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Solutions to Problems in Condensed Matter Physics 2(Textbook: Michael P. Marder, Condensed Matter Physics, Wiley, apterChapterChapterpage6 and 7 (“electrons”)11 (“phonons”)12 (“elasticity”)17 (“transport phenomena”)22 and 23 (“optical properties”)24 (“ordering”)25 and 26 (“magnetism”)27 (“superconductivity”)1681016212633(Niels Bohr Institute, February 2011)

1Problems in Chapter 6 and 7 (“electrons”)Solutions to the problems in Chapter 66.3 Pressure of a Fermi gas at zero temperatureThe number of electrons N and the internal energy U , in the volume V , are N V0 D(ε)f (ε)dε,U VεD(ε)f (ε)dε,0(1)The Fermi distribution function f (ε) and the density of states (per unit volume)D(ε) are 12m3(2), D(ε) A ε, A 2 3 .f (ε) β(ε µ)π e 1At zero temperature 1/3 2 kF2, kF 3π 2 n(3)2m Introducing the step function and D(ε) A ε in (1), we may determine U (0) andan alternative expression for A: ε F3n3/25/2A εdε 23 V AεF A 3/2 ,U (0) V 25 AεF 35 N εF (4)N V02εFf (ε) θ(εF ε),εF i.e. U εF kF2 V 2/3 and the pressure at zero temperature is(3π 2 )2/3 n5/3 2 U (0)2U 25 nεF (5) V3V5min sharp contrast to a classical gas, where the zero-temperature pressure is zero.P 6.4 Density of states in low dimensionsThe density of states in k-space is D k 2/(2π)d in d dimensions (6.34) D(ε) [d k]δ(ε ε k ) D k δ(ε ε k )d k 2(2π)d δ(ε ε k )d k(1)Introducing “spherical coordinates” in d dimensions and k k , thend k 2 dk2πk dk 4πk2 dk, d 1, d 2, d 3(2)The factor 2 in the one-dimensional case appears because the one-dimensional wavevector may assume both positive and negative values, whereas k 0 per definition. 2 k2ε k εk 2mD(ε) k 2mεk, dk 1/2 2m 2 δ(ε εk )dεk 2π 02 2 εk 22 (2π) 2(2π)30 02π 1/2dεk (3) 2m 1/2, d 1επ 2mεkmδ(ε εk ) 2 2 εk2mεm4π 2 k δ(ε εk ) 2 2 εkm2 2 εk1/2m,π 2d 22m3 1/2ε ,π 2 3d 3dεk 1/2dεk

Condensed Matter Physics 226.5 Fermi pancakesThin layer of Ag: Lx Ly L 106 Å and Lz d.The density of electrons is the same as the density of atoms (one conductionelectron per Ag atom), i.e. n 5.86 · 1022 cm 3 0.0586 Å 3 , according to theproperties given in the periodic table on the front page of Marder. The wavefunction of the “free” electrons should vanish at the boundaries z 0 and z d.This condition is fulfilled if assuming the one-electron wave function to beψ(x, y, z) ei(kx x ky y) sin (pqz) ,q π,dp 1, 2, . . .(1)The electron states are characterized by the two-dimensional wave vector k (kx , ky , 0) and the “band index” p. The eigenenergies are (k2 kx2 ky2 ):εp k 2 2 2 q 2k p2 q 2 ε1 0 2m2m(2)At T 0 (T TF ), as assumed implicitly in the exercise, the occupied states areall those with energies smaller than εF . The number of electrons in the pth band,Np , is zero if εp 0 εF . In the opposite case:Np L2 kp0222 kp2πkdk L(2π)22π(3)where kp is the Fermi wave number of the pth band, i.e. the largest value of k ofoccupied states in the pth band, as determined byεpk p 2 2 kp p2 q 2 εF2mkp2 2mεF p2 q 2 q 2 2 εF p2ε1 0(4)(a) In the first case d 4.1 Å or q π/d 0.76624 Å 1 , andε1 0 (1.054572 · 10 27 0.76624 · 108 )2 2 q 2 eV 2.2369 eV2m2 9.109389 · 10 28 1.602177 · 10 12(5)Assuming the Fermi energy to be smaller than the lowest energy of the (p 2)band, εF ε2 0 4ε1 0 , then only the (p 1) states are occupied, in which case N1has to be equal the total number N of free electrons:N L2 d n N1 L2k122π k1 2πd n 1.22866 Å 1(6)Introducing this result in (3) we get the Fermi energy and the band width W ofthe occupied states:εF ε1k ε1 0 (k1 /q)2 1 7.99 eV,1W εF ε1 0 5.75 eV(7)which is in accordance with our starting assumption of εF ε2 0 8.948 eV. Theseresults may be compared with that obtained for bulk Ag:εF (bulk) W (bulk) 2 2 2/33π n 5.50 eV2m(8)

