Advanced Chemistry - Mono Mole

July 14, 2022January 30, 2026

Spin-orbit coupling

Spin-orbit coupling is the interaction between a particle’s spin angular momentum and orbital angular momentum. An electron orbiting around the nucleus ‘sees’ the nucleus circling it, just as a person on earth perceives the sun circling the earth while the latter orbits around the sun.

This apparent nuclear orbit creates a magnetic field $B$ that exerts a torque on the electron’s spin magnetic dipole moment $\mu_s$ , resulting in an additional term of $\hat{H}_{so}=\sum_{i=1}^{n}A_i\hat{\boldsymbol{\mathit{S}}}_i\cdot\hat{\boldsymbol{\mathit{L}}}_i$ (where $A_i=\frac{1}{2m_e^{\;2}c^{2}}\frac{1}{r_i}\frac{dV_i}{dr_i}$ ) in the multi-electron Hamiltonian. To derive this term, we consider a 1-electron atom.

Let $L=m_erv_{\perp}$ be the orbital angular momentum of the electron and $I$ be the proton’s current loop, which generates a magnetic field of magnitude $B=\frac{\mu_0I}{2r}$ given by the Biot-Savart law. Since $\frac{1}{t}=\frac{v_{\perp}}{2\pi r}$ , we have

$I=\frac{e}{t}=\frac{eL}{2\pi m_er^{2}}\; \; \; \; \; \; \; \; 259$

Substitute eq259 in $B$ , we have, $B=\frac{\mu_0e}{4\pi m_er^{3}}L$ or $\frac{\boldsymbol{\mathit{B}}}{\hat{\boldsymbol{\mathit{B}}}}=\frac{\mu_0e}{4\pi m_er^{3}}\frac{\boldsymbol{\mathit{L}}}{\hat{\boldsymbol{\mathit{L}}}}$ , where $\hat{\boldsymbol{\mathit{B}}}$ and $\hat{\boldsymbol{\mathit{L}}}$ are unit vectors. Since $\boldsymbol{\mathit{B}}$ and $\boldsymbol{\mathit{L}}$ point in the same direction, $\hat{\boldsymbol{\mathit{B}}}=\hat{\boldsymbol{\mathit{L}}}=\hat{\boldsymbol{\mathit{M}}}$ . Multiplying both sides of $\frac{\boldsymbol{\mathit{B}}}{\hat{\boldsymbol{\mathit{M}}}}=\frac{\mu_0e}{4\pi m_er^{3}}\frac{\boldsymbol{\mathit{L}}}{\hat{\boldsymbol{\mathit{M}}}}$ by $\hat{\boldsymbol{\mathit{M}}}$ , we have $\boldsymbol{\mathit{B}}=\frac{\mu_0e}{4\pi m_er^{3}}\boldsymbol{\mathit{L}}$ , which we substitute in eq65 (where $\boldsymbol{\mathit{\mu}}$ is the spin analogue of eq61) to give $U=-\frac{\gamma_e\mu_0e}{4\pi m_er^{3}}\boldsymbol{\mathit{S}}\cdot\boldsymbol{\mathit{L}}$ .

Substituting eq164 and $\frac{1}{c^{2}}=\mu_0\varepsilon_0$ in $U$ yields

$U=\frac{e^{2}}{4\pi\varepsilon_0 m_e^{\;2}c^{2}r^{3}}\boldsymbol{\mathit{S}}\cdot\boldsymbol{\mathit{L}}$

For a 1-electron atom, $V=-\frac{e^2}{4\pi\varepsilon_0r}$ and so

$U=\frac{1}{ m_e^{\;2}c^{2}r}\frac{dV}{dr}\boldsymbol{\mathit{S}}\cdot\boldsymbol{\mathit{L}}\; \; \; \; \; \; \; \; 260$

Eq260 can be written in terms of the Larmor frequency of the electron. From eq149, $\omega_L=\frac{eB}{m_e}=\frac{e}{m_e}\frac{\mu_0e}{4\pi m_er^{3}}L=\frac{1}{m_e^{\;2}c^{2}r}\frac{dV}{dr}L$ . So, $U=\frac{\omega_L}{L}\boldsymbol{\mathit{S}}\cdot\boldsymbol{\mathit{L}}$ . Swapping $\omega_L$ with the Thomas precession rate $\omega_T$ , we obtain the correction term of $U_{tp}=-\frac{1}{2 m_e^{\;2}c^{2}r}\frac{dV}{dr}\boldsymbol{\mathit{S}}\cdot\boldsymbol{\mathit{L}}$ .

