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

Chapter 5: Spin Hall Effect 89
1.0
1.0
0.5
0.5
ΣsH e 8Π
0.0
σsH/(e/8π)
0.0
0.5
0.5
1.0
4 2 0 2 4
EFermi
1.0
4 2 0 2 4
E/t
(a)
(b)
Figure 5.2: (a) SHC σ SH as a function of Fermi energy E F , in a clean system of size
L 2 = 170×170 with both Rashba and linear Dresselhaus SOC with α 2 > α 1 (blue curve)
and α 2 < α 1 (red curve). (b) SHC σ SH as a function of Fermi energy E F , in a clean system
of size V = 150×150 with only Rashba SOC of strength α 2 = 0.8t (blue/solid), α 2 = 1.4t
(red/dotted) and α 2 = 2t (yellow/dashed).
where we used Eq.(5.20). The level broadening η is introduced for regularization in finitesize
systems [NSSM05, MMF08]. It is chosen to be of the order of the level spacing δE,
vanishing in the thermodynamic limit. Integration over the Brillouin zone we finally arrive
at the clean solution for the SHC which is plotted in Fig.(5.2).
Results and Discussion
TheSHC shows electron-hole symmetry: TheSHCvanishes at half-filling, E F = 0,
and is an odd function of E F . From Eq.(5.18) one can see that the sign-change is due to
the term which is proportional to the SOC strength. It is worth noticing that an evaluation
of the commutator [r,H] in Eq.(5.13) will lead to
v = − i m e
∂ R½2×2 +α 2 (σ x e y −σ y e x )+α 1 (σ y e y −σ x e x ) (5.24)
(compare for pure Rashba case e.g. with [SCN + 04]). A straight forward tight binding
formulation of the velocity operator in this form would yield, in contrast to Eq.(5.16), onsite
matrix elements proportional to the spin-orbit interaction which could yield unphysical
results: The contribution of the velocity operator to the SHC would give 〈+|v y |−〉 =
iα 2 sin(k x )/∆(k). Obviouslythemissingcos(k y )factor leads toaneven andthereforewrong
SHC function of E F . Because the linear in momentum SO interactions can be interpreted