固体物理习题解答Chapter 4 problems

1、固体物理习题解答固体物理习题解答Chapter 4 Chapter 4 problemsproblems(b) The time-average total energy per atomSubstitute us = u cos(t sKa) into the expression)(2sin)cos1 (sin)(sinsin )(cos)cos1(21)(sin21 )sin(sin)cos()cos1(21 )(sin21 sin)sin( cos)cos()cos(21 )(sin21 ) 1(cos)cos(21 )sin(2122222222222222222222sKatKaK

2、asKatKasKatKaCusKatuMsKatKasKatKaCusKatuMKasKatKasKatsKatCusKatuMKastusKatuCsKatuMEssssssssThe total energy average over time)cos22(4 sin21)cos1 (212141 )(2sin)cos1 (sin )(sinsin)(cos)cos1(2121 )(sin2121 222222222222220220222/20KaCuuMNKaKaNCuuNMsKaxKaKasKaxKasKaxKadxCusKaxdxuMEdtEsswhere N is the nu

3、mber of the atoms.The time-average total energy per atom222)cos1 (2141 /uKaCuMNEFrom the dispersion relation, 2/sin422KaMCwe have).cos1 (212/sin4122KaCKaCM2222221 )cos1 (2141uMuKaCuMThus 2. Continuum wave equationWe have the equation of the motion)2(1122ssssuuuCdtudMIn the long wavelength limit, a,

4、the difference of the displacements of nearest atoms is very small. Hence us(t) could be treated as a continuous function u(x, t).22222 )(2)()(dxudCaadxduadxduCxuaxuaxuCdtudMxaxThen we haveThe solution of this equation is )(exp),(0tKxiutxuwith the dispersion relation, 2/sin422KaMCIn the long wavelen

5、gth limit, Ka 1,2222)2/(4KvKaMCwhereMCav22Therefore the equation of motion reduces to the continuum wave equation22222dxudvdtud3. Basis of two unlike atomsFrom the equation of motions)2()2(12221221ssssssvuuCdtvdMuvvCdtudMWe have the solutions)(exp)(exptKaivvtKaiuuss(1)(2)Substitute Eq(2) into Eq(1)C

6、veCuvMCueCvuMiKaiKa2) 1(2)1 (2212(3)(4)At the Brillouin zone boundary K = Kmax = /a, we haveCvvMCuuM2222120)2(0)2(2212vCMuCMi.e.122 when , 0MCvor222 when , 0MCui.e. these two lattices act as if decoupled: one lattice remains at rest while the other lattice moves4. Kohn anomalyConsidering the interac

7、tions between p nearest planes, we have the dispersion relation012)cos1 (2pjjjKaCMSupposingpaapkACp0sinwe have1012)cos1 (sin2 )cos1 (2ppppKapaapkMApKaCMwith A and k0 are constants and p runs over all integers,Then102sinsin2ppKaapkMAKWhen K = k01022sin2papkMAKinfinite is sin )/( 0sin 02002limppapkaka

8、pki.e. is infinite when K = k0.K2Thus a plot of 2 vs K (or vs K) have a vertical tangent at K = k0: there is a kink at k0 in the phonon dispersion relation (K).5. Diatomic chainWe have the equation of motions)(10)()()(10122122ssssssssssvuvuCdtvdMuvCuvCdtudMThe solutions are)(exp)(exptKaivvtKaiuussSu

9、bstitute the solutions into equation of motionsCveCuMvCueCvMuiKaiKa11)10(11)10(22The homogenous linear equations have a solution only if the determinant of the coefficients of the unknown u, v vanishes.011 )exp(10)exp(10 1122MCiKaCiKaCMCthe dispersion relation is)cos1 (20111122KaMCor0)cos1 (20222242

10、KaCCMM2/1MCK a022022Discussions:(1) K = 00 and ,2222MC(2) K = /aMCMC2 and ,2022Obviously, the acoustic branch indicates the interactions between molecules while the optical branch shows the interactions inside the molecules. 6. Atomic vibrations in metalrRSuppose the restoring force is due to the el

11、ectric charge within the sphere of the radius r centered to the equilibrium position.)()(rEqrFAs shown in the figure, the electric field in a sphere with the charge homogenous distributed isrRerreRrrrqrE323324/34/3)(The equation of motionrReFqdtrdM3222i.e.322222with , 0MRerdtrd(a) The frequency of a

12、 single ion oscillationFor a harmonic oscillation )exp(0tirrwhere 2/132MRe(b) Estimate the value of this frequency for sodiumSodium has a bcc structure with lattice constant a = 4.225 (p. 23).Then R = 31/2a/4 = 1.83 =1.83E8 cmM 23Mp 3.84E23 g 4.8E10/3.43E23(1.83E8)31/2 3.3E13 s1(c) Estimate the velo

13、city of the sound in metalSuppose the dispersion relation is = vgK for metal, where vg is constant.In estimation, we take K = /a = /4.225E8 1E8 cm-1vg = /K 3E13/1E8 =3E5 cm/s7. Soft phonon modes(a) Force constant of the Coulomb interactionThe Coulomb between atom s and atom s+p is spspsppspuurererU0

14、22) 1() 1()(Then 20232222221 ) 1(221) 1( ) 3(21)()(pCparppparparspspCUrepaeOrUrUparUrUwhere 332) 1(2apeCppC(b) The dispersion relation03200332203320102)cos1 () 1(21sin )cos1 () 1(21sin4 )cos1 () 1(22)cos1 (2 )cos1 (2)cos1 (2 )cos1 (22ppppppppCRpppKapKapKapaeKaMpKaapeMKaMpKaCMKaCMpKaCM132202)cos1 () 1(21sin i.e.ppppKaKa3220/ and /4 whereaeM(c) Discussions0313202) 12(21 )cos1 () 1(1nppnpp(1) At the first Brillouin zone boundary, Ka = 1313130387)2() 12( nnnnnnnn13202471nn2 is negative when 1374nn(2