Introduction

In quantum physics, the spin-orbit interaction (also called spin-orbit effect or spin-orbit coupling) is a relativistic interaction of a particle's spin with its motion inside a potential.

A key example of this phenomenon is the spin-orbit interaction leading to shifts in an electron's atomic energy levels, due to electromagnetic interaction between the electron's magnetic dipole, its orbital motion, and the electrostatic field of the positively charged nucleus. This phenomenon is detectable as a splitting of spectral lines, which can be thought of as a Zeeman effect product of two relativistic effects: the apparent magnetic field seen from the electron perspective and the magnetic moment of the electron associated with its intrinsic spin. A similar effect, due to the relationship between angular momentum and the strong nuclear force, occurs for protons and neutrons moving inside the nucleus, leading to a shift in their energy levels in the nucleus shell model. In the field of spintronics, spin-orbit effects for electrons in semiconductors and other materials are explored for technological applications. The spin-orbit interaction is one cause of magnetocrystalline anisotropy and the spin Hall effect.

Relativistic origin of the Spin-Orbit Interaction

According to the theory of relativity, a particle moving in an electrical field experiences an effective magnetic field, which is directed perpendicularly to the electrical field and the particle movement direction. The interaction of this effective magnetic field with the electron spin is called Spin-Orbit interaction. It is important to emphasize that the direction and magnitude of the effective magnetic field do not depend either on the particle charge or on the particle spin.

According to this theory, the electric and magnetic field mutually transformed into each other depending on the speed of an observer. For example, if in a coordinate system of static observer there is only a magnetic field, a movable observer will experience this field as both an electrical field and a magnetic field.

A particle moving in a static magnetic field experience an effective electric field. The effective electrical field acts on the particle charge (the Lorentz force, Hall effect) and forces the particle to move along this field.

A particle moving in a static electrical field experience an effective magnetic field. The effective magnetic field acts on the particle magnetic moment (spin-orbit interaction) and causes the precession of the magnetic moment around the direction of the effective magnetic field.

The electromagnetic field is a relativistic object and the Lorentz transformation equations are written as, $$\mathbf{E}_\mathrm{move} = \frac{\mathbf{E}_\mathrm{static} + \mathbf{v} \times \mathbf{H}_\mathrm{static}}{\sqrt{1 - \frac{v^2}{c^2}}} \text{ ;} \\ \mathbf{H}_\mathrm{move} = \frac{\mathbf{H}_\mathrm{static} - \frac{\mathbf{v}}{c^2} \times \mathbf{E}_\mathrm{static}}{\sqrt{1 - \frac{v^2}{c^2}}}$$ where, $\mathbf{E}_\mathrm{static}, \mathbf{H}_\mathrm{static}$ are the electric and magnetic field in the static coordinate system and $\mathbf{E}_\mathrm{move}, \mathbf{H}_\mathrm{move}$ are the elctric and magnetic field in the coordinate system, which moves with a constant velocity $\mathbf{v}$. As a result, an electron, which moves in a static magnetic field $\mathbf{H}_\text{static}$, experiences in own reference frame an effective electrical field $\mathbf{E}_\text{Hall}$, which is called the Hall field (Hall voltage). Similarly, when an electron moves in a static electrical field $\mathbf{E}_\text{static}$, it experiences in own reference frame an effective magnetic field $\mathbf{H}_\text{SO}$, which is called the effective spin-orbit magnetic field.

For example, when an electron moves in the $x$-direction, $$v_x \neq 0 \qquad \text{;} \qquad v_y = 0 \qquad \text{;} \qquad v_z = 0$$ the effective electric field and the effective magnetic field are, $$E_\text{Hall} = E_\text{y} = - \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} H_\text{z} \text{ ;} \\ H_\text{SO} = H_\text{y} = \frac{\frac{v}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} E_\text{z}$$ For the non-relativistic case $(v << c)$: $$\sqrt{1 - \frac{v^2}{c^2}} \simeq 1$$ The effective electric field and the effective magnetic field are then, $$E_\text{Hall} = E_\text{y} = - v H_\text{z} \text{ ;} \\ H_\text{SO} = H_\text{y} = \frac{v}{c^2} E_\text{z}$$ Question: An electron always moves along the electrical field (but not perpendicularly). Then, how it can experience the spin-orbit interaction?
Answer: The trajectory of an electron may be very different. There are many cases when an electron moves perpendicularly to an electrical field. For example, when the electron is orbiting around a nucleus.

Question: Which parameter characterizes the spin-orbit interaction? Is it a classical parameter or a quantum mechanical parameter?
Answer: There is only one cause of spin-orbit interaction. It is the magnetic field $\mathbf{H}_\text{SO}$. All other parameters are consequences of $\mathbf{H}_\text{SO}$. For example, the electron spin aligns itself along $\mathbf{H}_\text{SO}$. It changes the energy of the electron. The change in energy can be defined as the energy of the spin-orbit interaction $E_\text{SO}$. If an additional external magnetic field $\mathbf{H}_\text{ext}$ is applied, the electron spin is aligned along the total magnetic field $\mathbf{H}_\text{SO} + \mathbf{H}_\text{ext}$ and $E_\text{SO}$ has no physical meaning.

