p268-ps3solutions-v3 - Phys 268r Fall 2013 Problem Set 3 Solutions B Halperin Solutions to Problem Set 3 Problem 1 We start in a reference frame moving

# p268-ps3solutions-v3 - Phys 268r Fall 2013 Problem Set 3...

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Phys 268r Fall 2013 Problem Set 3 Solutions B. Halperin Solutions to Problem Set 3 Problem 1. We start in a reference frame moving with the lattice at a velocity ~v = d ~ R/dt , (so that the problem looks stationary), and mark it as the primed frame. In this frame, the current will be the standard quantized Hall current: ~ j 0 = ν e 2 h ~ E 0 × ˆ z. where ~ E 0 is the electric field in the moving frame. This may be obtained from the standard Lorentz Transformations (letting ~ β = ~v/c ): ~ E 0 = γ ~ E + ~ β × ~ B ~ B 0 = γ ~ B - ~ β × ~ E . We have already discarded higher order terms in β , which we assume small, so we approxi- mate γ = 1 Further, the assumption that v D is small implies E/B small. Thus ~ β × ~ E is small, and ~ B 0 = ~ B . Consequently, we may assume that ν is the same in the moving frame as it would have been for the system at rest. We assume here that ~ B is oriented along the z-axis, and we define ρ B = B/ Φ 0 = eB/hc . We thus find: ~ j 0 = eνρ B c ~ E 0 × ~ B B 2 (1.1) = eνρ B c ~ E + ~ β × ~ B × ~ B B 2 (1.2) = eνρ B ~v D + eνρ B ~v × ~ B × ~ B B 2 (1.3) = eνρ B ~v D - eνρ B ~v (1.4) where in the last step, we used the vector triple product identity, and the fact that ~ B · ~v = 0. Finally, we transform back to the lab frame. To do this, we note that: ~ j = ~ j 0 + el ~v Combining this with our previous result, and using ρ el = νρ B + ηρ V we see that: ~ j = eνρ B ~v D + eηρ V ~v as claimed. 1
Phys 268r Fall 2013 Problem Set 3 Solutions B. Halperin Problem 2. We proceed similar to the derivation of σ xy in class. We first calculate expectation value of the space-integrated current < J > , induced by a small time-dependent perturbation δH due to δX ( t ). To first order in the perturbation due to δX , we may write our states as: | ψ ( t ) > = | 0 > + | δψ > t where | δψ > t = - i X N 6 =0 | N > Z t -∞ < N | ∂H 0 ∂X | 0 > δX ( t 0 ) e i ( E N - E 0 ) t 0 dt 0 . We have: < J > = < 0 | J | 0 > + < 0 | J | δψ > + < δψ | J | 0 > + .... The first term on RHS is 0 as discussed in class, since there is no electric field in the unperturbed Hamiltonian. We next calculate the second term, noting that the third is just its complex conjugate (which we will add in later). We also note that the J operator can be written as ∂H/∂k . Let us now assume that δX ( t ) = Re[ λe iωt + γt ], where γ is a positive infinitesimal, and λ is small. We thus have: < 0 | J | δψ > = - ie X N < 0 | ∂H ∂k | N >< N | ∂H ∂X | 0 > Z t -∞ δX ( t 0 ) e i ( E N - E 0 ) t 0 dt 0 = - ie X N < 0 | ∂H ∂k | N >< N | ∂H ∂X | 0 > Z t -∞ λe - iωt 0 + λ * e iωt 0 2 e i ( E N - E 0 ) t 0 dt 0 = - ie X N < 0 | ∂H ∂k | N >< N | ∂H ∂X | 0 > λe - iωt 2 i ( ω - E N + E 0 ) + λ * e iωt 2 i ( ω + E N - E 0 ) = e dX dt X N < 0 | ∂H ∂k | N >< N | ∂H ∂X | 0 > i ( E N - E 0 ) 2 , Using the fact that h δψ | J | 0 i = h 0 | J | δψ i * , we obtain an expression relating h J i to λ and λ * . In the limit of ω 0, we can simplify that expression by noting that λe - iωt 2 i ( ω - E N + E 0 ) + λ * e iωt 2 i ( ω + E N - E 0 ) - c.c. iωλe - iωt ( E N - E 0 ) 2 - c.c. as was noted in class. We also note that iωλe - iωt - c.c. = - 2 i dX/dt. 2
Phys 268r Fall 2013 Problem Set 3 Solutions B. Halperin We thus find < J > = ie dX dt X N 6 =0 < 0 | ∂H ∂k | N >< N | ∂H ∂X | 0 > - ( k X ) ( E N - E 0 ) 2 !

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• Fall '13
• BertrandI.Halperin
• Physics, Trigraph, Chern class, B. Halperin