Hall Viscosity of Hierarchical Quantum Hall States

Transcription

1 M. Fremling, T. H. Hansson, and J. Suorsa ; Phys. Rev. B Mikael Fremling Fysikum Stockholm University, Sweden Nordita 9 October 2014

2 Outline 1 Introduction Fractional Quantum Hall Eect 2 Quantum Hall Viscosity 3 Wave Functions on the Torus 4 Summary

3 The Hall Eect Edwin Hall: 1879 Transverse resistance: R H = V H I B.

4 The Quantum Hall Eect 2D interface, low temperatures, clean samples, high magnetic elds Klitzing, Dorda & Pepper 80' Stormer 92' Plateaus at R H = 1 ν h e 2, with a rational number ν = p q

5 Electrons in a Magnetic Field: Landau Levels Landau Hamiltonian H = 1 2m (ˆp j ea j ) 2 j=x,y Harmonic oscillator solution H = ω(a a ) ω = eb, l mc B = eb Each energy level is called a Landau Level (LL)

6 Many-body Hamiltonian for the FQHE Fractional Quantum Hall Eect: Rational lling, ν = 1 3, 2 5, 3 7, Flat bands Kinetic energy zero. H Int = i<j V (r i r j)

7 Many-body Hamiltonian for the FQHE Fractional Quantum Hall Eect: Rational lling, ν = 1 3, 2 5, 3 7, Flat bands Kinetic energy zero. H Int = i<j V (r i r j) No small parameter! Numerically hard, as the electrons are strongly interacting

8 Trial Wave Function for the FQHE Use representative wave functions for particular ν. Laughlin: ν = 1 q Laughlin 83' ψ 1 q = e 1 4 i z i 2 i<j (z i z j ) q

9 Trial Wave Function for the FQHE Use representative wave functions for particular ν. Laughlin: ν = 1 q Laughlin 83' ψ 1 q = e 1 4 i z i 2 i<j (z i z j ) q Moore-Read: ν = 5/2 Moore & Read 91' ψ MR = e 1 4 i z i 2 ( ) 1 (z i z j ) 2 Pf z i z j i<j Typically fractionally charged excitations and anyonic statistics!

10 Trial Wave Function for the FQHE II A whole zoo of dierent wave functions. Two layer version of Moore-Read: ν = 5/2 Cappelli, Georgiev & Todorov 01' ψ 5 2 = A e 1 4 i<j i( z i 2 + w i 2 ) i,j (z i z j ) 3 i<j (z i w j ) (w i w j ) 3

11 Trial Wave Function for the FQHE II A whole zoo of dierent wave functions. Two layer version of Moore-Read: ν = 5/2 Cappelli, Georgiev & Todorov 01' ψ 5 2 = A e 1 4 i( z i 2 + w i 2 ) i,j (z i w j ) i<j (z i z j ) 3 i<j (w i w j ) 3 Composite fermions: ν = 2 5 Jain 89' ψ 2 5 = A e 1 4 i<j i( z i 2 + w i 2 ) i (z i z j ) 3 i<j zi (z i w j ) 2 i,j (w i w j ) 3

12 Testing Wave Functions What kind of test can I perform on my trial wave functions? Compute energy expectation value E = d N r ψ (r)v (r)ψ(r) V

13 Testing Wave Functions What kind of test can I perform on my trial wave functions? Compute energy expectation value E = d N r ψ (r)v (r)ψ(r) Compute overlap with exact ground state φ 0 ψ φ 0 = d N r ψ (r)φ 0 (r) V V

14 Testing Wave Functions What kind of test can I perform on my trial wave functions? Compute energy expectation value E = d N r ψ (r)v (r)ψ(r) Compute overlap with exact ground state φ 0 ψ φ 0 = d N r ψ (r)φ 0 (r) V But, two trial wave functions Ψ A and Ψ B can not necessarily be compared directly. V

15 The Shift Filling fraction is a thermodynamic quantity ν = N e N Φ N e, N Φ For nite systems ν (N φ + S) = N e Shift S!

16 The Shift Filling fraction is a thermodynamic quantity ν = N e N Φ N e, N Φ For nite systems ν (N φ + S) = N e Shift S! Charged particles, curved space extra contribution to the eective magnetic eld

17 The Shift Filling fraction is a thermodynamic quantity ν = N e N Φ N e, N Φ For nite systems ν (N φ + S) = N e Shift S! Charged particles, curved space extra contribution to the eective magnetic eld The shift is a topological property of a FQH state. Filled lowest LL ν = 1 S = 1 Laughlin ν = 1 q S = q Composite Fermions ν = m 2m+1 S = 2 + m

18 The Sphere and Shift II All wave functions do not have the same shift. Shift can distinguish dierent wave functions!

19 The Sphere and Shift II All wave functions do not have the same shift. Shift can distinguish dierent wave functions! but ψ A and ψ B at dierent shifts can not be compared directly, since N φ and N e do not match!

