 (2.28) or M ∂ 2 u ( x, t ) ∂ 2 u ( x, t ) , ] = C (Δx) 2 2 ∂t ∂x 2 (2.29) or 2 2 ∂ 2 u ( x, t ) C C 2 ∂ 2 u ( x, t ) 2 ∂ u ( x, t ) 2 ∂ u ( x, t ) , ( x ) a = v = Δ = ∂t 2 M ∂x 2 M ∂x 2 ∂x 2 (2.30) which is the continuum elastic wave equation with the velocity of sound given by C v= a. (2.31) M 3. Eigenvalue-problem; solution using Mathematica We solve the eigenvalue-problem given by (3.1) ω 2us = K (−us +1 + 2us − us −1 ) , where s = 1, 2, …, N, K = C/M. These equations can be expressed using a matrix M (for example, N x N) and column matrix (1xN); MU = ω2U. (3.2) 2 Here ω is the eigenvalue. The number of eigenvalues is N. For example, the matrix M and the column matrix U for N = 12 (for example), are given by ⎛ 2K ⎜ ⎜− K ⎜ 0 ⎜ ⎜ 0 ⎜ 0 ⎜ ⎜ 0 M =⎜ ⎜ 0 ⎜ 0 ⎜ ⎜ 0 ⎜ 0 ⎜ ⎜ 0 ⎜ 0 ⎝ −K 2K −K 0 0 0 0 0 0 0 0 0 0 −K 2K −K 0 0 0 0 0 0 0 0 0 0 −K 2K −K 0 0 0 0 0 0 0 0 0 0 −K 2K −K 0 0 0 0 0 0 0 0 0 0 −K 2K −K 0 0 0 0 0 0 0 0 0 0 −K 2K −K 0 0 0 0 and 8 0 0 0 0 0 0 −K 2K −K 0 0 0 0 0 0 0 0 0 0 −K 2K −K 0 0 0 0 0 0 0 0 0 0 −K 2K −K 0 0 0 0 0 0 0 0 0 0 −K 2K −K 0 ⎞ ⎟ 0 ⎟ 0 ⎟ ⎟ 0 ⎟ 0 ⎟⎟ 0 ⎟ ⎟ ,(3.3) 0 ⎟ 0 ⎟ ⎟ 0 ⎟ 0 ⎟ ⎟ −K⎟ 2 K ⎟⎠ ⎛ u1 ⎞ ⎜ ⎟ ⎜ u2 ⎟ ⎜u ⎟ ⎜ 3⎟ ⎜ u4 ⎟ ⎜u ⎟ ⎜ 5⎟ ⎜u ⎟ U = ⎜ 6 ⎟ , for each eigenvalue. (3.4) u ⎜ 7⎟ ⎜ u8 ⎟ ⎜ ⎟ ⎜ u9 ⎟ ⎜u ⎟ ⎜ 10 ⎟ ⎜ u11 ⎟ ⎜u ⎟ ⎝ 12 ⎠ Using the Mathematica (program called Eigenvalues), we can calculate the dispersion relation of ωk vs k = (π/a) (n/N) with N = 1, 2, 3,…, N-1, N. For simplicity, here we choose N = 100. For comparison we also make a plot of the solution given by ω = 4C / M | sin(ka / 2) | . We find that there is a good agreement between these two curves. K = C/M. ((Mathematica-1)) The mathematica program is attached to this lecture note. Phonon dispersion relation for one atom in the unit cell. The Mathematica program is presented in the Appendix. w 2 K 1.0 0.8 0.6 0.4 0.2 p n 20 40 60 80 100 a 100 Fig.8 The dispersion relation with red (from Mathematica calculation) and blue ( ω = 4 K sin(ka / 2) with K = C/M). The x axis is k = (π/a) (n/N) with n = 1, 2, …, N (= 100). The y xais is ω /(2 K ) . The agreement between these two results are excellent. 9 ((Mathematica-2)) The Mathematica program used is attached to this lecture note. Here we use the Mathematica program (Eigensystem) to obtain the eigenvalues and the corresponding eigenfunction U. For each eigenvalue, we can calculate the deviation (us) of the position of the atom from the equilibrium position, as a function of the equilibrium position xs = s a, where s = 1, 2, 3,….. ,N. Here we consider the eigenvalue problem (N = 50). We show the plot of Un=(u1, u2, u3, …, uN) as a function of the location (xs = s a), where n = 1, 2, …, N. u = 1 . We find that the wavelength λn is given by λn = 2Na/n. The corresponding wavenumber qn is equal to kn = (2π/λn) = (π/a)(n/N). The deviation us is well described by the form sin[πs(n/N)] = sin[kn(sa)] ( ≈ eik n sa ). u 0.2 0.1 x 10 20 30 40 50 a -0.1 -0.2 Fig.9 Plot of u = (u1, u2, u3, …., u50) as a function of xs = s a for the eigenvalues ωi (i = 1, 2, 3, …, 10). ω1<ω2<…<ω10. Note that u is normalized ( u = 1 ). N = 50. 10 81 < 82 < u u 0.20 0.2 0.1 0.0 -0.1 -0.2 0.15 0.10 x 10 20 30 40 50 a 83 < u 0.2 0.1 0.0 - 0.1 - 0.2 85 < u x 10 20 30 40 50 a 86 < u 0.2 0.1 0.0 - 0.1 - 0.2 x 10 20 30 40 50 a 87 < u x 10 20 30 40 50 a 88 < u 0.2 0.1 0.0 -0.1 -0.2 x 10 20 30 40 50 a 89 < u 0.2 0.1 0.0 - 0.1 - 0.2 84 < u 10 20 30 40 50 a 0.2 0.1 0.0 - 0.1 - 0.2 10 20 30 40 50 a 0.2 0.1 0.0 -0.1 -0.2 x 0.2 0.1 0.0 - 0.1 - 0.2 x x 10 20 30 40 50 a 810< u 0.2 0.1 0.0 -0.1 -0.2 x 10 20 30 40 50 a x 10 20 30 40 50 a Fig.10 Plot of u = (u1, u2, u3, …., u50) as a function of xs = s a for the eigenvalues ωi (i = 1, 2, 3, …, 10). ω1<ω2<…<ω10. The wavenumber kn = (π/Na)n. The value of n is denoted in each figure. 11 85 < 810< u u 0.2 x 0.1 0.0 - 0.1 1020 304050 a - 0.2 815< 0.2 x 0.1 0.0 -0.1 10 203040 50 a -0.2 820< u u 0.2 x 0.1 0.0 - 0.1 1020 304050 a - 0.2 825< 0.2 x 0.1 0.0 -0.1 10 203040 50 a -0.2 830< u u 0.2 x 0.1 0.0 - 0.1 1020 304050 a - 0.2 835< 0.2 x 0.1 0.0 -0.1 10 203040 50 a -0.2 840< u u 0.2 x 0.1 0.0 - 0.1 1020 304050 a - 0.2 845< 0.2 x 0.1 0.0 -0.1 10 203040 50 a -0.2 850< u 0.2 x 0.1 0.0 - 0.1 1020 304050 a - 0.2 u 0.2 x 0.1 0.0 -0.1 10 203040 50 a -0.2 Fig.11 Plot of u = (u1, u2, u3, …., u50) as a function of xs = s a for the eigenvalues ωn (n = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50). ω5<ω10<…<ω50. u = 1 . N = 50. The wavenumber kn = (π/Na)n. The value of n is denoted in each figure. 12 4. 4.1 First Brillouin zone and group velocity Definition of the group velocity The transmission velocity of a wave pocket is the group velocity ∂ω vg = = ∇ k ω (k ) . ∂k This is the velocity of energy propagation in the medium. vg = ∂ω Ca 2 ⎛1 ⎞ = cos⎜ ka ⎟ , ∂k M ⎝2 ⎠ v g = 0 at k = (4.1) (4.2) π a The wave is a standing wave: zero net transmission velocity for a standing wave. Note that the phase velocity is defined by v p = ω / k . Long wavelength limit When ka « 1, ⎡ 1 ⎤ 1 1 − cos ka = 1 − ⎢1 − (ka) 2 ⎥ ≅ (ka) 2 . ⎣ 2! ⎦ 2 (4.3) Then ω2 = 2C 1 2 2 C 2 2 k a = k a M 2 M or ω= Ca 2 k, M (4.4) ∂ω Ca 2 = = v0 at ka = 0. (4.5) ∂k M The velocity of sound is independent of frequency in this limit. Thus ω = vk , exactly as in the continuum theory of elastic waves --- in the continuum limit a = 0 and thus ka = 0. vg = 4.2 The physical meaning of the first Brillouin zone We discuss the physical meaning of the first Brillouin zone. To this end, we consider π π 2π 7π and k = k0 + G = + = . the case of us with k = k0 = 3a 3a a 3a (i) k = k0 = π . 