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

Chapter 4: Direction Dependence of Spin Relaxation and Diffusive-Ballistic Crossover 69
as presented in Eq.(3.41). Expanding the Cooperon to second order in (Q+2eA+2m e âS)
and performing the angular integral which is for 2D diffusion (elastic mean free path l e
smaller than wire width W) continuous from 0 to 2π, yields:
Ĉ(Q) =
1
D e (Q+2eA+2eA S ) 2 +H γD
. (4.4)
The effective vector potential due to SO interaction is A S = m eˆαS/e, where ˆα = 〈â〉 is
averaged over angle. The SO term H γD , which cannot be rewritten as a vector potential, is
in our case due to the appearance of cubic Dresselhaus SOC.
4.1.1 Example
Toget anideaoftheprocedurewerecall thesituation presentedinRef.[Ket07] and
Chapter3 and take up the remark about the additional term due to cubic Dresselhaus SOC,
Eq.(3.45) concerning the width dependence: Starting with the Cooperon Hamiltonian, in
the case of Rashba and lin. and cubic Dresselhaus SOC,
with the effective vector potential
with ˜α 1 = α 1 −m e γ D E F /2,
H c := Ĉ−1
= (Q+2eA S ) 2 +(m 2 e
D E Fγ D ) 2 (Sx 2 +S2 y ), (4.5)
e
A S = m e
e ˆαS = m e
e
⎛
⎝ −˜α 1 −α 2 0
α 2 ˜α 1 0
⎛ ⎞
⎞ S x
⎠⎜
⎝ S y
⎟
⎠ , (4.6)
S z
it can be easily shown that the Hamiltonian Eq.(4.5) has only non vanishing eigenvalues
due to the last term in Eq.(4.5), which is due to cubic Dresselhaus SOC, if we assume no
boundaries except the lateral confinement. This term is neither reduced by reason of the
boundaryin the diffusive case, as has been shown in the previous chapter for quantum wires
in [100] direction and will be extended now to other directions, too. However two triplet
eigenvalues of this term depend on the wire width,
E QD1 = q2 s3
2 , (4.7)
(
E QD2,3 = q2 s3 3
2 2 ± sin(Q )
SOW)
, (4.8)
2Q SO W
with q 2 s3 /2 = (m2 e E Fγ D ) 2 . In the following we are going to diagonalize the whole Hamiltonian
and change the direction of the wire in the (001) plane.