3.4 Wave Functions: Analytical Form

As SO couples two opposite spin projections on two different valleys, one has to use the unitary transformation as described below in order to decouple the spins with opposite direction in those valleys. The four-component wave functions in the two-valley two spin projection basis can thus be straightforwardly obtained as described in [72148].

The four basis functions X1, X1, X2, and X2 for the two [001] valleys with spin up, spin down are transformed acccording to the steps in [129]. After these transformations, the spin quantization is oriented along the axis of the spin-orbit field. One assumes

      ∘ --------
kxy =   k2x + k2y.
(3.7)

       [                                      ]
ψ  =  1- (X   + X  ′) + (X    + X ′ ) ⋅ kx --iky ,
  1   2    1↑     2↑      1↓     2↓     kxy
(3.8a)

       [                                      ]
      1                               kx - iky
ψ2 =  -- (X1↑ + X2 ′↑) - (X1 ↓ + X2′↓) ⋅-------- ,
      2                                 kxy
(3.8b)

       [                                      ]
ψ  =  1- (X   - X  ′) + (X    - X ′ ) ⋅ kx --iky ,
  3   2    1↑     2↑      1↓     2↓     kxy
(3.8c)

       [                                      ]
      1                               kx - iky
ψ4 =  2- (X1 ↑ - X2 ′↑) - (X1 ↓ - X2 ′↓) ⋅--k----- .
                                         xy
(3.8d)

            (  )          (   )
X  =  ψ cos   γ- -  iψ  sin  γ- ,
  1    1      2       3     2
(3.9a)

            ( γ)          ( γ )
X2 =  ψ2cos   -- +  iψ4 sin  -- ,
              2             2
(3.9b)

            (  )          (   )
X3 =  ψ3cos   γ- -  iψ1 sin  γ- ,
              2             2
(3.9c)

            (  )          (   )
X4 =  ψ4cos   γ- +  iψ2 sin  γ- .
              2             2
(3.9d)

Here, X1, X2, X3, and X4 are the new rotated basis, and the angle γ is given by,

            △SO  ⋅ kxy
tan (γ ) = --------ℏ2kxky.
          D εxy -  M
(3.10)

Under these conditions, the new Hamiltonian can be written as

     [          ]
H  =   H1   H3   ,
       H3   H2
(3.11)

with

         [ 2 2     2  2    2                 ]
Hj=1,2 =  ℏ-kz-+  ℏ-(kx +-k-y)+ (- 1)jδ + ˜U (z ) I,
          2ml        2mt
(3.12)

      ⌊   2              ⌋
         ℏ-k0kz-    0
H   = |⌈    ml            |⌉ .
  3               ℏ2k0kz-
            0      ml
(3.13)

The wave functions corresponding to the Hamiltonian Equation 3.11 can be expressed as Φ1, Φ1, Φ2, and Φ2.

Wave Functions with Arbitrary Spin Orientation


     

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Figure 3.5: Sketch showing the spin injection into a (001) thin silicon film of thickness t in an arbitrary direction, described by the polar angle Θ.



PIC

Figure 3.6: The large component of the wave function of the lowest unprimed subband in an unstrained film located in the valley centered at k0 is shown. Θ = π
3 (c.f. Figure 3.5) is maintained.



PIC

Figure 3.7: The absolute value of the large (small) component of the spin wave functions reduces (increases), when the spin injection changes gradually from OZ- to OX-direction. kx, ky, and εxy are set to be 0.4nm-1, 0.4nm-1, and 0.5% respectively.


By considering Θ as the polar angle (c.f. Figure 3.5, and also the azimuthal angle by Φ), a set of linear transformations can be performed to obtain the wave functions Ψ1, Ψ1, Ψ2, and Ψ2 with spin along the injection orientation. This transformation is described below [149].

       (                                )
         cos Θ-  sin Θ-
Ψn ↑ =   -√--2-+ -√--2 ⋅ exp (- i(ϕ1 - Φ)) ⋅ Φn↑
            2       2
       (     Θ       Θ                  )
         cos-2-  sin-2-
     +    √ 2  -  √ 2  ⋅ exp(- i(ϕ1 - Φ)) ⋅ Φn ↓.
(3.14)

       (       Θ-      Θ-                 )
Ψn ↓ =   --s√in-2-+ co√s-2-⋅ exp (- i(ϕ1 - Φ )) ⋅ Φn↑
             2        2
       (                                  )
         --sin-Θ2-  cos-Θ2-
     +     √2--  -  √2-- ⋅ exp (- i(ϕ1 - Φ)) ⋅ Φn↓.
(3.15)

Here, tan ϕ1 = -ky
kx. Thus, when Θ=0 and the spin is orientated along the OZ-axis and the up(down)-spin wave function in each subband consists of majority and minority components [150] respectively. Their absolute values depend on εxy. The small components of the wave functions are the result of the SO term [150]. When k=0, the up(down)-spin states become eigenstates, which means the small component will be completely absent. Only when kx0 or ky0 the small component will be non-vanishing. As depicted in Figure 3.6 and for Θ = π
3, the large components of the four-components’ wave functions (Ψ1,1) can be described by the envelope quantization function

       exp(ik0z)-sin(πzt )
Ψ1,1 =        √t        ⋅ sin(Θ ),
(3.16)

where t signifies the sample thickness.

The small component is considerably suppressed by the application of the shear strain εxy [150]. The vanishing values of the small components decrease the spin mixing between the states with opposite spin projections, which causes the increment of the spin lifetime with εxy. Indeed, the large components of the wave functions do not change significantly with shear strain [150].

The spin injection orientation also has a strong effect on the components of the wave functions. For a fixed in-plane wave vector (kx, ky) and at every stress point, the absolute value of the majority (minority) component reduces (increases), when the spin injection direction is changed gradually from perpendicular (OZ-axis) to in-plane (say, OX-axis). This phenomenon is depicted in Figure 3.7, and has an impact on the spin relaxation as will be explained later.