3a At t = 0, we have u s = ue i ( ksa −ωt ) at t = 0. We make a plot of π ⎛π ⎞ ] = cos⎜ s ⎟ , ⎝3 ⎠ as a function of s, where s = 0, 1, 2, 3, …, where u = 1 in Fig.17. π 2π 7 + = π (ii) k = k0 + G = 3a a 3a Re[us ] = Re[e i ( s − ωt ) 3 We also make a plot of Re[us ] = Re[e ⎛ 7π ⎞ i⎜ s⎟ ⎝ 3 ⎠ ⎛ 7π ⎞ s ⎟ at t = 0 in Fig.17. ] = cos⎜ ⎝ 3 ⎠ 13 0.5 s Re[u /u] 1 0 -0.5 -1 0 1 2 3 4 5 6 7 s Fig.12 Plot of Re[ei ( ksa −ωt ) ] at t = 0 for k = k0 + G = 7π/3 (the red solid line) and k = k0 = π/3a (the dotted blue line). As shown in Fig.12, the wave represents by the solid curve (k = 7π/3a) conveys no information not give by the dashed curve (k = π/3a). Because the wave displacement is defined only at lattice points, the propagation of a large wavenumber (k0 + G) lying outside the first Brillouin zone is identical to a short wavenumber k0 lying inside the first Brillouin zone. In a continuum, the amplitudes would have a value everywhere, as represented by the continuous lines, so that both wavenumbers would be distinguishable. 4.3 Standing wave At the zone boundary k = ± π / a , the group velocity is equal to zero, implying that no energy is propagated. The amplitude Re[us] at time t is described [ ] [ ] Re[us ] = Re uei ( ksa −ωt ) = Re u (−1) s e − iωt = u (−1) s cos ωt (4.1) s The alternate atoms oscillate in opposite phases, because (-1) = -1 for odd integers s and (-1)s = 1 for even integers s. Note that the wave is a standing wave. A standing wave is an example of wave motion with zero group velocity. The wave moves neither to the right nor to the left. This situation is equivalent to Bragg reflections of x-ray. Fig.13 The condition for the Bragg reflection in the reciprocal lattice. k − k ' = 2π / a (or k = π / a and k ′ = −π / a ) The Bragg reflection arises when k = ± π / a . Even if we excite only the state k = π/a. we should obtain k = -π/a through the Bragg reflection. The superposition of these two wave leads to a standing wave, 1 i ( ksa −ωt ) 1 −iωt isπ e e e ± e −isπ ± e i ( k ′sa −ωt ) = 2 2 (4.6) 1 −iωt ⎧ 2 cos sπ ⎫ −iωt ⎧cos sπ e ⎨ = ⎬ = 2e ⎨ 2 ⎩2i sin sπ ⎭ ⎩i sin sπ [ ] [ ] 14 This implies that the wave cannot propagate in a lattice, but through successive reflections back and forth, a standing wave is set up. 4.4 General property of the group velocity There are two main features of the phonon dispersion relation; (i) E = ħωk is an even function of k; ω− k = ωk or E− k = Ek , where ħ is the Dirac’s constant (ħ = h/2π) and h is the Planck’s constant. (ii) Periodicity in Ek + G = Ek with G [= (2π/a) x integer] is the reciprocal lattice. Fig.14 Values of k in the first Brillouin zone related by the symmetrical relation (Ek+G = E k and E-k = E k). k = ±π/a is the boundary of the first Brillouin zone (|k|≤π/a). We now consider the value of the group velocity at the Brillouin zone boundary. From the condition Ek = E− k , we have E(3) = E(4). From the condition Ek = Ek + G , we have E(3) = E(5). Therefore, we have E(4) = E(5). On taking δ → 0, the group velocity at the boundary of Brillouin zone is defined as [ E (5) − E (4)] / 2δ , which reduces to zero (dΕk/dk→0). ((Note)) It follows that from the condition ( Ek = E− k ), we have E(1) = E(2). On taking δ→0, the group velocity defined by [ E (2) − E (1)] / 2δ reduces to zero (dΕk/dk→0). On applying the periodicity condition Ek = Ek + G this result can immediately be extended as follows. dΕk/dk→0 at k = 0, ±2π/a, ±4π/a,….. This prediction (only from the symmetry consideration) that the group velocity is equal to zero at k = 0, may be inconsistent with the result derived from the linear chain model above described. In the linear chain model, it is predicted that there is a discontinuous jump in the group velocity at k = 0, from -v0 (v0 = a C / M ) at k = 0- to –v0 at k = 0+. We note that this motion at k = 0 is a translation of the crystal as a whole and it is therefore not property of a vibration. 5. 5.1 Determination of force constants The system with the nearest neighbor interaction We can make a statement about the range of the forces from the observed dispersion relation for ω. The generalization of the dispersion relation to p nearest planes is found to be 15 ω k2 = 2 M ∑ [1 − cos( pka )]C p >0 π /a M ∫ dkω π 2 k − /a p , cos(rka) = 2∑ C p p >0 (5.1) π /a ∫π dk [1 − cos( pka)]cos(rka) . (5.2) − /a Fig.15 Linear chain having the nearest-neighbor coupling (C1), the second nearestneighbor coupling (C2), and the third nearest-neighbor coupling (C3), and the Here π /a π /a ⎡ sin(rka) ⎤ dk cos(rka) = ⎢ = 0, ∫ ⎣ ra ⎥⎦ −π / a −π / a (5.3) and π /a π /a − /a − /a 1 ∫ dk cos( pka) cos(rka) = π∫ dk 2 [cos{( p + r )ka} + cos{( p − r )ka}] π π /a ⎧ 1 2π 1 ⎪ = dk cos[( p − r )ka] = ⎨ 2 a ∫ 2 −π / a ⎪⎩ 0 for p = r , otherwise (5.4) Thus π /a M ∫π dkω 2 k cos(rka) = −2π − /a Cr , a (5.5) or π /a Ma (5.6) dkω k2 cos( pka) , ∫ 2π −π / a gives the force constant at range pa, for a structure with a monatomic basis. Cp = − 5.2 System with long-ranged interactions Fig.16 Linear chain with the nearest-neighbor and the next nearest-neighbor interactions. Here we assume that there are the second-, third,…nearest neighbor interactions in the system in addition to the nearest neighbor-interaction. Then the Lagrangian L is given by 16 L = T −V ∞ 1 1 = ∑ Mu s2 − ∑∑ C p (u s − u s + p ) 2 s 2 s p =1 2 . { (5.6) } ∞ 1 1 = ∑ Mu s2 − ∑ C p (u s − u s + p ) 2 + (u s − u s − p ) 2 − s 2 p =1 2 The Lagrange equation is d ∂L ∂L = , dt ∂u s ∂u s or ∞ ∞ p =1 p =1 (5.7) Mu s = −∑ C p (u s − u s + p ) − ∑ C p (u s − u s − p ) (5.8) ∞ = ∑ C p (u s + p − 2u s + u s − p ) p =1 Here we assume a traveling wave given by u s = ue i ( ksa −ωt ) . Then we have ( ∞ ) − Mω 2 u s = ∑ C p e ipka − 2 + e −ipka u s , (5.9) p =1 or ( ∞ ) − Mω 2 = ∑ C p e ipka − 2 + e −ipka , (5.10) p =1 or ω2 = 2 M ∞ C C ∑ C (1 − cos pka ) = M [1 − cos(ka)] + M [1 − cos(2ka)] + ... , p =1 p 1 2 (5.11) which includes the higher harmonics terms through long-range spring constants. The value of Cp can be determined experimentally from the phonon dispersion relation. 