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

HQD = Un(n − 1) − enVg

U = e2 Cg kBT

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 H0

Q

= ǫ(|LL| − |RR|) Left-Right coherent coupling HQ = Ω(|LR| + |RL|)

Ω

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Eigen-energies

Zeeman splitting H0

Z

= bzσz, bz = gµB(BzL + BzR)

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Eigen-energies

Zeeman splitting H0

Z

= bzσz, bz = gµB(BzL + BzR) Unperturbed Hamiltonian H0

DQD

= ǫdz + Ωdx + bzσz

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Eigen-energies

Zeeman splitting H0

Z

= bzσz, bz = gµB(BzL + BzR) Unperturbed Hamiltonian H0

DQD

= ǫdz + Ωdx + bzσz

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Coherent spin rotation

HZ = δbxσxdz, δbx = gµB(BxL − BxR)/2

• B0
• BR
• BL

e-

z x

δbx

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Double quantum dot

HDQD = ǫdz + Ωdx + bzσz + δbxσxdz

VL VR

BR BL

nL nR

Ω

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Effective Hamiltonian

˜ H = eSHe−S ≈ ǫdz + bzσz + Ωdx + Ωδbx bz σxdx [bzσz, S] = δbxσx ⇒ S ∝ δbx bz σydz

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Effective Hamiltonian

˜ H = eSHe−S ≈ ǫdz + bzσz + Ωdx + Ωδbx bz σxdx [bzσz, S] = δbxσx ⇒ S ∝ δbx bz σydz tunneling assisted spin-flip processes (ǫ ∼ bz)

E

Ω Ω

δbx δbx

Ω Ω

−δbx Bz

• 1

1 3 3 2 2 4 4

−δbx

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

A B C (0,0) (0,1) (1,0) (1,1) Manipulate Readout

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Spinless case

˙ ρa = ΓLρ0 + iΩ(ρab − ρba) ˙ ρb = −ΓRρb − iΩ(ρab − ρba) ˙ ρab = −ΓR 2 ρab + iǫρab + iΩ(ρa − ρb) ˙ ρba = −ΓR 2 ρba − iǫρba − iΩ(ρa − ρb)

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Stationary Current: spinless case

¯ I = Ω2ΓR ǫ2 + (ΓR/2)2 + Ω2(2 + ΓR/ΓL) ΓL = ΓR = 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

bz = 5 δbx = Ω = 0.5 ΓL = ΓR = 0.01 ΓR = 103ΓL = 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?

bz = 5, Ω = δbx = 0.5, ΓR = ΓL = 0.01

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Analytical expression for satellite peak

¯ I = 4Ω2δb2

x

b2

z

ΓR ǫ2 + 4Ω2δb2

x

b2

z

(3 + ΓR/ΓL) , ΓR, ΓL ≪ δbx ∼ Ω ≪ ǫ ∼ bz bz = 5, Ω = δbx = 0.5, ΓR = ΓL = 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) = ΓR/(2 + ΓR/ΓL) ¯ I(ǫ = bz) = ΓR/(3 + ΓR/ΓL) ¯ I(0) ∼ ¯ I(ǫ = bz)

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Optimal point

Lorentzian Heights ¯ I(0) = ΓR/(2 + ΓR/ΓL) ¯ I(ǫ = bz) = ΓR/(3 + ΓR/ΓL) ¯ I(0) ∼ ¯ I(ǫ = bz) Lorentzian Widths δǫ|ǫ=0 ∼ Ω

• (2 + ΓR/ΓL)

δǫ|ǫ=bz ∼ Ωδbx

bz

• (3 + ΓR/ΓL)

best results Ω

• (2 + ΓR/ΓL) ∼ bz which gives δǫ|ǫ=bz ∼ δbx.

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 ↑}) ˙ ρ1 = ΓLρ0 + iδbx(ρ12 − ρ21) + iΩ(ρ13 − ρ31) ˙ ρ2 = ΓLρ0 − iδbx(ρ12 − ρ21) + iΩ(ρ24 − ρ42) ˙ ρ3 = −ΓRρ3 − iδbx(ρ34 − ρ43) − iΩ(ρ13 − ρ31) ˙ ρ4 = −ΓRρ4 + iδbx(ρ34 − ρ43) − iΩ(ρ24 − ρ42) Stationary solution ( ˙ ¯ ρ = 0) 2ΓL¯ ρ0 = 2Ωℑ(¯ ρ13) + 2Ωℑ(¯ ρ24) ΓR ¯ ρ3 = 2δbxℑ(¯ ρ34) + 2Ωℑ(¯ ρ13) ΓR ¯ ρ4 = −2δbxℑ(¯ ρ34) + 2Ωℑ(¯ ρ24) Sanity check: ¯ IR = ΓR(¯ ρ3 + ¯ ρ4) = 2ΓL¯ ρ0 = ¯ IL

Farzad Qassemi Spin readout of a single electron in a double quantum dot

Single electron in DQD

Spinful Case

Off-diagonal elements (Eij = Ei − Ej) ˙ ρ12 = iE12ρ12 + iδbx(ρ1 − ρ2) − iΩρ∗

23 + iΩρ14

˙ ρ13 = iE13ρ13 + iΩ(ρ1 − ρ3) − iδbxρ23 + iδbxρ14 − ΓR 2 ρ13 ˙ ρ14 = iE14ρ14 + iΩ(ρ12 − ρ34) + iδbx(ρ13 − ρ24) − ΓR 2 ρ14 ˙ ρ23 = iE23ρ23 + iΩ(ρ∗

12 − ρ∗ 34) − iδbx(ρ24 + ρ13) − ΓR

2 ρ23 ˙ ρ24 = iE24ρ24 + iΩ(ρ2 − ρ4) − iδbxρ23 + iδbxρ14 − ΓR 2 ρ24 ˙ ρ34 = iE34ρ34 + iΩρ∗

23 − iωρ14 − iδbx(ρ3 − ρ4) − ΓRρ34

Farzad Qassemi Spin readout of a single electron in a double quantum dot