Solutions to Problems in Condensed Matter Physics 2

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1 Solutions to Problems in Condensed Matter Physics Textbook: Michael P. Marder, Condensed Matter Physics, Wiley, Contents page Chapter 6 and 7 electrons 1 Chapter 11 phonons 6 Chapter 1 elasticity 8 Chapter 17 transport phenomena 1 Chapter and 3 optical properties 16 Chapter 4 ordering 1 Chapter 5 and 6 magnetism 6 Chapter 7 superconductivity 33 Niels Bohr Institute, February 11

2 1 Problems in Chapter 6 and 7 electrons Solutions to the problems in Chapter Pressure of a Fermi gas at zero temperature The number of electrons N and the internal energy U, inthevolumev,are N = V Dεfεdε, U = V εdεfεdε, 1 The Fermi distribution function fε and the density of states per unit volume Dε are 1 m fε = e βε µ +1, Dε =A 3 ε, A = π 3. At zero temperature fε =θε F ε, ε F = kf 1/3 m, k F = 3π n 3 Introducing the step function and Dε =A ε in 1, we may determine U and an alternative expression for A: N = V εf A εdε = 3 VAε3/ F A = 3n ε 3/ F, U = V 5 Aε5/ F = 3 5 Nε F 4 i.e. U ε F kf V /3 and the pressure at zero temperature is P = U = U V 3 V = 5 nε F = 3π /3 n 5/3 5 5m in sharp contrast to a classical gas, where the zero-temperature pressure is zero. 6.4 Density of states in low dimensions The density of states in k-space is D k =/π d in d dimensions 6.34 Dε = [d k]δε ε k D k δε ε k d k = π d δε ε k d k 1 Introducing spherical coordinates in d dimensions and k = k, then d dk, d =1 k = πk dk, d = 4πk dk, d =3 The factor in the one-dimensional case appears because the one-dimensional wave vector may assume both positive and negative values, whereas k per definition. ε k = ε k = k 1/ m k = mεk m, dk = dε ε k 3 k 1/ dε k = Dε = m δε ε π k ε k π π 3 mεk π δε ε k 4π mε k δε ε k m 1/ dε k = ε k 1/ m dε k = ε k m π ε 1/, d =1 m π, d = m 3 π 3 ε1/, d =3

3 6.5 Fermi pancakes Condensed Matter Physics Thin layer of Ag: L x = L y = L =1 6 ÅandL z = d. The density of electrons is the same as the density of atoms one conduction electron per Ag atom, i.e. n = cm 3 =.586 Å 3, according to the properties given in the periodic table on the front page of Marder. The wave function of the free electrons should vanish at the boundaries z =andz = d. This condition is fulfilled if assuming the one-electron wave function to be ψx, y, z e ik x x+k y y sin pqz, q = π, p =1,,... 1 d The electron states are characterized by the two-dimensional wave vector k = k x,k y, and the band index p. The eigenenergies are k = k x + k y : ε p k = k + p q ε m 1 = q m At T =T T F, as assumed implicitly in the exercise, the occupied states are all those with energies smaller than ε F. The number of electrons in the pth band, N p, is zero if ε p >ε F. In the opposite case: N p = L kp π πk dk = L k p π 3 where k p is the Fermi wave number of the pth band, i.e. the largest value of k of occupied states in the pth band, as determined by ε pkp = kp + p q = ε m F kp = mε F p q = q εf p 4 ε 1 a In the first case d =4.1 Åorq = π/d =.7664 Å 1,and ε 1 = q m = ev =.369 ev Assuming the Fermi energy to be smaller than the lowest energy of the p = band, ε F <ε =4ε 1, then only the p = 1 states are occupied, in which case N 1 has to be equal the total number N of free electrons: N = L dn = N 1 = L k 1 π k 1 = πd n =1.866 Å 1 6 Introducing this result in 3 we get the Fermi energy and the band width W of the occupied states: ε F = ε 1 k1 = ε 1 [ ] k 1 /q +1 =7.99 ev, W = ε F ε 1 =5.75 ev 7 which is in accordance with our starting assumption of ε F <ε =8.948 ev. These results may be compared with that obtained for bulk Ag: ε F bulk = W bulk = 3π /3 n =5.5 ev 8 m

4 3 Problems in Chapter 6 and 7 electrons b In the case of d =8. Å, q = π/d =.3831 Å 1 and ε 1 =.559 ev. In order to determine ε F in this case, we need to include more bands than the lowest one. The number of bands turns out to be 3, and [ ] N = N 1 +N +N 3 πd n = q εf 1 εf + εf ε 1 ε 1 ε 1 Solving this equation with respect to ε F /ε 1,wegetε F /ε =11.53, which is 1 larger than 3 but smaller than 4, in accordance with the assumption that all electrons are found in the three lowest bands. Hence the results are: ε F =6.44 ev, W = ε F ε 1 =5.89 ev 1 16 p Band energy ev Band energy ev ε F Wave number k [r.l.u.] ε F k p The two first figures show the energy bands as functions of k in the cases a and b, where the unit of k is the length of a reciprocal lattice vector π/a [r.l.u.]. The last figure shows the Fermi energy and the band width as a function of the number of atomic layers of Ag. The crystal structure of Ag is fcc with the lattice parameter a = 4.9 Å. This is very nearly the thickness d assumed in a, i.e. this case corresponds to a film with two atomic layers of Ag atoms. Fermi energy and band width ev k Wave number k [r.l.u] ε F W Number of Ag layers d/a k k 1 Bulk value

5 Solutions to the problems in Chapter Normals to surfaces Condensed Matter Physics 4 r =x 1,x,x 3 = st is the parametrization of a curve lying within the surface defined by f r =ε. Since f st is a constant ε, the derivative of this function is : d dt f st = α f ds α x α dt = f d s dt = 1 Because st may be any arbitrary curve lying within the surface, the same is true for the curve tangent d st/dt, and 1 is only generally valid if f is normal to the surface. 7.3 Van Hove singularities a The problem becomes the same as the one considered in problem 7.1 if making the replacements k r and ε n f r, hence k k ε n is perpendicular to the k energy surface defined by ε n k = ε. b The energy is assumed to be ε n k = ε max k, and in the two-dimensional case Dε = [d k]δε ε n k = π πk δε ε max + k dk = 1 δε ε π max + k dk = 1 1 π θε max ε The density of states is zero if ε>ε max and 1/π whenε<ε max. c In the three dimensional case, the result is Dε = [d k]δε ε n k = π 3 4πk δε ε max + k dk = 1 π kδε ε max + k dk = 1 εmax π εθε max ε Solution to HS s problem 1 Heat capacity of a two-dimensional electron gas GaAs/AlGaAs heterostructure see Marder Section 19.5: n =1 11 cm, ε = k m, m =.67m e 1 The present situation corresponds to the case a of the previous problem 6.5 in Marder, i.e. the gas is purely two-dimensional in the sense that only the p =1 band needs to be considered, and k is a two-dimensional vector with the length k = kx + ky. The most important quantity is the Fermi energy, which is determined by evaluating N at zero temperature: n = N A = D k d k = kf k k F π πk dk = k F π k F = πn a

