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1 Δσ = 2e 2 2π D e Q α,β =± C αββα,ω =0 (Q), α, β = ± ˆ C ǫ F τ 1 ǫ F ˆ C ω =E E (Q = p + p )= τ 1 q E, p + q E , p q 1 . ˆ C Q ˆ C ω =0 (Q) ˆ C (Q)=(D e (Q +2eA +2eA S ) 2 + H γ ) 1 (001) α 2 ˜ α 1 = α 1 m e γ D E F /2 A S = m e e ˆ a.S = m e e ˜ α 1 α 2 0 α 2 ˜ α 1 0 .S, H γ =(m 2 e γǫ F ) 2 (S 2 x + S 2 y ). H c := ˆ C 1 D e physical meaning? ˆ C H c = U HU (v F |∇ r s |) 10= t s + 1 ˆ τ s s D e 2 s + γ (B 2τ (v F )B SO (k)) × s 0= t s + D e H s B =0 1 τ sij = τγ 2 B SO (k) 2 δ ij −〈B SO (k) i B SO (k) j . n · j s i | ± W 2 =0, j s i = τ v F (B (k) × s) i D e s i . U U A 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 K x E/DQ 2 SO (d) QW=30 SO E min,2D k=0 U A τ D e n ·〈v F ((B (k) · S))〉− in U A U A CU A | =0 U A (n · 2eA S in ) U A ˜ C | =0, in ˜ C | =0 H c n θ =0: n e y E 1/2,min = 3 2 q 2 s3 2 + q 2 sm q 2 s3 2 ( α 2 x1 + α 2 x2 q 2 2 ) 2 24q 2 sm W 2 , E 3,min = q 2 s3 2 + q 2 s3 2 + q 2 sm ( α 2 x1 + α 2 x2 q 2 2 ) 2 12q 2 sm W 2 q sm = (α x2 q 2 ) 2 + α 2 x1 α x1 = 1 2 m e γ D cos(2θ )((m e v ) 2 4k 2 z ), α x2 = 1 2 m e γ D sin(2θ )((m e v ) 2 4k 2 z ), q 2 =2m e α 2 ,q 2 s3 /2=(m 2 e ǫ F γ D ) 2 . θ =0 q 2 W = 30 K x θ W α x1 (θ = 0) = q 2 θ = 1 2 arcsin 2k 2 z (m e γ D ) 2 ((m e v ) 2 2k 2 z ) q 2 2 ( m 3 e v 2 γ D 4k 2 z m e γ D ) q 2 . W 2 0 Π 4 Π 2 3 Π 4 Π 0.0 0.5 1.0 1.5 2.0 0 Π 4 Π 2 3 Π 4 Π 0 1 θ q 1 /q 2 W 2 (θ ) α x1 (θ = 0) = q 2 q 1 /q 2 =1.63 θ = (1/4+ n)π, n q 1 < (qs3/ 2) q 1 =(qs3/ 2) θ = (3/4+ n)π, n q s3 =0.9 H c (θ ) E 1 (k x = 0) = q 2 s3 + q 2 sm (α 2 x1 + α 2 x2 q 2 2 ) 2 + qs3 2 2 q 2 s 12 W 2 , |S = 1; m = 0= (| ↑↓〉 + | ↓↑〉)/ 2 | ˆ =(0, 1, 0) T 0 Π 4 Π 2 3 Π 4 Π 5 Π 4 3 Π 2 7 Π 4 2 Π 1.0 0.5 5.0 10.0 50.0 0.4 0.3 0.2 0.1 0. θ T 2 D e q 2 2 T 2 [001] (D e q 2 2 ) q s3 /q 2 W =0.4/q 2 q s3 =0 T 2 θ = (1/4+ n)π, n T 2 =1/(4q 2 2 D e ) z k z = k 3 z =0 k 2 z = |∇φ| 2 dz H [110] = γ D σ z k x 1 2 k 2 z 〉− 1 2 (k 2 x 2k 2 y ) . q 2 =2m e α 2 C 1 D e = ( Q x ˜ q 1 S z q 2 S y ) 2 +(Q y + q 2 S x ) 2 + ˜ q 2 3 2 S 2 z , ˜ q 1 =2m e γ D 2 k 2 z 〉− γ D 2 m e E F 2 , ˜ q 3 = (3m e E 2 F (γ D /2)). (i∂ y +2m e α 2 S x )C x, y = ± W 2 =0, x k x = ± Δ 24 (24 (q 2 W ) 2 ) 1 D e τ s = q 2 1 + q 2 2 ) 24 (q 2 W ) 2 . = k x =0 W L L l e < W l 1D l 1D N = k F W/π m e 1 C 1 D e =2f 1 Q y +2α 2 S x +2 a 1 γ D v 2 f 3 f 1 S y 2 +2f 2 Q x 2α 2 S y 2 a 1 γ D v 2 f 3 f 2 S x 2 +8γ 2 D v 4 f 4 f 2 3 f 2 S 2 x + f 5 f 2 3 f 1 S 2 y , f i 5 10 15 20 25 30 0.0 0.2 0.4 0.6 0.8 1.0 3. 2.5 2. 1.5 1. 0.5 0. N E min 2 F γ 2 D Q =0 N = k F W/π α 2 /q s q s = (2α 2 ) 2 +(γv 2 F /2 2α 1 ) 2