The total spin-orbit Hamiltonian is

$\hat{H}_{so}=U+U_{tp}=U_{tp}=\frac{1}{2 m_e^{\;2}c^{2}r}\frac{dV}{dr}\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}\; \; \; \; \; \; \; \; 261$

The spin-orbit energy $E_{so}$ corresponds to the expectation value $\langle\psi\vert\hat{H}_{so}\vert\psi\rangle$ , where $\psi$ is the hydrogenic wavefunction. Since $\hat{J}^2=(\boldsymbol{\mathit{J}}+\boldsymbol{\mathit{S}})\cdot(\boldsymbol{\mathit{J}}+\boldsymbol{\mathit{S}})=\hat{L}^2+\hat{S}^2+2\boldsymbol{\mathit{J}}\cdot\boldsymbol{\mathit{S}}$ , we have $\boldsymbol{\mathit{J}}\cdot\boldsymbol{\mathit{S}}=\frac{1}{2}(\hat{J}^2-\hat{L}^2-\hat{S}^2)$ . Substituting this and $V=-\frac{e^2}{4\pi\varepsilon_0r}$ into $\langle\psi\vert\hat{H}_{so}\vert\psi\rangle$ gives:

$E_{so}=\frac{e^2\hbar^2}{8\pi\varepsilon_0m_e^2c^2}\langle\psi\vert\frac{1}{r^3}\vert\psi\rangle[J(J+1)-L(L+1)-S(S+1)]$

For a given $L$ , we usually express the above equation as:

$E_{so}=\frac{1}{2}A\hbar^2[J(J+1)-L(L+1)-S(S+1)]\;\;\;\;\;\;\;\;261a$

where $A=\frac{e^2\hbar^2}{8\pi\varepsilon_0m_e^2c^2}\langle\psi\vert\frac{1}{r^3}\vert\psi\rangle$ is regarded as a constant.

For a multi-electron system,

$\hat{H}_{so}=\sum_{i=1}^n\frac{1}{2m_e^{\;2}c^2r_i}\frac{dV_i}{dr_i}\hat{\boldsymbol{\mathit{L}}}_i\cdot\hat{\boldsymbol{\mathit{S}}}_i\; \; \; \; \; \; \; \; 261b$

The spin-orbit energy expression for a multi-electron system is very similar to eq261a if we assume Russell-Saunders coupling, where spin-orbit interactions are weak compared to the Coulomb interaction ( $\hat{H}_{Coulomb}\gg\hat{H}_{so}$ ). Here, the electrons’ orbital angular momenta couple to form the total orbital angular momentum $\boldsymbol{\mathit{L}}=\sum_i\boldsymbol{\mathit{L}}_i$ separately from their spin angular momenta, which couple to form $\boldsymbol{\mathit{S}}=\sum_i\boldsymbol{\mathit{S}}_i$ . Each electron in the system experiences a force due to its Coulomb interaction with the other electrons $\sum_{j\neq i}\frac{e^2}{4\pi\varepsilon_0\vert\boldsymbol{\mathit{r}}_i-\boldsymbol{\mathit{r}}_j\vert}$ .

Consider electron $i$ at $\boldsymbol{\mathit{r}}_i=(R,0)$ along the laboratory $x$ -axis and electron $j$ at $\boldsymbol{\mathit{r}}_j=(0,R)$ on the $y$ -axis. The Coulomb force $\boldsymbol{\mathit{F}}_{ij}$ between electrons $i$ and $j$ acts in the direction $\boldsymbol{\mathit{r}}_{i}-\boldsymbol{\mathit{r}}_j$ , pointing downwards and to the right. The component of $\boldsymbol{\mathit{F}}_{ij}$ perpendicular to the $x$ -axis produces a torque $\boldsymbol{\mathit{\tau}}_{i}$ on electron $i$ , where $\boldsymbol{\mathit{\tau}}_{i}=\boldsymbol{\mathit{r}_{i}\times\boldsymbol{\mathit{F}}_{ij}$ . Because the electrons are not symmetrically arranged at every instant, the resultant torque on electron $i$ due to all other electrons is generally non-zero. Since $\boldsymbol{\mathit{\tau}}_{i}=\frac{d\boldsymbol{\mathit{L}}_{i}}{dt}$ (see eq71), the individual orbital angular momenta are not conserved. By Newton’s third law, the torque exerted by electron $i$ on electron $j$ is equal and opposite to the torque exerted by $j$ on $i$ . When summed over all electrons, these internal torques cancel, giving

