Exam (elaborations) TEST BANK FOR The Oxford Solid State Basics By Steven H. Simon (Solution Manual)

Exam (elaborations) TEST BANK FOR The Oxford Solid State Basics By Steven H. Simon (Solution Manual) The Oxford Solid State Basics Solutions to Exercises Steven H. Simon Oxford University CLARENDON PRESS . OXFORD 2015 Contents 1 About Condensed Matter Physics 1 2 Specific Heat of Solids: Boltzmann, Einstein, and Debye 3 3 Electrons in Metals: Drude Theory 15 4 More Electrons in Metals: Sommerfeld (Free Electron) Theory 21 5 The Periodic Table 35 6 What Holds Solids Together: Chemical Bonding 39 7 Types of Matter 47 8 One-Dimensional Model of Compressibility, Sound, and Thermal Expansion 49 9 Vibrations of a One-Dimensional Monatomic Chain 55 10 Vibrations of a One-Dimensional Diatomic Chain 71 11 Tight Binding Chain (Interlude and Preview) 81 12 Crystal Structure 95 13 Reciprocal Lattice, Brillouin Zone, Waves in Crystals 99 14 Wave Scattering by Crystals 111 15 Electrons in a Periodic Potential 125 16 Insulator, Semiconductor, or Metal 135 17 Semiconductor Physics 139 18 Semiconductor Devices 149 19 Magnetic Properties of Atoms: Para- and Dia-Magnetism 159 vi Contents 20 Spontaneous Magnetic Order: Ferro-, Antiferro-, and Ferri-Magnetism 167 21 Domains and Hysteresis 175 22 Mean Field Theory 179 23 Magnetism from Interactions: The Hubbard Model 191 About Condensed Matter Physics 1 There are no exercises for chapter 1. Specific Heat of Solids: Boltzmann, Einstein, and Debye 2 (2.1) Einstein Solid (a) Classical Einstein (or “Boltzmann”) Solid: Consider a three dimensional simple harmonic oscillator with mass m and spring constant k (i.e., the mass is attracted to the origin with the same spring constant in all three directions). The Hamiltonian is given in the usual way by H = p2 2m + k 2 x2  Calculate the classical partition function Z = Z dp (2π~)3 Z dx e−H(p,x) Note: in this problem p and x are three dimensional vectors.  Using the partition function, calculate the heat capacity 3kB.  Conclude that if you can consider a solid to consist of N atoms all in harmonic wells, then the heat capacity should be 3NkB = 3R, in agreement with the law of Dulong and Petit. (b) Quantum Einstein Solid: Now consider the same Hamiltonian quantum mechanically.  Calculate the quantum partition function Z = X j e−Ej where the sum over j is a sum over all eigenstates.  Explain the relationship with Bose statistics.  Find an expression for the heat capacity.  Show that the high temperature limit agrees with the law of Dulong of Petit.  Sketch the heat capacity as a function of temperature. (See also exercise 2.7 for more on the same topic) (a) H = p2 2m + k 2 x2 Z = Z dp (2π~)3 Z dx e−βH(p,x) Since, Z ∞ −∞ dy e−ay2 = p π/a in three dimensions, we get Z = h 1/(2π~) p π/(β/2m) p π/(βk/2)) i3 = (~ωβ)−3 with ω = p k/m. From the partition function U = −(1/Z)∂Z/∂β = 3/β = 3kBT 4 Specific Heat of Solids: Boltzmann, Einstein, and Debye Thus the heat capacity ∂U/∂T is 3kB. (b) Quantum mechanically, for a 1d harmonic oscillator, we have eigenenergies En = ~ω(n + 1/2) with ω = p k/m. The partition function is then Z1d = X n≥0 e−β~ω(n+1/2) = e−β~ω/21/(1 − e−β~ω) = 1/[2 sinh(β~ω/2)] The expectation of energy is then 00 1 2 0.25 0.5 0.75 1 Fig. 2.1 Heat capacity in the Einstein model (per atom) in one dimension. Units are kb on vertical axis and kbT/ω on horizontal. In three dimensions, the heat capacity per atom is three times as large. hEi = −(1/Z)∂Z/∂β = (~ω/2) coth(β~ω/2) = ~ω(nB(β~ω) + 1 2 ) where nB is the boson occupation factor nB(x) = 1/(ex − 1) (hence again the relationship with free bosons). The high temperature limit gives nB(x) → 1/(x+x2/2) = 1/x−1/2 so that hEi → kBT . More generally, we obtain C = kB(β~ω)2 eβ~ω (eβ~ω − 1)2 In 3D, En1,n2,n3 = ~ω[(n1 + 1/2) + (n2 + 1/2) + (n3 + 1/2)] and Z3d = X n1,n2,n3≥0 e−βEn1,n2,n3 = [Z1d]3 and correspondingly hEi = 3~ω(nB(β~ω) + 1 2 ) So the high temperature limit is hEi → 3kBT and the heat capacity C = ∂hEi/∂T = 3kB. More generally we obtain C = 3kB(β~ω)2 eβ~ω (eβ~ω − 1)2 Plotted this looks like Fig. 2.1. 5 (2.2) Debye Theory I (a)‡ State the assumptions of the Debye model of heat capacity of a solid.  Derive the Debye heat capacity as a function of temperature (you will have to leave the final result in terms of an integral that cannot be done analytically).  