3Problems in Chapter 6 and 7 (“electrons”)(b) In the case of d 8.2 Å, q π/d 0.38312 Å 1 and ε1 0 0.5592 eV. In orderto determine εF in this case, we need to include more bands than the lowest one.The number of bands turns out to be 3, and N N1 N2 N3 2πd n q2εF 12 ε1 0 εF 22 ε1 0 εF 32ε1 0 (9)Solving this equation with respect to εF /ε1 0 , we get εF /ε1 0 11.523, which islarger than 32 but smaller than 42 , in accordance with the assumption that allelectrons are found in the three lowest bands. Hence the results are:W εF ε1 0 5.89 eVεF 6.44 eV,20(10)p2Band energy (eV)16121εF84k100. number k [r.l.u.]1.0520p4Band energy (eV)16312218εF4k300.00.2k2k10.40.60.8Wave number k [r.l.u]1.0Fermi energy and band width (eV)12εF1086W4Bulk value200481216Number of Ag layers (2d/a)20The two first figures show the energy bands as functions of k inthe cases (a) and (b), where theunit of k is the length of a reciprocal lattice vector 2π/a [r.l.u.].The last figure shows the Fermienergy (and the band width) as afunction of the number of atomiclayers of Ag. The crystal structure of Ag is fcc with the latticeparameter a 4.09 Å. This isvery nearly the thickness d assumed in (a), i.e. this case corresponds to a film with two atomiclayers of Ag atoms.

Condensed Matter Physics 24Solutions to the problems in Chapter 77.1 Normals to surfaces r (x1 , x2 , x3 ) s(t) is the parametrization of a curve lying within the surfacedefined by f ( r ) ε. Since f ( s(t)) is a constant ε, the derivative of this functionis 0: f dsαd sdf ( s(t)) f · 0(1)dt xdtdtααBecause s(t) may be any arbitrary curve lying within the surface, the same is truefor the curve tangent d s(t)/dt, and (1) is only generally valid if f is normal tothe surface.7.3 Van Hove singularities(a) The problem becomes the same as the one considered in problem 7.1 if makingthe replacements k r and ε f ( r), hence ε is perpendicular to thek nknkenergy surface defined by εn k ε.(b) The energy is assumed to be εn k εmax k2 , and in the two-dimensional case D(ε) [d k]δ(ε εn k ) 1 2π 02(2π)2 02πk δ(ε εmax k2 ) dk(1)1θ(εδ(ε εmax k )d(k ) ε)2π max22The density of states is zero if ε εmax and 1/(2π) when ε εmax .(c) In the three dimensional case, the result is [d k]δ(ε εn k ) D(ε) 12π 2 02(2π)3 04πk2 δ(ε εmax k2 ) dkk δ(ε εmax k2 )d(k2 ) 1 εmax ε θ(εmax ε)2π 2(2)Solution to HS’s problem 1Heat capacity of a two-dimensional electron gasGaAs/AlGaAs heterostructure (see Marder Section 19.5):n 1011 cm 2 ,ε 2 k2,2m m 0.067me(1)The present situation corresponds to the case (a) of the previous problem 6.5 inMarder, i.e. the gas is purely two-dimensional in the sense that only the (p 1)bandneeds to be considered, and k is a two-dimensional vector with the length k 2kx ky2 . The most important quantity is the Fermi energy, which is determinedby evaluating N at zero temperature:n N A k kFD k d k 2(2π)2 kF02πk dk kF22π kF 2πn(2a)

