Rashba spin-orbit coupling and spin relaxation in silicon quantum wells
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m
at/0401615v2 [cond-mat.mes-hall] 27 Jul 2004
T
1T
2
k·p
Step graded SiGeSi CapSi Cap
Doped SiGe donor layer
(P or Sb)
Strained Si 2DEG
SiGe virtual substrateStrained Si QW (empty)28
Strained Si 2DEG backgate
SiGe barrier/spacer
Donor layer
SiGe virtual substrateStep graded SiGe
electron
qubit
induced
charge12-201025204 (nm)
4
< 60
6-10
< 30 (nm)SiGe barrier/spacer
E
ESchottky gates
Device 1 (D1)Device 2 (D2)SiGe barrier/spacer
E
z∼106
E
z>105
E
z≈en
s
4πǫ
0ǫ
Sid2=3×105,
d=20
3×107
g
0=2.00232
∆g
≈−0.003∆g
⊥≈−0.004∆g
g
Ge=1.4
∆gV
H
SO=zz
H2D
R=α(p
xσ
y−p
yσ
x)∝E
z(σ×p)
z,
E
c
E
z
α
α=2PP
z∆
d
2E
v1E
v2
1
E
v2
eE
z,
zP=X|p
x|S/imP
z=Z|p
z|S/im
∆
d=0.044E
v1=3.1
E
v2=7
m
PP
z
PP
z
2mP2
(z)/2
≈22
α≈1.66×10−6
E
z
k=0
α≈5.94αe/=
0.55/√
4πns
02040608022.533.544.55x 10-7
Time (s)
Magnetic field angle, θT
1
T
2
n
s=3x10 m-2
µ=9 m2
/Vs
E=2.3 MV/m
α=3.8 m/s(a)
0204060801.41.61.822.22.42.62.833.2x 10-7
Time (s)
Magnetic field angle, θT
1
T
2
n
s=4x10 m-2
µ=5 m2
/Vs
E=3.1 MV/mα=5.1 m/s(b)
1515
0204060800.511.52x 10-7
Time (s)
Magnetic field angle, θT
1
T
2n
s
µ=20 m2
/Vs
E=3.1 MV/m
α=5.1 m/s(c)
02040608000.20.40.60.811.21.41.6x 10-7
Time (s)
Magnetic field angle, θT
1
T
2n
s
µ=50 m2
/Vs
E=3.8 MV/m
α=6.4 m/s(d)
=4x10 m-215
=4x10 m-215
θ=0
α
H2D
R=α(p
xσ
y−p
yσ
x)
pε
p=ε
Fp′ε
p′=ε
F,ε
F
T
2,T
1,2×2ρσ
i=Tr{σ
iρ}.
H,dρ/dt=i
n =
9x10
8
7
6
5
13
24
Mobility, µ (m /Vs)T (µsec)-2s
α = 1 m/s
2
2010203040500102030m15
T
2α=1
µn
sα2T
2(α)=T
2(α=1)/α2
B=0.33
H
1
ρ
int=exp(−iH
0t/)ρexp(iH
0t/),
dρ
int
ρ
int,Hint
1(t)
,
Hint
1(t)=exp(iH
0t/)H
1exp(−iH
0t/).
ρ
int(t)=ρ
int(0)+i
dt=i
2
t
0dt′
ρ
int(0),Hint
1(t′
)
,Hint
1(t)
.
zH
0=ω
cσ
z/2.
Hint
1(t)
ss′′=
i=x,yhi
(t)σi
ss′′
i
dt
2=
s′′
s′′′ρ
ss′′(0)
Hint
1(t′
)
s′′s′′′
Hint
1(t)
s′′′s′ei(s′′
−s′′′
)t′
ei(s′′′
−s′
)t
+
s′′s′′′
Hint
1(t)
ss′′ei(s−s′′
)t
Hint
1(t)
s′′s′′′ei(s′′
−s′′′
)t′
ρ
s′′′s′(0)
−
s′′s′′′
Hint
1(t′
)
ss′′ei(s−s′′
)t′
ρint
s′′s′′′(0)
Hint
1(t)
s′′′s′ei(s′′′
−s′
)t
−
s′′s′′′
Hint
1(t)
ss′′ei(s−s′′
)t
ρint
s′′s′′′(0)ei(s′′′
−s′
)t′
Hint
1(t′
)
s′′′s′,est
=eǫ
st
.
hi(t)hj(t′)σi
ss′′σj
s′′s′′′
=
i=x,y,zχi
(t−t′
)σi
ss′′σi
s′′s′′′,
τ=t−t′χi
(τ)=
−2
dρ
++
−2
dρ
+−
dt=d[Tr(σz
ρ)]
dt(ρ
++−ρ
−−)
=2
i
2
[χx
(ω
c)+χy
(ω
c)]σ
z
1/T
1=2[χx
(ω
c)+χy
(ω
c)]/2
.
dσx
dt=d1/T
2(θ)=
cos2
θχx
(ω
L)+χy
(ω
L)+2sin2
θχx
(0)
/2
.
xt=0hx
,
hxexp(−t/τ
p)
2α2
p2
Fe−t/τ
p,
χx,y
(ω)=α2
p2
Fτ2pTBz
2=420
TB⊥z
2=140
TBz
2=
TB⊥z
2=
µ∼92
n
s∼3×1015−2TBz
2TB⊥z
2
TBz
1TB⊥z
1
TBz
2TB⊥z
2
TBz
1TB⊥z
1
TBz
2=µ
TB⊥z
2=
TBz
2=
TB⊥z
2=
µ∼52
n
s∼4×1015−2TBz
2TB⊥z
2
TBz
2TB⊥z
2
TBz
1TB⊥z
1
TBz
2=105
TB⊥z
2=49
TBz
2=
TB⊥z
2=
T
2=2/√