6 5 Problems in Chapter 6 and 7 electrons The same result is obtained by using that, according to problem 6.4 or equation 6.35, Dε =m /π in the two-dimensional case: n = εf Dε dε = m ε F π ε F = nπ m or k F = πn b Introducing the numbers, and using that k F =1Å 1 corresponds to a Fermi energy ε F =3.81 ev when the mass is m e,thenweget k F = Å 1, ε F = ev = 3.58 mev 3.67 This Fermi energy corresponds to a Fermi temperature T F = ε F /k B =41.5 K. 1 T = 1 K is much smaller than the Fermi temperature and the heat capacity may be determined by the leading order expression 6.77 c V = π 3 Dε Fk BT = π 6 T T F k Fk B, as Dε F = m π = In the case of a sample with the area A =1cm the result is k F πε F = k F πk B T F 4 C V el = Ac V = A π cm J/K = J/K 5 The explicit result is C V el = Aπm kb T/3. Hence, the small value of C V is not due to the low electron density n but to the two-dimensionality of the system and the small effective mass. In order to estimate the phonon contribution to the heat capacity we shall use the Debye model Section in Marder. The Debye temperature of GaAs is Θ D = 344 K which value is not changed much when some of the Ga ions are replaced by Al ions. Using 13.7, in Marder.Ed. and x 4 e x 4π4 e x dx = c V = 1π4 T 5 n Ak B T Θ Θ D 6 D According to Table.5 page 7 in Marder, GaAs has the zincblende structure diamond structure with the lattice parameter a = 5.63 Å. In this structure there are 8 atoms per unit cell, and the density of atoms is n A =8/a 3 = cm 3. [The mean atomic mass periodic table is / = 7.3 u, implying a mass density ρ =5.38 g cm 3 ]. Assuming V =1cm 3 and T = 1 K, the phonon contribution becomes C V ph = 1π J/K = J/K which is much larger than the electronic contribution. Utilizing that C V el T and C V ph T 3, the temperature T at which the two contributions are equal, is determined by T in K T = T 3 T = K 8

7 Condensed Matter Physics 6 A more fair comparison would be to consider a film of thickness 1 µm, i.e. A =1cm and V =1 cm 3,inwhichcaseT =1.8 mk,atemperaturewithin an accessible range however, the reduction of the size of the sample makes it more difficult to determine the heat capacity. Solution to the problem in Chapter Zinc in copper Copper is a monovalent fcc and zinc a divalent hcp not fcc metal. At a small concentration of Zn ++, the crystal structure is fcc until the Fermi surface touches some 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 = D k d k = k k F π 3 kf 4πk dk = 1 1/3 3π k3 F k F = 3π n 1 Defining c to be the concentration of Zn ++ ions and a to be the lattice parameter of the fcc lattice, then the electron density is n = 4 a 3 [1 c+c] = 4 a c [ ] 1π 1/3 k F = 1 + c = c 1/3 a3 a b The primitive unit vectors of the face-centered cubic lattice are,.: a 1 = a 1, 1,, a = a 1,, 1, a 3 = a, 1, 1 3 and the corresponding primitive vectors of the reciprocal lattice are b1 =π a a 3 a 1 a a 3 = π a 1, 1, 1, b = π a 1, 1, 1, b3 = π a 1, 1, 1 4 The boundaries of the 1. Brillouin zone are established by the planes perpendicular to ± b i at the distance k 1 = b i / from the origin. Hence the shortest distance from the origin to the zone boundaries is k 1 = b i = π 3 a = a 5 k F c = is smaller than k 1,andk F c becomes equal to k 1 at the zinc concentration c = c c 1 1/3 = c 1 =.36 6 The experimental phase diagram see Fig. 4 in Chapter 1 of Kittel shows a transition from fcc to bcc, for increasing c values, at about the Zn concentration determined here.

8 7 Problems in Chapter 11 phonons c The primitive unit vectors of the body-centered cubic lattice are,.3: a 1 = a 1, 1, 1, a = a 1, 1, 1, a 3 = a 1, 1, 1 7 and the corresponding primitive vectors of the reciprocal lattice are b1 = π a 1, 1,, b = π a 1,, 1, b3 = π a, 1, 1 8 Notice, that the reciprocal of the bcc lattice is fcc, and equivalently, that the reciprocal lattice of fcc is the bcc lattice. In the case of bcc there are atoms per cubic unit cell, hence n = [ ] 6π 1/3 a c, k F = 1 + c = c 1/3 9 a3 a The shortest distance from the origin to the boundaries of the 1. Brillouin zone is k = b i = π a = a and k = k F when c = c =.48 1 Once again, this is about the right Zn concentration at which the Cu-Zn alloy system 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, the Fermiwavevectoristhesameoneachsideofthephaseline,butthelatticeparameter a is changed, a bcc =1/ 1/3 a fcc =.794 a fcc. The cubic lattice parameter a also changes increases gradually with the Zn concentration, in between the phase lines, since n Cu =1.33 n Zn, however, all the possible variations of a have no influence on the arguments above. β-brass is the alloy with c =.5, and it shows an order-disordered phase transition at about 47 C, which we are going to 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 contact with the Brillouin zone? We recall that the energy bands split into two at the region of contact on the zone boundary [Marder, Chapter 8, 8.4]. If we add more electrons to the alloy at this stage, they will have to be accommodated in the 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 it may be energetically favorable for the crystal structure to change to one which can contain a Fermi surface of larger volume more electrons before contact is made with the zone boundary. In this way H. Jones made plausible the sequence of structures fcc, bcc, γ, hcp with increasing electron concentration.