5Problems in Chapter 6 and 7 (“electrons”)The same result is obtained by using that, according to problem 6.4 or equation(6.35), D(ε) m /π 2 in the two-dimensional case:n εF0D(ε) dε m εFπ 2 εF nπ 2m 1Introducing the numbers, and using that kF 1 ÅεF 3.81 eV when the mass is me , then we getkF 0.793 · 10 2 1ÅkF 2πn(2b)corresponds to a Fermi energy ,or 20.793 · 10 2εF 3.81 eV 3.58 meV0.067(3)This Fermi energy corresponds to a Fermi temperature TF εF /kB 41.5 K.1) T 1 K is much smaller than the Fermi temperature and the heat capacitymay be determined by the leading order expression (6.77)cV π2π T 22D(εF )kBT k k ,36 TF F BasD(εF ) m kF2kF2 (4)π 22πεF2πkB TFIn the case of a sample with the area A 1 cm2 the result isπ 1(0.793·106 cm 1 )2 1.38066·10 23 J/K 1.1 · 10 13 J/K6 41.5(5) 22The explicit result is CV (el) Aπm kB T /(3 ). Hence, the small value of CV isnot due to the low electron density n but to the two-dimensionality of the systemand the small effective mass.CV (el) A cV A2) In order to estimate the phonon contribution to the heat capacity we shall usethe Debye model (Section 13.3.2 in Marder). The Debye temperature of GaAs isΘD 344 K (which value is not changed much when some of the Ga ions arereplaced by Al ions). Using (13.70), (13.75) in Marder (2.Ed.) and 04π 4x4 exdx (ex 1)215 12π 4cV n k5 A B TΘD3(T ΘD )(6)According to Table 2.5 (page 27) in Marder, GaAs has the zincblende structure ( diamond structure) with the lattice parameter a 5.63 Å. In this structure thereare 8 atoms per unit cell, and the density of atoms is nA 8/a3 4.48 · 1022 cm 3 .[The mean atomic mass (periodic table) is (69.72 74.92)/2 72.3 u, implying amass density ρ 5.38 g cm 3 ]. Assuming V 1 cm3 and T 1 K, the phononcontribution becomes CV (ph) 12π 41 4.48 · 1022 1.38066 · 10 235344 3J/K 3.55 · 10 6 J/K (7)which is much larger than the electronic contribution. Utilizing that CV (el) Tand CV (ph) T 3 , the temperature T0 at which the two contributions are equal,is determined by (T0 in K)1.10 · 10 13 T0 3.55 · 10 6 T03 T0 1.8 · 10 4 K(8)