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Dire tion Dependen e of Spin Relaxation and Diffusive-Ballisti CrossoverP. Wenk∗, S. Kettemann†

∗S hool of Engineering and S ien e, Ja obs University Bremen, Bremen 28759, Germany†S hool of Engineering and S ien e, Ja obs University Bremen, Bremen 28759, Germany, and Asia Pa i� Center for Theoreti al Physi s and Division of Advan ed MaterialsS ien e Pohang University of S ien e and Te hnology (POSTECH) San31, Hyoja-dong, Nam-gu, Pohang 790-784, South Korea

Abstra tThe dependen e of spin relaxation on the dire tion of the quantumwire under Rashba and Dresselhaus (linear and ubi ) spin orbit ou-pling (SOC) is studied using the Cooperon equation. Comprising thedimensional redu tion of the wire in the di�usive regime, the lowestspin relaxation and dephasing rates for (001) and (110) systems arefound. The analysis of spin relaxation redu tion is then extended tonon-di�usive wires where it is shown that, in ontrast to the theoryof dimensional rossover from weak lo alization to weak antilo aliza-tion in di�usive wires, the relaxation due to ubi Dresselhaus spinorbit oupling is redu ed and the linear part shifted with the numberof transverse hannels.[1�3℄We set ~ ≡ 1.

Cooperon and Spin Di�usionThe weak lo alization orre tion to the ondu tivity is given by

∆σ = −2e2

DeVol.∑Q

α,β=±Cαββα,ω=0(Q), (1)

where α, β = ± are the spin indi es, and the Cooperon propagator Cis for ǫF τ ≫ 1 (ǫF , Fermi energy), given byCω=E−E ′(Q = p + p′) = τ

1−∑

q

PSfrag repla ementsE,p + q

E′,p′−q

−1

. (2)Expanding C to lowest order in the generalized momentum Q leads to

Cω=0(Q) ≡ C(Q) = (De(Q + 2eA + 2eAS)2 +Hγ)

−1 (3)e.g. in GaAs (001), with the Rashba parameter α2 and the shiftedlinear Dresselhaus oupling α1 = α1 −meγDEF/2,AS =

me

ea.S =

me

e

(−α1 −α2 0α2 α1 0

)

.S, (4)Hγ = (m2

eγǫF )2(S2

x + S2y). (5)

2D spe trum of Hc :=C−1

Desplits in gapless singlet and triplet modes

physical

meaning?

The lo al orre tions given by C an be related to spin relaxation (re-quiring time reversal symmetry)Hc = UCDHSDU†CDwith the spin di�usion equation for (vF | ∇rs |) ≪ 1/τ

0 = ∂ts +1

τss

︸︷︷︸spin relax.−De∇2s + γ(B− 2τ〈(∇vF )BSO(k)〉)× s︸ ︷︷ ︸spin pre ession

0 = ∂ts +DeHSDs (6)with the D'yakonov-Perel' Spin Relaxation Tensor, B = 0

1

τsij= τγ2

(

〈BSO(k)2〉δij − 〈BSO(k)iBSO(k)j〉

)

.