9 Solutions to the problems in Chapter Elastic constants Condensed Matter Physics 8 a The application of a uniform external gas pressure no shear stress implies a uniform dilations of an isotropic solid, where only e xx = e yy = e zz are non-zero, or e αβ = δ αβ V V 3V = V 3V 1 where V is the zero-pressure equilibrium volume. energy expression for an isotropic solid 1.3 Introducing this in the free [ F = 1 ] V λ e αα +µ e αβ d r = 1 V [λ + V +µ 3 ] V α V αβ 9 V the bulk modulus is found to be B = V F V = λ + 3 µ 3 when V V. This result is also obtained from the bulk modulus in the cubic case 1.19, B =[c 11 +c 1 ]/3, by replacing the elastic constants with the Lamé constants of the isotropic solid, c 1 = λ, c 44 = µ, andc 11 = c 1 +c 44 = λ +µ. b According to 1.31, or Fig. 1., Young s modulus Y is defined to be Y = σ zz /e zz, when applying a uniform stress σ zz in the z direction. In the cubic case, the diagonal elements of stress and strain tensors are related by, , σ xx σ yy σ zz = c e xx e yy e zz c 11 c 1 c 1, c = c 1 c 11 c 1 4 c 1 c 1 c 11 [the off-diagonal elements relations are σ α = c 44 e α,whereα =4, 5, 6. In the case of α = 4 this equation reads σ yz =c 44 e yz ]. In order to determine e zz when σ zz is non-zero, we need to determine the inverse matrix. The determinant of c is D = c c3 1 3c 11 c 1 =c 11 +c 1 c 11 c 1, and the inverse matrix is found to be A B B c 1 = B A B, A = B B A showing that e zz = Aσ zz, hence c 11 + c 1 c 11 +c 1 c 11 c 1, B = c 1 c 11 + c 1 A 5 Y = A 1 = c 11 +c 1 c 11 c 1 c 11 + c 1 6 [In order to complete the cubic case, then Poisson s ratio 1.34 is B/A or c ν = 1, and the shear modulus is G = c c 11 + c 44 ]. 1

10 9 Problems in Chapter 1 elasticity 1.5 Waves in cubic crystals a Combining 1.14 and 1.16 we may write the free energy of a cubic solid as F = 1 d r [c 11 e xx + e yy + e zz +c 1 e xx e yy + e xx e zz + e yy e zz + c 44 e xy + e yx + e xz + e zx + e yz + ezy ] = 1 d r 1 σ αβ e αβ αβ which shows that the stress-strain relations are σ αα = c 11 e αα + c 1 e ββ, σ αβ =c 44 e αβ α β, e αβ 1 β α Introducing these relations in the equations of motion, 1.5, we get ρü α = β σ αβ r β = c 11 u α r α + β α [ c 1 uα + u β r β r α u β uα + c r α r 44 + u ] β 3 β r β r β r α b Introducing a plane-wave solution u r, t = u e i k r ωt into these equations 3, we get ρω u α = c 11 kαu α c 1 k α k β u β c 44 β α β α kβu α + k α k β u β 4 5 leading to the following matrix equation for determining u and ω c 11 kα + c 44 kβ ρω u α +c 1 + c 44 k α k β u β = 6 β α β α c In the case of waves propagation along [1], the k-vector is k =k,, and the matrix equation is diagonal: c 11 k ρω u 1 =, The sound velocities are c 44 k ρω u =, c 44 k ρω u 3 = 7 c l = ω k = c11 ρ, u k longitudinal c t = ω k = c44 ρ, u k transverse 8 i.e. one longitudinal and two degenerate transverse sound waves. Table 1.1 shows that c 11 = 165 GPa, c 44 =79.4 GPa, and c 1 = 64 GPa in the case of silicon. From the periodic table: the atomic density is n = cm 3 and the atomic mass is 8.9 u implying ρ = [kg] [m 3 ] = 38 kg/m 3. Introducing these numbers in 8, we get c l = 84 m/s, c t = 583 m/s 9

11 Condensed Matter Physics 1 d Waves propagating along [111], in which case we assume k =k/ 31, 1, 1, and the eigenvalue equation for determining λ = ρω is found to be A λ B B B A λ B u 1 u = 1a B B A λ where λ = ρω, A = 1 3 c 11 +c 44 k, B = 1 3 c 1 + c 44 k 1b The characteristic determinant is A λ A λ B B A λ B =A+B λa B λ = 11 Introducing the solution λ = A +B in 1a, we get u 1 = u = u 3 or u k, whereas the two other degenerate solutions λ = A B imply u 1 + u + u 3 =or u k =. Hence the sound velocities are u 3 c l = ω k = c11 +c 1 +4c 44 3ρ c t = ω k = c11 c 1 + c 44 3ρ = 934 m/s, u k longitudinal = 58 m/s, u k transverse 1 Solutions to the problems in Chapter AC conductivity In the presence of a time-dependent uniform electrical field E = E e iωt 1 we may use the general solution of the Boltzmann equation given by 17.4, before the integration with respect to t is performed: t g = f dt e t t/τ ε e iωt v k E e ft µ Considering only effects which are linear in the applied field the linearized solution of the Boltzmann equation, then v k = v k t andft within the integral may be replaced by their time-independent equilibrium values at zero field, and the time integration may be performed straightforwardly g = f τ ε v 1 iωτ k E e f τ ε µ e iωt = f ε f v 1 iωτ k Ee ε µ Introducing this expression into we get the frequency-dependent conductivity valid in the limit of E σ αβ ω =e [d τ k] ε f v 1 iωτ α v β 4 ε µ or in the case of a cubic or an isotropic free electron system: σω = ne m τ 1 iωτ = ne τ m 3 1+iωτ 1+ωτ 5

12 11 Problems in Chapter 17 transport phenomena 17.5 Current driven by thermal gradient We shall consider a metal subject to a constant temperature gradient T = T x,, and ε k = ε k = 1 m v k 1 According to 17.6, 17.6, and the electrical current, in the case of G =,is j = L 1 T, L 1 = 1 T e L1 = 1 π e 3 k B T σ ε F The assumption of an isotropic mass, 1, implies σε to be diagonal, and according to the diagonal element is σ αα ε =e τ d kd k v kα δε ε k =e τ dε k Dε k 1 3 v k δε ε k 3 The integration of v kα over all solid angles, at a constant k, is 1/3 of the result deriving from Tr v kα = v k and using v k =ε k /m,weget σ αα ε =e τdε ε 3m σ ααε = e τ 3m [ Dε+εD ε ] 4 This result is introduced in j = 1 π e 3 k BT e τ 3m Dε D ε F 1+ε F T F Dε F T 5 In terms of the heat capacity c V =π /3kB TDε F, 6.77, we finally get j x = eτc V 3m T D ε 1+ε F x F Dε F = eτc V m T x 6 where the last equality sign is valid only if Dε ε. 17.8Halleffect elementaryargument The Hall effect geometry: The applied field E is along the x axis leading to a current j in this direction, i.e. an electron hole current in the minus plus x direction. The magnetic part of the Lorenz force F j B is in the minus y direction, when B is along z, leading to opposite signs of the resulting charge distributions in the electron and hole cases.

