Spin–orbit interaction

The energy of a magnetic moment in a magnetic field is given by

$${\displaystyle \Delta H=-{\boldsymbol {\mu }}\cdot \mathbf {B} ,}$$

where μ is the magnetic moment of the particle, and B is the magnetic field it experiences.

We shall deal with the magnetic field first. Although in the rest frame of the nucleus, there is no magnetic field acting on the electron, there is one in the rest frame of the electron (see classical electromagnetism and special relativity). Ignoring for now that this frame is not inertial, we end up with the equation

$${\displaystyle \mathbf {B} =-{\frac {\mathbf {v} \times \mathbf {E} }{c^{2}}},}$$

where v is the velocity of the electron, and E is the electric field it travels through.[lower-alpha 1] Here, in the non-relativistic limit, we assume that the Lorentz factor ${\displaystyle \gamma \backsimeq 1}$. Now we know that E is radial, so we can rewrite ${\textstyle \mathbf {E} =\left|E\right|{\frac {\mathbf {r} }{r}}}$. Also we know that the momentum of the electron ${\displaystyle \mathbf {p} =m_{\text{e}}\mathbf {v} }$. Substituting this in and changing the order of the cross product gives

$${\displaystyle \mathbf {B} ={\frac {\mathbf {r} \times \mathbf {p} }{m_{\text{e}}c^{2}}}\left|{\frac {E}{r}}\right|.}$$

Next, we express the electric field as the gradient of the electric potential ${\displaystyle \mathbf {E} =-\nabla V}$. Here we make the central field approximation, that is, that the electrostatic potential is spherically symmetric, so is only a function of radius. This approximation is exact for hydrogen and hydrogen-like systems. Now we can say that

$${\displaystyle |E|=\left|{\frac {\partial V}{\partial r}}\right|={\frac {1}{e}}{\frac {\partial U(r)}{\partial r}},}$$

where ${\displaystyle U=-eV}$ is the potential energy of the electron in the central field, and e is the elementary charge. Now we remember from classical mechanics that the angular momentum of a particle ${\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }$. Putting it all together, we get

$${\displaystyle \mathbf {B} ={\frac {1}{m_{\text{e}}ec^{2}}}{\frac {1}{r}}{\frac {\partial U(r)}{\partial r}}\mathbf {L} .}$$

It is important to note at this point that B is a positive number multiplied by L, meaning that the magnetic field is parallel to the orbital angular momentum of the particle, which is itself perpendicular to the particle's velocity.

The spin magnetic moment of the electron is

$${\displaystyle {\boldsymbol {\mu }}_{S}=-g_{\text{s}}\mu _{\text{B}}{\frac {\mathbf {S} }{\hbar }},}$$

where ${\displaystyle \mathbf {S} }$ is the spin angular-momentum vector, ${\displaystyle \mu _{\text{B}}}$ is the Bohr magneton, and ${\displaystyle g_{\text{s}}=2.0023...\approx 2}$ is the electron-spin g-factor. Here ${\displaystyle {\boldsymbol {\mu }}}$ is a negative constant multiplied by the spin, so the spin magnetic moment is antiparallel to the spin angular momentum.

The spin–orbit potential consists of two parts. The Larmor part is connected to the interaction of the spin magnetic moment of the electron with the magnetic field of the nucleus in the co-moving frame of the electron. The second contribution is related to Thomas precession.

The Larmor interaction energy is

$${\displaystyle \Delta H_{\text{L}}=-{\boldsymbol {\mu }}\cdot \mathbf {B} .}$$

Substituting in this equation expressions for the spin magnetic moment and the magnetic field, one gets

$${\displaystyle \Delta H_{\text{L}}={\frac {g_{\text{s}}\mu _{\text{B}}}{\hbar m_{\text{e}}ec^{2}}}{\frac {1}{r}}{\frac {\partial U(r)}{\partial r}}\mathbf {L} \cdot \mathbf {S} \approx {\frac {2\mu _{\text{B}}}{\hbar m_{\text{e}}ec^{2}}}{\frac {1}{r}}{\frac {\partial U(r)}{\partial r}}\mathbf {L} \cdot \mathbf {S} .}$$

Now we have to take into account Thomas precession correction for the electron's curved trajectory.