Condensed Matter Physics 26A more fair comparison would be to consider a film of thickness 100 µm, i.e.A 1 cm2 and V 10 2 cm3 , in which case T0 1.8 mK, a temperature withinan accessible range (however, the reduction of the size of the sample makes it moredifficult to determine the heat capacity).Solution to the problem in Chapter 1111.2 Zinc in copperCopper is a monovalent fcc and zinc a divalent hcp (not fcc) metal. At a smallconcentration of Zn , the crystal structure is fcc until the Fermi surface touchessome point on the edge of the Brillouin zone:(a) In the nearly free electron approximation, the Fermi wave vector is (nearly)determined as in the free electron case, n k kFD k d k 2(2π)3 kF04πk2 dk 1 3k3π 2 F kF 3π 2 n 1/3(1)Defining c to be the concentration of Zn ions and a to be the lattice parameterof the fcc lattice, then the electron density is44n 3 [(1 c) 2c] 3 (1 c)aa 1/312π 2kF (1 c)a34.9109(1 c)1/3a(2)(b) The primitive unit vectors of the face-centered cubic lattice are, (2.2): a1 a(1, 1, 0),2 a2 a(1, 0, 1),2 a3 a(0, 1, 1)2(3)and the corresponding primitive vectors of the reciprocal lattice are b 2π a2 a3 2π (1, 1, 1),1 a1 · a2 a3a b 2π (1, 1, 1),2a b 2π ( 1, 1, 1)3a(4)The boundaries of the 1. Brillouin zone are established by the planes perpendicularto bi at the distance k1 bi /2 from the origin. Hence the shortest distance fromthe origin to the zone boundaries is π 35.4414 bi k1 2aa(5)kF (c 0) is smaller than k1 , and kF (c) becomes equal to k1 at the zinc concentration c c14.9109(1 c1 )1/3 5.4414 c1 0.36(6)The experimental phase diagram (see Fig. 4 in Chapter 21 of Kittel) shows atransition from fcc to bcc, for increasing c values, at about the Zn concentrationdetermined here.

7Problems in Chapter 11 (“phonons”)(c) The primitive unit vectors of the body-centered cubic lattice are, (2.3): a1 a(1, 1, 1),2 a2 a(1, 1, 1),2 a3 a( 1, 1, 1)2(7)and the corresponding primitive vectors of the reciprocal lattice are b 2π (1, 1, 0),1a b 2π (1, 0, 1),2a b 2π (0, 1, 1)3a(8)Notice, that the reciprocal of the bcc lattice is fcc, and equivalently, that thereciprocal lattice of fcc is the bcc lattice.In the case of bcc there are 2 atoms per cubic unit cell, hence2n 3 (1 c),a 1/36π 2kF (1 c)a3 3.8979(1 c)1/3a(9)The shortest distance from the origin to the boundaries of the 1. Brillouin zone is π 24.4429 bi k2 2aaand k2 kFwhenc c2 0.48(10)Once again, this is about the right Zn concentration at which the Cu-Zn alloysystem shows a change of crystal structure (from bcc to a complex γ structure).If the atomic density is assumed to stay constant at the transition fcc bcc, theFermi wave vector is the same on each side of the phase line, but the lattice parameter a is changed, abcc (1/2)1/3 afcc 0.794 afcc . The cubic lattice parametera also changes (increases) gradually with the Zn concentration, in between thephase lines, since nCu 1.33 nZn , however, all the possible variations of a haveno influence on the arguments above. β-brass is the alloy with c 0.5, and itshows an order-disordered phase transition at about 470 C, which we are goingto discuss later on.Citation from Kittel: “Why is there a connection between the electron concentrations at which a new phase appears and at which the Fermi surface makes contactwith the Brillouin zone? We recall that the energy bands split into two at theregion of contact on the zone boundary [Marder, Chapter 8, (8.24)]. If we addmore electrons to the alloy at this stage, they will have to be accommodated inthe upper band or in states of high energy near the zone corners of the lower band.Both options are possible, and both involve an increase of energy. Therefore itmay be energetically favorable for the crystal structure to change to one whichcan contain a Fermi surface of larger volume (more electrons) before contact ismade with the zone boundary. In this way H. Jones made plausible the sequenceof structures fcc, bcc, γ, hcp with increasing electron concentration.”