13 Condensed Matter Physics 1 a The equation of motion, when assuming the Drude model, is m v = e E + v c B m v τ 1 in the case of electrons with charge e. UsingB =,,B, then we get v B = v y B, v x B, and since the current, by geometry, is constrained to be along x, the steady state is characterized by v x being constant and v y =. These conditions imply m v y = e E y v x c B = E y = v x c B b The current is j =j x,, with j x = nev x,andthehallcoefficientis R = E y Bj x = v x c B 1 B nev x = 1 nec, E y = RBj x 3 The electric field in the x direction may be determined from v x =,or m v x = ee x mv x τ = E x = me τ v x = me j x τ ne = mj x ne τ = j x σ 4 i.e. E x is determined by the Drude resistivity σ 1 as in the case of B = Hall effect Boltzmann equation The Boltzmann equation in the relaxation-time approximation is given by and dg dt = g t + r g r + g k k = g f 1 τ When the state is uniform in space and steady in time, this equation reduces to k g k = g f τ The semiclassical equation of motion is k = e 1 E + c v k B, v k = r = 1 ε k k = k m 3 where the second equation expresses that the mass tensor is assumed to be isotropic for simplicity. Like in Problem 17.4 we are only interested in the response current which is linear in the electric field. This means that g may be replaced by f in products on the left hand side of which already involve E. The linearized version of the Boltzmann equation is therefore when using that v k B f e E f k e c v k B g f = g f k τ k = because f k is parallel with v k,as 4 f k = f ε k ε k k = f µ v k 5

14 13 Problems in Chapter 17 transport phenomena a The geometry is the same as applied in problem 17.8, hence we define B =,,B and assume the resulting E =E x,e y, to be perpendicular to B. In this geometry, we guess that the solution has the form Introducing this in eq. 4 and utilizing 5, we get g = f + ak x + bk y 6 e f E m µ x k x + E y k y eb ak m c y bk x = ak x + bk y τ 7 Since k x and k y are independent variables, this equation leads to two independent conditions, which determine a and b to be a = E x ω c τe y 1+ω c τ e τ m f µ, b = E y + ω c τe x 1+ω c τ e τ m f µ where we have introduced the cyclotron frequency ω c = eb m c. The α component of the current density j is, according to Marder s eq , j α = e = em [d k] v kα g k f k = e [d k] v kα ak x + bk y 8 [d 9 k] av kx δ αx + bv ky δ αy where the last equality sign follows because the off-diagonal terms vanish, when the mass tensor is assumed to be diagonal isotropic. Using the same procedure as in Marder s eqs , we have em [d k] v kα e τ m f µ = e τ [d k] v f kα µ = ne τ m σ 1 and combining the three equations 8-1, we finally get j x = E x ω c τe y 1+ω c τ σ, j y = E y + ω c τe x 1+ω c τ σ 11 In the case where E is assumed to be along the z axis parallel to the field, we have to add a term c z k z to the trial function g in 6. In this situation, the magnetic field does not contribute to the Boltzmann equation, and we get j z = σ E z. Hence, for a system with an isotropic mass m, the total conductivity tensor is found to be 1 ω c τ σ = ω c τ 1, 1 σ 1+ω c τ 1+ω c τ when the magnetic field B is applied along the z axis. ω c = eb m c This result may also be expressed in terms of the Hall coefficient R, where ω c τ = RBσ, or R = 1 nec 13

15 Condensed Matter Physics 14 The resistivity tensor ρ, defined by the relation E = ρ j, is the inverse of the conductivity tensor and is particularly simple ρ = σ 1 = 1 1 ω c τ ω σ c τ 1 = 1 1 σ RB σ RB σ 1 which is in perfect agreement with the results derived from the Drude model problem The calculations in section assume ω c τ 1, in which case the diagonal elements of σ may be neglected in comparison with the off-diagonal ones, σ xx = σ yy andσ xy = σ yx ω c τ 1. Solution to HS s problem 3 Hall effect of a two-dimensional electron gas A two-dimensional electron gas with an anisotropic dispersion ε = ak x + k y+bk 4 x + k 4 y, a >, b > 1 Introducing the polar angle θ in the k x,k y -coordinate system of the reciprocal lattice, the dispersion relation may be written ε = ak b 3 + cos 4θk4, k x = k cos θ, k y = k sin θ reflecting directly the four-fold, cubic symmetry of the dispersion. 1 The equation determining the constant energy contour is obtained by solving with respect to k k = k ε = [ a 1+ bε ] 1/ b3 + cos 4θ a 3 + cos 4θ 1 3 In the case of bε a, the square root may be expanded, and to second order 1+x =1+ 1 x 1 8 x the result is k ε ε a 1 3bε 4a bε cos 4θ 4 4a3 In the figure below to the left I show a constant energy contour, which differs visible from a circle. It is obtained by a numerical evaluation of 3 Mathematica program in the case of a = b =1andε = assuming dimensionless quantities. The thin line shows the average length of kε. The figure to the right show the corresponding k ε k as a function of the angle θ. The gradient, and hence the velocity v k = k ε k /, is perpendicular to the constant energy contour. Notice that v k is smallest along the 11 directions, where kε has its maxima.

16 15 Problems in Chapter 17 transport phenomena The Boltzmann equation in the relaxation-time approximation, , dg dt = g t + r g r + g k k = g f τ g k k = g f τ when considering the steady state of a uniform system. The fields are assumed to be E =E,, and B =,,B, and the semiclassical equation of motion is k = e 1 E + c v k B, v k = r = 1 ε k k Introducing this in the right-hand part of 5 we get where e E + 1 c v k B g f k + f = g f k τ f k = f ε k ε k k = v k f ε k 8 This gradient is perpendicular to v k B, and the linearized version of 7 is ee f v k e ε k c v k B g f = g f k τ Since g f is going to scale with E, the term neglected, e E g f,isof k second order in E. Inserting E =E,, and B =,,B in 9, we finally get ev x E f ε k = 9 [ eb v c y v k x 1 ] g f 1 x k y τ 3 With the assumption of g f = k y F ε, we find g f g f df ε df ε v y v k x = v x k y k y v y dε k x F v x k y x dε k y df = v y k y dε v df x v x F v x k y dε v y = v x F 11

