Why s orbital does split up in spin orbit coupling?
Answers
Hso=2/3Δ0L⋅σ
where L and σ are the angular momentum and Pauli spin operators, respectively.
Also given in Fig. 3 is the double group notation for the spin–orbit split bands as well as the lowest conduction bands [Γ2′(Γ1)] in diamond- and zinc blende–type semiconductors. Because of the J = 3/2 degeneracy, the valence bands have warped energy surfaces.
The fundamental direct gap, that is, the energy difference between the Γ6−Γ6conduction and Γ8−Γ8 valence bands, is denoted as E0.·The spin–orbit splitting is Δ0, and E0 + Δ0 is the transition energy between the conduction band and the spin–orbit split Γ7+Γ7 bands.
A strain with a uniaxial component splits the J = 3/2 multiplet into a pair of degenerate Kramers doublets [1–6]. The three valence bands for the case of a compressive uniaxial strain are shown schematically in Fig. 3, where the bands are labeled v1, v2, and v3. In addition, the hydrostatic component of the strain will shift the energy gap between the valence bands and the lowest-lying conduction band. The transition between the conduction band and the vi valence band is denoted E0(i), where i = 1, 2, 3. The removal of the J = 3/2 degeneracy produces energy surfaces that are ellipsoids of revolution around the uniaxial strain axis.
It has been shown [1–6] that the strain Hamiltonian Hє(v) for a p-like multiplet can be expressed as
(2)Hєν=−aνtrє−3bLx2−13L2єxx+cp−3dLxLy+LyLxєxy+cpwhere tr(є) is the trace of the strain matrix є, єij denotes the components of the strain tensor, and cp denotes cyclic permutation with respect to the indices x, y, and z. The quantity av represents the intraband (absolute) shift of the orbital valence bands due to the hydrostatic component of the stress (intraband or absolute hydrostatic deformation potential), while b and d are uniaxial deformation potentials appropriate to strains of tetragonal and rhombohedral symmetries, respectively.
At k = 0 the conduction band minima for the diamond- and zinc blende–type solids (except for Si and “zero-bandgap” materials such as α-Sn and HgTe) are an antibonding s-state with symmetry, Γ2′(Γ1). The effect of a strain is to produce a hydrostatic shift given by
(3)Hєc=actrєwhere ac is the intraband (absolute) hydrostatic deformation potential of the Γ2′(Γ1) conduction band [1–6].
We take our wave functions in the |J, MJ〉 representation that makes Hso diagonal. For the s-like Γ2′(Γ1) conduction band [3–5],
(4a)1/2,1/2c=S↑(4b)1/2,−1/2c=S↓whereas the p-like Γ8+Γ8 and Γ7+Γ7valence bands can be written as
(4c)3/2,3/2=1/2X+iY↑(4d)3/2,1/2=1/6X+iY↓−2/3Z↑(4e)3/2,−1/2=−1/6X−iY↑−2/3Z↓(4f)3/2,−3/2=−1/2X−iY↓(4g)1/2,1/2=1/3X+iY↓+1/3Z↑(4h)1/2,−1/2=1/3X−iY↑−1/3Z↓At the BZ center, the matrix elements connecting the Γ2′(Γ1) conduction band with the Γ8+Γ8 and Γ7+Γ7 valence bands are zero.
The strain-dependent Hamiltonian matrix of the valence band (neglecting the small strain-dependent spin–orbit splitting terms) is
(5)3/2,±3/2m3/2,±1/2m1/2,±1/2m−avtrє¯–δES,vm000−avtrє¯–δES,vm2δES,vm02δES,vm−avtrє¯–Δ0where m denotes either the [001] or [111] direction [1–5]. Since the strain does not remove the Kramers degeneracy, there is a similar expression for the |S ↓ 〉, |J, –MJ〉m manifold.
Solving the Hamiltonian in Eq. (5) yields the following for the stress dependence of the v2, v1, and v3 bands:
(6a)δEv2=δEH,vm+δES,vm(6b)δEv1=δEH,vm+Δ0−δES,vm/2−1/2Δ02+2Δ0δES,vm+9δES,vm21/2(6c)δEv3=δEH,vm+Δ0−δES,vm/2+1/2Δ02+2Δ0δES,vm+9δES,vm21/2The band v2 corresponds to |3/2, ± 3/2〉m, whereas v1 and v3 are straininduced linear combinations of |3/2, ± 1/2〉m and |1/2, ± 1/2〉m [3–5].
Combining Eqs. (3), (4a), (4b), (6a), (6b), and (6c), one can write the following for the stress dependence of the E0(2), E0(1), and E0(3) bandgaps:
(7a)E02=E0+δEHm+δES,vm(7b)E01=E0+δEHm+Δ0−δES,vm/2−1/2Δ02+2Δ0δES,vm+9δES,vm21/2(7c)E01=E0+δEHm+Δ0−δES,vm/2+1/2Δ02+2Δ0δES,vm+9δES,vm21/2