Physics 211B : Solution Set #1

Transcription

1 Physics B : Solution Set # [] Drude formula Consider a hypothetical monovalent s-band metal with a simple cubic crystal structure he valence band dispersion is given by the tight binding result, εk) = t { cosk x a) + cosk y a) + cosk z a) } Compute the DC conductivity tensor σ αβ Show that σ αβ = σ δ αβ is diagonal, and obtain an expression for σ Numerically evaluate any integrals he following result may prove useful: π π du π π dv δcos u + cos v + λ) = 4 K λ ) Θ λ ), where Kx) is the complete elliptic integral of the second kind Compare your result with the Drude value you would obtain by approximating the band as parabolic, based on its curvature at the zone center Solution: We must evaluate σ αβ = e τ d k π) vα v β f ), where the dispersion is he velocity is εk) = t { cosk x a) + cosk y a) + cosk z a) } vk) = k = ta ) sink x a), sink y a), sink z a) As a consequence of the cubic symmetry, σ αβ = σ δ αβ is diagonal, and we may write σ = e τ ta ) d k π) sin k z a) f ) he monovalency condition means that there is one electron per unit cell, which in turn means the chemical potential lies at µ = 0 so the s-band is half-filled Changing variables to u = k x a, etc, σ = e τ t π a π π π π du dv dw sin w) δcos u + cos v + cos w) π π

2 We now use Iλ) = π π du = 4 π π λ dv δcos u + cos v + λ) ) x x + λ ) = 4 λ ) ds ) ) + λ + λ s + λ ) ) s 4 ) 8 λ = + λ K = 4 K ) λ 4) + λ ) hus, where C = Numerical integration gives C σ = 8C π e τ t a 5) dx ) x K 4 x Expanding the dispersion about the zone center, we find hence the effective mass is given by he Drude conductivity is then εk) = 6t + ta k + Ok 4 ), 6) ta σ = ne τ m m = m = ta 7) = ne τ ta = x e τt a, 8) where x na is the dimensionless density, equal to the average number of electrons per unit cell 0 x ) For a monovalent metal, the band is half filled, and x =, and the prefactor x is he exact value of the prefactor is 8C/π 0668 [] hermal transport in a magnetic field Consider a metal with a parabolic band εk) = k /m in the presence of a uniform magnetic field B Use the Boltzmann equation* to compute a) the resistivity tensor ρ, b) the thermal conductivity tensor κ, c) the thermopower tensor Q, and d) the Peltier tensor Assume is small, and work to lowest nontrivial order in the temperature Also assume a constant relaxation time τ Does the Wiedemann-Franz law hold for the matrices κ and ρ?

3 Solution: We begin with the Boltzmann equation, [ v ee + ε µ ] f ) = e c v B δf k δf τ We take εk) = k /m, and we write δfk) = k Aε) v x δf k y v y δf k x = m k xa y k y A x ), from which we obtain the linear relations +ω c τ 0 A x ω c τ 0 A y = τ f 0 m 0 0 A z he solution is easily obtained: A x A y = τ/m f 0 + ω A z cτ ee x + ε µ x ee y + ε µ y ee z + ε µ z ω c τ 0 ee x + ε µ x +ω c τ 0 ee y + ε µ ωcτ y ee z + ε µ z he electrical and thermal currents are given by j α = e dε gε) ε A α jα q = dε ε µ) gε) ε A α We now read off the transport coefficients from the relations We find ρ = e dε gε) ε f ) m j = ρ E ρ Q ρ Q = e εε µ) m dε gε) j q = ρ E κ + ρ Q) τ + ωcτ f ) ρ = e m dε ε µ) gε) ε f ) κ + ρ Q = ε ε µ) m dε gε) f ) ω c τ 0 +ω c τ ωcτ ω τ c τ 0 + ωcτ +ω c τ ωcτ ω τ c τ 0 + ωcτ τ + ωcτ +ω c τ ωcτ ω c τ 0 +ω c τ ωcτ