17 Condensed Matter Physics 16 When introducing this result in 1 and neglecting 1/τ, we get ev x E f = eb ε k c [ v xf ε] F ε = ce f 1 B ε k The current in the y direction is then j y = e [d k]v y g f = e ce [d B f k]v y k y ε k = e ce ε k dk B x dk y D k k y k f y = ece ε k B = ece B dk x dk y D k f = ece B dk x dk y D k k y f k y [d k]f = nece B = 1 RB E, R = 1 nec when performing the y integration by parts, where D k =/π is a constant and the boundaries of the integral is the boundaries of the first Brillouin zone. 13 is the usual high-field result for the off-diagonal conductivity σ yx = 1/RB. 13 Solutions to the problems in Chapter and 3. Faraday rotation The combination of the semiclassical equation of motion and the Drude relaxationtime model leads to the following equation of motion for the electrons when k = m v = m r r = e m c r B e m E r r or r + τ τ + e m c r B = e m E 1 The applied magnetic field is assumed along ẑ. The electrical field vector, due to an incident light wave, is assumed to be circular polarized in the plane perpendicular to z B =,,B, E = E ± = E ˆx ± iŷ e iωt If the light wave propagates in the positive z direction, i.e. E e iqz ωt, then the plus minus sign in corresponds to the left right circular polarization. In the case of the right circular polarization, the end point of the E-vector is making a right-handed screw line in the direction of propagation at a certain time t see for instance Griffiths, Introduction to electrodynamics, p a Introducing the expressions for the fields in 1, the equation may be solved, with respect to r, by assuming r = A ± ˆx ± iŷ e iωt 3 Because ˆx ± iŷ ẑ = ŷ ± iˆx = ±iˆx ± iŷ, we get from 1 [ iω+ 1 τ ± i eb ] m A ± ˆx ± iŷe iωt = e c m E ˆx ± iŷe iωt 4 or A ± = ee [ m iω+ 1 τ ± i eb ] 1 m = ee τ c m 1 iωτ ± iω c τ, ω c = eb m c 5

18 17 Problems in Chapter and 3 optical properties where ω c is the cyclotron frequency section 1.. The current density and the conductivity tensor are determined from j = ne r, j = σ E 6 The calculations show that r =A ± /E E and defining j ± = j ˆx ± iŷ, then j + = σ ++ E + + σ + E = σ + E + and j = σ + E + + σ E = σ E. The current has the same circular polarization as the electric field, i.e. the conductivity tensor is diagonal with respect to the choice of sign, σ + = σ + =. The diagonal components are σ ++ σ + and σ σ σ ± = ne A± E = σ 1 iωτ ± iω c τ, σ = ne τ m 7 b In the case of ωτ 1andω ω c,weget σ ± = σ 1 iωτ ± iω c τ = σ [ 1+ i iωτ ωτ ω ] 1 c iσ 1 i ω ωτ ωτ ± ω c ω 8 c The dielectric constant,.14, is ɛ ± =1+ 4πi ω σ± 1 ω p ω 1 i ωτ ± ω c ω 9 where ω p is the plasma frequency introduced by.3 or 3.7 ωp = 4πne m = 4πσ τ 1 The index of refraction,.18, is ñ ± = n ± + iκ ± = ɛ ± 1 ω p ω 1 i ωτ ± ω c, ω ω ω p 11 The Faraday rotation θ of the polarization vector per unit length along the sample is then found to be θ = ω c n n + = ω c ω p ω ω c ω = ω pω c cω 1 Notice, that the sign of θ is determined by the sign of ω c,orb, i.e. if B is along the negative z direction then θ is negative. If the active carriers are holes rather than electrons, we may perform exactly the same calculation except that e is replaced by e in all expressions m is then the positive hole mass, hence the sign of θ, like the Hall voltage, depends on the kind of carriers. Derivation of the expression for the Faraday rotation angle: Assuming the incident light wave to be linearly polarized along the x axis at z =, then we want to find the angle of rotation of the polarization vector in the xy plane as a function of z. Ez/E = 1 ˆx + iŷeiq + z ωt + 1 ˆx iŷeiq z ωt 13

19 Condensed Matter Physics 18 At z =, the polarization vector is parallel to ˆx, E/E =ˆxe iωt. Introducing ψ = 1 q + + q z and ψ = 1 q + q z, orq + z = ψ + ψ and q z = ψ ψ, then we get [ Ez/E = 1 ˆx + iŷei ψ + 1 ˆx iŷe i ψ] e iψ ωt 14 The polarization vector at z is determined by the factor in the square bracket, which is found to be 1 ˆx + iŷcos ψ + i sin ψ+ 1 ˆx iŷcos ψ i sin ψ =ˆx cos ψ ŷ sin ψ This expression shows that the polarization vector at z is making the angle θz = ψ = 1 q + q z with the x axis. According to.18 q ± = ωn ± /c, andthe rotation angle per unit length, along the direction of light propagation, is θ = 1 q + q = ω n n c + 15 In the case where the carriers are electrons, then θ>, which sign corresponds to a counter-clockwise or left-handed rotation of the polarization vector along z. 3.3 Helicon waves a From the Maxwell equations 3.71 and 3.7 we find c E = B t = 1 E c t + 4π c j t = 1 E c t + 4πσ E c t 1 when utilizing the linear relation jω =σω Eω. Assuming E = E e i k r ωt, k =,,k, k ω c we get from the left- and right-hand sides of 1 c i k i k E = 1 c iω E 4πσ + c iω E = ω 1+i 4πσ E = ω c ω c ɛ E 3 when introducing the dielectric tensor ɛ,.14. Leaving out the common phase factor, we finally get k k E + k ɛ E = 4 b Symmetry analysis of the relation j = σ E: The conducting sample is assumed to be isotropic in the shape of a sphere, if surface effects are of importance. In the presence of a magnetic field B = Bẑ, the sample, or alternatively the coordinate axes, may be rotated any angle around the z axis without changing the coordinate representation the matrix equation for the tensor relation between j and E. If the coordinate axes are rotated by an angle of 9 degrees around the z axis, then j x,j y,j z is renamed j y,j x,j z ande x,e y,e z is replaced by E y,e x,e z, whereas the components σ αβ determined by the sample are kept unchanged: j x = σ xx E x + σ xy E y + σ xz E z j y = σ yx E x + σ yy E y + σ yz E z j z = σ zx E x + σ zy E y + σ zz E z j y = σ xx E y + σ xy E x + σ xz E z j x = σ yx E y + σ yy E x + σ yz E z 5 j z = σ zx E y + σ zy E x + σ zz E z