20 The Sphere and Shift II All wave functions do not have the same shift. Shift can distinguish dierent wave functions! but ψ A and ψ B at dierent shifts can not be compared directly, since N φ and N e do not match! Solution: Choose geometry where S = 0. Shift is manifest though the quantum Hall viscosity η

21 A Flat Geometry Sphere:

22 A Flat Geometry Sphere: Curved geometry S 0

23 A Flat Geometry Sphere: Curved geometry S 0 Plane/Disc:

24 A Flat Geometry Sphere: Curved geometry S 0 Plane/Disc: Compactication at sphere same problem

25 A Flat Geometry Sphere: Curved geometry S 0 Plane/Disc: Compactication at sphere same problem Torus: Flat geometry S = 0 Easy access to geometry parameter τ

26 Ordinary Viscosity A reminder about ordinary viscosity: Viscosity thickness of a uid

27 Ordinary Viscosity A reminder about ordinary viscosity: Viscosity thickness of a uid Honey: η = 2 10 Pa s is thicker than water: η = mpa s.

28 Ordinary Viscosity A reminder about ordinary viscosity: Viscosity thickness of a uid Honey: η = 2 10 Pa s is thicker than water: η = mpa s. Small η Liquid nitrogen: (at 77 K) η = mpa s.

29 Ordinary Viscosity A reminder about ordinary viscosity: Viscosity thickness of a uid Honey: η = 2 10 Pa s is thicker than water: η = mpa s. Small η Liquid nitrogen: (at 77 K) η = mpa s. Large η Pitch tar: η = Pa s.

30 Ordinary Viscosity Viscosity η and elastic modulus λ, relate stress tensor σ to the strain u, and the strain rate u. σ α,β = γ,δ λ α,β,γ,δ u γ,δ + γ,δ η α,β,γ,δ u γ,δ α, β, γ, δ = 1,..., d in d dimensions. The viscosity tensor may be divided as η = η S + η A where η S α,β,γ,δ = η S γ,δ,α,β η A α,β,γ,δ = η A γ,δ,α,β

31 Normal Viscosity and Dissipation Normally one associates viscosity with dissipation,... and in ordinary three dimensions (or higher) that is natural.

32 Normal Viscosity and Dissipation Normally one associates viscosity with dissipation,... and in ordinary three dimensions (or higher) that is natural. The familiar viscous dissipation comes from the symmetric tensor η S. Force and displacement are parallel dissipation

33 Anti-symmetric Viscosity In all other dimensions than d = 2, isotropic uids will have η A = 0. However in two dimensions there is more to the story...

34 Anti-symmetric Viscosity In all other dimensions than d = 2, isotropic uids will have η A = 0. However in two dimensions there is more to the story... In d = 2 dimensions η A may be non-zero.... but time reversal symmetry has to be broken.... like in the quantum Halls system.

35 Anti-symmetric Viscosity In all other dimensions than d = 2, isotropic uids will have η A = 0. However in two dimensions there is more to the story... In d = 2 dimensions η A may be non-zero.... but time reversal symmetry has to be broken.... like in the quantum Halls system. Now, force and displacement are orthogonal, and no dissipation may occur. η A carries information about the shift S.

36 Analytic Quantum Hall Viscosity Viscosity has been computed for: The lled lowest Landau level: η = 1 n 4 Avron, Seiler & Zograf 95' S = 1

37 Analytic Quantum Hall Viscosity Viscosity has been computed for: The lled lowest Landau level: η = 1 4 n Avron, Seiler & Zograf 95' S = 1 The Laughlin ν = 1 q wave function: η = 1 4 nq Read 09' S = q.

38 Analytic Quantum Hall Viscosity Viscosity has been computed for: The lled lowest Landau level: η = 1 4 n Avron, Seiler & Zograf 95' S = 1 The Laughlin ν = 1 q wave function: η = 1 4 nq Read 09' S = q. The (conjectured) general relation is Read 09' η = 1 4 ns Viscosity can distinguish between dierent states.