$\sum_i\boldsymbol{\mathit{\tau}}_{i}=\frac{d\boldsymbol{\mathit{L}}}{dt}=0$

Thus, while the total orbital angular momentum $\boldsymbol{\mathit{L}}$ is conserved, the individual $\boldsymbol{\mathit{L}}_i$ are not. To satisfy $\frac{d\boldsymbol{\mathit{L}}_i}{dt}\neq 0$ , $\frac{d\boldsymbol{\mathit{L}}}{dt}= 0$ and $\boldsymbol{\mathit{L}=\sum_i\boldsymbol{\mathit{L}}_i$ at all times, the individual orbital angular momenta can change only in direction, leading to their precession about the fixed total orbital angular momentum $\boldsymbol{\mathit{L}}$ . By the same reasoning, the individual spin angular momenta precess around $\boldsymbol{\mathit{S}}$ . Since $\hat{H}_{Coulomb}\gg\hat{H}_{so}$ , these precessions occur on a much shorter timescale than the subsequent precessions of $\boldsymbol{\mathit{L}}$ and $\boldsymbol{\mathit{S}}$ around the total angular momentum $\boldsymbol{\mathit{J}}$ .

To continue the derivation of the spin-orbit energy expression for a multi-electron system, we decompose the individual orbital and spin angular momenta into components parallel and perpendicular to $\boldsymbol{\mathit{L}}$ and $\boldsymbol{\mathit{S}}$ :

$\boldsymbol{\mathit{L}}_i=\boldsymbol{\mathit{L}}_{i\parallel}+\boldsymbol{\mathit{L}}_{i\perp}\;\;\;\;\boldsymbol{\mathit{S}}_i=\boldsymbol{\mathit{S}}_{i\parallel}+\boldsymbol{\mathit{S}}_{i\perp}$

Under rapid precession, the time-averaged value of $\boldsymbol{\mathit{L}}_{i\perp}$ and $\boldsymbol{\mathit{S}}_{i\perp}$ vanish. So, the average value of $\boldsymbol{\mathit{L}}_i\cdot\boldsymbol{\mathit{S}}_i$ is

$\boldsymbol{\mathit{L}}_i\cdot\boldsymbol{\mathit{S}}_i=\boldsymbol{\mathit{L}}_{i\parallel}\cdot\boldsymbol{\mathit{S}}_{i\parallel}$

Since $\boldsymbol{\mathit{L}}_{i\parallel}=(\boldsymbol{\mathit{L}}_{i}\cdot\hat{\boldsymbol{\mathit{L}}})\hat{\boldsymbol{\mathit{L}}}=\biggr$\boldsymbol{\mathit{L}}_i\cdot\frac{\boldsymbol{\mathit{L}}}{\vert\boldsymbol{\mathit{L}}\vert}\biggr$\frac{\boldsymbol{\mathit{L}}}{\vert\boldsymbol{\mathit{L}}\vert}=\frac{\boldsymbol{\mathit{L}}_i\cdot\boldsymbol{\mathit{L}}}{L^2}\boldsymbol{\mathit{L}}$ and correspondingly $\boldsymbol{\mathit{S}}_{i\parallel}=\frac{\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{S}}}{S^2}\boldsymbol{\mathit{S}}$ , where $\hat{\boldsymbol{\mathit{L}}}$ and $\hat{\boldsymbol{\mathit{S}}}$ are unit vectors,

$\sum_i\boldsymbol{\mathit{L}}_i\cdot\boldsymbol{\mathit{S}}_i=\Lambda\boldsymbol{\mathit{L}}\cdot\boldsymbol{\mathit{S}}\;\;\;\;\;\;\;\;261c$

where $\Lambda=\sum_i \frac{(\boldsymbol{\mathit{L}}_i\cdot\boldsymbol{\mathit{L}})(\boldsymbol{\mathit{S}}_i\cdot\boldsymbol{\mathit{S}})}{L^2S^2}$ .