6. Vibration of square lattice4 Fig.17 Square array of lattice constant a. The displacements are normal to the plane of the lattice. 17 We consider transverse vibrations of a planar lattice of rows and columns of identical atoms, and let u(l, m) denote the displacement normal to the plane of the lattice of the atom in the l-th column and m-th row. The mass of each atom is M, and C is the force constant for nearest neighbor atoms. The equation of motion is expressed by d 2u (l , m) M = C{[u (l + 1, m) + u (l − 1, m) − 2u (l , m)] + [u (l , m + 1) + u (l , m − 1) − 2u (l , m)] dt 2 (6.1) We assume that the solution of u(l, m) is given by u (l , m) = u (0) exp[i (lk x a + mk y a − ωt ) , (6.2) where a is the spacing between nearest-neighbor atoms. The equation of motion is satisfied only if 2C ω= 2 − cos(k x a ) − cos(k y a ) , (6.3) M where the first Brillouin zone is − π a ≤ kx ≤ π a and − π a ≤ kx ≤ π a . (6.4) Fig.18 Brillouin zone of the two-dimensional square lattice with a lattice constant a. Bragg reflections occur at the zone boundary of the first Brillouin zone. 18 Fig.19 Three-dimensional plot of the dispersion relation (of the 2D square lattice. The height (ω) is plotted as a function kxa vs kya (in the 2D wavevector-plane). 7. Two atoms per primitive basis We consider a cubic crystal where atoms of mass M1 lie on one set of planes and atoms of mass M2 lie on planes interleaved between those of the first set. Fig.20 (a) Diatomic linear chain, There are two atoms with masses M1 and M2 per unit cell (lattice constant a). (b) The plane of atoms stacked alternatively (lattice constant a). The Lagrangian of the system is given by 1 1 L = M 1u s2 + M 2 v s2 + 2 2 1 − C (u s +1 − v s ) 2 + (v s − u s ) 2 + (u s − v s −1 ) 2 + 2 The Lagrange’s equations are as follows. d ∂L ∂L = , dt ∂u s ∂u s or [ 19 . ] (7.1) (7.2) 1 M 1u s = − C [− 2(v s − u s ) + 2(u s − v s −1 )], 2 (7.3) or M 1u s = −C (− v s + 2u s − v s −1 ) = C (v s + v s −1 − 2u s ) . (7.4) Similarly d ∂L ∂L = , (7.5) dt ∂v s ∂v s 1 M 2 v s = − C [− 2(u s +1 − v s ) + 2(v s − u s )] = C (u s +1 − 2v s + u s ) . (7.6) 2 Let a denote the repeat distance of the lattice in the direction normal to the lattice plane. Equation of motion is given by ⎧ d 2u s = C (v s + v s −1 − 2u s ) ⎪⎪M 1 dt 2 , (7.7) ⎨ 2 ⎪M d v s = C (u + u − 2v ) s +1 s s ⎪⎩ 2 dt 2 Here we assume that each plane interacts only with its nearest-neighbor and that the force constants are identical between all pairs of nearest-neighbor planes. Suppose that us and vs have the forms of ⎧⎪u s = ue i ( ska −ωt ) . (7.8) ⎨ ⎪⎩v s = ve i ( ska −ωt ) Then − ω 2 M 1u = Cv (1 + e − ika ) − 2Cu . (7.9) − ω 2 M 1v = Cu (e ika + 1) − 2Cv When the determinant of the coefficients of u and v vanishes, the homogeneous linear equations have a non-trivial solution. 2C − M 1ω 2 − C 1 + e − ika (7.10) = 0, − C 1 + e ika 2C − M 2ω 2 or M 1 M 2ω 4 − 2C ( M 1 + M 2 )ω 2 + 2C 2 (1 − cos ka ) = 0 . (7.11) We examine this equation in the limiting cases (i) ka « 1 (ii) ka = ±π at the zone boundary ( ) ( ) (i) ka « 1 1 2 2 k a 2 M 1 M 2ω 4 − 2C ( M 1 + M 2 )ω 2 + C 2 k 2 a 2 = 0 , 1 − cos ka ≈ Or 20 (7.12) ω = 2 C (M 1 + M 2 ) ± C 2 (M 1 + M 2 ) 2 − M 1M 2C 2 k 2 a 2 M 1M 2 C (M 1 + M 2 ) ⎡ M 1M 2C 2 k 2 a 2 = ⎢1 ± 1 − 2 M 1M 2 C (M 1 + M 2 ) 2 ⎢⎣ ≈ ⎤ ⎥ ⎥⎦ . C (M 1 + M 2 ) ⎡ ⎧ M 1 M 2 C 2 k 2 a 2 ⎫⎤ 1 1 ± − ⎢ ⎨ 2 2 ⎬⎥ M 1M 2 ⎢⎣ ⎩ 2C ( M 1 + M 2 ) ⎭⎥⎦ Then we have the two roots for ka « 1 ⎛ 1 1 ⎞ ⎟⎟ (optical branch), + ω 2 = 2C ⎜⎜ ⎝ M1 M 2 ⎠ 1 C 2 2 ω = k 2 a 2 (acoustic branch). M1 + M 2 (*) The extent of the first Brillouin zone is − π a (7.14) (7.15) ≤k≤ π (ii) ka = ±π M 1 M 2ω 4 − 2C ( M 1 + M 2 )ω 2 + 4C = 0 M 1ω 2 − 2C M 2ω 2 − 2C = 0 or 2C 2C ω2 = ; ω2 = . M1 M2 ( (7.13) )( ) a . (7.16) (7.17) ω optical phonon branch acoustic phonon branch -π/a π/a 0 k for M1 > M2 Fig.21 The dispersion curve (ω vs k) for a diatomic linear chain with nearest neighbor atoms interacting with interaction C only. The masses of the atoms are M1 and M2; The first Brillouin zone is the segment between –π/a and π/a. There is an acoustic mode (lower branch) and an optical mode (upper branch). 21 We consider the particle displacement in TA and TO branches. For the optical branch at k = 0, we have − ω 2 M 1u = 2Cv − 2Cu M 2C 2C u =− 2 : out-of-phase (7.18) = = 2 v 2C − M 1ω M1 ⎛ 1 1 ⎞ ⎟⎟ 2C − M 1 2C ⎜⎜ + ⎝ M1 M 2 ⎠ For the acoustical branch at k = 0, we have − ω 2 M 1u = 2Cv − 2Cu . (7.19) 2 Since ω = 0 , then we have u = v . : in-phase Fig.22 Nature of the vibration in the acoustical and optical branch of a vibrational spectrum. u s = ue i ( ska −ωt ) , v s = ve i ( ska −ωt ) . (7.20) Acoustical waves in a diatomic linear lattice ( k = π 2a ) Fig.23 Transverse acoustic waves in a diatomic linear lattice. u cos( ska) = v cos( ska) . (7.21) 22 Optical waves in a diatomic linear lattice ( k = π 2a ) Fig.24 Transverse optical waves in a diatomic linear lattice. Figure represents a snapshot of the wave motion. M 1u + M 2 v = 0 ⇒ The velocity of center of mass does not move. M1 + M 2 The atoms vibrate against each other, but their center of mass is fixed. If the two atoms carry opposite charges, we may excite a motion of this type with the electric field of a light wave. vc ≡ Fig.25 Origin of the name of optical mode So that the branch is called the optical branch. The atoms (and their center of mass) move together, as in long wavelength acoustical vibrations, whence the term acoustical branch. Fig.26 Origin of the name of acoustic mode This is a frequency gap at k max = ± π a of the first Brillouin zone. ((Note)) If we look for solutions in the gap with ω real, then k will be complex, so that the wave is damped in space. ((Mathematica-3)) phonon dispersion for the system with two atoms in unit cell 23 wêw0 1.5 1.0 0.5 -2 -1 1 2 kaêp Fig.27 The dispersion curve of the diatomic linear chain with two atoms in a unit cell. The ratio M2/M1 is varied as a parameter between 2, 3, 4, … and 10. 