Spin Relaxation Anisotropy in the (001) SystemConsider spe ular s atteringfrom spin- onserving bound-aries:

n · jsi|±W2

= 0,

jsi = − τ〈vF (BSO(k)× s)i〉−De∇si.Using UCD we �nd the BC forthe Cooperon and simplifythem by applying a se ondtransformation UA to simpleNeumann BC: 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Kx

E/DQ2 SO

(d)QW=30SO

Emin,2D

k=0

UA

Den · 〈vF ((BSO(k) · S))〉 − i∂n

)

U†AUACU

†A|Surfa e = 0

UA (n · 2eAS − i∂n)U†AC|Surfa e = 0,

−i∂nC|Surfa e = 0

Solving the Cooperon Hamiltonian Hc with BC for di�erent dire tionsn, (θ = 0 : n = ey), in the (001) plane gives the minima

E1/2,min =3

2

q2s32

+

(

q2sm − q2s32

) (α2x1 + α2x2 − q22

)2

24q2smW 2, (7)

E3,min =q2s32

+

(q2s32 + q2sm

) (α2x1 + α2x2 − q22

)2

12q2smW 2 (8)

where we set qsm =√

(αx2 − q2)2 + α2x1 andαx1 =

1

2meγD cos(2θ)((mev)

2 − 4〈k2z〉),

αx2 = − 1

2meγD sin(2θ)((mev)

2 − 4〈k2z〉),q2 = 2meα2, q2s3/2 = (m2

eǫFγD)2.In the Fig. above, the spe trum for θ = 0 at q2W = 30 is plotted. Theabsolute minimum at �nite wave ve tors Kx leads to long persistingspin heli es as shown in the inset.For a general θ we an dedu e about the minimal spin-relaxation ratethat Eq. (7) is independent of the widthW if αx1(θ = 0) = −q2 and/orthe dire tion of the wire is pointing in

θ =1

2arcsin

(

2〈k2z〉(meγD)2((mev)

2 − 2〈k2z〉)− q22(m3

ev2γD − 4〈k2z〉meγD

)q2

)

. (9)Analyzing the prefa tor of W 2 in the Eq. (8) gives the optimal angle asplotted in Fig. 1.

0 ���Π

4���Π

2 �������3 Π4

Π

0.0

0.5

1.0

1.5

2.0

0 ���Π

4���Π

2�������3 Π4 Π

0

1

θ

q 1/q 2

Figure 1: Dependen e of the W 2 oe� ient in Eq. (8) on the lateral rotation (θ).The absolute minimum is found for αx1(θ = 0) = −q2 (here: q1/q2 = 1.63) andfor di�erent SO strength we �nd the minimum at θ = (1/4 + n)π, n ∈ Z if q1 <(qs3/

√2) (dashed line: q1 = (qs3/

√2)) and at θ = (3/4 + n)π, n ∈ Z else. Herewe set qs3 = 0.9. The s aling is arbitrary.� Spin DephasingThe eigenve tor of Hc(θ) whi h has the eigenvalue

E1(kx = 0) = q2s3 + q2sm − (α2x1 + α2x2 − q22)2 + qs32

2 q2s12

W 2, (10)is the triplet state |S = 1;m = 0〉 = (| ↑↓〉 + | ↓↑〉)/

√2 ≡

| ⇉〉=(0, 1, 0)T . This is equal to the z- omponent of the spin densitywhose evolution is des ribed by the spin di�usion equation, Eq. (6).This gives an analyti al des ription (Fig. 2) of numeri al al ulationdone by J.Liu et al., Ref. 4.

0����

Π

4����

Π

2 ��������

3 Π

4

Π

��������

5 Π

4��������

3 Π

2��������

7 Π

4

2 Π

1.0

0.5

5.0

10.0

50.0

0.4

0.3

0.2

0.1

0.

θ

T2D

eq2 2

Figure 2: The spin dephasing time T2 of a spin initially oriented along the [001]dire tion in units of (Deq2

2) for the spe ial ase of equal Rashba and lin. DresselhausSOC. The di�erent urves show di�erent strength of ubi Dresselhaus in units of

qs3/q2. In the ase of �nite ubi Dresselhaus SOC we set W = 0.4/q2. If qs3 = 0:T2 diverges at θ = (1/4+n)π, n ∈ Z (dashed verti al lines). The horizontal dashedline indi ated the 2D spin dephasing time, T2 = 1/(4q2

2De).

