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

76 Chapter 4: Direction Dependence of Spin Relaxation and Diffusive-Ballistic Crossover
[HPB + 97]
( 1
H [110] = −γ D σ z k x
2 〈k2 z〉− 1 )
2 (k2 x −2ky)
2 . (4.35)
Including the Rashba SOC (q 2 ), noting that its Hamiltonian does not depend on the orientation
of the wire,[HPB + 97] we end up with the following Cooperon Hamiltonian
C −1
D e
= (Q x − ˜q 1 S z −q 2 S y ) 2 +(Q y +q 2 S x ) 2 + ˜q2 3
2 S2 z , (4.36)
with
˜q 1 = 2m e
γ D
2 〈k2 z〉− γ D
2
m e E F
2
, (4.37)
q 2 = 2m e α 2 , (4.38)
and ˜q 3 = (3m e E 2 F(γ D /2)). (4.39)
We see immediately that in the 2D case states polarized in the z-direction have vanishing
spin relaxation as long as we have no Rashba SOC. Compared with the (001) system the
constant term due to cubic Dresselhaus does not mix spin directions. Here we set the
appropriate Neumann boundary condition as follows:
(
(−i∂ y +2m e α 2 S x )C x,y = ± W )
= 0, ∀x. (4.40)
2
The presence of Rashba SOC adds a vector potential proportional to S x . Applying a nonabelian
gauge transformation as before to simplify the boundary condition, we diagonalize
the transformed Hamiltonian (App.(D.1)) up to second order in q 2 W in the 0-mode approximation.
4.3.1 Special case: without cubic Dresselhaus SOC
The spectrum is found to be
E 1 = k 2 x + 1
12 ∆2 (q 2 W) 2 , (4.41)
E 2,3 = k 2 x + 1
24 ∆2( 24−(q 2 W) 2)
± ∆ 24√
∆ 2 (q 2 W) 4 +4k 2 x (24−(q 2W) 2 ) 2 , (4.42)
with the lowest spin relaxation rate found at finite wave vectors k x min
= ± ∆ 24 (24−(q 2W) 2 ),
We set ∆ = √˜q 2 1 +q2 2 .
1
D e τ s
= ∆2
24 (q 2W) 2 . (4.43)