Spin readout of a single electron in a double quantum dot Farzad - PowerPoint PPT Presentation

Single electron in DQD Spin readout of a single electron in a double quantum dot Farzad Qassemi EPIQ, Univ de Sherbrooke Nov 5, 2013 Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Outline Single and double quantum dots Coherent spin manipulation in DQD Stationary current for spinless and spinful charge Readout using satellite peaks Collaboration: Julien Camirand Lemyre, Michel Pioro-Ladriere Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Single electron in double quantum dot 200 nm Figure: SEM picture of DQD in Michel’s lab Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Coulomb Blockade H QD = Un ( n − 1 ) − enV g U = e 2 � k B T C g Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Single electron in double dot Left-Right charging energy Ω H 0 = ǫ ( | L �� L | − | R �� R | ) Q Left-Right coherent coupling H Q = Ω( | L �� R | + | R �� L | ) Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Eigen-energies Zeeman splitting H 0 = b z σ z , b z = g µ B ( B zL + B zR ) Z Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Eigen-energies Zeeman splitting H 0 = b z σ z , b z = g µ B ( B zL + B zR ) Z Unperturbed Hamiltonian H 0 = ǫ d z + Ω d x + b z σ z DQD Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Eigen-energies Zeeman splitting H 0 = b z σ z , b z = g µ B ( B zL + B zR ) Z Unperturbed Hamiltonian H 0 = ǫ d z + Ω d x + b z σ z DQD Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Coherent spin rotation H Z = δ b x σ x d z , δ b x = g µ B ( B xL − B xR ) / 2 � δ b x B 0 x z � � B R B L e- Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Double quantum dot H DQD = ǫ d z + Ω d x + b z σ z + δ b x σ x d z B L B R Ω n L n R V R V L Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Effective Hamiltonian e S He − S ≈ ǫ d z + b z σ z + Ω d x + Ω δ b x ˜ H = σ x d x b z [ b z σ z , S ] = δ b x σ x ⇒ S ∝ δ b x σ y d z b z Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Effective Hamiltonian e S He − S ≈ ǫ d z + b z σ z + Ω d x + Ω δ b x ˜ H = σ x d x b z [ b z σ z , S ] = δ b x σ x ⇒ S ∝ δ b x σ y d z b z tunneling assisted spin-flip processes ( ǫ ∼ b z ) 1 1 Ω δ b x B z Ω δ b x E 2 3 2 3 � − δ b x Ω Ω − δ b x 4 4 Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Transport regime Double quantum dot is coupled to the source and drain (1,1) (1,0) A C B (0,1) (0,0) Manipulate Readout Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Spinless case ρ a ˙ = Γ L ρ 0 + i Ω( ρ ab − ρ ba ) ρ b ˙ = − Γ R ρ b − i Ω( ρ ab − ρ ba ) − Γ R ρ ab ˙ = 2 ρ ab + i ǫρ ab + i Ω( ρ a − ρ b ) − Γ R ρ ba ˙ = 2 ρ ba − i ǫρ ba − i Ω( ρ a − ρ b ) Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Stationary Current: spinless case Ω 2 Γ R ¯ I = ǫ 2 + (Γ R / 2 ) 2 + Ω 2 ( 2 + Γ R / Γ L ) Γ L = Γ R = 0 . 