8. 8.1 The number of modes; degree of freedom One-dimensional case We consider the degree of freedom for N atoms in the linear lattice chain. There are N atoms (each unit cell has one atom). Each atom has three degrees of freedom. One for each of the x, y, z directions. Then we have 3N degree of freedoms. This indicates that there are 3N modes in the system. This implies that the number of allowed k values in a single branch is just N for each Brillouin zone; 2N transverse acoustic (TA) modes and N longitudinal acoustic (LA) mode. 8.2 by Three-dimensional case We consider the lattice waves in the 3D system. The displacement vector us is given u s ~ e ik ⋅ R n , (7.1) where Rn is the position vector of the atom located in the equilibrium positions of the lattice. R n = n1a1 + n2 a 2 + n3 a 3 , (7.2) a1, a2, and a3 are the primitive lattice vectors along the x, y, and z directions. From the boundary condition, we have (7.3) eik ⋅( R n + N1a1 ) = eik ⋅R n , or e ik x N1a1 = 1 , leading to the selected values of wavenumber kx 2π 2π kx = l1 = l1 for a1 = a2 = a3 = a. (7.4) N1a1 N1a Similarly, we have 2π 2π 2π 2π ky = l2 = l2 , and kz = l3 = l3 (7.5) N 2 a2 N 2a N 3a3 N 3a Here l1, l2, and l3 are integers given by 24 ⎧ ⎪ N N N ⎪ l1 = − 1 ,− 1 + 1, , 1 2 2 2 ⎪ ⎪ N1 ⎪ N2 N2 N (7.6) ,− + 1, , 2 . ⎨l2 = − 2 2 2 ⎪ N2 ⎪ ⎪ N3 N3 N3 ⎪ l3 = − 2 ,− 2 + 1, , 2 ⎪ N3 ⎩ This means that there are N (=N1N2N3 ) modes, where N is the number of unit cells in the system. Each modes has 3 degrees of freedom (1 longitudinal mode and 2 transverse modes). Then the total number of modes is 3 × N = 3N. (a) In the case of one atom per unit cell, we have 3N degree of freedom, with N longitudinal acoustic mode and transverse acoustic mode 2N (b) We consider the number of degrees of freedom of the atoms. With p atoms in the primitive cell and N primitive cells, there are pN atoms. Each atom has three degrees of freedom, one for each of the x, y, z directions, making a total of 3pN degrees of freedom for the crystal. The number of allowed k values in a single branch is just N for one Brillouin zone. Thus, the one LA and two TA branches have a total of 3N modes. The remaining (3p - 3) x N degrees of freedom are accommodated by the optical branches. 3 acoustical branches 1 longitudinal acoustical (LA) mode 2 transverse acoustical (TA) mode 3p - 3 optical branches (p - 1) longitudinal optical (LO) mode 2(p - 1) transverse optical (TO) mode For p = 2, for example, we have 1 LA, 1 LO modes, and 2 TA and 2 TO modes. 9. 9.1 Classical Model Theory of the transverse wave in a string 25 Fig.28 Oscillation of one-dimensional continuum Suppose that a traveling wave is propagating along a string that is under a tension Ts. Let us consider one small element of length Δx. The ends of the element make small angle θA and θB with the x axis. The net force acting on the element along the y-axis is ∑ Fy = T sin θ B − T sin θ A , (9.1) = T (sin θ B − sin θ A ) ≈ T (tan θ B − tan θ A ) or ⎛ ∂y ⎞ ⎛ ∂y ⎞ (9.2) Fy = T [⎜ ⎟ − ⎜ ⎟ ] . ⎝ ∂x ⎠ B ⎝ ∂x ⎠ A We now apply the Newton’s second law to the element, with the mass of the element given by m = μΔx , Fy = ma y = μΔx ∂2 y . ∂t 2 (9.3) Then we have ∂2 y ⎛ ∂y ⎞ ⎛ ∂y ⎞ μΔx 2 = Ts [⎜ ⎟ − ⎜ ⎟ ] ∂t ⎝ ∂x ⎠ B ⎝ ∂x ⎠ A , ⎛ ∂y ⎞ ⎛ ∂y ⎞ ⎜ ⎟ −⎜ ⎟ 2 2 μ ∂ y ⎝ ∂x ⎠ B ⎝ ∂x ⎠ A ∂ y = = 2 Δx ∂x Ts ∂t 2 which leads to a wave equation given by ∂2 y μ ∂2 y 1 ∂2 y , = = ∂x 2 Ts ∂t 2 v 2 ∂t 2 where v is the velocity of the sound, T v= s . (9.4) (9.5) (9.6) μ Here we use the Taylor expansion, 26 ∂2 y ∂ ⎛ ∂y ⎞ ⎛ ∂y ⎞ ⎛ ∂y ⎞ ⎛ ∂y ⎞ ⎛ ∂y ⎞ − ⎜ ⎟ = Δx ⎜ ⎟ = Δx 2 . ⎜ ⎟ −⎜ ⎟ =⎜ ⎟ ∂x ∂x ⎝ ∂x ⎠ ⎝ ∂x ⎠ B ⎝ ∂x ⎠ A ⎝ ∂x ⎠ x + Δx ⎝ ∂x ⎠ x 9.2 (9.6) Energy density of the elastic wave 2 ⎛ ∂y ⎞ When the length of element changes to Δx to Δx 1 + ⎜ ⎟ , the work done in the ⎝ ∂x ⎠ system (= Wc) of the conserved system is given by 2 2 2 1 ⎛ ∂y ⎞ T ⎛ ∂y ⎞ ⎛ ∂y ⎞ − Ts Δx 1 + ⎜ ⎟ + Ts dΔ = −Ts dΔ[1 + ⎜ ⎟ − 1] = − s Δx⎜ ⎟ . (9.7) 2 ⎝ ∂x ⎠ 2 ⎝ ∂x ⎠ ⎝ ∂x ⎠ Since the potential energy ΔU is related to Wc by ΔU = −Wc , the potential energy ΔU is given by 2 T ⎛ ∂y ⎞ ΔU = s Δx⎜ ⎟ . 2 ⎝ ∂x ⎠ The kinetic energy contribution ΔK is given by (9.8) 2 1 ⎛ ∂y ⎞ ΔK = μΔx ⎜ ⎟ . 2 ⎝ ∂t ⎠ Then the energy density is given by (9.9) 2 ΔE = ΔK + ΔU = Δx[ 2 Ts ⎛ ∂y ⎞ μ ⎛ ∂y ⎞ ⎜ ⎟ + ⎜ ⎟ ]. 