In 1926 Llewellyn Thomas relativistically recomputed the doublet separation in the fine structure of the atom.[1] Thomas precession rate ${\displaystyle {\boldsymbol {\Omega }}_{\text{T}}}$ is related to the angular frequency of the orbital motion ${\displaystyle {\boldsymbol {\omega }}}$ of a spinning particle as follows:[2][3]

$${\displaystyle {\boldsymbol {\Omega }}_{\text{T}}=-{\boldsymbol {\omega }}(\gamma -1),}$$

where ${\displaystyle \gamma }$ is the Lorentz factor of the moving particle. The Hamiltonian producing the spin precession ${\displaystyle {\boldsymbol {\Omega }}_{\text{T}}}$ is given by

$${\displaystyle \Delta H_{\text{T}}={\boldsymbol {\Omega }}_{\text{T}}\cdot \mathbf {S} .}$$

To the first order in ${\displaystyle (v/c)^{2}}$, we obtain

$${\displaystyle \Delta H_{\text{T}}=-{\frac {\mu _{\text{B}}}{\hbar m_{\text{e}}ec^{2}}}{\frac {1}{r}}{\frac {\partial U(r)}{\partial r}}\mathbf {L} \cdot \mathbf {S} .}$$

The total spin–orbit potential in an external electrostatic potential takes the form

$${\displaystyle \Delta H\equiv \Delta H_{\text{L}}+\Delta H_{\text{T}}={\frac {(g_{\text{s}}-1)\mu _{\text{B}}}{\hbar m_{\text{e}}ec^{2}}}{\frac {1}{r}}{\frac {\partial U(r)}{\partial r}}\mathbf {L} \cdot \mathbf {S} \approx {\frac {\mu _{\text{B}}}{\hbar m_{\text{e}}ec^{2}}}{\frac {1}{r}}{\frac {\partial U(r)}{\partial r}}\mathbf {L} \cdot \mathbf {S} .}$$

The net effect of Thomas precession is the reduction of the Larmor interaction energy by factor of about 1/2, which came to be known as the Thomas half.

Thanks to all the above approximations, we can now evaluate the detailed energy shift in this model. Note that L_z and S_z are no longer conserved quantities. In particular, we wish to find a new basis that diagonalizes both H₀ (the non-perturbed Hamiltonian) and ΔH. To find out what basis this is, we first define the total angular momentum operator

$${\displaystyle \mathbf {J} =\mathbf {L} +\mathbf {S} .}$$

Taking the dot product of this with itself, we get

$${\displaystyle \mathbf {J} ^{2}=\mathbf {L} ^{2}+\mathbf {S} ^{2}+2\,\mathbf {L} \cdot \mathbf {S} }$$

(since L and S commute), and therefore

$${\displaystyle \mathbf {L} \cdot \mathbf {S} ={\frac {1}{2}}\left(\mathbf {J} ^{2}-\mathbf {L} ^{2}-\mathbf {S} ^{2}\right)}$$

It can be shown that the five operators H₀, J², L², S², and J_z all commute with each other and with ΔH. Therefore, the basis we were looking for is the simultaneous eigenbasis of these five operators (i.e., the basis where all five are diagonal). Elements of this basis have the five quantum numbers: ${\displaystyle n}$ (the "principal quantum number"), ${\displaystyle j}$ (the "total angular momentum quantum number"), ${\displaystyle \ell }$ (the "orbital angular momentum quantum number"), ${\displaystyle s}$ (the "spin quantum number"), and ${\displaystyle j_{z}}$ (the "z component of total angular momentum").