Condensed Matter Physics 28Solutions to the problems in Chapter 1212.4 Elastic constants(a) The application of a uniform external gas pressure (no shear stress) implies auniform dilations of an isotropic solid, where only exx eyy ezz are non-zero, oreαβ δαβ VV V0 3V03V0(1)where V0 is the zero-pressure equilibrium volume. Introducing this in the freeenergy expression for an isotropic solid (12.23)F 12 λeαα 2 2µα e2αβd r αβ12 (V0 V V ) λV0 2 3 V 2µ9 V0 2 (2)the bulk modulus is found to beB V 2F λ 23 µ V 2(3)(when V V0 ). This result is also obtained from the bulk modulus in the cubiccase (12.19), B [c11 2c12 ]/3, by replacing the elastic constants with the Laméconstants of the isotropic solid, c12 λ, c44 µ, and c11 c12 2c44 λ 2µ.(b) According to (12.31), or Fig. 12.2, Young’s modulus Y is defined to be Y σzz /ezz , when applying a uniform stress σzz in the z direction. In the cubic case,the diagonal elements of stress and strain tensors are related by, (12.14)-(12.17), σxxexx σ c e ,yyyyσzzezz c11c c12c12c12c11c12 c12c12 c11(4)[the “off-diagonal elements relations” are σα c44 eα , where α 4, 5, 6. In thecase of α 4 this equation reads σyz 2c44 eyz ]. In order to determine ezz whenσzz is non-zero, we need to determine the inverse matrix. The determinant of c isD c311 2c312 3c11 c212 (c11 2c12 )(c11 c12 )2 , and the inverse matrix is foundto be c 1A BBBAB BB ,AA c11 c12,(c11 2c12 )(c11 c12 )B c12Ac11 c12(5)showing that ezz Aσzz , henceY A 1 (c11 2c12 )(c11 c12 )c11 c12(6)[In order to complete the cubic case, then Poisson’s ratio (12.34) is B/A orc12, and the shear modulus is G c44 ].ν c11 c12

9Problems in Chapter 12 (“elasticity”)12.5 Waves in cubic crystals(a) Combining (12.14) and (12.16) we may write the free energy of a cubic solid asF 12 d r c11 e2xx e2yy e2zz 2 c12 exx eyy exx ezz eyy ezz 2 c44e2xy e2yx e2xz e2zx e2yze2zy 12 d r (1)σαβ eαβαβwhich shows that the stress-strain relations areσαα c11 eαα c12 β αeββ ,σαβ 2 c44 eαβ (α β),eαβ1 2 uβ uα rβ rα(2)Introducing these relations in the equations of motion, (12.25), we getρüα σαβ rββ 2 uβ 2 uα uα uβ c11 c12 c44 rα2 rα rβ rβ rβ rαβ α(3)(b) Introducing a plane-wave solution u( r , t) u0 ei(k· r ωt)(4)into these equations (3), we get ρω 2 u0α c11 kα2 u0α c12 kα kβ u0β c44β α β αkβ2 u0α kα kβ u0β (5)leading to the following matrix equation for determining u0 and ω c11 kα2 c44 β αkβ2 ρω2 u0α (c12 c44 ) β αkα kβ u0β 0(6)(c) In the case of waves propagation along [100], the k-vector is k (k, 0, 0) andthe matrix equation is diagonal: c11 k2 ρω 2 u01 0, c44 k2 ρω 2 u02 0, c44 k2 ρω 2 u03 0(7)The sound velocities are ωc11 ,k ρωc44,ct kρcl u0 k u0 k(longitudinal)(8)(transverse)i.e. one longitudinal and two degenerate transverse sound waves. Table 12.1 showsthat c11 165 GPa, c44 79.24 GPa, and c12 64 GPa in the case of silicon.From the periodic table: the atomic density is n 4.99 · 1022 cm 3 and the atomicmass is 28.09 u implying ρ 28.09 1.66054 · 10 27 [kg] 4.99 · 1028 [m 3 ] 2328kg/m3 . Introducing these numbers in (8), we getcl 8420 m/s ,ct 5830 m/s(9)