Question: Since the spin-orbit magnetic field $\mathbf{H}_\text{SO}$ is proportional to $1/c^2$, then it should be negligibly small. Should we care about such a tiny effect?
Answer: Yes. It is small, except for when $\mathbf{E}_\text{static}$ is huge. For example, near the nucleus $\mathbf{H}_\text{SO}$ is large. Any substantial spin-orbit interaction is induced only by an electrical field of a nucleus. Other realistic sources of the electrical field in a solid induce very weak spin-orbit interaction. This field is very symmetric, and it cancels itself out in a nucleus. External fields may break this symmetry.

Question: Is spin-orbit interaction is a quantum mechanical effect?
Answer: No. The spin-orbit interaction affects both small and large objects. It exists in the macro world as well.

Energy of Spin-Orbit interaction

There is a precession of electron spin $\mathbf{S}$ around magnetic field $\mathbf{H}_\text{SO}$ of spin-orbit interaction until it is aligned parallel to $\mathbf{H}_\text{SO}$. After the electron spin is aligned, the energy of spin-orbit interaction becomes, $$E_\text{SO} = \mu_\text{B} H_\text{SO}$$ where, $\mu_\text{B}$ is the Bohr magneton.

The $\mathbf{H}_\text{SO}$ only interacts with the spin magnetic moment, not with the orbital magnetic moment.

Question: Why $\mathbf{H}_\text{SO}$ does not induce the Lorentz force and cannot interact with the orbital magnetic moment?
Answer: $\mathbf{H}_\text{SO}$ has relativistic origin. It appears only in the coordinate system which moves together with the electron. The Lorentz force has the relativistic origin as well and it is originated from the electrical field, which the electron experiences when moves in a magnetic field. In the moving coordinate system, where the electron experiences $\mathbf{H}_\text{SO}$, the electron does not move and therefore experience no Lorentz force. The interaction of the orbital moment with a magnetic field is due to the Lorentz force.

Electron spin is aligned along due to the spin precession damping. The spin precession damping is a complex mechanism, which involves an external particle with a non-zero spin (e.g. a photon, a magnon). It could take a relatively long time until the full alignment of electron spin along $\mathbf{H}_\text{SO}$.

Question: Is there any cases when the electron spin is not aligned along $\mathbf{H}_\text{SO}$?
Answer: There are many such cases. For example, the conduction electrons in a metal. There are many conduction electrons, which simultaneously overlap with each other. For this reason, the scattering between quantum states of conduction electrons is very frequent. The time between two consequent scattering is very short $(\sim 1 \mathrm{\: ps})$. This time is not enough for the complete alignment.

Magnitude of the Spin-Orbit interaction

Except for a few weak effects, all spin-orbital effects are induced by an electrical field of an atomic nucleus and the electron movement (rotation) in the close proximity of the nucleus.

For a realistic spin-orbit interaction, the maximum value of drift velocity of electron in a solid is $1 \times 10^7 \mathrm{\: cm/s}$ (GaAs, Si) [Ref]. The electron cannot go faster, because above the saturation velocity the electron intensively illuminates phonons. The maximum breakdown voltage applicable to a semiconductor is $5 \times 10^5 \mathrm{\: cm/s}$ (a oxide). For higher voltage, the avalanche breakdown occurs.

For these values, the effective magnetic field of the spin-orbit interaction is only $0.5 \mathrm{\: Gauss}$. For comparison, Earth's magnetic field is about $0.25-0.65 \mathrm{\: Gauss}$.

For an electron rotating around the nucleus, the speed of the electron is about $\sim 2.1 \times 10^6 \mathrm{\: m/s}$. The Coulomb electrical field in H atom at the first orbital $(r = 0.053 \mathrm{\: nm})$ is $5.1 \times 10^9 \mathrm{\: V/cm}$.

For these values, the effective magnetic field of the spin-orbit interaction is $125 \mathrm{\: kGauss} = 12.5 \mathrm{\: T}$.

Spin-Orbit interaction and orbital moment

Electron orbital can be represented as 2D electron rotation around a rotating axis. Question: Is the spin-orbit interaction proportional to the orbital moment?
Answer: No. Even though there are common tendencies between the spin-orbit interaction and the orbital moment. For example, when the orbital moment is zero, the spin-orbit interaction is zero. When the orbital moment changes its sign, the spin-orbit interaction changes its sign as well.
Even though "orbit" is a part of the name of the spin-orbit interaction, the relation between orbital moment and $\mathbf{H}_\text{SO}$ is complex and not straightforward.

Question: How is it possible that an electron, while rotating around a nucleus, does not experience the spin-orbit interaction?
Answer: It is because for the spherical orbit an electron makes an equal number of rotations in two opposite directions. Since for opposite rotation directions, the directions of the effective magnetic field of the spin-orbit interaction are opposite, an electron does not experience any spin-orbit interaction.