From the final result, obtain the high and low temperature limits of the heat capacity analytically. You may find the following integral to be useful Z ∞ 0 dx x3 ex − 1 = ∞X n=1 Z ∞ 0 x3e−nx = 6 ∞X n=1 1 n4 = π4 15 By integrating by parts this can also be written as Z ∞ 0 dx x4ex (ex − 1)2 = 4π4 15 . (b) The following table gives the heat capacity C for potassium iodide as a function of temperature. T (K) C(J K−1mol−1) 0.1 8.5 × 10−7 1.0 8.6 × 10−4 5 .12 8 .59 10 1.1 15 2.8 20 6.3  Discuss, with reference to the Debye theory, and make an estimate of the Debye temperature. (a) The key assumption of Debye theory is that the dispersion curve is linear (ω = vk) up to a cut-off frequency ωDebye determined by the requirement that the total number of vibrational modes is correct. For a crystal containing N atoms, the low temperature limiting form is C = 12NkBπ4 5  T TD 3 (2.1) and the high temperature limit is 3NkB. Here, TD = ~ωDebye/kB. The full derivation goes as follows. For oscillators with frequency ω(k) a system has a full energy E = L3 Z d3k(2π)3~ω(k)[nB(β~ω(k)) + 1/2] One includes also a factor of 3 out front to account for the three different sound modes (two transverse and one longitudinal) and we cut off the integral at some cutoff frequency ωcutoff . We use the assumption that ω = v|k| although it is not much harder to consider three different velocities for the three different modes. We thus obtain E = Z ωcutoff 0 dωg(ω)[nB(β~ω) + 1/2]~ω where g(ω) = N  12πω2 (2π)3nv3  = N 9ω2 ω3 d and we have replaced nL3 = N where n is the density of atoms. Here ω3 d = 6π2nv3 is the Debye frequency, and ~ωd = kBTDebye defines the Debye temperature. Note that there is no dependence of g(ω) on the density n (it cancels). This shows that until the cutoff is imposed, there 6 Specific Heat of Solids: Boltzmann, Einstein, and Debye is actually no knowledge of the underlying lattice — only the overall volume and sound velocity. We should choose the cutoff frequency such that we have the right number of modes in the system, thus we have 3N = Z ωcutoff 0 dωg(ω) performing this integral, we find that the proper value of ωcutoff is exactly the Debye frequency ωd that we just defined. The general heat Debye theory heat capacity will then be C = dhEi/dT = kB (kBT )2 Z ωd 0 dωg(ω)(~ω)2 eβ~ω (eβ~ω − 1)2 Defining x = ~ω/kBT we obtain C = dhEi/dT = NkB  T TDebye 3 9 Z ~ωd/kBT 0 dxx4 ex (ex − 1)2 This integral is known as the Debye integral. In the low temperature limit, we can extend the integral out to infinity whereupon it just gives the constant 4π4/15 recovering the above claimed result Eq. 2.1. In the high temperature limit, the exponents can be expanded such that the Debye integral becomes Z ~ωd/kBT 0 dxx4 ex (ex − 1)2 = Z ~ωd/kBT 0 dxx2 = (1/3)(~ωd/kBT )3 which then recovers the law of Dulong-Petit C = 3NkB (b) Given the heat capacity and the temperature, in the low T limit we should have (from Eq. 2.1) TD =  12Rπ4T 3 5C 1/3 The table of heat capacity looks like T (K) 0.1 1.0 5 8 10 15 20 C (J K −1 mol −1) 8.5 × 10−7 8.6 × 10−4 1.2 × 10−1 5.9 × 10−1 1.1 2.8 6.3  12Rπ4T3 5C 1/3 (K) 132 135 So TDebye is about 130K. The fact that the T 3 fit is not perfect is a reflection of (a) that Debye theory is just an approximation (in particular that phonons have a nonlinear spectrum!) and (b) that one needs to be in the low T limit to obtain perfect T 3 scaling. (Note that at low enough T , the T 3 scaling does indeed work). 7 (2.3) Debye Theory II Use the Debye approximation to determine the heat capacity of a two dimensional solid as a function of temperature.  State your assumptions. You will need to leave your answer in terms of an integral that one cannot do analytically.  At high T , show the heat capacity goes to a constant and find that constant.  At low T , show that Cv = KT n Find n. Find K in terms of a definite integral. If you are brave you can try to evaluate the integral, but you will need to leave your result in terms of the Riemann zeta function. In 2d there should be 2N modes. So high T heat capacity should be C = 2kbN (Law of Dulong-Petit). Assume longitudinal and transverse sound velocities are equal. 