Condensed Matter Physics 210 (d) Waves propagating along [111], in which case we assume k (k/ 3)(1, 1, 1),and the eigenvalue equation for determining λ ρω 2 is found to be A λBBu010 BA λB u02 0 u03BBA λ0whereλ ρω 2 ,A 13 (c11 2c44 )k2 ,The characteristic determinant is B 13 (c12 c44 )k2(10a)(10b) (A λ) (A λ)2 B 2 2B (A λ) B 2 (A 2B λ)(A B λ)2 0 (11)Introducing the solution λ A 2B in (10a), we get u01 u02 u03 or u0 k,whereas the two other degenerate solutions λ A B imply u01 u02 u03 0 or u0 · k 0. Hence the sound velocities areωc11 2c12 4c44 9340 m/s ,cl k3ρωc11 c12 c44 5080 m/s ,ct k3ρ u0 k(longitudinal)(12) u0 k(transverse)Solutions to the problems in Chapter 1717.4 AC conductivityIn the presence of a time-dependent (uniform) electrical field E e iωtE0(1)we may use the general solution of the Boltzmann equation given by (17.24), beforethe integration with respect to t is performed:g f t 0 e f (t )dt e(t t)/τε e iωt v k · E µ (2)Considering only effects which are linear in the applied field (the linearized solutionof the Boltzmann equation), then v k v k (t ) and f (t ) within the integral may bereplaced by their time-independent equilibrium values at zero field, and the timeintegration may be performed straightforwardlyg f τετε e f e iωt f e f v k · E v k · E01 iωτε µ1 iωτε µ(3)Introducing this expression into (17.43)-(17.44) we get the frequency-dependent 0)conductivity (valid in the limit of E0σαβ (ω) e2 [d k]τε fvα vβ1 iωτε µ(4)or in the case of a cubic or an isotropic (free electron) system:σ(ω) ne2ne2 τ 1 iωττ m 1 iωτm 1 (ωτ )2(5)

11Problems in Chapter 17 (“transport phenomena”)17.5 Current driven by thermal gradientWe shall consider a metal subject to a constant temperature gradient T T, 0, 0 x ε k εk 12 m v k2and(1)According to (17.60), (17.62), and (17.68) the electrical current, in the case ofG 0, is j L12 TT ,11 π2L12 L(1) (k T )2 σ (εF )ee 3 B(2)The assumption of an isotropic mass, (1), implies σ(ε) to be diagonal, and according to (17.64) the diagonal element is2σαα (ε) e τ 2d kD k v kαδ(ε ε k ) e2 τ dεk D(εk ) 13 v k2 δ(ε εk )(3)2 over all solid angles, at a constant k , is 1/3 of the resultThe integration of v kα222 deriving from Tr v v and using v 2εk /m , we getkαkσαα (ε) e2 τ D(ε)k2ε3m σαα(ε) 2e2 τ D(ε) εD(ε)3m (4)This result is introduced in (2) 22 j 1 π (kB T )2 2e τ D(εF ) 1 εF D (εF )e 33m D(εF ) TT(5)2 T D(ε ), (6.77), we finally getIn terms of the heat capacity cV (π 2 /3)kBF2eτ cV Tjx 3m x 1 εFD (εF )D(εF )where the last equality sign is valid only if D(ε) eτ cV Tm x(6)ε.17.8 Hall effect – elementary argument is along the x axis leading to a current j inThe Hall effect geometry: The applied field Ethis direction, i.e. an electron (hole) current in the minus (plus) x direction. The magnetic is in the minus y direction, when B is along z, leadingpart of the Lorenz force F j Bto opposite signs of the resulting charge distributions in the electron and hole cases.

Condensed Matter Physics 212(a) The equation of motion, when assuming the Drude model, is m v v Bm v e Ecτ(1) (0, 0, B), then we get v B in the case of electrons with charge e. Using B(vy B, vx B, 0) and since the current, by geometry, is constrained to be along x,the steady state is characteri

Solutions to Problems in Condensed Matter Physics 2 (Textbook: MichaelP.Marder, Condensed Matter Physics,Wiley,2000) Contents page Chapter6and7(“electrons”) 1 Chapter11(“phonons”) 6