- Spin-Orbit Interaction in Hydrogen-like Atoms

Total Angular Momentum and Commutation Relations #

For a single electron with orbital angular momentum $\mathbf{L}$ and spin $\mathbf{S}$, we define the total angular momentum as

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

The square of $\mathbf{J}$ is given by

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

Hence,

$$\mathbf{L} \cdot \mathbf{S} \ =\ \frac{1}{2}\Bigl(J^2 - L^2 - S^2\Bigr).$$

Because

$$[J_z,\ L^2] = [J_z,\ S^2] = [J_z,\ J^2] = 0,$$

it follows that

$$[J_z,\ \mathbf{L} \cdot \mathbf{S}] = 0.$$

Therefore, $\mathbf{L}\cdot\mathbf{S}$ commutes with all of $L^2$, $S^2$, $J^2$, and $J_z$. In other words, one can choose simultaneous eigenkets of

$${\ H_0,\ L^2,\ S^2,\ J^2,\ J_z},$$

where $H_0$ is the unperturbed Hamiltonian.

Effective Potential and Spin-Orbit Hamiltonian #

In hydrogen-like atoms, the valence electron experiences an effective central potential $V_c(r)$, which may not be a pure Coulomb potential of the form $-Ze^2/r$ if there are additional effects from the electron cloud.

Nevertheless, the electron moving in this central potential $V_c(r)$ “feels” an effective magnetic field due to its motion in an electric field. In semi-classical terms, an electron with velocity $\mathbf{v}$ in an electric field $\mathbf{E}=-\nabla V_c(r)/e$ experiences an effective magnetic field

$$\mathbf{B}_\mathrm{eff} \ \propto\ -\ \frac{1}{c}\ \mathbf{v} \times \mathbf{E}.$$

More precisely, using $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ and identifying $\mathbf{v} \sim \mathbf{p}/m_e$, one obtains an approximate form

$$\mathbf{B}_\mathrm{eff} \ =\ -\ \frac{\mathbf{L}}{m_e\ e\ c\ r}\ \frac{dV_c(r)}{dr}.$$

The spin-orbit interaction can be written as

$$H_{LS} = V_{LS} = -\ \boldsymbol{\mu} \cdot \mathbf{B}_\mathrm{eff},$$

where the electron magnetic moment is

$$\boldsymbol{\mu} = \frac{g_e\ e\ \mathbf{S}}{2\ m_e\ c} \ \approx\ \frac{e\ \mathbf{S}}{m_e\ c} \quad (\text{with } g_e \approx 2).$$

In a more precise relativistic treatment (from the Dirac equation), the coefficient becomes

$$V_{LS} \ =\ \frac{1}{2\ m_e^2\ c^2}\ \frac{1}{r}\ \frac{dV_c(r)}{dr}\ \mathbf{L}\cdot \mathbf{S}.$$

Hence, the total Hamiltonian is

$$H \ =\ H_0 + V_{LS}, \quad H_0 \ =\ \frac{p^2}{2\ m_e} + V_c(r).$$

First-Order Energy Corrections #

Since $[\mathbf{L}\cdot \mathbf{S},\ H_0]=0$ (for a central $V_c(r)$) and $\mathbf{L}\cdot \mathbf{S}$ also commutes with $L^2,\ S^2,\ J^2,\ J_z$, we may choose $H_0$ eigenkets that simultaneously diagonalize $V_{LS}$. In the standard notation, these kets are labeled by quantum numbers $\ket{n,\ l,\ j,\ m}$ (or $\ket{n,\ l}\otimes\ket{j,\ m}$), where $j=l\pm \tfrac{1}{2}$.

We use

$$\mathbf{L}\cdot \mathbf{S} = \frac{1}{2}\Bigl(J^2 - L^2 - S^2\Bigr)$$

$$= \frac{\hbar^2}{2}\Bigl[j(j+1) - l(l+1) - s(s+1)\Bigr],$$

with $s=\tfrac{1}{2}$. Thus,

$$\bra{j=l+\frac{1}{2},\ m}\mathbf{L}\cdot\mathbf{S}\ket{j=l+\frac{1}{2},\ m}$$

$$= \frac{\hbar^2}{2}\ l,$$

$$\bra{j=l-\frac{1}{2},\ m}\mathbf{L}\cdot\mathbf{S}\ket{j=l-\frac{1}{2},\ m}$$

$$= -\ \frac{\hbar^2}{2}\ (l+1).$$

The first-order energy shift in perturbation theory is

$$\Delta_{n l j}^{(1)} = \bra{n,l,j,m} V_{LS} \ket{n,l,j,m}$$

$$= \frac{1}{2\ m_e^2\ c^2}\ \bra{nl}\frac{1}{r}\frac{dV_c(r)}{dr}\ket{nl} \ \bra{j,m}\mathbf{L}\cdot \mathbf{S}\ket{j,m}.$$

Hence, explicitly,

$$\Delta_{n l,\ j=l+\frac{1}{2}}^{(1)} = \frac{\hbar^2\ l}{4\ m_e^2\ c^2} \ \bra{nl}\frac{1}{r}\frac{dV_c(r)}{dr}\ket{nl},$$

$$\Delta_{n l,\ j=l-\frac{1}{2}}^{(1)} = -\ \frac{\hbar^2\ (l+1)}{4\ m_e^2\ c^2} \ \bra{nl}\frac{1}{r}\frac{dV_c(r)}{dr}\ket{nl}.$$

Estimation and Coulomb Case #

For rough scaling estimates, we often note that

$$\bra{nl}\frac{1}{r}\frac{dV_c(r)}{dr}\ket{nl} \sim \frac{e^2}{a_0^3},$$

where $$a_0 = \frac{\hbar^2}{m_e\ e^2} \quad(\text{the Bohr radius}).$$

Thus,

$$\Delta_{n l j}^{(1)} \sim \frac{e^2}{a_0^3}\ \frac{\hbar^2}{m_e^2\ c^2}.$$

In the pure Coulomb potential $V_c(r) = -\frac{Z e^2}{r}$, one has

$$\frac{1}{r}\ \frac{dV_c(r)}{dr} = \frac{Z e^2}{r^3},$$

and more specific radial matrix elements can be computed via the hydrogenic wavefunctions. For instance, one finds relations such as

$$\left\langle\frac{Z e^2}{r^3}\right\rangle_{n l} = \frac{m_e}{\hbar^2\ l(l+1)}\ \left\langle\frac{(Z e^2)^2}{r^2}\right\rangle_{n l},$$

using commutator techniques with $p_r$ and $[H_0,\ p_r]=0$ for the hydrogenic Hamiltonian.