39 Recapitulation Sphere: Curved geometry S 0 Plane/Disc: Curved geometry S 0

40 Recapitulation Sphere: Curved geometry S 0 Plane/Disc: Curved geometry S 0 Torus: Flat geometry S = 0 Easy access to geometry parameter τ

41 The Mathematical Torus The torus is periodic in two directions: L x and τl x = L + ıl y. Torus area: 2πN s l 2 = L x L y. Torus geometry: τ = τ 1 + ıτ 2 = L + ıl y L x τ 2 : Aspect ratio τ 1 : Skewness ratio N s : Number of ux quanta

42 The Complex τ-plane

43 The Complex τ-plane Fremling 13' τ 2 = 1 Low ψ 2 High

44 The Complex τ-plane Fremling 13' τ 2 = 1 τ 1 = Low ψ 2 High

45 Computing Viscosity Viscosity is a response to strain rate, such as a velocity gradient. Is related to a Berry phase under changes in geometry. F = ı τ A τ ı τ A τ η H = 2τ 2 2 A Torus F A Torus = L x L y = 2πN s l 2 B. Berry connection A µ = ı ϕ µ ϕ

46 Computing Viscosity Viscosity is a response to strain rate, such as a velocity gradient. Is related to a Berry phase under changes in geometry. F = ı τ A τ ı τ A τ η H = 2τ 2 2 A Torus F A Torus = L x L y = 2πN s l 2 B. Berry connection A µ = ı ϕ µ ϕ Numerically: Discretizing path in τ-space W = e ıa Ω F = e ı A µ(λ) dλ µ j ϕ j ϕ j+1 giving A Ω : Area of circle. η H = 2τ 2 2 A Torus I (W ) A Ω

47 Computing Viscosity from Diagonalization W j ϕ j ϕ j+1 ϕ n = k α (n) k k ϕ n+1 ϕ n = k ᾱ (n+1) α (n) k k

48 Computing Viscosity from Diagonalization W j ϕ j ϕ j+1 ϕ n = k α (n) k k ϕ n+1 ϕ n = k ᾱ (n+1) α (n) k k Extra piece with Fock basis η H = 2τ 2 2 I (W ) + 1 A Torus A Ω 4 n

49 Viscosity from Exact Diagonalization Numerically computed viscosity s = S/2 for the exact ground state Laughlin ν = 1/2, ν = 1/3 Read & Rezayii 11'

50 Viscosity from Exact Diagonalization Numerically computed viscosity s = S/2 for the exact ground state Laughlin ν = 1/2, ν = 1/3 Read & Rezayii 11' Moore - Read ν = 1, ν = 5/2 Read & Rezayii 11'

51 Viscosity from Exact Diagonalization Numerically computed viscosity s = S/2 for the exact ground state Laughlin ν = 1/2, ν = 1/3 Read & Rezayii 11' Moore - Read ν = 1, ν = 5/2 Read & Rezayii 11' Our contribution ν = 2/5 Fremling, Hansson & Suorsa 14'

52 Viscosity from Exact Diagonalization Numerically computed viscosity s = S/2 for the exact ground state Laughlin ν = 1/2, ν = 1/3 Read & Rezayii 11' Moore - Read ν = 1, ν = 5/2 Read & Rezayii 11' Our contribution ν = 2/5 Fremling, Hansson & Suorsa 14' But what about the trial wave functions?

53 Wave Function on the Torus Next, compute viscosity of trial wave function. Problem: Only a few known on the torus.

54 Wave Function on the Torus Next, compute viscosity of trial wave function. Problem: Only a few known on the torus. Simplest approach: Begin with Laughlin ν = 1 q ψ 1 ({z}) = e 1 4 q j z j 2 i<j on the plane (z i z j ) q

55 Wave Function on the Torus Next, compute viscosity of trial wave function. Problem: Only a few known on the torus. Simplest approach: Begin with Laughlin ν = 1 q ψ 1 ({z}) = e 1 4 q j z j 2 i<j on the plane (z i z j ) q Generalize to the torus by letting z ϑ 1 (z τ), New piece F s (Z), xed by boundary conditions. Haldane 85'

56 Wave Function on the Torus Next, compute viscosity of trial wave function. Problem: Only a few known on the torus. Simplest approach: Begin with Laughlin ν = 1 q ψ 1 ({z}) = e 1 4 q j z j 2 i<j on the plane (z i z j ) q Generalize to the torus by letting z ϑ 1 (z τ), New piece F s (Z), xed by boundary conditions. Haldane 85' ψ 1 ({z}) = e 1 4 q Z = i j z j 2 i<j z i F s (Z) = ϑ[ s q 0 ϑ 1 (z i z j τ) q F s (Z) ] (qz qτ)

57 The Conformal Field Theory (CFT) Construction I Simplest approach works for Laughlin, Moore-Read, but then dicult. The center of mass piece F is hard to guess. Idea: Use CFT to construct Laughlin wave function instead On the plane ψ 1 ({z}) = q N e O V (z j ) j=1 Electron operator V (z) =: e ı qφ(z) : O : Background operator φ(z) : Scalar boson

58 The Conformal Field Theory (CFT) Construction II Works also on the torus, but with some more work. Both chiralities in φ V (z, z) =: e ı qφ(z, z) : N e O V (z j, z j ) = F j=1 Ψ F ({z}) Ψ F ({ z}) The Laughlin wave function is a linear combination ψ 1 ({z}) = q F e ı2πλ F Ψ F ({z}) Yes, I'm hiding a lot of ugly details!

