This section presents a relatively simple and quantitative description of the spin–orbit interaction for an electron bound to a
hydrogen-like atom, up to first order in
perturbation theory, using some
semiclassical electrodynamics and non-relativistic quantum mechanics. This gives results that agree reasonably well with observations. A rigorous calculation of the same result would use
relativistic quantum mechanics, using the
Dirac equation, and would include
many-body interactions. Achieving an even more precise result would involve calculating small corrections from
quantum electrodynamics.
Energy of a magnetic moment The energy of a magnetic moment in a magnetic field is given by \Delta H = -\boldsymbol{\mu}\cdot\mathbf{B}, where is the
magnetic moment of the particle, and is the
magnetic field it experiences.
Magnetic field We shall deal with the
magnetic field first. Although in the rest frame of the nucleus, there is no magnetic field acting on the electron, there
is one in the rest frame of the electron (see
classical electromagnetism and special relativity). Ignoring for now that this frame is not
inertial, we end up with the equation \mathbf{B} = -\frac{\mathbf{v} \times \mathbf{E}}{c^2}, where is the velocity of the electron, and is the electric field it travels through. Here, in the non-relativistic limit, we assume that the Lorentz factor \gamma \backsimeq 1. Now we know that is radial, so we can rewrite \mathbf{E} = \left| E \right| \frac{\mathbf{r}}{r} . Also we know that the momentum of the electron \mathbf{p} = m_\text{e} \mathbf{v} . Substituting these and changing the order of the cross product (using the identity \mathbf{A} \times \mathbf{B} = -\mathbf{B} \times \mathbf{A}) gives \mathbf{B} = \frac{\mathbf{r} \times \mathbf{p}}{m_\text{e} c^2} \left| \frac{E}{r} \right|. Next, we express the electric field as the gradient of the
electric potential \mathbf{E} = -\nabla V. Here we make the
central field approximation, that is, that the electrostatic potential is spherically symmetric, so is only a function of radius. This approximation is exact for hydrogen and hydrogen-like systems. Now we can say that |E| = \left|\frac{\partial V}{\partial r}\right| = \frac{1}{e} \frac{\partial U(r)}{\partial r}, where U = -eV is the
potential energy of the electron in the central field, and is the
elementary charge. Now we remember from classical mechanics that the
angular momentum of a particle \mathbf{L} = \mathbf{r} \times \mathbf{p}. Putting it all together, we get \mathbf{B} = \frac{1}{m_\text{e} ec^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \mathbf{L}. It is important to note at this point that is a positive number multiplied by , meaning that the
magnetic field is parallel to the
orbital
angular momentum of the particle, which is itself perpendicular to the particle's velocity.
Spin magnetic moment of the electron The
spin magnetic moment of the electron is \boldsymbol{\mu}_S = -g_\text{s} \mu_\text{B} \frac{\mathbf{S}}{\hbar}, where \mathbf{S} is the spin (or intrinsic angular-momentum) vector, \mu_\text{B} is the
Bohr magneton, and g_\text{s} = 2.0023... \approx 2 is the electron-spin
g-factor. Here \boldsymbol{\mu} is a negative constant multiplied by the
spin, so the
spin magnetic moment is antiparallel to the spin. The spin–orbit potential consists of two parts. The Larmor part is connected to the interaction of the spin magnetic moment of the electron with the magnetic field of the nucleus in the co-moving frame of the electron. The second contribution is related to
Thomas precession.
Larmor interaction energy The Larmor interaction energy is \Delta H_\text{L} = -\boldsymbol{\mu} \cdot \mathbf{B}. Substituting in this equation expressions for the spin magnetic moment and the magnetic field, one gets \Delta H_\text{L} = \frac{g_\text{s}\mu_\text{B}}{\hbar m_\text{e} e c^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \mathbf{L} \cdot \mathbf{S} \approx \frac{2\mu_\text{B}}{\hbar m_\text{e} e c^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \mathbf{L} \cdot \mathbf{S}. Now we have to take into account
Thomas precession correction for the electron's curved trajectory.