20 19 Problems in Chapter and 3 optical properties A comparison of the two sets of relations then shows that σ xx = σ yy, σ xy = σ yx, and σ zx = σ zy = σ xz = σ yz =. [Extension of the symmetry arguments: At zero field, the isotropic sample is insensitive to a rotation around any arbitrary direction and σ αβ = σ xx δ αβ,whichresult shows that σ xy whenb. A reflection of the system with respect to, for instance, the yz plane implies that the axial vector B along ẑ is reversed, whereas it is the x components of the polar vectors j and E which change sign. Because of this different behaviour of the polar and axial vectors it is straightforwardly shown that σ xy changes sign if B is reversed, whereas nothing happens with the diagonal elements. This symmetry element is in accordance with the general Onsager relation σ αβ B =σ βα B. Hence, in the case of an isotropic sample the diagonal components may only involve terms quadratic even powers in B, whereasthe off-diagonal ones are linear of odd powers in B.] These symmetry relations are consistent with the results obtained in the previous problems 17.9 and., and they apply as well for the dielectric tensor ɛ, or ɛ xx ɛ xy ɛ = ɛ xy ɛ xx 6 ɛ zz Introducing this in 4 in combination with k k E = k E + k k E, the result is k + ɛ xx k ɛ xy k E ɛ xy k k + ɛ xx k x Ey = 7 ɛ zz k Ez which only has a non-zero solution for E, if the characteristic determinant vanishes, or if [ ] ɛ zz k/k ɛ xx + ɛ xy = 8 We are going to utilize the solution of problem. in the following. Here σ ± is defined by the vector relation j ± = σ ± E ± in the xy plane, or j x + ij y = { σ + E x + ie y σ xx E x + σ xy E y + i σ xy E x + σ xx E y =σ xx iσ xy E x + ie y 9 showing that σ + = σ xx iσ xy and correspondingly, 9 in the solution of., ɛ ± = ɛ xx iɛ xy ɛ xx = 1 ɛ+ + ɛ, ɛ xy = 1 ɛ+ ɛ i 1 The latter relations reduce 8 to the diagonal form ɛ zz [k/k ɛ +][ k/k ɛ ] = 11 c The situation is exactly the one considered in problem., and the result found there is [combining 8 and 9 in the solution of.] ɛ ± =1+ 4πi ω =1+ ω p ω σ 1 iωτ ± iω c τ =1+iω p ω 1 1 iτ ω ± ω c =1 ω p ω 1 ω ω c, τ 1 iωτ ± iω c τ ω = ω + i τ 1

21 Condensed Matter Physics or finally ɛ ± =1 ω p ω ω 1 ± ω c / ω 1 ω c / ω, ω c = eb m c, ω p = 4πσ τ = 4πne m 13 which result agrees with the expression for ɛ xx =ɛ + + ɛ / given by 3.8. d The atomic density front page of aluminium metal is 6. 1 cm 3. According to Table 6.1, the number of conduction electrons per atom is 3, i.e. the electron density in Al 3+ -metal is n = cm 3. The cyclic plasma frequency is ω p = rad/s, when n =1 cm 3 and m = m, see 3.3. [Incidentally, the unit of a cyclic frequency is not Hz but rad/s or just s 1. The unit Hz hertz is reserved for the case of cycles per second, ν = ω/π.] In the case of Al-metal, ω p = s 1, ν p = Hz, ω p = 16 ev 14 The cyclotron frequency may be expressed in terms of the fundamental magnetic quantity, the Bohr magneton: µ B = e mc = ev/g ω c = eb m c =m m µ BB 15 Assuming m = m and B =1T=1kG=1 4 G, we get ω c =.1 mev, ν c = Hz, ω c = s 1 16 The numbers show that ω c 1 5 ω p ω p.inthecaseofω ω c, the fraction in the expression for ɛ ± is of the order of 1 1, and the adding of one to this fraction has no effect at all. e Assuming ω ω p we get from equation 1 ω p ω p ɛ ± = ωω ω c + i τ ωω ω c, ω c τ 1 17 In most cases ɛ ± <, which implies a purely imaginary refraction index ñ = ɛ and therefore no light propagation. However, there is one situation, where electromagnetic waves with ω ω p are capable to propagate through the sample, which one is determined by the solution ɛ +,whenω<ω c or ɛ if the carriers are holes. Hence, when placing a metal like aluminium in a strong magnetic field, it is possible for low-frequency light to propagate through the metal, but low temperature and high purity is demanded in order to secure that the damping is small, ω c τ 1. The dispersion relation for these helicon waves is determined by 11 and 17: ω ω k = k p ɛ+ = ω = ω cck c ωω c ω ωp +ck ω c ωp ck, k ω p c 18 showing that the dispersion is not linear but quadratic in k in the long wavelength limit. The name of the waves is a derivative of helix and refers to the circular polarization of the waves ɛ + corresponds to left circular polarization the polarization vector of the helicon wave is forming a helix along the z direction.

22 1 Problems in Chapter 4 ordering The helicon resonance is related to the Hall effect, and the result may be written in a simple form in terms of the Hall coefficient R = nec 1 : ω = B R 4π ck cgs-gaussian units 19 In the experiments of, for instance, R. Bowers, C. Legendy and F. Rose, Phys. Rev. Lett. 7, , wavelengths of the order of 1 cm are considered. In the case of Al-metal, λ = 1 cm corresponds to a helicon frequency of ν =1.77 Hz at B = 1 kg. In sodium n =.54 1 cm 3 this frequency is 1.3 Hz. Bowers et al. observed a standing wave resonance in a Na sample of length l = 4 mm at 3 Hz and 1 kg. The resonance condition is l = λ/, and our simple model predicts λ/ to be about 3.1 mm at the same frequency and field. Solutions to the problems in Chapter Mean field theory The antiferromagnetic two-dimensional Ising model on a square lattice. In this lattice, the number of nearest neighbours is z = 4. The Hamiltonian is supposed to be H = Jσ R σ Hµ R B σ R = 1 Jσ R R R R σ Hµ R B σ R, J < 1 R R where H = E in the usual representation of σ R = ±1. Because J is negative, the ground state is not the ferro- but the antiferromagnet. In the sums, R and R are nearest neighbours, and the antiferromagnetic ground state is assumed to be determined by the mean field values { σ,σ σ R, σ R =, R sublattice A σ,σ, R sublattice B R The two sublattices of the square lattice: A shown by filled and B by open circles. The figure shows the case, where the site at R belongs to the A-sublattice, in which case σ R = σ. Its four nearest neighbours at positions R,then all belong to the B sublattice and σ = σ R. a In the mean-field MF approximation we get, when inserting in 1, E MF = 1 J σ R σ R + σ R σ R σ R σ R Hµ B R R R = J σ σ R R Hµ B σ R + 1 NzJσ σ R R = N/ R A R zjσ + µ B H σ R N/ R B σ R zjσ + µ B H σ R + 1 NzJσ σ 3