Spin Relaxation Anisotropy in the (110) SystemThe Dresselhaus Hamiltonian with the on�nement in z ≡[110℄ dire -tion (〈kz〉 = 〈k3z〉 = 0, and 〈k2z〉 =

∫|∇φ|2dz) has the following form

H[110] = −γDσzkx

(1

2〈k2z〉 −

1

2(k2x − 2k2y)

)

. (11)In luding the Rashba SOC q2 = 2meα2, noting that its Hamiltoniandoes not depend on the orientation of the wire, we end up with

the following Cooperon HamiltonianC−1

De=(Qx − q1Sz − q2Sy

)2+ (Qy + q2Sx)

2 +q232S2z , (12)

with q1 = 2meγD2〈k2z〉 −

γD2

meEF

2, (13)

q3 = (3meE2F (γD/2)). (14)

⇒ In 2D states polarized in the z-dire tion have vanishing spin relax-ation as long as we have no Rashba SOC.� Spin Relaxation in the WireAgain we apply appropriate Neumann boundary ondition(−i∂y + 2meα2Sx)C

(

x, y = ±W

2

)

= 0, ∀x (15)and solve the Cooperon equation:� Spe ial ase: without ubi Dresselhaus SOCThe lowest spin relaxation rate is found at �nite wave ve torskxmin = ±∆

24(24− (q2W )2),1

Deτs=

(q21 + q22)

24(q2W )2. (16)

As in the 2D ase the spin relaxation rate vanishes for vanishingRashba SOC =⇒ there is no width dependen e of weak lo aliza-tion/weak antilo alization in the ase without Rashba SOC.If ubi Dresselhaus SOC annot be negle ted, the absolute minimumof spin relaxation an also shift to kxmin = 0 (see solution in Ref. 3).Di�usiv-Ballisti Crossover

For every dire tion in the di�usive (001) system there is stilla �nite spin relaxation, Eq. (7) and (8), due to ubi Dressel-haus SOC, at wire widths W ≪ LSO (spin pre ession lengthLSO). Experiments, e.g. the work by Kunihashi et al., Ref.5, however, show that in wires whi h do not ful�ll the ondi-tion le < W , the ubi Dresselhaus term is suppressed, too.In the right Fig. we show widthdependen e of the spin relaxationlength l1DSO of di�erent arrier den-sity[5℄. Solid lines and dashed linesshow the l1DSO al ulated from thetheory by S.K.[1℄, with negle ting ubi Dresselhaus term and tak-ing into a ount full SOIs, respe -tively.To explain this one has to assume a �nite number of transverse han-nels N = kFW/π in the q spa e, Eq. (2). The Cooperon to diagonalizereads then (me ≡ 1)

C−1

De= 2f1

(

Qy + 2α2Sx + 2

(

a1 − γDv2f3f1

)

Sy

)2

+ 2f2

(

Qx − 2α2Sy − 2

(

a1 − γDv2f3f2

)

Sx

)2

+ 8γ2Dv4

[(

f4 −f23f2

)

S2x +

(

f5 −f23f1

)

S2y

]

, (17)with fun tions fi whi h depend on N. The suppression of the ubi Dresselhaus term, also depending on Rashba SOC, is plotted in Fig. (3).

5 10 15 20 25 300.0

0.2

0.4

0.6

0.8

1.0

3.

2.5

2.

1.5

1.

0.5

0.

N

Em

in/ǫ2 F

γ2 D

Figure 3: The lowest eigenvalues of the on�ned Cooperon Hamiltonian Eq. (17),equivalent to the lowest spin relaxation rate, are shown for Q = 0 for di�erentnumber of modes N = kFW/π. Di�erent urves orrespond to di�erent values ofα2/qs, qs =√(2α2)2 + (γv2F/2− 2α1)2.Referen es[1℄ S. Kettemann, Phys. Rev. Lett. 98, 176808 (2007).[2℄ P. Wenk and S. Kettemann, Phys. Rev. B 81, 125309 (2010).[3℄ P. Wenk and S. Kettemann, Phys. Rev. B 83, 115301 (2011).[4℄ J. Liu, T. Last, E. Koop, S. Denega, B. van Wees, and C. van der Wal, J. Super ond. Nov.Magn. 23, 11 (2010).[5℄ Y. Kunihashi, M. Kohda, and J. Nitta, Phys. Rev. Lett. 102, 226601 (2009).