Substituting eq261c into eq261b gives:

$\hat{H}_{so}\propto\hat{\boldsymbol{\mathit{L}}}\cdot\hat{\boldsymbol{\mathit{S}}}$

It follows that

$E_{so}=\frac{1}{2}A'\hbar^2[J(J+1)-L(L+1)-S(S+1)]\;\;\;\;\;\;\;\;261d$

If spin-orbit interactions become strong compared to the Coulomb interaction ( $\hat{H}_{Coulomb}<\hat{H}_{so}$ ), $jj$ -coupling replaces Russell-Saunders coupling, and the spin-orbit Hamiltonian must be treated in the form of eq261b.

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July 14, 2022December 29, 2024

Thomas precession

Thomas precession is a relativistic kinematic correction to the spin-orbit coupling Hamiltonian.

Consider an electron orbiting counter-clockwise around a nucleus at O along the path of a regular $n$ -sided polygon in the $xy$ -plane (see diagram above). At $t=0$ , the nucleus sees the electron making a turn at $(x_0,y_0)$ at an angle $\theta=tan^{-1}\frac{y_1}{x_1}$ . At subsequent vertices, the nucleus sees the electron turning at the same angle before arriving back to $(x_0,y_0)$ .

In the rest frame of the electron (denoted by blue axes), $(x_1,y_1)$ becomes $\left ( \frac{x_1}{\gamma},y_1 \right )$ due to length contraction along the direction of travel, and the electron sees itself turning at angle of $\theta'=tan^{-1}\left ( \gamma\frac{y_1}{x_1} \right )$ at $(x_0,y_0)$ , where $\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ . At subsequent vertices, the electron sees the same turning angle before arriving back to $(x_0,y_0)$ . Substituting $tan\theta=\frac{y_1}{x_1}$ in $tan\theta'=\gamma\frac{y_1}{x_1}$ , we have $tan\theta'=\gamma tan\theta$ . In the limit of large $n$ , the polygon approaches a circle with each turning angle being very small. Using the small-angle approximation, $\theta'=\gamma\theta$ . According to the nucleus and the electron, each turn corresponds to $\theta=\frac{2\pi}{n}$ and $\theta'=\frac{2\pi\gamma}{n}$ respectively. So, after completing a full orbit, the electron sees itself turning an angle of $\theta'_T=2\pi\gamma$ .

In the rest frame of the nucleus (denoted by red axes), the orientation of the spin axis of the electron $S$ remains unchanged (just like the spin axis of the earth being constant from the sun’s perspective). However, in the rest frame of the electron, the spin axis of the electron appears to rotate clockwise. After a completing a full orbit, the change in the orientation of $S$ is $\Delta\theta'=2\pi\gamma-2\pi$ . Since the rate of precession is defined as the rate of change in orientation of the rotational axis of a rotating body, the rate of $\Delta\theta'$ , called the Thomas precession, is

$\omega_T=\omega(\gamma-1)\; \; \; \; \; \; \; \; 255$

where $\omega_T=\frac{\Delta\theta'}{t}$ , $\omega=\frac{2\pi}{t}$ and $t$ is the period of the orbit.

Substitute $a=\frac{v_{\perp}^{\;2}}{r}$ and eq58, where $\omega=\frac{v_{\perp}}{r}$ , in eq255 (note that the radius is perpendicular to the path of travel and is not contracted, and therefore common to both nucleus and electron), $\omega_T=\frac{v_{\perp}}{r}(\gamma-1)=\frac{v_{\perp}a}{v_{\perp}^{\;2}}(\gamma-1)$ . Since $v\ll c$ , we have $\left | -\frac{v_{\perp}^{\;2}}{c^{2}} \right |<1$ . Let $x=-\frac{v_{\perp}^{\;2}}{c^{2}}$ in the binomial series $(1+x)^{-\frac{1}{2}}=1-\frac{x}{2}+\frac{3x^{2}}{8}-\cdots$ . We have $\omega_T\approx \frac{v_{\perp}a}{v_{\perp}^{\;2}}\left ( 1+\frac{v_{\perp}^{\;2}}{2c^{2}}-1 \right )$ or

$\omega_T= \frac{v_{\perp}a}{2c^{2}}\; \; \; \; \; \; \; \; 256$

Substitute $F=m_ea=-\frac{dV}{dr}$ and eq59, where $L=rm_ev_{\perp}$ , in eq256

$\omega_T= -\frac{1}{2m_e^{\;2}c^{2}r}\frac{dV}{dr}L\; \; \; \; \; \; \; \; 257$

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