23 Condensed Matter Physics Because of the MF-approximation, the total energy E = R E R is the sum of the individual contributions of each site, and the partition function of the total system Z = all E e βe = R all E R e βe R = R Z R 4 is reduced so to become the product of the partition functions Z R for the individual sites, see for instance 6.4. For the sites, where R belongs to the sublattice A, we get, when omitting the constant energy term 1 zjσ σ, Z = σ=±1 which implies that e β[ zjσ +µ B Hσ] = σ R = σ = 1 Z σ=±1 σ=±1 [ ] e βzjσ +µ B Hσ =cosh zjσ + µ B H [ ] σe βzjσ +µ B Hσ =tanh βzjσ + µ B H 5 6 Notice that the omission of a constant energy term does not affect the calculation of a thermal expectation value. Z and σ are determined equivalently [ ] [ ] Z =cosh zjσ + µ B H, σ =tanh βzjσ + µ B H 7 and the total free energy is F = N k B T ln Z +lnz + N 1 zjσ σ = N { [ ] [ ]} k BT ln coshzjσ + µ B H +ln coshzjσ + µ B H + N zjσ σ 8 The constant energy term neglecting in the partition functions adds to F. b The self-consistent equations determining σ and σ are derived above. c In the paramagnetic phase, T > T N, σ and σ vanish in the limit H. Expanding tanh to leading order, tanh x x, we get from 6 and 7 [ ] σ =tanh βzjσ + µ B H [ ] σ =tanh βzjσ + µ B H βzjσ + µ B H βzjσ + µ B H σ = σ = µ B Hβ 1 zjβ The expectation values σ and σ are equal and are proportional to the field. The magnetization is M =N/V µ B σ + σ / and the susceptibility is defined as χ = M/H in the limit of zero field, and we find 9 χ = N V µ B k B T zj N V µ B k B T Θ, Θ= z J k B = T N 1 The paramagnetic Curie, or Curie Weiss, temperature Θ is defined by 4.41, χ T Θ 1. In the present system Θ is negative J is negative, and Θ is, as we shall see, the same as the Néel temperature, T N, below which the system is antiferromagnetically ordered at zero field.

24 3 Problems in Chapter 4 ordering d The Néel temperature is the temperature below which the self-consistent equations for σ and σ have a non-zero solution at zero field. The solution is obtained by assuming σ = σ,andh =,inwhichcaseweget [ ] σ =tanh β z J σ =tanhβz J σ = tanhασ 11 This equation is the same as considered in the case of the ferromagnet, 4.56, and it has a non-zero solution, when the coefficient α 1. The Néel temperature T N is the temperature at which α = z J β =1,orT N = z J /k B. The paramagnetic susceptibility, which diverges at the transition, is obtained by applying a field with the same symmetry as the ordered phase. The application of a staggered field, where H R =+H at the A sites and H at the B sites determines straightforwardly using σ = σ the staggered susceptibility to be χ stag = N V µ B σ σ H = N V µ B k B T T N 1 which diverges at T N. e Spin-flip transition: In the ordered phase, the application of a small field along z, in the up direction, does not change σ =1orσ = 1. However, if the field becomes sufficiently large, the spins antiparallel to the field are going to flip so to become parallel to the field. This first-order spin-flip transition happens, when the Zeeman-energy gain is able to compensate for the loss of exchange energy. Introducing dimensionless quantities, then σ + h σ =tanh, σ t =tanh σ + h t, h = µ B H z J, t = T T N 13 In the limit of t, then the introduction of σ = 1 in the second equation implies σ = 1 if1>h>andσ =1ifh>1. The first equation then predicts σ = 1 in both cases, hence assuring that this is the self-consistent solution. Using ln[ cosh x] =ln[e x + e x ] ln e x = x for x ±, the free energy is found to be determined by f = F Nz J = t t { σ + h + t { [ ln cosh ] [ σ + h σ + h +ln cosh t t } { 1 1 σ σ = ]} 1 σ σ σ + h, <h<1 t h + 1, h > 1 14 Hence, the free energies of the antiferromagnetic phase f = 1/ and of the spinflipped phase f = h +1/ are equal at the transition at h = 1. At a finite temperature the spin-flip transition is smeared out due to thermal effects. The figure below shows the numerical calculated magnetization, σ + σ /, as a function of the field h at various values of temperature t.

25 Condensed Matter Physics 4 Averaged spin t Magnetic field h 4.6 Superlattices Model of Cu 3 Au In the fcc lattice, the sites in the centers of the six faces are named A s and the 8 corners are named B s as indicated on the figure by, respectively, black and white spheres. There is 8/8 =1 B s site and 6/ =3A s sites per unit cell. The distance between nearest neighbours is a/, and the coordination number is z = 1. The kind of nearest neighbours of the two kind of sites are A s -site: 8 A s sites and 4 B s sites. B s -site: 1 A s sites. The correct solution to the equations 4.6 and 4.63 is not 4.64, but fσ R,σ R = ɛ BB ɛ AA 4 σ R + σ R ɛ AB ɛ AA ɛ BB σ 4 R σ R 4.64 when leaving out the constant term C 1 =ɛ AB +ɛ AA +ɛ BB /4. This modification affects the definition of the effective parameters in 4.67, which should read µ B H = µ ɛ BB ɛ AA z, J = ɛ AB ɛ AA ɛ BB a In the present system we have a 3:1 mixture of A and B atoms. The effective Hamiltonian equals the energy minus the chemical potential times the difference between the numbers of B and A atoms, is leaving out the constant contribution H = E µ σ R = J σ R σ µ R B H σ 1 R R R R R We are going to use the same choice as Marder, that σ R =1or 1 signifies a site occupied by, respectively, a B or an A atom. With this choice, the effective energy parameters are those given by 4.67 above. In the mean-field approximation H MF = J R R σ R σ + σ R R σ R σ R σ µ R B H R σ R = R H R

26 5 Problems in Chapter 4 ordering where σ σ R +σ R R σ R = 1 σ σ R + σ R R σ R = σ R σ R R R R = N B s σ Bs R R 3N 1 σ As + σ As 8 σ As +4 σ Bs A s Here N is the number of unit cells or B s sites, and σ R = σ As R nn 3 σ Bs, if R = A s B s is the position of an A s B s site. The grand partition function of the total system is the product of the individual partition functions for each site, because H = R H R,andinthecaseofanA s site omitting the constant energy term Z As = e β[j8 σ +4 σ As Bs +µ B H]σ σ=±1 4 implying that and equivalently σ As = 1 σe β[j8 σ +4 σ As Bs +µ B H]σ Z As σ=±1 { [ ]} =tanh β 8 J σ As +4J σ Bs + µ B H { [ ]} σ B =tanh β 1 J σ s A + µ B H s 5 6 b σ R =+1fortheB atoms and the number of these atoms is equal the number of unit cells N, whereas the number of A atoms is 3N. These conditions imply N 3N = N σ R = N 3N σ R Bs + σ As = N σ Bs +3N σ As B s A s σ Bs = 3 σ As 7 c The chemical potential, or effectively µ B H, has to be adjusted so that the two equations 5 and 6 are in accordance with the relation σ Bs = 3 σ As. The effective Zeeman term may be eliminated: [ ] tanh 1 σ A tanh 1 σ s B = β 8 J σ s A +4J σ s B + µ B H [ ] s 8 β 1 J σ As + µ B H =4βJ σ Bs σ As = 8βJ 1+ σ As Introducing an effective order parameter Q and J = J, we finally get Q +1 fq tanh 1 3Q 1 +tanh 1 =8β J Q, Q = 1 σ A 9 s In the disordered phase both the A s and the B s sites are occupied by an A atom with the probability 3 4 or σ As = σ Bs = = 1. This result is in accordance with 7 and Q = 1 σ As =. In the completely ordered phase at