2 ⎝ ∂x ⎠ 2 ⎝ ∂t ⎠ (9.10) 10. Quantum mechanical approach: phonon 10.1 Annihilation and creation operators1,3 We assume U = (u1, u2, ,uN) for the eigenvalue ω = ωk. The displacement u(x) is expresses using a Dirac delta function, u ( x) = ∑ usδ ( x − sa ) , (10.1) s with L = Na . The Fourier transform of u(x) is given by 1 1 Uk = dxu ( x)e − ikx = us e − iksa . ∑ ∫ N N s The inverse Fourier transform of Uk is 1 1 1 eiksaU k = ∑ ∑ ∑ us 'e−iks 'aeiksa N k N k N s' 1 = ∑ us ' ∑ eik ' a ( s − s ') N s' k' N = ∑ usδ s , s ' N s = us or 27 (10.2) (10.3) 1 e iksaU k . ∑ N k The Lagrangian of the system is given by L = T −V us = (10.4) 1 ⎡M 2⎤. = ∑ ⎢ u s2 − C (u s +1 − u s ) ⎥ 2 ⎦ s ⎣ 2 We use the notation 1 us = U k eiksa . ∑ N k Since us is real, it is required that * U −k = U k . Using these relations, we have 1 U kU k 'ei ( k + k ') sa = ∑∑U kU − k . ∑s us2 = N ∑∑∑ k k' s k s Here we use N ∑e i ( k + k ') sa s =1 = Nδ k ', − k . (10.5) (10.6) (10.7) (10.8) (10.9) Similarly ∑ (u s +1 s 1 2 − us ) = 2∑U kU − k (1 − cos ka) = ∑ Mωk2U kU − k , k k k (10.10) where 2C (10.11) (1 − cos ka) . M Then L can be rewritten as ⎛M ⎞ Mωk2 (10.12) L = ∑ ⎜⎜ U kU − k − U kU − k ⎟⎟ . 2 k ⎝ 2 ⎠ The linear momentum Pk conjugate to Uk is defined by ∂L Pk = = MU − k . (10.13) ∂U k Then Hamiltonian H is obtained as 1 ⎛ 1 1 ⎞ H = ∑ PkU k − L = ∑ ⎜ Pk P− k + Mωk2U kU − k ⎟ . (10.14) 2 k ⎝M 2 ⎠ k Note that we define the Fourier transform of the linear momentum by 1 ps = e −iksa Pk ∑ N k (10.15) 1 iksa Pk = ∑ e ps N s + with P− k = Pk . (10.16) Let us work it out by the operator technique, starting from the commutation relations, [us , ps ' ] = i δ s , s ' . (10.17) ω k2 = 28 Then the commutation relation is preserved in [U k , Pk ' ] = i δ k , k ' , since [U k , Pk ' ] = [ 1 N ∑e −iksa us , s 1 N ∑p −ik 'sa s ] s 1 e −iksa [u s , ps ' ]e ik 's 'a ∑∑ N s s' . (10.18) i −iksa ik 's 'a = ∑∑ e e δ s ,s ' N s s' i = ∑ e −i ( k −k ') sa = i δ k ,k ' N s Thus our new displacements and momenta are canonically conjugate, and noncommuting, if they are of the same wavenumber; otherwise they are dynamically independent operators. The Hamiltonian is rewritten as 1 ⎛ 1 + 1 ⎞ + H = ∑ ⎜ Pk Pk + Mωk2U k U k ⎟ . (10.19) 2 k ⎝M 2 ⎠ The final step is the introduction of annihilation and creation operators defined by 1 Mωk + 1 ak+ = ( Uk − iPk ) 2 M ωk , (10.20) 1 Mωk 1 + ak = ( Uk + iPk ) 2 M ωk = where ak and ak+ act to destroy, and create a phonon of waveuumber, k and energy ωk , respectively. One can get the expressions for Pk and Uk from the above equations for ak and ak + Pk = i M ωk + ( ak − a− k ) , 2 Uk = ( a k + a− k ) , (10.21) + 2 Mωk The annihilation and creation operators satisfy the commutation [a , a ] k + k' = δ k ,k ' , (10.22) (10.23) [a k , a k ' ] = [a k+ , a k+' ] = 0, (10.24) The transformed Hamiltonian is 1⎞ 1⎞ ⎛ ⎛ H = ∑ ωk ⎜ ak+ ak + ⎟ = ∑ ωk ⎜ N k + ⎟ , (10.25) 2⎠ k 2⎠ ⎝ ⎝ k where N k = ak+ ak . So that each phonon may be regarded as possessing an energy ωk . The total Hamiltonian is the sum of the Hamiltonian of independent linear oscillators of angular frequency ωk. The various properties of the operators and eigenstates of the Hamiltonian are seen in the Appendix A. 10.2 Symmetry of lattice and translation operator13,15 29 We now consider the displacement defined by u ( x) = x u = ∑ us x s . (10.26) s When we use the Dirac notation x s = δ ( x − sa ) , the ket vector is described by u = ∑ us s . (10.27) s We introduce the translation operator given by T(a). The Hamiltonian H is invariant under the translation of the system by the lattice constant a. In other words, T (a) commutes with the Hamiltonian H. The eigenstate of H should be simultaneously the eigenstate of T (a) . T(a) is a unitary operator, but not a Hermite operator. Then the eigenvalue of T(a) is a complex number (see the Appendix B for more detail). Since T (a ) s = s + 1 , the ket s is not an eigenstate of T(a). Suppose that us has the form of exp(isak), where k is a real parameter. u is a linear combinationof s with s = 0, 1, …, N-1. u = ∑ eisak s . (10.28) s When T(a) is applied to the eigen ket u , we get T (a ) u = ∑ eisak T (a ) s = ∑ eisak s + 1 s =e −ika ∑e s i ( s +1) ak s + 1 = e −ika u , (10.29) s which means that u is the eigenket of T(a). Similarly, we repeat this process N times, [T (a )]N u = e − ikNa u = u , (10.30) since [T (a )]N =1 (we use the periodic boundary condition). Then we have e − iNka =1, or 2π 2π k= n where n =0, 1, 2,…, N-1 ( 0 ≤ k < ). Alternatively we chose the value of k Na a as the N states for − π ≤k< π (the first Brillouin zone). a a It follows from this discussion that when u = uk are introduced, the N harmonic oscillators (N being the number of atoms in the crystal) becomes uncoupled and that to each uk corresponds to one separate oscillator with the angular frequency ωk. This motion is called a normal mode of vibration or a normal mode. This motion does not describe the motion of a single atom in the crystal but, rather, of all atoms in it. A few examples will show how this works. We take a normal mode for which k equals the reciprocal lattice G (=2π/a). In this case, the coefficient us equals unity. This motion is a translation of the crystal as a whole and it is therefore not property a vibration. The case just considered corresponds of course to the center (k = 0) of the Brillouin zone. In order to have another example, we consider the case where k equal to a point at the zone boundary of the Brillouin zone. The displacement vector is out of phase in going from one cell to the next, whether the successive lines show the evolution of the motion in time, the harmonic function depicted corresponding to the particular frequency of this normal mode. 30 11. Crystal momentum A phonon of k will interact with particles such as photons, neutrons, and electrons as if it had a momentum k . However, a phonon does not carry physical momentum. The physical momentum of a crystal is given by ∂L Pk = = MU − k ∂U k , (11.1) 1 iksa =M ∑ us e N s where 1 1 Uk = dxu ( x)e − ikx = (11.2) ∑ use−iksa . ∫ N N s We assume that u s is independent of s; us = u . Then the momentum Pk is evaluated as M Pk = u ∑ eiksa N s M = u[1 + eika + e 2ika + .... + eika ( N −1) ] , (11.3) N = M N uδ k ,0 2πr where k = (r = 0, ±1, ±2,…, ±N/2). The mode k = 0 represents a uniform translation Na of the crystal as a whole. Such a translation does carry momentum. For most practical purpose, a phonon acts as if its momentum were ħk, sometimes called the crystal momentum. 