01, Ω = 0 . 1 Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Stationary current: spinful case b z = 5 δ b x = Ω = 0 . 5 Γ R = 10 3 Γ L = 0 . 01 Γ L = Γ R = 0 . 01 Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Can we use it as a readout? b z = 5 , Ω = δ b x = 0 . 5 , Γ R = Γ L = 0 . 01 Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Analytical expression for satellite peak 4 Ω 2 δ b 2 Γ R x b 2 ¯ I = , Γ R , Γ L ≪ δ b x ∼ Ω ≪ ǫ ∼ b z z ǫ 2 + 4 Ω 2 δ b 2 ( 3 + Γ R / Γ L ) x b 2 z b z = 5 , Ω = δ b x = 0 . 5 , Γ R = Γ L = 0 . 01 Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Optimal point Lorentzian Heights ¯ I ( 0 ) = Γ R / ( 2 + Γ R / Γ L ) ¯ I ( ǫ = b z ) = Γ R / ( 3 + Γ R / Γ L ) ¯ I ( 0 ) ∼ ¯ I ( ǫ = b z ) Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Optimal point Lorentzian Heights ¯ I ( 0 ) = Γ R / ( 2 + Γ R / Γ L ) ¯ I ( ǫ = b z ) = Γ R / ( 3 + Γ R / Γ L ) ¯ I ( 0 ) ∼ ¯ I ( ǫ = b z ) Lorentzian Widths � δǫ | ǫ = 0 ∼ Ω ( 2 + Γ R / Γ L ) δǫ | ǫ = b z ∼ Ω δ b x � ( 3 + Γ R / Γ L ) b z � best results Ω ( 2 + Γ R / Γ L ) ∼ b z which gives δǫ | ǫ = b z ∼ δ b x . Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Summary and Outlook Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Summary and Outlook Single spin readout 1 We have theoretically analyzed the possibility of new spin readout 2 Extending our model to include the effect of decoherence and relaxation 3 Realizing our theoretical prediction in real experiment (undergoing in Michel’s lab) Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Spinful case Diagonal elements ( {| 1 � , | 2 � , | 3 � , | 4 �} ≡ {| L ↓� , | L ↑� , | R ↓� , | R ↑�} ) ρ 1 ˙ = Γ L ρ 0 + i δ b x ( ρ 12 − ρ 21 ) + i Ω( ρ 13 − ρ 31 ) ρ 2 ˙ = Γ L ρ 0 − i δ b x ( ρ 12 − ρ 21 ) + i Ω( ρ 24 − ρ 42 ) ρ 3 ˙ = − Γ R ρ 3 − i δ b x ( ρ 34 − ρ 43 ) − i Ω( ρ 13 − ρ 31 ) ρ 4 ˙ = − Γ R ρ 4 + i δ b x ( ρ 34 − ρ 43 ) − i Ω( ρ 24 − ρ 42 ) Stationary solution ( ˙ ρ = 0) ¯ 2 Γ L ¯ ρ 0 = 2 Ω ℑ (¯ ρ 13 ) + 2 Ω ℑ (¯ ρ 24 ) Γ R ¯ ρ 3 = 2 δ b x ℑ (¯ ρ 34 ) + 2 Ω ℑ (¯ ρ 13 ) Γ R ¯ ρ 4 = − 2 δ b x ℑ (¯ ρ 34 ) + 2 Ω ℑ (¯ ρ 24 ) Sanity check: ¯ ρ 0 = ¯ I R = Γ R (¯ ρ 3 + ¯ ρ 4 ) = 2 Γ L ¯ I L Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD Spinful Case Off-diagonal elements ( E ij = E i − E j ) ρ 12 ˙ = iE 12 ρ 12 + i δ b x ( ρ 1 − ρ 2 ) − i Ω ρ ∗ 23 + i Ω ρ 14 iE 13 ρ 13 + i Ω( ρ 1 − ρ 3 ) − i δ b x ρ 23 + i δ b x ρ 14 − Γ R ρ 13 ˙ = 2 ρ 13 iE 14 ρ 14 + i Ω( ρ 12 − ρ 34 ) + i δ b x ( ρ 13 − ρ 24 ) − Γ R ρ 14 ˙ = 2 ρ 14 34 ) − i δ b x ( ρ 24 + ρ 13 ) − Γ R ρ 23 ˙ = iE 23 ρ 23 + i Ω( ρ ∗ 12 − ρ ∗ 2 ρ 23 iE 24 ρ 24 + i Ω( ρ 2 − ρ 4 ) − i δ b x ρ 23 + i δ b x ρ 14 − Γ R ρ 24 ˙ = 2 ρ 24 ρ 34 ˙ = iE 34 ρ 34 + i Ω ρ ∗ 23 − i ωρ 14 − i δ b x ( ρ 3 − ρ 4 ) − Γ R ρ 34 Farzad Qassemi Spin readout of a single electron in a double quantum dot