59 The Normalization Good news, can be done for more complicated states than Laughlin and Moore-Read. Eg. positive Jain series ν = 1 3, 2 5, 3 7, 4 9,... Essential Bonus: Gives control over the τ-dependence of the normalization constant. For Laughlin N 0 : τ-independent (assumption). η(τ) : Dedekind's η-function. N = N 0 ( τ 2 η(τ) 2) qn e/2

60 Wave Function for ν = 2 on the Torus 5 We have applied the CFT construction to the ν = 2 wave function 5 ψ 2 = A e 1 4 i( z i 2 + w i 2 ) zi (z i w j ) 2 5 i i,j i<j (z i z j ) 3 i<j (w i w j ) 3

61 Wave Function for ν = 2 on the Torus 5 We have applied the CFT construction to the ν = 2 wave function 5 ψ 2 = A e 1 4 i( z i 2 + w i 2 ) zi (z i w j ) 2 5 i i,j i<j (z i z j ) 3 i<j (w i w j ) 3 On the torus becomes ψ 2 5 = A m,n i<j D (z,τ) m,n e 1 4 i( z i 2 + w i 2 ) i,j ϑ 1 (z i z j τ) 3 i<j ϑ 1 (z i w j τ) 2 ϑ 1 (w i w j τ) 3 F (Z, W )

62 Wave Function for ν = 2 on the Torus 5 We have applied the CFT construction to the ν = 2 wave function 5 ψ 2 = A e 1 4 i( z i 2 + w i 2 ) zi (z i w j ) 2 5 i i,j i<j (z i z j ) 3 i<j (w i w j ) 3 On the torus becomes ψ 2 5 = A m,n i<j D (z,τ) m,n e 1 4 i( z i 2 + w i 2 ) i,j ϑ 1 (z i z j τ) 3 i<j ϑ 1 (z i w j τ) 2 ϑ 1 (w i w j τ) 3 F (Z, W ) D m,n : Translation operator on z F (Z, W ) : Center of mass function S = 4, η = 1 n 4 4

63 Overlap with Coulomb: The Whole τ-plane Dierent terms D m,n are dominant in dierent regions Red lines: Fundamental SL(2, Z) domain Black lines: Boundaries for the dominating D m,n The anzats for very well in the entire τ-plane. Above 99% overlap by taking the 8 most dominant terms. Fantastic!

64 Computing Viscosity from Real-space Wave Functions W j ϕ j ϕ j+1 ϕ n+1 ϕ n = Ω ϕ n+1(z)ϕ n (z) η H = 2τ 2 2 A Torus I (W ) A Ω

65 Computing Viscosity from Real-space Wave Functions W j ϕ j ϕ j+1 ϕ n+1 ϕ n = Ω ϕ n+1(z)ϕ n (z) Evaluated using Monte Carlo as ϕ n+1 ϕ n = 1 Z N m ϕ n (zm)ϕ n+1 (z m) p m Z N = m 1 p m η H = 2τ 2 2 A Torus I (W ) A Ω

66 Computing Viscosity from Real-space Wave Functions W j ϕ j ϕ j+1 ϕ n+1 ϕ n = Ω ϕ n+1(z)ϕ n (z) Evaluated using Monte Carlo as ϕ n+1 ϕ n = 1 Z N m ϕ n (zm)ϕ n+1 (z m) p m Z N = m 1 p m All MC integrals use the same set of zm for numerical stability η H = 2τ 2 2 A Torus I (W ) A Ω

67 Viscosity at ν = 2/5 We then numerically calculate s for a trial wave function we believe describes the ground state

68 Viscosity at ν = 2/5 We then numerically calculate s for a trial wave function we believe describes the ground state Again, s is not constant in the τ-plane, but converges on s = 2 as N φ.

69 Summary We have constructed torus wave functions for ν = 2/5 (among many other) The construction is valid for all τ We have qualitative understanding of with terms D m,mn are dominant We have computed the viscosity for these state and it coincided with the Coulomb ground state, and theoretical predictions.

70 Summary We have constructed torus wave functions for ν = 2/5 (among many other) The construction is valid for all τ We have qualitative understanding of with terms D m,mn are dominant We have computed the viscosity for these state and it coincided with the Coulomb ground state, and theoretical predictions. Thank you!