4 We evaluate the integrals using the Sommerfeld expansion, and invoking the density of states m ) / gε) = ε π With an energy-independent relaxation time τ, we obtain ω ϱ = σ = ne τ c τ 0 m + ωcτ +ω c τ ωcτ he quantity ρ Q is proportional to the same matrix as is ρ, and one readily finds that Q is a multiple of the unit matrix, Q = π k B eε F I Here we have used the results dε gε) ε ε µ) f ) = π k B ) gµ) + O 4 ) We also find gε F ) = m k F π = n ε F = Q t B) = π k B ) I eε F Finally, the thermal conductivity tensor is ω c τ 0 κ = π kb nτ/m + ωcτ +ω c τ ωcτ hus, the Wiedemann-Franz law, holds at the matrix level as well κ = π e k B ρ [] wo bands Calculate the frequency-dependent conductivity tensor for a direct gap semiconductor in the presence of a magnetic field B = Bẑ You should begin with the Boltzmann equation in the relaxation time approximation f 0 f 0, δf δ f for holes), δf t e v E f 0 e c v B δf k = δf τ, and the conduction and valence band dispersions, ε v k) = ε v 0 m v αβ k α k β ε c k) = ε c 0 + m c αβ k α k β 4

5 Assume the two bands behave independently, and solve the two Boltzmann equations for the conduction electrons and valence holes In each case, try a solution of the form δfk, t) = k µ A µ εk)) e iωt he currents are { jc j v } = e d k π) { } vc δf c + v v δ f v Compute σ αβ along principal axes of the effective mass tensors You may assume that m v and m c commute, ie they have the same eigenvectors You should further assume that B lies along a principal axis Solution: his problem is essentially solved in the notes in section 7 All that is left to do is to sum the contributions from the valence and conduction bands: σ xx ω) = n ce τ c m c,x iωτ c iωτ c ) + ω c, τ c + p ve τ v m v,x iωτ v iωτ v ) + ω c, τ v σ xy ω) = n ce τ c ω c, τ c m c,x m c,y iωτ c ) + ωc, τ c σ zz ω) = n ce τ c m c,z iωτ c + p ve τ v m v,z iωτ v + p ve τ v m v,x m v,y ω v, τ v iωτ v ) + ω c, τ v [4] Spin disorder resistivity for the brave only!) Consider an isolated trivalent b impurity ion in a crystal field Application of Hund s rules gives a total angular momentum J = 6 A cubic crystal field splits this -fold degenerate multiplet into six levels: two singlets, one doublet, and three triplets he ground state is a singlet Using the first Born approximation, calculate the temperature-dependent resistivity in a free electron model with a scattering Hamiltonian H imp = A g ) N imp j= δr R j ) S J j /, where r and S are the conduction electron position and spin operators R j and J j are the impurity position and angular momentum of the j th b impurity A is the strength of the exchange interaction, and g = is the gyromagnetic factor a) In general the relaxation time is energy-dependent: τ = τε) Show that the resistivity is given by ρ = m/ne τ, where the average is with respect to the weighting function ε gε) f 0 /) Show also that τ τ, which provides an upper bound for ρ which can often be computed 5

6 b) Use the results of a) to derive the approximate expression for the resistivity ρ ρ 0 p ij Q ji, where E i E i )/k p ij = B e E i E j )/k B Q ij = i J + j + i J j + i J z j, e E i/k B k e E k/k B where the ionic energy levels are denoted by E i and where the summations run over the J + ) crystal field states Show that ρ 0 = πm g ) A n imp 8e ε F c) Show that the high temperature limiting value of ρ is JJ + ) ρ 0 his is often called the spin-disorder resistivity Solution: he collision term in Boltzmann equation is, from Fermi s golden rule, I kσ [f] = π jkσ H imp ik σ δ E j + ε k E i ε k ) fk σ f kσ) π ij k σ P i ij k σ P i fk σ H imp ikσ δ E j + ε k E i ε k ) fkσ f k σ ), where P i is the Boltzmann weight for the ion in state i: P i = exp E i/k B ) n exp E n/k B ) Using plane wave states ψ k r) = V / expik r), the matrix element is obtained: jkσ H imp ik σ = V A g ) e ik k ) R jσ S J iσ We assume the impurity positions are uncorrelated, so that jkσ H imp ik σ = A g ) V jσ S J iσ R,R e ik k ) R R ) R = N imp A g ) jσ V 4 S J iσ ) + Nimp )δ kk he term proportional to δ kk cancels when inserted into the collision integral We are left with I kσ [f] = π 5 A g ) d k n imp π) δ ) E j + ε k E i ε k jσ S J iσ ijσ } {P j f k σ f kσ) P i f kσ f k σ ) 6

