h a p 1 a p 1 i\u03b3 \u0394 2 h X p pa p 1 a p 1 hc 121 Notice that this reduces

# H a p 1 a p 1 iγ δ 2 h x p pa p 1 a p 1 hc 121

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h a p 1 a p 1 + Δ 0 2 h X p> 0 [ pa p 1 a - p 1 + h.c.] . (1.21) Notice that this reduces precisely to the form of a spinless p-wave superconductor, which we have previously studied! On diagonalizing the above Hamiltonian, we get, E ± ( p ) = ± s p 2 2 m 1 - γ 2 m h - μ - h 2 + γ 2 Δ 2 0 p 2 4 h 2 . (1.22) We see that E ± ( p ) vanishes at μ * = - h at p = 0. In the vicinity of this point, if we expand μ = μ * + δμ , we obtain, E ± ( p ) = ± C | p | , where (1.23) C 2 = γ 2 Δ 2 0 4 h 2 - 1 - γ 2 m h δμ m . (1.24) 4
Phys 268r Fall 2013 Problem Set 5 Solutions - Corrected v2 B. Halperin ********************************************************************************* If you wish, you can also study the full pairing problem, H 1 + H 2 given by (1.18) . After using the explicit forms of Δ αβ p , it is straightforward to diagonalize H p , which yields the following four bands ( i = 1 , .., 4): E i ( p ) = ξ 2 p 1 + ξ 2 p 2 + | Δ 0 | 2 q ( ξ 2 p 1 - ξ 2 p 2 ) 2 + | Δ 0 | 2 [( ξ p 1 - ξ p 2 ) 2 + | Δ 0 | 2 ] 2 1 / 2 . (1.25) This can vanish iff ξ 2 p 1 + ξ 2 p 2 + | Δ 0 | 2 = q ( ξ 2 p 1 - ξ 2 p 2 ) 2 + | Δ 0 | 2 [( ξ p 1 - ξ p 2 ) 2 + | Δ 0 | 2 ] . (1.26) In the particular limit that we are interested in, ξ p 1 = p 2 2 m 1 - γ 2 m h - μ - h (= A - μ ) , (1.27) ξ p 2 = p 2 2 m 1 + γ 2 m h - μ + h (= B - μ ) . (1.28) Therefore, we need to find roots of the following equation for μ : ( A - μ ) 2 + ( B - μ ) 2 + | Δ 0 | 2 = p | Δ 0 | 4 + | Δ 0 | 2 ( A - B ) 2 + ( A - B ) 2 ( A + B - 2 μ ) 2 . (1.29) This yields 4 solutions: μ * = 1 2 A + B ± q ( A - B ) 2 - 2( | Δ 0 | 2 ± p | Δ 0 | 4 - ( A - B ) 2 | Δ 0 | 2 ) . (1.30) Problem 3. a) We solve this problem using the same logic we used to find the charge density for the Laughlin wave function with one species, and the (332) generalization for the ν = 2 / 5 state in a system with two species. We have now four species, with N particles of each, and a (non-normalized) trial wavefunc- tion: Ψ 0 { z jt } = Y j<k Y t ( z jt - z kt ) 3 ! × Y j,k Y t<t 0 ( z jt - z kt 0 ) 2 ! × Y j,t e -| z jt | 2 / 4 l 2 0 . Taking the absolute square of this wave function gives a probability distribution P 0 { z it } = Z - 1 e - H 0 (1.31) 5
Phys 268r Fall 2013 Problem Set 5 Solutions - Corrected v2 B. Halperin where Z is a normalization constant, and H 0 = 6 X j<k X t ln | z jt - z kt | + 4 X j,k X t<t 0 ln | z jt - z kt 0 | - πρ B X jt | z jt | 2 , (1.32) where ρ B = (2 πl 2 0 ) - 1 is the density of magnetic flux quanta. P 0 is the equilibrium distribution function for set of classical particles with a Hamiltonian H 0 at a temperature T = 1. Let ρ t ( ~ r ) be the probability density for finding a particle of species t at point ~ r in this equilibrium distribution. We may get a good estimate of this by using mean field theory (i.e., classical Hartree approximation or Debye-Huckel theory) ρ t ( ~ r ) = C t e - φ t ( ~ r ) (1.33) where C t is a normalization constant and φ t is the average potential felt by a particle of type t : φ t ( ~ r ) = - πρ B r 2 + Z d 2 r 0 " 6 ln[ z t - z 0 t ] ρ t ( ~ r 0 ) + 4 X t 0 6 = t ln[ z 1 - z 0 t ] ρ t 0 ( ~ r 0 ) # . (1.34) Since we are considering a large system, and φ t enters in an exponential, we see that φ t must be, to a good approximation, independent of position in any region where ρ t 6 = 0, that is φ t 0 in such regions.

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