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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 cqj /
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 )
FSOZ
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 )} eiZ 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
xp 
Interacting with a wide range
of phonon
Δ𝑝1 ∼ ℏ/2Δ𝑥1
Phonon frequency
2
Interacting with a small range
of phonon,
Decoherence is suppressed.
xp 
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 cqj /
Displacement operator, adding momentum “𝑞”
Creating (annihilating) a phonon with momentum “−𝑞” (“𝑞”)
Momentum conservation
xp 
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.