7 Note that the Fermi-Dirac distribution fkσ 0 is annihilated by I: P j f 0 k σ f 0 kσ ) = e βe j n e βen e βε k µ) + eβεk µ) = e βe i n e βen = P i fkσ 0 f k 0 σ ) e βε k µ) + e βε k µ) e βεk µ) + δ ) E j + ε k E i ε k e βεk µ) + δ ) E j + ε k E i ε k We therefore write and find f kσ = f 0 kσ + δf kσ, f 0 kσ = f 0 k = e βε k µ) + P j f 0 k σ + δf k σ ) f 0 kσ δf kσ) P i f 0 kσ + δf kσ) f 0 k σ δf k σ ) = [ P j f 0 kσ ) + P if 0 kσ] δfk σ [ P j f 0 k σ + P i f 0 k σ ] δfkσ he Boltzmann equation then takes the form ev k E f ) = π 5 A g ) n imp jσ S J iσ ijσ d k { π) δ ) [Pj E j + ε k E i ε k fkσ 0 ) + P ifkσ] 0 δfk σ [ } P j fk 0 σ + P i fk 0 σ )] δf kσ We now sum both sides on σ and divide by two Using jσ S J iσ = j J α i i J β j σ S α σ σ S β σ σ σ = SS + )δ αβ j J α i i J β j = 4 j J i, we obtain ev k E f ) = π 5 A g ) n imp j J i ij d k π) δ ) E j + ε k E i ε k { [Pj fkσ 0 ) + P ifkσ 0 ] δfk σ [ } P j fk 0 σ + P i fk 0 ] σ δfkσ It is convenient to define the structure factor Su) ij P i j J i δu Ej + E i ) 9) 7

8 Note that S u) = e βu Su) We assume a solution of the form δf k = eτε k ) E f 0 and plug this into our Boltzmann equation Further assuming an isotropic Fermi surface, the δf k term integrates to zero since it is proportional to v k, aside from energy-dependent hence rotationally isotropic) factors We then have τε k ) = π A g ) n imp Furthermore, since u = ε k ε k, we have du Su) d k π) δu + ε k ε k) [ ] fk 0 + e βu fk 0 giving τε k ) = π A g ) n imp f 0 k + e βu f 0 k = f 0 k f 0 k du Su), d k π) δu + ε k ε k) f 0 k f 0 k a) he conductivity is determined from j = e which, for free electrons, gives d k π) eτε k) v k E ) v k f ) σ = ne τ m τ = dε ε gε) τε) f ) / dε ε gε) f ) Now the triangle inequality requires for any real functions fε) and gε) [ [ ][ ] dε fε) hε)] dε fε) dε hε),, so taking fε) hε) [ ε gε) f [ ε gε) f ) ] / τε) ) ] / τε) 8

9 we conclude hence ρ = τε) τε) m ne τ m ne τ b) Let us compute τ Noting that f ) = β f 0 f 0), we have τ = m n π A g ) βn imp = πm A g ) n π) 6 4 β n imp d k π) d k π) du Su) δu + ε k ε k ) fk 0 f 0 k ) v k du Su) dε dε δu + ε ε) f 0 ε) f 0 ε ) ) ds ε v ds ε v he factor f 0 ε) f 0 ε u) varies on a scale k B If k B µ and if u µ then we can approximate ds ε v ds ε v 4πkF), and the energy integral becomes dε f 0 ε) f 0 ε) ) = hus, and τ ω e βω = m π n A g ) k F / ) 4 βu n imp du e βu Su), ρ m ne = π τ 8 ma g ) e ε F since n = k F/π and ε F = k F/m Note that βu du e βu Su) = ij n imp βe j E i ) e βe j E i ) βu du e βu Su) e βe i n e βen j J i ij p ij Q ji, 9

10 where p ij = βe j E i ) e βe j E i ) e βe i n e βen Q ji = j J i = j J + i + j J i i + j J z i herefore, the upper bound on the resistivity is ρ + = ρ 0 p ij Q ji, with ρ 0 = πm 8 A g ) ε F n imp c) At high temperatures, βu/e βu ) and therefore p ij P i, independent of j In this limit, then, p ij Q ji = ij i e βe i = i = JJ + ) j J i / n j i J i / Hence, ρ + = JJ + )ρ 0 at high temperatures n e βen e βen [5] Cyclotron resonance in Si and Ge Both Si and Ge are indirect gap semiconductors with anisotropic conduction band minima and doubly degenerate valence band maxima In Si, the conduction band minima occur along the 00 ΓX ) directions, and are six-fold degenerate he equal energy surfaces are cigar-shaped, and the effective mass along the ΓX principal axes the longitudinal effective mass) is m l 0 m e, while the effective mass in the plane perpendicular to this axis the transverse effective mass) is m t 00 m e he valence band maximum occurs at the unique Γ point, and there are two isotropic hole branches: a heavy hole with m hh 049 m e, and a light hole with m lh 06 m e In Ge, the conduction band minima occur at the fourfold degenerate L point along the eight directions) with effective masses m l 6 m e and m t 008 m e he valence band maximum again occurs at the Γ point, where the hole masses are m hh 04 m e and m lh 0044 m e Use the following figures to interpret the cyclotron resonance data shown below Verify whether the data corroborate the quoted values of the effective masses in Si and Ge Solution: We found that σ αβ = ne Γ αβ, with Γ αβ τ iω) m αβ ± e c ɛ αβγ Bγ τ iωτ)m x ±eb z /c eb y /c = eb z /c τ iω)m y ±eb x /c ±eb y /c eb x /c τ iω)m z 0

