(a) What is the origin of the weak localization effect?

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1 1 Problem 1: Weak Localization a) Wat is te origin of te weak localization effect? Weak localization arises from constructive quantum interference in a disordered solid. Tis gives rise to a quantum mecanical correction to te classical teory of conduction, te Drude formula. Te origin is te enanced quantum mecanical probability for an electron to return to its initial position. Tis is because a particle can return to its origin by following a clock-wise or counter-clock-wise loop. Te quantum mecanical pases acquired along tese two loops are identical. Terefore, te quantum mecanical probability for returning is twice te classical probability for return. As a consequence, weak localization enances te resistivity and makes te system sligtly more localized. b) Wy is te effect called weak localization? Te effects is a small correction to te resistivity tat is a precursor to complete, or strong [Anderson], localization. It is terefore called weak localization. c) Wy will a magnetic field affect te weak localization effect? Weak localization is caused by an increased probability for an electron to be back scattered. Tis is because a particle tat returns to its origin can originate in two time-reversed pats tat ave exactly te same quantum mecanical pases. Te quantum mecanical probability of returning is terefore twice te classical probability of returning. A magnetic field breaks time reversal symmetry, induces a pase difference between te two time-reversed pats, reduces te return probability, and enances te conductivity. Problem 2: Te Quantum Hall Effect Consider a two-dimensional electron gas in te x-y plane, were tere is a transverse armonic potential of te form V y) = mω 2 0 y2 /2 and a magnetic field B = Bz applied along te z-direction. Te Scrödinger equation is [ 2 + iē ) 2 2m A + 1 ] 2 mω20y 2 ψx, y) = Eψx, y). 1) Coose te Landau gauge were te electromagnetic vector potential is A = yb, 0, 0).

2 2 a) Use te Scrödinger equation 1) and te ansatz ψ k,n x, y) = ϕ n y) exp ikx to sow tat te equation for ϕ n is [ ] d2 du 2 + u K)2 + R 2 u 2 ϕ n u) = ϵ k,n ϕ n u), 2) were we ave introduced te dimensionless variables u = y/l b, R = ω 0 /ω c, ϵ = E/ ω c /2), K = kl B. Additionally, l B = /eb)) 1/2 is te magnetic lengt, and ω c = eb/m is te cyclotron frequency. By using te Landau gauge, te Scrödinger equation 1) can be written as [ 2 2m x iey ) 2 B 2 2 2m y ] 2 mω2 0y 2 ψx, y) = Eψx, y). 3) Wit ψ k,n x, y) = ϕ n y) exp ikx, / x ik so tat x iey ) 2 B ik iey ) 2 B = k ey ) 2 B. 4) We also make use of 2 1 2m l 2 B = 1 2 eb m = 1 2 ω c. 5) Te first term in te Hamiltonian appearing in Eq. 3 is ten Te second term is and te tird term is 2 k ey ) 2 2m B = 2 2m l 2 B 1 l 2 B kl B y ) 2 = 1 l 2 ω c kl B y B l B 2 ) 2. 6) 2 1 2m y/l B ) 2 = 1 2 ω c y/l B ) 2 7) mω2 0y 2 = 1 2 mω2 c l 2 b ) y 2 ω0 2 l B ωc 2 = 2 ω 1 ) y 2 ω0 2 c l B ωc 2. 8) Wit y = l B u and k = K/l B we find Eq. 2 wit ϵ = E/ ω c /2), qed. b) Demonstrate tat Eq. 2 can be re-written into te form of an equation for a particle in a armonic potential and tat te solution in terms of te original variables is ψ k,n x, y) = e ikx ϕ n y L 2 Bk), 9) were ϕ n y) is a armonic oscillator function tat satisfies 2 d 2 2m dy ) 2 mω2 y 2 ϕ n y) = Ω n + 1 ) ϕ n y), 10) 2

3 3 L 2 B = ω2 c ωc 2 + ω0 2 lb 2 11) and te eigenenergy for te eigenstate ψ k,n x, y) is E k,n = Ωn ) + 2 k 2 were Ω = ω 2 c + ω 2 0 )1/2, and M B = mω 2 /ω M B, 12) Te potential energy increases at most as a quadratic function in te transverse coordinate u, so we can re-write te potential terms as u K) 2 + R 2 u 2 = 1 + R 2) u K ) R 2 + K 2 K2 1 + R 2 13) Tis corresponds to a displaced armonic oscillator potential term quadratic in coordinate) wit a re-defined energy: [ d2 du R2 )u K ] 1 + R 2 )2 ϕ n u) = ϵ k,n R2 K 2 ) 1 + R 2 ϕ n u) 14) Let us now return to te original variables: u 1 + R 2 = 1 + ω2 0 ω 2 c K 1 + R 2 = y kl B l B Ω 2 ω2 c = 1 y l B 1 2 ω c ϵ ω2 0 ω 2 ) c ωc 2 Ω 2 l Bk 2 = E 1 2 eb m Tis implies tat te energy levels are eb ω2 c = Ω2 ω 2 c ω ω2 c ω 2 0 kl 2 B, 15) ) ω0 2 + k 2 = E 2 ω2 c 2m ω0 2 + ω2 c = 1 ) y L 2 l Bk, 16) B ω 2 0 k 2 = E 2 k 2 2M B. 17) E k,n = Ω n + 1 ) + 2 k 2, 18) 2 2M B were n = 0, 1, 2, and k is a continuum number. Te armonic oscillator functions are displaced and ave arguments y L 2 B k. c) Compute te particle current density j = ) ψ m Im ψ + e m A ψ 2 19) for te eigenstate ψ k,n tat we found above. Wat is te sign of te current density along te x-direction, j x?