Types of Spin-Orbit interaction

It is convenient to divide effects related to the spin-orbit interaction into 3 classes depending on the source of the electrical field and the source of breaking of the time-reversal symmetry.

• Type 1: Weak spin-orbit interaction
• Type 2: Strong spin-orbit interaction
• Type 3: Moderate spin-orbit interaction

Type 1: Weak spin-orbit interaction

Only the conduction electrons experience this type of spin-orbit effect. The electrical field at an interface or electrical field of a Schottky barrier is the sources of the electrical field for this type. An electrical current flowing along with the interface and perpendicularly to the interface electrical field breaks the time-reversal symmetry.

This type of effect includes:

• spin Hall effect
• anomalous spin Hall effect
• anomalous magneto-resistance
• rashba effect

Type 2: Strong spin-orbit interaction

Only localized electrons (e.g. d- or f- electrons) experience this type of spin-orbit effect. The centrosymmetric electrical field of the atomic nucleus is the source of this type of spin-orbit effect. An external magnetic field is a source of breaking the time-reversal symmetry.

This type of effect includes:

• perpendicular magnetic anisotropy
• g-factor
• magneto-elastic effect
• voltage-controlled magnetic anisotropy
• interface sensing
• magnetic anisotropy

In absence of external magnetic field, a localized electron does not experience any $\mathbf{H}_\text{SO}$. However, when an external magnetic field $\mathbf{H}_\text{ext}$ is applied, it induces strong $\mathbf{H}_\text{SO}$ parallel to $\mathbf{H}_\text{ext}$ and the electron experiences a stronger total magnetic field $\mathbf{H}_\text{total} = \mathbf{H}_\text{ext} + \mathbf{H}_\text{SO}$.

The orbital moment of the localized electrons is zero. Any 3D orbital can be represented as the sum of two 2D orbitals of clockwise and anti-clockwise electron rotation. Clockwise and anti-clockwise orbitals are identical. As a result, the electron experience the same but opposite $\mathbf{H}_\text{SO}$ for the clockwise and anti-clockwise orbitals.

When an external magnetic field is applied, the time-reversal symmetry is broken. Since electron rotations in the clockwise and anti-clockwise orbitals are opposite, the Lorentz force is the opposite. As a result, the clockwise and anti-clockwise orbitals are deformed differently in an external magnetic field, $\mathbf{H}_\text{SO}$ becomes different for clockwise and anti-clockwise orbitals and in total the electron experiences a non-zero $\mathbf{H}_\text{SO}$.

Type 3: Moderate spin-orbit interaction

Only conduction electrons experience this type of effect. The centrosymmetric electrical field of the atomic nucleus is the source of the electrical field. Electrical current breaks the time-reversal symmetry.

This type of effect includes:

• spin Hall effect
• spin pumping
• inverse spin Hall effect
• spin damping
• spin-orbit torque

Spin-Orbit interaction from the Dirac equations

Einstein's relativistic equation for the energy is, $$E^2 = p^2c^2 - m_o^2c^4$$ which should describe the electric field as well. The quantum-mechanical operator for the energy and the momentum are, $$\bar{p} \rightarrow - i \hbar \nabla \text{ ;} \\ \bar{E} = i \hbar \frac{\partial}{\partial t}$$ Now, the relation between relativistic and quantum mechanics can found that, $$\left( \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \frac{m_o^2 c^2}{\hbar^2}\right) \Psi(r,t) = 0$$ This is called the Klein-Gordon equation.

The wave equation should be 1st order differential equation with respect to time and space. Dirac found that the Klein-Gordon equation can be represented as a product of a 1st order differential equation and its conjugate. Therefore, the Dirac equations describes the electrical field. $$\left( A \partial_x + B \partial_y + C \partial_z + \frac{i}{c} D \partial_t - \frac{m_o c}{\hbar}\right) \Psi = 0$$ Setting, $$D = \gamma^0 \text{ ;} \qquad A = i \gamma^1 \text{ ;} \\ B = i \gamma^2 \text{ ;} \qquad C = i \gamma^3$$ $$\partial_0 = \frac{1}{c} \partial_t \text{ ;} \qquad \partial_1 = \partial_x \text{ ;} \\ \partial_2 = \partial_y \text{ ;} \qquad \partial_3 = \partial_z$$ where, $\gamma$ denotes the $(2 \times 2)$ sub-matrices taken from the Pauli matrices. The Dirac equation, $$i \hbar \gamma^\mu \partial_\mu \Psi - m_o c \Psi = 0$$ where, $\mu = 0,1,2,3$

The Dirac equation, which includes the gauge potential, is $$\gamma^\mu (i \hbar \partial_\mu - e A_\mu) \Psi - m_o c \Psi = 0$$

References

Dr. Vadym Zayets, "Spin-Orbit Interaction" - https://staff.aist.go.jp/v.zayets/spin3_32_SpinOrbit.html