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.

Spin-orbit coupling

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