To evaluate the energies, we note that

$${\displaystyle \left\langle {\frac {1}{r^{3}}}\right\rangle ={\frac {2}{a^{3}n^{3}\;\ell (\ell +1)(2\ell +1)}}}$$

for hydrogenic wavefunctions (here ${\displaystyle a=\hbar /(Z\alpha m_{\text{e}}c)}$ is the Bohr radius divided by the nuclear charge Z); and

$${\displaystyle \left\langle \mathbf {L} \cdot \mathbf {S} \right\rangle ={\frac {1}{2}}{\big (}\langle \mathbf {J} ^{2}\rangle -\langle \mathbf {L} ^{2}\rangle -\langle \mathbf {S} ^{2}\rangle {\big )}={\frac {\hbar ^{2}}{2}}{\big (}j(j+1)-\ell (\ell +1)-s(s+1){\big )}.}$$

We can now say that

$${\displaystyle \Delta E={\frac {\beta }{2}}{\big (}j(j+1)-\ell (\ell +1)-s(s+1){\big )},}$$

where

$${\displaystyle \beta =\beta (n,l)=Z^{4}{\frac {\mu _{0}}{4\pi }}g_{\text{s}}\mu _{\text{B}}^{2}{\frac {1}{n^{3}a_{0}^{3}\;\ell (\ell +1/2)(\ell +1)}}.}$$

For the exact relativistic result, see the solutions to the Dirac equation for a hydrogen-like atom.

The derivation above calculates the interaction energy in the (momentaneous) rest frame of the electron and in this reference frame there's a magnetic field that's absent in the rest frame of the nucleus.

Another approach is to calculate it in the rest frame of the nucleus, see for example George P. Fisher: Electric Dipole Moment of a Moving Magnetic Dipole (1971).[4] However the rest frame calculation is sometimes avoided, because one has to account for hidden momentum.[5]

Hole bands of a bulk (3D) zinc-blende semiconductor will be split by ${\displaystyle \Delta _{0}}$ into heavy and light holes (which form a ${\displaystyle \Gamma _{8}}$ quadruplet in the ${\displaystyle \Gamma }$-point of the Brillouin zone) and a split-off band (${\displaystyle \Gamma _{7}}$ doublet). Including two conduction bands (${\displaystyle \Gamma _{6}}$ doublet in the ${\displaystyle \Gamma }$-point), the system is described by the effective eight-band model of Kohn and Luttinger. If only top of the valence band is of interest (for example when ${\displaystyle E_{\text{F}}\ll \Delta _{0}}$, Fermi level measured from the top of the valence band), the proper four-band effective model is

$${\displaystyle H_{\text{KL}}(k_{\text{x}},k_{\text{y}},k_{\text{z}})=-{\frac {\hbar ^{2}}{2m}}\left[\left(\gamma _{1}+{{\tfrac {5}{2}}\gamma _{2}}\right)k^{2}-2\gamma _{2}\left(J_{\text{x}}^{2}k_{\text{x}}^{2}+J_{\text{y}}^{2}k_{\text{y}}^{2}+J_{\text{z}}^{2}k_{\text{z}}^{2}\right)-2\gamma _{3}\sum _{m\neq n}J_{m}J_{n}k_{m}k_{n}\right]}$$

where ${\displaystyle \gamma _{1,2,3}}$ are the Luttinger parameters (analogous to the single effective mass of a one-band model of electrons) and ${\displaystyle J_{{\text{x}},{\text{y}},{\text{z}}}}$ are angular momentum 3/2 matrices (${\displaystyle m}$ is the free electron mass). In combination with magnetization, this type of spin–orbit interaction will distort the electronic bands depending on the magnetization direction, thereby causing magnetocrystalline anisotropy (a special type of magnetic anisotropy). If the semiconductor moreover lacks the inversion symmetry, the hole bands will exhibit cubic Dresselhaus splitting. Within the four bands (light and heavy holes), the dominant term is

$${\displaystyle H_{{\text{D}}3}=b_{41}^{8{\text{v}}8{\text{v}}}[(k_{\text{x}}k_{\text{y}}^{2}-k_{\text{x}}k_{\text{z}}^{2})J_{\text{x}}+(k_{\text{y}}k_{\text{z}}^{2}-k_{\text{y}}k_{\text{x}}^{2})J_{\text{y}}+(k_{\text{z}}k_{\text{x}}^{2}-k_{\text{z}}k_{\text{y}}^{2})J_{\text{z}}]}$$