12 Semiclassical approach 12.1 Simple case The energy of a lattice vibration is quantized. The quantum of energy is called a phonon. Elastic waves in crystals are made up of phonons. Thermal vibrations in crystals are thermally excited phonons. The energy of an elastic mode of angular frequency ωk is 1⎞ ⎛ (12.1) ε k = ⎜ nk + ⎟ ω k , 2⎠ ⎝ when the mode is excited to quantum number n k , where the mode is occupied by n 1 ω is the zero pint energy of the mode. We consider the wave of the 2 mode k with the amplitude. u = u k cos(ω k t − kx) , (12.2) where u is the displacement of a volume element from its equilibrium positions at x in the crystal. uk = u0 cos(ωt − kx) = u0 (cos ωt cos kx + sin ωt sin kx) . (12.3) The energy in the mode is εk = [ K + P ] = 2 K (∵ K = P ) (12.4) photons. The term 31 when it is averaged over time. 2 1 ⎛ ∂u ⎞ K = ρ⎜ ⎟ , (12.5) 2 ⎝ ∂t ⎠ ∂u (12.6) = u 0ω k [− sin ω k t cos kx + cos ω k t sin kx ] , ∂t where ρ is the mass density. 1 2 2 2 2 2 2 ∫ KdV = 2 ρu 0 ω k ∫VdV [− sin ω k t cos kx + cos ω k t sin kx . (12.7) − 2 sin ω k t cos ω k t sin kx cos kx] Here we have ⎧ x = 0 ~ Na (= L) ⎪ ⎪ 2π N⎞ π ⎛ (12.8) = ⎜ = 0,±1, ,± ⎟ , ⎨k = N aN 2 ⎛ ⎞ ⎝ ⎠ ⎪ a⎜ ⎟ ⎪⎩ ⎝2⎠ from the boundary condition: cos(kNA) = 1 ⇒ kNa = 2π ). Then we obtain Na Na 0 0 2 ∫ dx cos kx = ∫ dx 1 (1 + cos 2kx ) = 1 Na + [k sin 2kx]0Na = 1 Na . 2 2 2 ⎛ 2π ⎞ (∵ 2kNa = 2⎜ ⎟ Na = 4π ). ⎝ aN ⎠ Similarly, we have Na Na 1 2 ∫0 dx sin kx = 2 Na , ∫0 dx sin kx cos kx = 0 . Then we have V 2 2 V 2 2 2 2 ∫ KdV = 4 ρu0ω (sin ωt + cos ωt ) = 4 ρu0ω . The time average kinetic energy is T V V 2 2 1 K = ρu 0 ω dt = ρu 02ω 2 . ∫ 4 4 T 0 1 ε , we have 2 1 1⎛ 1⎞ V ρu 02ω k2 = ε = ⎜ nk + ⎟ ω k , 4 2 2⎝ 2⎠ (12.9) (12.10) (12.11) (12.12) Since K = (12.13) or 1⎞ ⎛ 2 ⎜ nk + ⎟ 1 2⎠ u0 . ⇒ u eff = u 02 = ⎝ ρVω k 2 Since ρV = NM , 32 (12.15) 1⎞ ⎛ 2 ⎜ nk + ⎟ 2⎠ . (12.16) u 02 = ⎝ NMω k This relates the displacement in a given mode to the phonon occupancy n of the mode. An optical mode with ω close to zero is called a soft mode. 12.2 General case The Lagrangian L is given by L = T −V 1 ⎡M 2⎤ = ∑ ⎢ us2 − C (us +1 − us ) ⎥ . 2 ⎦ s ⎣ 2 = L(u1 , u2 ,.., u N −1 , u1 , u2 ,.., u N −1 ) The linear momentum conjugate to us, is given by ∂L ps = = Mu s . ∂u s Then the Hamiltonian H can be derived as H = ∑ ps us − L (12.17) (12.18) s 1 M 2 2 u s + ∑ C (u s +1 − u s ) . 2 s s s 2 1 M 2 = ∑ u s2 + ∑ C (u s +1 − u s ) 2 s s 2 Now we calculate the total energy for the case of u s = u cos( ska − ωt ) . = ∑ Mu s2 − ∑ u s = u (−1)(−ω ) sin( ska − ωt ) , (12.19) (12.20) or E= 1 Mu 2ω 2 ∑ [1 − cos 2(ωt − ska)] 4 s , (12.21) 1 2 1 ⎡ ⎤ + Cu (1 − cos ka)∑ ⎢1 − cos 2(ωt − ska − ka)⎥ 2 2 ⎦ s ⎣ NM (mass density), V = Na. with ρ = V The dispersion relation is given by 2C ω2 = (1 − cos ka) . (12.22) M Then the total energy is ⎧ 1 1 ⎡ ⎤⎫ E = Mu 2ω 2 ⎨∑ [1 − cos 2(ωt − ska)] + ∑ ⎢1 − cos 2(ωt − ska − ka)⎥ ⎬ . 4 2 ⎦⎭ s ⎣ ⎩ s (12.23) The time-average is 33 T E = 1 1 1 Edt = Mu 2ω 2 ∑1 = Mu 2ω 2 N ∫ T0 2 2 s 1 = ρVu 2ω 2 2 where T= 2π ω , ρ= , (12.24) NM NM , NM = ρV = V Na In general, we use u s = ∑ u k cos( ska − ωt ) (12.25) k Then we have 1 ρV ∑ u k2ω k2 (12.26) 2 k This energy is compared with the result derived from the quantum mechanics. 1 1⎞ ⎛ ρV ∑ u k2ω k2 = ∑ ω k ⎜ nˆ k + ⎟ . 2 2⎠ ⎝ k k or (12.27) 1 1⎞ ⎛ ρVu k2ω k2 = ω k ⎜ nˆ k + ⎟ , (12.28) 2 2⎠ ⎝ or 1⎞ 1⎞ ⎛ ⎛ 2 ⎜ nˆ k + ⎟ 2 ⎜ nˆ k + ⎟ 2⎠ 2⎠ , (12.29) u k2 = ⎝ = ⎝ ρVω k NMω k where ρV = NM . Here we define the effective amplitude as 1 (12.30) ukeff = uk , 2 where 1⎞ ⎛ ⎜ nˆk + ⎟ (ukeff )2 = ⎝ NMω 2 ⎠ . (12.31) k E = eff If the root-mean square of the average displacement (= < (uk ) 2 > ) is comparable to the lattice constant a, the system may melt (Lindeman criterion).8 13. 13.1 Inelastic neutron scattering by crystal with lattice vibration Scattering cross section 34 Fig.29 The neutron scattering by phonons in the system. ki is the wavevector of the incident neutron and kf is the wavevector of the outgoing neutron. Q is the scattering vector. Q = kf – ki. |ki| = |kf| = 2π/λ, λ is the wavelength of neutron. We consider the neutron diffraction by a crystal with lattice vibration. The scattering form factor is given by − iQ ⋅ R j , (13.1) F (Q) = ∑ be j where Q (=kf – ki) is the scattering vector, kf and ki are the wave vectors of incident and outgoing neutrons, and b is scattering amplitude (which is independent of Q). The vector Rj is the actual position of the atom that ought to have been at the lattice Rj0. For Rj = Rj0 (periodic configuration), we have a Bragg condition − iQ ⋅ R (13.2) ∑ e j = Nδ (Q − G ) . j When Rj = Rj(0) + uj, we have (0) F (Q) = b∑ exp[−iQ ⋅ (R j + u j )] j = b∑ exp[−iQ ⋅ (R j )] − ibQ ⋅ ∑ u j exp(−iQ ⋅ R j ) (0) j (0) . (13.3) j The first term is a Bragg reflection. We assume that the displacement vector uj is given by u j = e(q)[exp(iq ⋅ R (j0 ) ) + exp(−iq ⋅ R (j0) )]u (q) , (13.4) where q is the wavevector of phonon (k is not used in this section to avoid confusion), e(q) is the polarization vector and u(q) is the displacement amplitude. Then the second term is rewritten as F2 (Q) = −ib[Q ⋅ e(q)]∑{exp[i (q − Q) ⋅ R (j0 ) ] + exp[−i (q + Q) ⋅ R (j0 ) ]}u (q) j . (13.5) = −iNb[Q ⋅ e(q)]u (q){δ (Q + q − G ) + δ (Q − q − G )} Then the scattering intensity S(Q) is proportional to S (Q) = N 2b 2 [Q ⋅ e(q)]2 | u (q) |2 , (13.6) where Q = G ± q . 