2N = 2A Z |k|=kDebye 0 d2k (2π)2 = 2(πk2D ebye (2π)2 with A the area. So kDebye = √4πn with n = N/A the density. So Debye = ~kDebyec with c the sound velocity. Since phonons obey bose statistics we have E = 2A Z |k|=kDebye 0 d2k (2π)2 ǫknB(βǫk) = 2A Z |k|=kDebye 0 d2k (2π)2 ~ck 1 eβ~ck − 1 = 2A 2π (2π)2 Z |k|=kDebye 0 k dk ~ck 1 eβ~ck − 1 = A π Z Debye 0 dǫ ~c ǫ ~c ǫ 1 eβǫ − 1 Let z = βǫ = ǫ/(kbT ) and we get E = A(kbT )3 π~2c2 Z Debye/(kbT) 0 z2dz ez − 1 For large T , /T is small so z is small, so z2dz ez − 1 = z so we get Z Debye/(kbT) 0 zdz = (Debye/(kbT ))2/2 so in this limit E = A(kbT )2 Debye 2π~2c2 = Ak2D ebyekbT 2π = A (4πN/A)kbT 2π = 2NkbT 8 Specific Heat of Solids: Boltzmann, Einstein, and Debye which gives C = dE/dT = 2Nkb as expected. For small T , the upper limit of the integral goes to infinity and we have E = A(kbT )3 π~2c2 Z ∞ 0 z2dz ez − 1 So Cv = KT 2 where K = 3Ak3 b π~2c2 Z ∞ 0 z2dz ez − 1 To evaluate the integral we have Z ∞ 0 z2dz ez − 1 = Z ∞ 0 z2dz e −z ∞X n=0 e−nz = ∞X n=1 Z ∞ 0 dzz2e−nz = ∞X n=1 2/n3 = 2ζ(3) Thus we obtain K = 6Ak3 b ζ(3) π~2c2 (2.4) Debye Theory III Physicists should be good at making educated guesses: Guess the element with the highest Debye temperature. The lowest? You might not guess the ones with the absolutely highest or lowest temperatures, but you should be able to get close. Largest Debye temperature should be the one with the highest speed of sound which is probably the hardest element (ie., highest spring constant) and/or smallest mass. Diamond is the obvious guess (and indeed it does have the highest Debye temperature). Debye = 2230K. The lowest is harder to guess. One presumably wants a soft material of some sort – also possibly a heavy material. Material Debye Neon 75 K Argon 92 K Krypton 64 K Xenon 64 K Radon 64 K Mercury 69 K Potassium 91 K Rubidium 56 K Cesium 32 K Some Low Debye Temperatures Soft and heavy metals like mercury are good guesses. (in fact mercury is liquid at room temperature and one has to go to low T to measure a Debye temperature). Also good guesses are Noble gases where the spring constant is very low (weak interaction between the atoms). Also heavy soft group 1 metals are good guesses. Many of these are gas or liquid at room T and a Debye temeperature can only be measured at low T . 9 (2.5) Debye Theory IV From Fig. 2.3 (main text) estimate the Debye temperature of diamond. Why does it not quite match the result listed in Table 2.2 (main text)? Extracting the slope from the figure gives C/T 3 ≈ 1.9×10−7J/(mol − K4) Then using the formula C = 12NkBπ4 5  T TD 3 We obtain TD ≈ 2200K The reason that it does not match the Debye temperature given in the figure caption has to do with the comment in the caption. Debye theory predicts the heat capacity at all possible temperatures. The Debye temperature quoted in the text is chosen so as to give a good fit over the full temperature range. The Debye temperature measured here is chosen to give a good fit at the lowest temperatures (where Debye theory can actually be exact). (2.6) Debye Theory V* In the text we derived the low temperature Debye heat capacity assuming that the longitudinal and transverse sound velocities are the same and also that the sound velocity is independent of the direction the sound wave is propagating. (a) Suppose the transverse velocity is vt and the longitudinal velocity is vl. How does this change the Debye result? State any assumptions you make. (b) Instead suppose the velocity is anisotropic. For example, suppose in the ˆx, ˆy and ˆz direction, the sound velocity is vx, vy and vz respectively. How does this change the Debye result? (a) This is actually quite simple. The derivation of the heat capacity follows the text (or exercise 2.1). The only difference is in the density of states. In the isotropic calculation we use g(ω) = N  12πω2 (2π)3nv3  Recall the origin of these factors. Really we had (See Eq. 2.3 of the main text) g(ω) = 3L3 4πω2 (2π)3v3 where the factor of 3 out front is for the three polarizations of the sound waves. One could just as well have written it as g(ω) = L3 4πω2 (2π)3  1 v3 + 1 v3 + 1 v3  10 Specific Heat of Solids: Boltzmann, Einstein, and Debye separating out the three different polarizations. Now, if the three polarizations have three different velocities, we have g(ω) = L3 4πω2 (2π)3  1 v3 1 + 1 v3 2 + 1 v3 3  this is true since the density of states of the three different excitation modes simply add. In an isotropic solid, the two transverse mode have the same velocity vt and the one longitudinal mode has velocity vl and we would have g(ω) = L3 4πω2 (2π)3  2 v3 t + 1 v3 l  The remainder of the derivation