11 Figure : Constant energy surfaces near the conduction band minima in silicon here are six symmetry-related ellipsoidal pockets whose long axes run along the 00 directions he valence band maxima are isotropic in both cases, with m hh Si) 049 m e m hh Ge) 04 m e m lh Si) 06 m e m lh Ge) 0044 m e With isotropic bands, the absorption is peaked at ω = ω c = eb/m c, assuming ω c τ Writing ω = πf, the resonance occurs at a field Bf) = πf m c e = hc e m hf m e πa B e /a B ) = G m m e f[hz] = 8590 G m m e,

12 Figure : Cyclotron resonance data in Si G Dresselhaus et al, Phys, Rev, 98, )) he field lies in a 0) plane and makes an angle of 0 with the [00] axis where we have used hc e = G cm a B = m e e = 059 Å h = ev s e a B = 7 ev = Ry f = Hz hus, we predict B hh Si) 40 G B hh Ge) 90 G B lh Si) 70 G B lh Ge) 78 G All of these look pretty good

13 Figure : Constant energy surfaces near the conduction band minima in germanium here are eight symmetry-related half-ellipsoids whose long axes run along the directions, and are centered on the midpoints of the hexagonal zone faces With a suitable choice of primitive cell in k-space, these can be represented as four ellipsoids, the half-ellipsoids on opposite faces being joined together by translations through suitable reciprocal lattice vectors Now let us review the situation with electrons near the conduction band minima: Si: 6-fold degenerate minima along 00 Ge: 4-fold degenerate minima along at L point) m l Si) 0 m e m l Ge) 6 m e m t Si) 00 m e m t Ge) 008 m e he resonance condition is that σ αβ =, which for τ > 0 occurs only at complex frequencies, ie for real frequencies there are no true divergences, only resonances he location of the resonance is determined by det Γ = 0 aking the determinant, one finds det Γ = τ iω) m l {τ iω) m t + e c B z + m t m l e B c x + By ) }

14 Figure 4: Cyclotron resonance data in Ge G Dresselhaus et al, Phys, Rev, 98, )) he field lies in a 0) plane and makes an angle of 60 with the [00] axis Assuming ωτ, the location of the resonance is given by ) ω eb = m t c + m t m l ) eb m t c, where B B z and B B x ˆx + B y ŷ Let the polar angle of B be θ, so B = B cos θ and B = B sin θ We then have ω = ) eb { m t c cos θ + m t m l } sin θ m )/ Bf) = 8600 G t cos m θ + m t e m l sin θ, where again we take f = ω/π = Hz 4

15 According to the diagrams, the field lies in the 0) plane, which means we can write ˆB = sin χ ê sin χ ê + cos χ ê, where χ is the angle ˆB makes with ê = [00] Ge We have and we are told χ = 60, so m t m e = 008 ˆB = 8 ê m t m l = 005, 8 ê + ê he conduction band minima lie along, which denotes a set of directions in real space: ±[] : ˆn = ± ê + ê + ê ) cos θ = ˆB ˆn) = B = 950 G ±[ ] : ˆn = ± ê + ê ê ) cos θ = ˆB ˆn) = B = 950 G ±[ ] : ˆn = ± ê + ê + ê ) cos θ = ˆB ˆn) = 7 6 B = 50 G ±[ ] : ˆn = ± ê ê + ê ) cos θ = ˆB ˆn) = 7+ 6 B = 70 G All OK! Si Again, B lies in the 0) plane, this time with χ = 0, so ˆB = 8 ê 8 ê + he conduction band minima lie along 00, so 4 ê ±[00] : ˆn = ±ê cos θ = ˆB ˆn) = 4 B = 80 G ±[00] : ˆn = ±ê cos θ = ˆB ˆn) = 8 B = 980 G ±[00] : ˆn = ±ê cos θ = ˆB ˆn) = 8 B = 980 G hese also look pretty good 5