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APS March Meeting 2015 Phonon mediated spin relaxation in a moving quantum dot: Doppler shift, Cherenkov radiation, and spin relaxation boom Xinyu Zhao1 Peihao Huang1,2 Xuedong Hu1 Department of Physics, University at Buffalo, SUNY 1 Department of Physics, California State University, Northridge 2 Supported by NSA/LPS through ARO For details, see arXiv:1503.00014 Moving spin qubit Spin qubits in double quantum dots. Figure is from Wiki R. P. G. McNeil et al., Nature 477, 439 (2011). Experimental realization of moving spin qubit – playing pingpong between two quantum dots. P. Huang and X. Hu, PRB 88, 075301 (2013) Spin qubit relaxation in a moving quantum dot In this work, we study the spin relaxation caused by the phonon environment Moving QD, Doppler effect Subsonic regime, Doppler shift Figure is from Internet Breaking the sound barrier, shock wave Figure is from Internet Can we observe these phenomenon in spin relaxation? Model Model H tot H d H Z H SO U ph (r ) U ph (r ) qj iq ·r F ( q z )e (e qj iq qj )(b†qj bqj ) 2 cqj / 2 1 * 2 H d * m d (r r0 ) 2 2m 2 1 H Z g B B0 · 2 H SO y x x y Total relaxation rate is an integration over 𝑑𝜃𝑑𝜙 2 1 d d f ( wz , , ), 0 T1 0 Kernel function (angular distribution in 𝑑θ𝑑𝜙 ) FSOZ f * 2 2 2 5 wz4 sin 3 cos 2 Cep Fz Fxy , j ( m d ) 8 c s j 𝑧 𝑑𝜃𝑑𝜙 𝑦 𝜃 𝜙 𝑥 For details, see arXiv:1503.00014 Subsonic to supersonic regime Kernel function (angular distribution in 𝑑θ𝑑𝜙) vs. 𝑣0 and 𝜙. 𝒗𝟎 > 𝒗𝒔 𝒗𝟎 = 𝒗𝒔 𝒗𝟎 < 𝒗𝒔 Transition from subsonic regime to supersonic regime. At 𝑣0 ≈ 𝑣1 , the phonon distribution is suddenly concentrated in one direction and then split into two branches. Supersonic regime: Cherenkov effect 𝑣0 = 6000 m/s 𝑧 𝑑𝜃𝑑𝜙 𝑦 𝜃 𝜙 𝑥 Angular distribution of the kernel function 𝑓(𝜃, 𝜙) Only the phonon from certain directions make obvious contribution to the relaxation. This is a characteristic feature of Cherenkov radiation. Quantum confinement Without quantum confinement Cherenkov angle is the same as the classical case 𝐴𝐶 vs C arccos v0 𝑣 𝑐𝑜𝑠𝜙𝐶 = 𝐴𝐵 = 𝑣𝑠 0 With quantum confinement Confinement causes phonon bottleneck effect The Cherenkov angle is slightly shifted 𝝓𝑪 = 𝐜𝐨𝐬 −𝟏 𝒗𝒔 = 𝟑𝟖∘ 𝒗𝟎 𝝓′𝑪 = 𝐜𝐨𝐬 −𝟏 𝒗𝒔 𝝎𝒁 − = 𝟒𝟎∘ 𝒗𝟎 𝟐𝒗𝟎 vs Z arccos v0 2v0 C Quantum correction on Cherenkov angle Spin relaxation boom Sonic boom vs. Spin relaxation boom Figure is from Wiki 1. For a single type of phonon, the relaxation curve is quite similar to the Prandtl-Glauert singularity. 2. The total relaxation (black, dash-dotted) is the combination of all types of phonon. 3. The position where the peak occurs is slightly shifted due to the quantum confinement effect. Spin relaxation for moving QD Spin relaxation rate as a function of magnetic field and moving velocity. For a strong 𝐵 field, decoherence is lower at high velocity. 1. A moving QD may have even lower relaxation rate than a static QD. 2. Decoherence can be suppressed by increasing moving velocity. In the right figure, the partial derivative with respect to 𝑣0 is plotted. In the white region (negative derivative), decoherence can be suppressed by increasing 𝑣0 . Summary In this work, we study the spin relaxation in a moving quantum dot. Several interesting features caused by Doppler effect are observed. Subsonic regime: Frequency shift, Doppler effect. Transonic regime: Breaking the sonic barrier, formation of the shock wave, spin relaxation boom vs. sonic boom. Supersonic regime: Cherenkov radiation of the phonons, A quantum correction on Cherenkov angle is given explicitly. arXiv:1503.00014 Cherenkov radiation Applications Phonon Detector Measuring environment is important in many quantum error correction schemes. Our results: 1. Phonon is concentrated in certain directions 2. Small correction of Cherenkov angle click These results are quite useful in measuring the environment. Feedback control operation Fluctuation of the Cherenkov angle reflects the fluctuation of the moving velocity. 