Spin-Orbit Interaction

Although neglected up to this lecture, the interaction between the electron-spin and the orbital angular momentum must also be included in the atomic Hamiltonian. Such interaction is described according to the spin-orbit Hamiltonian defined as follows,

$\displaystyle \hat{H}_{SO} = \frac{1}{2m_e c^2} \frac{1}{r} \Bigg(\frac{\partial V}{\partial r}\Bigg) \hat{L} \cdot \hat{S} = \xi \hat{L} \cdot \hat{S},$ (50)

where $ V$ is the Coulombic potential of the electron in the field of the atom. Note that the spin-orbit interaction is proportional to $ \hat{L} \cdot \hat{S}$. A proper derivation of Eq. (50) requires a relativistic treatment of the electron which is beyond the scope of these lectures.

Note: A classical description of such interaction also gives a perturbation proportional to $ \hat{L} \cdot \hat{S}$. This is because from the reference frame of the electron, the nucleus is a moving charge that generates a magnetic field $ B$, proportional to $ \hat{L}$. Such magnetic field interacts with the spin magnetic moment  $ m_s = -e/m_e \hat{S}$. Therefore, the interaction between $ B$ and $ m_s$ is proportional to $ \hat{L} \cdot \hat{S}$. Unfortunately, however, the proportionality constant predicted by such classical model is incorrect, and a proper derivation requires a relativistic treatment of the electron as mentioned earlier in this section.

In order to compute the spin-orbit Hamiltonian of a many-electron atom, it is necessary to compute first an approximate effective potential $ V_i$ of each electron $ i$ in the total electric field of electrons and nuclear charges. Then, we can compute the sum over all electrons as follows,

$\displaystyle \hat{H}_{SO} \approx \frac{1}{2m_e c^2} \sum_i \frac{1}{r_i} \fra...
...artial r_i} \hat{L}_i \cdot \hat{S}_i = \sum_i \xi_i \hat{L}_i \cdot \hat{S}_i.$ (51)

The correction of eigenfunctions and eigenvalues, due to the spin-orbit coupling, is usually computed according to perturbation theory after solving the atomic eigenvalue problem in the absence of the spin-orbit interaction. For example, the spin-orbit correction to the eigenvalue of state $ \mid \Psi \rangle$ for a one-electron atom is,

$\displaystyle E_{S.O.}^{(1)} \approx \langle \Psi \mid \xi \hat{L} \cdot \hat{S} \mid \Psi \rangle.$ (52)

Note that the $ L\cdot S$ product can be written in terms of $ J^2, L^2$ and $ S^2$ as follows, $ L\cdot S= \frac{1}{2}(J^2 - L^2 -S^2),$ because, $ J^2= J\cdot J =(L+S)(L+S)= L^2 +S^2 +2 L\cdot S,$ and, since the unperturbed wave function is an eigenfunction of $ L^2$, $ S^2$ and $ J^2$,

$\displaystyle L \cdot S\vert\psi> = \frac{1}{2} \hbar^2(J(J+1) -L(L+1) -S(S+1))\vert \psi>.$


$\displaystyle E_{S.O.} \approx \frac{1}{2} \hbar^2 <\xi> \left [ J(J+1)- L(L+1) -S(S+1) \right].$

It is important to note that, due to the spin-orbit coupling, the total energy of a state depends on the value of the total angular momentum quantum number $ J$. Furthermore, each of the energy levels is (2J+1) times degenerate, as determined by the possible values of $ M_J$. For example, when L=1, and S=1/2, then the possible values of J are 1/2 and 3/2, since (J=L+S, L+S-1, ..., L-S).

The spin orbit interaction is, therefore, responsible for splitting of spectroscopic lines in atomic spectra.

It is possible to remove the degeneracy of energy levels by applying an external magnetic field that perturbs the system as follows, $ H_B= -m\cdot B,$ where $ m=m_L +m_S$, with $ m_L=-\frac{e}{2m_e}L$, and $ m_S= -\frac{e}{m_e} S.$ The external perturbation is, therefore, described by the following Hamiltonian,

$\displaystyle H_B=- \frac{e}{2m_e}(L+2S) \cdot B= -\frac{e}{2m_e}(J+S) \cdot B.$

The energy correction according to first-order perturbation theory is:

$\displaystyle E_B= -\frac{e}{2m_e} B (\hbar M_J +<S_z>) = A B M_J,$

where $ <S_z> = \hbar M_J \frac{J(J+1) -L(L+1) +S(S+1)}{2J(J+1)}$ and $ A$ is a proportionality constant. Therefore, the perturbation of an external magnetic field splits energy level characterized by quantum number J into 2J+1 energy sub-levels. These sub-levels correspond to different possible values of $ M_J$, as described by the following diagram:

...{$ ^{-1}$}}
\put(130,4){\makebox(0,0)[t]{$ ^{-2}$}}

Exercise 45: (A). Calculate the energy of the spectroscopic lines associated with transitions 3S $ \rightarrow$ 3P for Na in the absence of an external magnetic field. (B). Calculate the spectroscopic lines associated with transitions 3S $ \rightarrow$ 3P for Na atoms perturbed by an external magnetic field $ B_z$ as follows:

$\displaystyle \hat{H}_B = -\hat{m} \cdot B = \beta_e B \hbar^{-1} (\hat{J_z} +\hat{S_z}),$

and $ E_B = <\psi \vert \hat{H}_B \vert \psi> = \beta_e B M_J g,$ with $ g=1+ \frac{J(J+1)-L(L+1) +S(S+1)}{2J(J+1)}.$