27 Condensed Matter Physics 6 zero temperature, all A B atoms are placed on the A s B s sites implying that σ As = 1 and σ Bs = 1 and hence Q =1. The numerical solution of 9 is discussed in a Mathematica program. The system orders at a first-order transition, when fq =Qf Q [the line y = αq with α = f Q c is parallel to the tangent of fq atq = Q c, and this line just touches fq atq = Q c,whenfq c = αq c. This then becomes a solution to 9 if choosing α = 8β J ]. The solution is Q = Q c =.3455 and defining the effective temperature scale t =8β J 1 then Q jumps from zero to Q c at the temperature t = t c =1/f Q c = The calculated variation of Q as a function of t is shown in the figure. Solutions to the problems in Chapter 5 and 6 5. Classical electrons in a magnetic field The spin degree-of-freedom does not occur in classical physics. The elementary particles do neither exist, but accepting their existence, they only interact with the electromagnetic fields because of their charges, and with the gravitational field because of their masses. The classical Hamiltonian H of a many-body system see for instance Goldstein is a function of the independent canonical variables r i, p i of all the particles. The canonical impulse p i, independent of r i, is defined in terms of the Lagrangian L = L r i, r i, as p i = L/ r i or, in terms of the Hamiltonian theory, as the generator of an infinitesimal translation r i r i + δ r i. a If the external constraints fields are independent of time, the Hamiltonian is equal the total kinetic plus potential energy of the system. In this case the classical partition function is Z = e βf = 1 h n d r 1 d p 1 d r n d p n e βh r 1, p 1,, r n, p n 1 b In the presence of a time-independent electromagnetic field, E only contributes to the potential energy as determined by the scalar potential. The magnetic field B only affects the velocities of the particles via the vector potential A = A r, where B = A.Fortheith particle with mass m i and charge q i, the canonical impulse is p i = m i ri + q i A c T i = 1 m i r i = 1 p m i q i A i c [The magnetic field needs not to be uniform in space, but only to be constant in time the case of a spatial uniform field is discussed in the next problem 5.4].

28 7 Problems in Chapter 5 and 6 magnetism c The canonical impulse p i appears in the Hamiltonian only via the kinetic energy T i. Assuming H in 1 to be the Hamiltonian function of the system, when the magnetic field is zero but including the potential energy contributions of the electric field, then the introduction of the magnetic field implies that H r 1, p 1,, r n, p n H r 1, p 1 q 1 A r c 1,, r n, p n q n A r c n 3 The partition function Z in the presence of the magnetic field is then Z B= 1 h n d r 1 d p 1 d r n d p n e βh r 1, p 1 q 1 A r1 c,, r n, p n q n A rn c = 1 4 h n d r 1 d p 1 d r n d p n e βh r 1, p 1,, r n, p n = Z The integration with respect to the canonical impulse p i may be performed before the r i -integrations. Introducing the following change of variables p i = p i q i c A ri, then d p i = d p i,since A r i is independent of p i. This transformation of variables therefore leads to the same integral as in 1. The elimination of B in the partition function means, for instance, that the magnetic susceptibility of a classical equilibrium system is zero. This result is called the Bohr van Leeuwen theorem. Notice, that it is important for the argumentation that the system is in equilibrium the basis for the use of the partition function. 5.4 Quantum electrons in a magnetic field The Hamiltonian of an electron with charge e in a magnetic field B is H = 1 m p + e c A 1 The choice of the Landau gauge A =,Bx, B = A =,,B, A = i.e. this vector potential corresponds to a spatial uniform field B in the z direction. Since the divergence of A is zero, the two quantum mechanical operators A and p commute, p = /i in the r-representation. Introducing this choice of gauge in 1, we get H = 1 [ p x + p m y + eb ] c x + p z 3 The wave function of the electron is assumed to be ψ r =ux e iyk y +zk z 4 and the eigenvalue equation is H ψ = e iyk y +zk z 1 m [p x + k y + ebc x + k z ] ux 5 When the common phase factor is deleted, this eigenvalue equation is reduced to one, which only involves a determination of ux, and the equation may be written: [ ] p H ux = x m + 1 mω c x x + ε z ux 6

29 where ω c = eb mc, Condensed Matter Physics 8 x = k y mω c, ε z = k z m Equation 6 is the eigenvalue equation for the one-dimensional harmonic oscillator, displaced to be centered around x = x and shifted in energy by the constant with respect to x energyε z. Hence, the final eigenstates are classified by k y,k z and the occupation number ν, and the energy eigenvalues are ε ν,ky,k z = ν + 1 ω c + kz, ν =, 1,, 8 m This is the result 5.48 utilized in the analysis of the Landau diamagnetism Ferromagnetic ground state We are going to discuss the ferromagnetic ground state of the Heisenberg model for an arbitrary integer/half-integer value of S. The Heisenberg Hamiltonian is H = J ll S l S l = 1 J ll S l S l 1 ll l l according to 6.1 in Marder. The possibility of J ll may be excluded, since such a term would contribute only by a constant term J ll SS + 1 to the energy. The components of S l fulfill the usual spin-commutator relations [S lx,s l y ]=iδ ll S lz, [S ly,s l z ]=iδ ll S lx, [S lz,s l x ]=iδ ll S ly a Defining the following operator: S z = l S lz 3 then the commutator of this operator with a single term S l S l in 1 is [ S z, S l S l ]=[S z, S l ] S l + S l [ S z, S l ] =[S lz,s lx ] S l x +[S lz,s ly ] S l y + S lx [S l z,s l x ]+S ly [S l z,s l y ] 4 = is ly S l x is lx S l y + is lx S l y is ly S l x = showing that [ S z, H ] =. The S z -representation for each site is defined by S lz m l = m l m l,wherem l = S, S+1,...,S,andS z l m l = l m l l m l. Most of these eigenstates are highly degenerated except for the two singlets, l m l = ±S = l + S or l S. Because S z and H commute the two non-degenerate eigenstates of S z are also eigenstates to H, and H l m l = ±S = 1 NJ S l m l = ±S, J k= l J ll e i k r l r l 5 b Defining S = S R + S R and using S = S R + S R = S R + S R + S R S R { Ψ S SS +1+SS +1+ Ψ S Ψ = R S Ψ R 6 SS +1, S =, 1,, S which shows that Ψ S R S R Ψ attains its maximum value, when S =S, and Max{ Ψ S R S R Ψ } = 1 [SS +1 SS +1]=S 7