Itinerant Spin Dynamics in Structures of ... - Jacobs University

70 Chapter 4: Direction Dependence of Spin Relaxation and Diffusive-Ballistic Crossover
4.2 Spin Relaxation anisotropy in the (001) system
4.2.1 2D system
We rotate the system in-plane through the angle θ (the angle θ = π/4 is equivalent
to [110]). This does not effect the Rashba term but changes the Dresselhaus one to[CWd07,
WJW10]
1
γ D
H D[001] = σ y k y cos(2θ)(〈k 2 z 〉−k2 x )−σ xk x cos(2θ)(〈k 2 z 〉−k2 y )
−σ y k x
1
2 sin(2θ)(k2 x −k2 y −2〈k2 z 〉)
1
+σ x k y
2 sin(2θ)(k2 x −k2 y +2〈k2 z 〉), (4.9)
with the wave vectors k i . The resulting Cooperon Hamiltonian, including Rashba and
Dresselhaus SOC, reads then
H c = (Q x +α x1 S x +(α x2 −q 2 )S y ) 2 +(Q y +(α x2 +q 2 )S x −α x1 S y ) 2 + q2 s3
2 (S2 x +S 2 y),
where we set
(4.10)
q 2 s3
2 = ( m 2 eE F γ D
) 2, (4.11)
α x1 = 1 2 m eγ D cos(2θ)((m e v) 2 −4〈kz 2 〉), (4.12)
α x2 = − 1 2 m eγ D sin(2θ)((m e v) 2 −4〈kz〉) 2 (4.13)
( )
=
q 1 −√
q
2
s3
2
sin(2θ) (4.14)
= 2m e˜α 1 sin(2θ), (4.15)
with q 1 = 2m e α 1 , q 2 = 2m e α 2 . We see that the part of the Hamiltonian which cannot be
written as a vector field and is due to cubic Dresselhaus SOC does not depend on the wire
direction in the (001) plane.
Special case: Only lin. Dresselhaus SOC equal to Rashba SOC
As a special example for the 2D case we set q s3 = 0 and q 1 = q 2 . To simplify
the search for vanishing spin relaxation we go to polar coordinates. Applying free wave