35 Fig.30 The relation between ki, kf, Q (= kf – ki), q and G. ki and kf are the wavevectors of incident and outgoing neutrons. ki and kf are on the Ewald sphere with radius (|ki| = |kf| = 2π/λ, λ is the wavelength of neutron). Q is the scattering vector. q is the wavevector of phonon. G is the reciprocal lattice vector. Q = G + q. When we use 1⎞ ⎛ ⎜ nq + ⎟ 2⎠ | u (q) |2 = ⎝ , (13.7) NMωq the Intensity S(Q) can be rewritten as 1⎞ ⎛ ⎜ nq + ⎟ 2⎠ . (13.8) S (Q) = Nb 2 [Q ⋅ e(q)]2 ⎝ Mωq Here we need to take into account of the energy conservation. The kinetic energy of the incident neutron and the outgoing neutrons is given by ε and ε’. The energy conservation law is valid; ε ' = ε + ωq for the absorption of phonon with ωq and ε ' = ε − ωq for the emission of phonon with ωq. Finally we get the dynamic structure factor given by [Q ⋅ e(q)]2 ωq 2 S (Q, ω ) = Nb [( nq + 1)πδ (ω − ωq ) + nq πδ (ω + ωq )] , (13.9) 2 Mωq where ω = (ε '−ε ) / , + aˆq nq = nq + 1 nq + 1 for the creation of phonon, (13.10) aˆq nq = nq nq − 1 for the destruction of phonon. (13.11) (see the Appendix A). Using the factor [Q ⋅ e(q)]2 , we can select the branch; if Q ⊥ e(q) , this branch does not contribute to the inelastic neutron scattering. We note that S(Q) is related to S(Q, ω) through 1 S (Q) = S (Q, ω )dω . (13.12) 2π ∫ 13.2 Energy and momentum conservation We consider the matrix element given by the integral over the entire space of the system, ∫ exp(−ik f ⋅ r) exp(iq ⋅ r) exp(ik i ⋅ r)dr = ∫ exp[−i(k f − k i − q) ⋅ r]dr . 36 (13.13) This integral is not equal to zero only when k f − ki − q = G . (13.14) The integral over the time is given by ∫ exp(−iωk f t ) exp(±iωqt ) exp(iωk i t )dt = ∫ exp[i(−ωk f ± ωq + ωk i )t ]dt . (13.15) is not equal to zero only when ω = ωk f − ωk i = ±ωq (energy conservation). (13.16) Conclusion We have shown that the well-defined lattice waves propagate over the crystal, forming a so-called phonon as a quantization of the lattice waves. Phonon has the dual characters of wave and particle, which is essential to the quantum mechanics. Phonon is one of bosons, obeying the Bose-Einstein statistics. Phonons will be seen to play an important role in any phenomena for which the energy of importance is comparable to ω , the energy of the phonon in question. In the BCS (Bardeen-Cooper-Schrieffer) model for the superconductivity, a specific interaction between electrons can lead to an energy gap separated from excited states by a energy gap. The formation of Cooper pairs is due to the electron-phonon interaction. The first electron interacts with the lattice and deforms it; the second electron sees the deformed lattice. Thus the second electron interacts with the first electron through the lattice deformation. REFERENCES 1. J.M. Ziman, Electrons and Phonons (Oxford at the Clarendon Press, 1960). 2. J.M. Ziman, Principles of the Theory of Solids (Cambridge University Press.1964). 3. J.M. Ziman, Elements of Advanced Quantum Theory (Cambridge University Press, Cambridge, 1969). 4. C. Kittel, Introduction to Solid State Physics, sixth edition (John Wiley & Sons, New York, 1986). 5. C. Kittel, Quantum Theory of Solids (John Wiley & Sons, New York, 1963). 6. G.H. Wannier, Elements of Solid State Theory (Cambridge at the University Press, 1960). 7. R.E. Peierls, Quantum Theory of Solids (Oxford at the Clarendon Press, 1964). 8. D. Pines, Elementary Excitation in Solids (W.A. Benjamin, Inc New York 1964). 9. W. Cochran, The dynamics of atoms in crystals (Edward Arnold (Publishers) Ltd., London,1973). 10. N.W. Ashcroft and N.D. Mermin, Solid State Physics (Holt, Rinehart and Winston, New York, 1976). 11. R.A. Levy, Principles of Solid State Physics (Academic Press, New York, 1968). 12. A.A. Maradudin, I.M. Lifshitz, A.M. Kosevich, W. Cochran, and M.J.P. Musgrave, Lattice Dynamics (W.A. Benjamin, Inc. New York, 1969). 13. S.L. Altman, Band Theory of Solids. An Introduction from the Point of View of Symmetry (Oxford University Press, New York 1991). 14. R. Huebener, Electrons in Action, Roads to Modern Computers and Electronics. (Wiley-VCH Verlag GmbH & Co. KGaA, 2005). 15. J.J. Sakurai, Modern Quantum Mechanics (Addison Wesley, New York, 1994). 37 APPENDIX Simple harmonics (1D) in quantum mechanics15 A (a) Annihilation and creation operator The commutation relation [xˆ, pˆ ] = i . The Hamiltonian of the simple harmonics is 1 2 mω02 2 ˆ H= pˆ + xˆ . 2m 2 The eigenvalue-problem of the simple harmonics Hˆ n = ε n n , (A.1) (A.2) (A.3) with 1⎞ ⎛ (A.4) 2⎠ ⎝ where n = 0, 1, 2, 3,.... In the { x } representation, the wave function of the simple harmonics can be described as ε n = ⎜ n + ⎟ ω0 , 2 ⎛ d 2 mω02 2 ⎞ ⎜⎜ − (A.5) + x ⎟⎟ x n = ε n x n . 2 2 ⎝ 2m dx ⎠ Here we introduce the creation operator and annihilation operators given by ipˆ ⎞ 1 ⎛⎜ mω0 ipˆ ⎞⎟ ⎜⎜ xˆ + ⎟⎟ = xˆ + , mω0 ⎠ m ω0 ⎟⎠ 2 ⎜⎝ 2⎝ β ⎛ ipˆ ⎞ 1 ⎛⎜ mω0 ipˆ ⎞⎟ ⎜⎜ xˆ − ⎟⎟ = aˆ + = xˆ − , mω0 ⎠ m ω0 ⎟⎠ 2 ⎜⎝ 2⎝ aˆ = β ⎛ (A.6) (A.7) with mω0 β= (A.8) ( ) 1 aˆ + aˆ + = 2β xˆ = (aˆ + aˆ ), + 2mω0 (A.9) 1 mω0 1 m ω0 ( (aˆ − aˆ + ), aˆ − aˆ + ) = i 2 2β i [xˆ, pˆ ] = 1 2 mω0 aˆ + aˆ + , aˆ − aˆ + = − aˆ, aˆ + , i i 2β pˆ = ( or ) [ ] [ ] [aˆ, aˆ ] = 1̂ + aˆ + aˆ = (A.11) (A.12) ⎞ ipˆ ⎞⎛ 1 ipˆ ⎞ β ⎛ 2 pˆ ⎜⎜ xˆ + 2 2 − i ⎟⎟ = ⎟⎟⎜⎜ xˆ + ⎜⎜ xˆ − [ pˆ , xˆ ]⎟⎟ 2 ⎝ mω0 ⎠⎝ mω0 ⎠ 2 ⎝ m ω0 mω0 ⎠ β ⎛ 2 (A.10) 2 38 2 (A.13) or aˆ + aˆ = 1 ⎛ ˆ 1 ⎞ ⎜ H − ω0 ⎟ 2 ω0 ⎝ ⎠ (A.14) or 1⎞ ⎛ Hˆ = ω0 ⎜ Nˆ + ⎟ , 2⎠ ⎝ (A.15) where Nˆ = aˆ + aˆ . The operator N̂ is Hermitian since (A.16) + Nˆ + = (aˆ + aˆ ) = aˆ + aˆ = Nˆ . (A.17) The eigenvectors of Ĥ are those of N̂ , and vice versa since [ Hˆ , Nˆ ] = 0 , [ Nˆ , aˆ ] = aˆ + aˆ , aˆ = aˆ + aˆaˆ − aˆaˆ + aˆ = aˆ + , aˆ aˆ = −aˆ , [ Nˆ , aˆ + ] = aˆ + aˆ , aˆ + = aˆ + aˆaˆ + − aˆ + aˆ + aˆ = aˆ + aˆ , aˆ + = aˆ + . Thus we have the relations [ Nˆ , aˆ ] = −aˆ , and [ Nˆ , aˆ + ] = aˆ + , 1⎞ ⎛ Nˆ n = ω0 ⎜ n + ⎟ n . 