4 4 Let us first consider te current along te y-direction. Since A y = 0, Im e ikx ϕ n y L 2 Bk) ) y eikx ϕ n y L 2 Bk) = 0, 20) and because te armonic oscillator functions are real, we find tat j y = 0. Second, we consider te current along te x-direction. In tis case we ave A x = yb and we make use of ) Im e ikx ϕ n x eikx ϕ n = kϕ 2 n 21) so tat j x = k m eyb ) ϕ 2ny) ) = ω c y l 2 m Bk ϕ 2 ny L 2 Bk) 22) Since ϕ 2 ny L 2 B k) is always positive, j x 0 wen y > l 2 B k and j x 0 wen y < l 2 B k irrespective of te quantum number n and te frequencies ω c and ω 0. Problem 3: Te Landauer-Büttiker formalism Te Landauer-Büttiker formula for te conductance is G = e2 T n, 23) n were T n is te transmission probability for transverse wave guide mode n and te sum is over transverse wave guide modes. a) Consider a one-dimensional system only one wave guide mode) wit a left and rigt reservoir wit cemical potentials µ L and µ R, respectively, suc tat ev = µ L µ R. Find arguments for ow te current sould be expressed in terms of te velocity vϵ) of an electron at energy ϵ, te density of states in one dimension Nϵ) = 2/[vϵ)], te transmission probability T ϵ), and te distribution functions in te left and te rigt reservoirs fϵ µ L ) and fϵ µ R ). Derive from tese arguments te Landauer-Büttiker conductance in te linear response regime te bias voltage is muc smaller tan te Fermi energy) at zero temperature, Eq. 23 wit only one transverse wave guide mode, G = e 2 /)T. Te probability of finding an electron at energy ϵ in te left rigt) reservoir is determined by te Fermi-Dirac distributiuon function fϵ µ L ) fϵ µ R )).

5 5 Te current consists of rigt-going and lef-going particles. Te probability tat an electron will move from te left reservoir to te rigt reservoir is P l r ϵ) = fϵ µ l ) [1 fϵ µ R )] T ϵ). 24) Similarly, te probability tat an electron will move from te rigt reservoir to te rigt reservoir is P r l ϵ) = fϵ µ r ) [1 fϵ µ l )] T ϵ). 25) Te net current produced by electrons wit energy ϵ is Iϵ) = Nϵ)evϵ) [P l r P r l ]. 26) were Nϵ) = 2/[vϵ)] is te density of states in one dimension. Te total curent is ten I = 2e dϵ fϵ µ r ) fϵ µ l )) T ϵ) 27) We consider te linear response regime were ev = µ l µ r is small. We may ten expand and fϵ µ l ) fϵ µ 0 ) + µ l µ 0 ) fϵ µ ) 0) ϵ fϵ µ r ) fϵ µ 0 ) + µ r µ 0 ) fϵ µ ) 0) ϵ 28) 29) were µ 0 is te equilibrium cemical potential. At low temperatures fϵ µ ) 0) = δϵ µ 0 ) 30) ϵ so tat te total current becomes I = 2e2 T µ 0)V 31) and te conductance G = I/V is G = 2e2 T µ 0) 32) q.e.d.

6 6 b) We consider a narrow quantum wire tat is created in a two-dimensional electron gas suc tat te system is infinite in te x-direction, but as a finite widt L in te y-direction. Te potential is assumed to be infinite outside te wire were te wave function vanises. We assume tere is no scattering in te wire and transport is ballistic. Sow tat te conductance is G = 2e2 [ ] LkF, 33) π were [ ] represents te integral part of te number, e.g. [2.1] = 2. Te solution can be found as follows. Te energy of te system as tree contributions arising from te motion along te x and y directions. Te particle is free to move along x and te energy associated wit tis motion is E x. Along te transverse direction y, te wave function must represent standing waves of te form ψy) = A sin nπy/l, were A is a normalization constants and n = 0, 1, 2, 3,. Te total energy of te system is ten ) E = E x + 2 π 2 n 2. 34) 2m L Te wave guide modes are only propagating wen E x > 0. Transport is governed by electrons at te Fermi energy E F modes are determined by = 2 kf 2 /2m). Tis implies tat te propagating ) E F 2 π 2 n 2 35) 2m L so tat ) LkF n. 36) π Since transport is ballistic, te transmission probability T n = 1 for all propagating modes. Keeping in mind tat te quantum number n is an integral number, te Landauer-Buttiker condutance is terefore Problem 4: Magnetoresistance G = 2e2 [ ] LkF. 37) π a) Wat do te abbrevations AMR, GMR, and TMR mean? Wat is te cause of AMR, GMR, and TMR?

7 7 AMR - Anisotropic magnetoresistance. GMR - giant magnetoresistance. TMR - tunnel magnetoresistance. Wen a current passes troug a ferromagnet, te resistance depends on te relative direction of te current and te magnetization. Tis anistropic magnetoresistance is caused by te spin-orbit coupling. Giant magnetoresistance occurs in metallic ybrid systems of ferromagnets and normal metals. Te resistance depends on te relative orientation of two or more ferromagnets and is caused by spin-dependent scattering in te ferromagnets. Electrons aligned or anti-aligned to te magnetization experience different potentials and ave different conductances. Tunnel magnetoresistance occurs in transport troug tunnel junctions between ferromagnets. Te current depends on te relative orientation of te ferromagnerts and is caused by spin-dependent tunneling.