𝛿𝜙 𝜙 e 𝑣0 + 𝛿𝑣 Direct detection of the decoherence rate !!! v arccos 1 Z v0 2v0 H eff 1 g B B B(t ) 2 B v 1 Re { B(0), B(t )} eiZ t dt T1 Schrieffer-Wolff transformation 2 1 * 2 2 H m ( r r ) H tot H d H Z H SO U ph (r ) d d 0 2m* 2 1 H Z g B B0 · H SO y x x y 2 [ H d H Z , S ] H SO H e S He S 1 H eff g B [ B0 B B (t )] , 2 2m* B ( v0 y , v0 x ,0), g B B(t ) 2 B0 (t ), (t ) | (r , t ) | , Eular rotation g 2 B2 Sij ( ) 4 2 𝑧 [ Bi (0), B j (t )] cos t dt 𝐵 𝑦 𝜃𝐵 𝜙𝐵 𝑥 1 S XX (Z ) SYY (Z ), T1 𝑧(𝑧 ′ ) 𝑧(z’) 𝐵 𝑦′ 𝜃𝐵 𝜙𝐵 𝑥 𝑦 𝑥′ 𝑦′ 𝐵 𝑦(𝑌) 𝜃𝐵 𝜃𝐵 𝜙𝐵 𝑥 𝑍 𝑋 𝜃𝐵 𝑥′ In or to make the correlation function in a diagonal form, an Eular rotation is needed 1. Rotate along z axis by angle 𝜙𝐵 𝑥𝑦𝑧 → 𝑥 ′ 𝑦 ′ 𝑧 ′ 2. Rotate along 𝑦′ axis by angle 𝜃𝐵 Finally, the 𝑍 axis of the reference frame after rotation (𝑋𝑌𝑍) is in the direction of the magnetic field 𝐵. Static QD Decoherence rate as a function of external magnetic field for static QD (𝑣0 = 0) Consistent with the results presented in Phys. Rev. Lett. 93, 016601 (2004). Quantum confinement Without quantum confinement Without considering cutoff function (quantum confinement), the kernel function is The Cherenkov angle is slightly shifted due to the quantum confinement effect. wz f wz4 Z 1- j 𝜔 The singularity occurs at 1−𝜉𝑍 → ∞, i.e., 𝑗 𝒗𝟏 𝝓𝑪 = 𝒂𝒓𝒄𝒄𝒐𝒔( ) 𝒗𝟎 This is just the angular relation in the Cherenkov radiation in optics With quantum confinement While considering cutoff function (quantum confinement) in x-y plane, the kernel function is wz2 2 f w exp( 2 / 2) vj 4 z 𝝓𝑪 = 𝒂𝒓𝒄𝒄𝒐𝒔 𝒗𝟏 = 𝟑𝟖∘ 𝒗𝟎 𝝓 = 𝒂𝒓𝒄𝒄𝒐𝒔 𝒗𝟏 𝝎𝒁 − = 𝟒𝟎∘ 𝒗𝟎 𝟐𝒗𝟎 𝜔 The “singularity” is slightly shifted from 1−𝜉𝑍 → ∞ to 𝑗 𝜕𝑓 𝜕𝑤𝑧 = 0, then, 𝑤𝑧 = 2𝑣1 𝜆 v arccos 1 Z v0 2v0 1. There is no singularity 2. The Cherenkov angle is slightly shifted My point: The spin relaxation depends on THREE major factors: 1. Moving velocity, reflected by Doppler effect 2. Magnetic field, determining the original Zeeman splitting 3. Quantum confinement, causing the phonon bottleneck effect Quantum confinement will affect the spin relaxation The position of spin relaxation boom depends on the B field There is a long tail for the LA phonon, it is better to plot the 3-D relaxation figure for single type of phonon separately. Quantum confinement will affect the spin relaxation boom (where the peak appears) wz Z 1- j My point: The spin relaxation depends on THREE major factors: 1. Moving velocity, reflected by Doppler effect 2. Magnetic field, determining the original Zeeman splitting 3. Quantum confinement, causing the phonon bottleneck effect Uncertainty relation Strong QD confinement high Weak QD confinement Δ𝑥1 is small Δ𝑝1 low xp Interacting with a wide range of phonon Δ𝑝1 ∼ ℏ/2Δ𝑥1 Phonon frequency 2 Interacting with a small range of phonon, Decoherence is suppressed. xp Δ𝑥2 is large Δ𝑝2 2 Δ𝑝2 ∼ ℏ/2Δ𝑥2 Uncertainty relation U ph (r ) Electron-phonon interaction qj Electron part Phonon part e iq ·r b†qj bqj iq ·r F ( q z )e (e qj iq qj )(b†qj bqj ) 2 cqj / Displacement operator, adding momentum “𝑞” Creating (annihilating) a phonon with momentum “−𝑞” (“𝑞”) Momentum conservation xp Strong confinement Δ𝑥1 is small Δ𝑝1 ∼ ℏ/2Δ𝑥1 low 2 Δ𝑝1 is large Interacting with a wide range of phonon high Phonon frequency Δ𝑝2 ∼ ℏ/2Δ𝑥2 Weak confinement Δ𝑥2 is large Interacting with a small range of phonon, Decoherence is suppressed. Δ𝑝2 is small Spin relaxation rate as a function of the size of the quantum dot in z direction.