where the material parameter ${\displaystyle b_{41}^{8{\text{v}}8{\text{v}}}=-81.93\,{\text{meV}}\cdot {\text{nm}}^{3}}$ for GaAs (see pp. 72 in Winkler's book, according to more recent data the Dresselhaus constant in GaAs is 9 eVų;[12] the total Hamiltonian will be ${\displaystyle H_{\text{KL}}+H_{{\text{D}}3}}$). Two-dimensional electron gas in an asymmetric quantum well (or heterostructure) will feel the Rashba interaction. The appropriate two-band effective Hamiltonian is

$${\displaystyle H_{0}+H_{\text{R}}={\frac {\hbar ^{2}k^{2}}{2m^{*}}}\sigma _{0}+\alpha (k_{\text{y}}\sigma _{\text{x}}-k_{\text{x}}\sigma _{\text{y}})}$$

where ${\displaystyle \sigma _{0}}$ is the 2 × 2 identity matrix, ${\displaystyle \sigma _{{\text{x}},{\text{y}}}}$ the Pauli matrices and ${\displaystyle m^{*}}$ the electron effective mass. The spin–orbit part of the Hamiltonian, ${\displaystyle H_{\text{R}}}$ is parametrized by ${\displaystyle \alpha }$, sometimes called the Rashba parameter (its definition somewhat varies), which is related to the structure asymmetry.

Above expressions for spin–orbit interaction couple spin matrices ${\displaystyle \mathbf {J} }$ and ${\displaystyle {\boldsymbol {\sigma }}}$ to the quasi-momentum ${\displaystyle \mathbf {k} }$, and to the vector potential ${\displaystyle \mathbf {A} }$ of an AC electric field through the Peierls substitution ${\textstyle \mathbf {k} =-i\nabla -{\frac {e}{\hbar c}}\mathbf {A} }$. They are lower order terms of the Luttinger–Kohn k·p perturbation theory in powers of ${\displaystyle k}$. Next terms of this expansion also produce terms that couple spin operators of the electron coordinate ${\displaystyle \mathbf {r} }$. Indeed, a cross product ${\displaystyle ({\boldsymbol {\sigma }}\times {\mathbf {k} })}$ is invariant with respect to time inversion. In cubic crystals, it has a symmetry of a vector and acquires a meaning of a spin–orbit contribution ${\displaystyle {\boldsymbol {r}}_{\text{SO}}}$ to the operator of coordinate. For electrons in semiconductors with a narrow gap ${\displaystyle E_{\rm {G}}}$ between the conduction and heavy hole bands, Yafet derived the equation[13][14]

$${\displaystyle {\mathbf {r} }_{\text{SO}}={\frac {\hbar ^{2}g}{4m_{0}}}\left({\frac {1}{E_{\rm {G}}}}+{\frac {1}{E_{\rm {G}}+\Delta _{0}}}\right)({\boldsymbol {\sigma }}\times {\mathbf {k} })}$$

where ${\displaystyle m_{0}}$ is a free electron mass, and ${\displaystyle g}$ is a ${\displaystyle g}$-factor properly renormalized for spin–orbit interaction. This operator couples electron spin ${\displaystyle \mathbf {S} ={\tfrac {1}{2}}{\boldsymbol {\sigma }}}$ directly to the electric field ${\displaystyle \mathbf {E} }$ through the interaction energy ${\displaystyle -e(\mathbf {r} _{\text{SO}}\cdot \mathbf {E} )}$.

Electric dipole spin resonance (EDSR) is the coupling of the electron spin with an oscillating electric field. Similar to the electron spin resonance (ESR) in which electrons can be excited with an electromagnetic wave with the energy given by the Zeeman effect, in EDSR the resonance can be achieved if the frequency is related to the energy band splitting given by the spin–orbit coupling in solids. While in ESR the coupling is obtained via the magnetic part of the EM wave with the electron magnetic moment, the ESDR is the coupling of the electric part with the spin and motion of the electrons. This mechanism has been proposed for controlling the spin of electrons in quantum dots and other mesoscopic systems.[15]