Thomas interaction energy In 1926
Llewellyn Thomas relativistically recomputed the doublet separation in the fine structure of the atom. Thomas precession rate \boldsymbol{\Omega}_\text{T} is related to the angular frequency of the orbital motion \boldsymbol{\omega} of a spinning particle as follows: \boldsymbol{\Omega}_\text{T} = -\boldsymbol{\omega} (\gamma - 1), where \gamma is the
Lorentz factor of the moving particle. The Hamiltonian producing the spin precession \boldsymbol{\Omega}_\text{T} is given by \Delta H_\text{T} = \boldsymbol{\Omega}_\text{T} \cdot \mathbf{S}. To the first order in (v/c)^2, we obtain \Delta H_\text{T} = -\frac{\mu_\text{B}}{\hbar m_\text{e} e c^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \mathbf{L}\cdot \mathbf{S}.
Total interaction energy The total spin–orbit potential in an external electrostatic potential takes the form \Delta H \equiv \Delta H_\text{L} + \Delta H_\text{T} = \frac{(g_\text{s}-1) \mu_\text{B}}{\hbar m_\text{e} e c^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \mathbf{L} \cdot \mathbf{S} \approx \frac{\mu_\text{B}}{\hbar m_\text{e} e c^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \mathbf{L} \cdot \mathbf{S}. The net effect of Thomas precession is the reduction of the Larmor interaction energy by factor of about 1/2, which came to be known as the
Thomas half.
Evaluating the energy shift Thanks to all the above approximations, we can now evaluate the detailed energy shift in this model. Note that and are no longer conserved quantities. In particular, we wish to find a new basis that diagonalizes both (the non-perturbed Hamiltonian) and . To find out what basis this is, we first define the
total angular momentum operator \mathbf{J} = \mathbf{L} + \mathbf{S}. Taking the
dot product of this with itself, we get \mathbf{J}^2 = \mathbf{L}^2 + \mathbf{S}^2 + 2\, \mathbf{L} \cdot \mathbf{S} (since and commute), and therefore \mathbf{L} \cdot \mathbf{S} = \frac{1}{2} \left(\mathbf{J}^2 - \mathbf{L}^2 - \mathbf{S}^2 \right) It can be shown that the five operators , , , , and all commute with each other and with Δ
H. Therefore, the basis we were looking for is the simultaneous
eigenbasis of these five operators (i.e., the basis where all five are diagonal). Elements of this basis have the five
quantum numbers: n (the "principal quantum number"), j (the "total angular momentum quantum number"), \ell (the "orbital angular momentum quantum number"), s (the "spin quantum number"), and j_z (the " component of total angular momentum"). To evaluate the energies, we note that \left\langle \frac{1}{r^3} \right\rangle = \frac{2}{a^3 n^3\; \ell (\ell + 1) (2\ell + 1)} for hydrogenic wavefunctions (here a = \hbar / (Z \alpha m_\text{e} c) is the
Bohr radius divided by the nuclear charge ); and \left\langle \mathbf{L} \cdot \mathbf{S} \right\rangle = \frac{1}{2} \big(\langle \mathbf{J}^2 \rangle - \langle \mathbf{L}^2 \rangle - \langle \mathbf{S}^2 \rangle\big) = \frac{\hbar^2}{2} \big(j (j + 1) - \ell (\ell + 1) - s (s + 1)\big).
Final energy shift We can now say that \Delta E = \frac{\beta}{2} \big(j(j+1) - \ell(\ell+1) - s(s+1)\big), where the spin-orbit coupling constant is \beta = \beta(n,l) = Z^4 \frac{\mu_0}{4\pi} g_\text{s} \mu_\text{B}^2 \frac{1}{n^3 a_0^3\;\ell(\ell+1/2)(\ell+1)}. For the exact relativistic result, see the
solutions to the Dirac equation for a hydrogen-like atom. The derivation above calculates the interaction energy in the (momentaneous) rest frame of the electron and in this reference frame there's a magnetic field that's absent in the rest frame of the nucleus. Another approach is to calculate it in the rest frame of the nucleus, see for example George P. Fisher:
Electric Dipole Moment of a Moving Magnetic Dipole (1971). However the rest frame calculation is sometimes avoided, because one has to account for
hidden momentum. == Scattering ==