2⎠ ⎝ [ Nˆ , aˆ ] n = −aˆ n , From the relation [ [ ] ] [ ] [ ] (A.18) (A.19) (A20) (A.21) (A.22) ( Nˆ aˆ − aˆNˆ ) n = −aˆ n , (A.23) Nˆ (aˆ n ) = (n − 1)aˆ n . (A.24) or â n is the eigenket of N̂ with the eigenvalue (n-1). aˆ n ≈ n − 1 . (A.25) From the relation [ Nˆ , aˆ + ] n = aˆ + n , ( Nˆ aˆ − aˆ Nˆ ) n = − aˆ n , or Nˆ (aˆ + n ) = (n + 1)aˆ + n . (A.26) (A.27) (A.28) aˆ + n is the eigenket of N̂ with the eigenvalue (n+1). aˆ + n ≈ n + 1 (A.29) Now we need to show that n should be either zero or positive integers: n = 0, 1, 2, 3,…. We note that n aˆ + aˆ n = n n n ≥ 0 , (A.30) 39 n aˆaˆ + n = n aˆ + aˆ + 1 n = (n + 1) n n ≥ 0 . (A.31) The norm of a ket vector is non-negative and the vanishing of the norm is a necessary and sufficient condition for the vanishing of the ket vector. In other words, n ≥ 0. If n = 0, aˆ n = 0. If n ≠ 0, â n is a nonzero ket vector of norm n n n . If n>0, one successively forms the set of eigenkets, aˆ n , aˆ 2 n , aˆ 3 n , …. aˆ p n , , belonging to the eigenvalues, n-1, n-2, n-3,….., n-p, This set is certainly limited since the eigenvalues of N̂ have a lower limit of zero. In other words, the eigenket aˆ p n ≈ n − p , or n-p = 0. Thus n should be a positive integer. Similarly, one successively forms the set of eigenkets, 2 3 p aˆ + n , aˆ + n , aˆ + n , …. aˆ + n , belonging to the eigenvalues, n+1, n+2, n+3,….., n+p, Thus the eigenvalues are either zero or positive integers: n = 0, 1, 2, 3, 4, . + The properties of â and â (A.32) 1. aˆ 0 = 0 , since 0 aˆ + aˆ 0 = 0 . 2. aˆ + n = n + 1 n + 1 , [ Nˆ , aˆ + ] n = aˆ + n , Nˆ aˆ + n = aˆ + Nˆ n + aˆ + n = (n + 1)aˆ + n . (A.33) (A.34) (A.35) aˆ + n is an eigenket of N̂ with the eigenvalue (n + 1). Then aˆ + n = c n + 1 . (A.36) Since 2 2 n aˆaˆ + n = c n + 1 n + 1 = c , (A.37) n aˆ + aˆ + 1 n = n + 1 = c , we obtain c = n +1 . 2 (A.39) aˆ n = n n − 1 , [ Nˆ , aˆ ] n = −aˆ n , (A.41) Nˆ aˆ n = aˆNˆ n − aˆ n = (n − 1)aˆ n , (A.43) or (A.40) 3. (A.42) â n is an eigenket of N̂ with the eigenvalue (n - 1) Then aˆ n = c n − 1 . (A.44) Since 40 2 (b) 2 n aˆ + aˆ n = c n − 1 n − 1 = c = n , (A.45) c= n (A.46) Basis vectors in terms of 0 We use the relation 1 = aˆ + 0 , 1 + 1 +2 (aˆ ) 0 , aˆ 1 = 2 2! 1 + 1 +3 (aˆ ) 0 , 3 = aˆ 2 = 3 3! ----------------------------------------------------1 +n 1 + (aˆ ) 0 , n = aˆ n − 1 = n n! The expression for x̂ n and p̂ n 2 = xˆ n = (aˆ + aˆ ) n + 2 mω 0 = 2mω0 (A.48) ( ) n +1 n +1 + n n −1 , ( (A.49) ) m ω0 + m ω0 i aˆ − aˆ n = i n +1 n +1 − n n −1 . (A.50) 2 2 Therefore the matrix elements of â , â + , x̂ , and p̂ operators in the { n } representation ( pˆ n = ) are as follows. n' aˆ n = nδ n ', n −1 , (A.51) n' aˆ + n = n + 1δ n ', n +1 , n' xˆ n = 2 mω 0 ( (A.52) ) n + 1δ n ', n +1 + nδ n ', n −1 , ( (A.53) ) m ω0 n + 1δ n ', n +1 − nδ n ', n −1 . (A.54) 2 Mean values and root-mean-square deviations of x̂ and p̂ in the state n . n' pˆ n = i n xˆ n = 0 , (A.55) n pˆ n = 0 , (A.56) 1⎞ ⎛ , (A.57) n xˆ 2 n = ⎜ n + ⎟ 2 ⎠ mω 0 ⎝ (Δp )2 = n pˆ 2 n = ⎛⎜ n + 1 ⎞⎟m ω0 . (A.58) 2⎠ ⎝ The product ΔxΔp is 1⎞ 1 ⎛ ΔxΔp = ⎜ n + ⎟ ≥ (Heisenberg’s principle of uncertainty), (A.59) 2⎠ 2 ⎝ Note that (Δx )2 = 41 (aˆ xˆ 2 = + + aˆ )(aˆ + + aˆ ) = (aˆ aˆ + + + aˆaˆ + aˆ + aˆ + aˆaˆ + ) , 2mω0 2mω0 m ω0 + m ω0 + + ( ( pˆ 2 = aˆ − aˆ )(aˆ + − aˆ ) = aˆ aˆ + aˆaˆ − aˆ + aˆ − aˆaˆ + ) , 2 2 and ( ) n aˆ + 2 n =0, (A.61) (A.62) n aˆ 2 n = 0 , (A.63) n aˆ + aˆ + aˆaˆ + n = n 2aˆ + aˆ + 1 n = 2n + 1 , (A.64) Mean potential energy 1 1 1 2 V = mω 2 n xˆ 2 n = mω 2 (Δx ) = ε n . 2 2 2 Mean kinetic energy 1 1 (Δp )2 = 1 ε n . K = n pˆ 2 n = 2m 2m 2 Thus we have V = K . (Virial theorem) B. (A.60) Translation operator in quantum mechanics15 Here we discuss the translation operator Tˆ (a ) in quantum mechanics, ψ ' = Tˆ (a ) ψ , (A.65) (A.66) (A.67) (B.1) or ψ ' = ψ ' Tˆ + (a) . (B.2) In an analogy from the classical mechanics, it is predicted that the average value of x̂ in the new state ψ ' is equal to that of x̂ in the old state ψ plus the x-displacement a under the translation of the system ψ ' xˆ ψ ' = ψ xˆ + a ψ , or ψ Tˆ + (a) xˆTˆ (a) ψ = ψ xˆ + a ψ , or Tˆ + (a) xˆTˆ (a) = xˆ + a1̂ . Normalization condition: ψ ' ψ ' = ψ Tˆ + (a)Tˆ (a) ψ = ψ ψ , or Tˆ + (a)Tˆ (a ) = 1 . [ Tˆ (a) is an unitary operator]. From Eqs.(B3) and (B4), we have xˆTˆ (a) = Tˆ (a)( xˆ + a) = Tˆ (a) xˆ + aTˆ (a) , 42 (B.3) (B.4) or the commutation relation: [ xˆ, Tˆ (a)] = aTˆ (a) . From this, we have xˆTˆ (a ) x = Tˆ (a) xˆ x + aTˆ (a) x = ( x + a)Tˆ (a) x . Thus, Tˆ (a) x is the eigenket of x̂ with the eigenvalue (x+a). (B.5) or Tˆ (a) x = x + a , (B.6) or Tˆ + (a)Tˆ (a) x = Tˆ + (a) x + a = x . When x is replaced by x-a in Eq.(B7), we get x − a = Tˆ + (a) x , or x − a = x Tˆ (a ) . Note that x ψ ' = x Tˆ (a) ψ = x − a ψ = ψ ( x − a) . (B.7) (B.8) (B.9) (B.10) The average value of p̂ in the new state ψ ' is equal to the average value of p̂ in the old state ψ under the translation of the system ψ ' pˆ ψ ' = ψ pˆ ψ , (B.11) or ψ Tˆ + (a) pˆ Tˆ (a ) ψ = ψ pˆ ψ , or Tˆ + (a) pˆ Tˆ (a) = pˆ . So we have the commutation relation [Tˆ (a ), pˆ ] = 0̂ . From this commutation relation, we have pˆ Tˆ (a) p = Tˆ (a) pˆ p = pTˆ (a) p . Thus, Tˆ (a) p is the eigenket of p̂ associated with the eigenvalue p. 43 (B.12)