# Fine structure

Interference fringes, showing fine structure (splitting) of a cooled deuterium source, viewed through a Fabry-Pérot étalon.

In atomic physics, the fine structure describes the splitting of the spectral lines of atoms due to quantum-mechanical (electron spin) and relativistic corrections.

The gross structure of line spectra is the line spectra predicted by the quantum mechanics of non-relativistic electrons with no spin. For a hydrogenic atom, the gross structure energy levels only depend on the principal quantum number n. However, a more accurate model takes into account relativistic and spin effects, which break the degeneracy of the energy levels and split the spectral lines. The scale of the fine structure splitting relative to the gross structure splitting is on the order of ()2, where Z is the atomic number and α is the fine-structure constant, a dimensionless number equal to approximately $1/137$.

The fine structure can be separated into three corrective terms: the kinetic energy term, the spin-orbit term, and the Darwinian term. The full Hamiltonian is given by

$H=H_{0}+H_{\mathrm{kinetic}}+H_{\mathrm{so}}+H_{\mathrm{Darwinian}}.\!$

This can be seen as a non-relativistic approximation of the Dirac equation.

## Kinetic energy relativistic correction

Classically, the kinetic energy term of the Hamiltonian is

$T=\frac{p^{2}}{2m},$

where $p$ is the momentum and $m$ is the mass of the electron.

However, when considering a more accurate theory of Nature viz. special relativity, we must use a relativistic form of the kinetic energy,

$T=\sqrt{p^{2}c^{2}+m^{2}c^{4}}-mc^{2},$

where the first term is the total relativistic energy, and the second term is the rest energy of the electron. ($c$ is the speed of light) Expanding this in a Taylor series ( specifically a Binomial series ), we find

$T=\frac{p^{2}}{2m}-\frac{p^{4}}{8m^{3}c^{2}}+\cdots.$

Then, the first order correction to the Hamiltonian is

$H_{\mathrm{kinetic}}=-\frac{p^{4}}{8m^{3}c^{2}}.$

Using this as a perturbation, we can calculate the first order energy corrections due to relativistic effects.

$E_{n}^{(1)}=\langle\psi^{0}\vert H'\vert\psi^{0}\rangle=-\frac{1}{8m^{3}c^{2}}\langle\psi^{0}\vert p^{4}\vert\psi^{0}\rangle=-\frac{1}{8m^{3}c^{2}}\langle\psi^{0}\vert p^{2}p^{2}\vert\psi^{0}\rangle$

where $\psi^{0}$ is the unperturbed wave function. Recalling the unperturbed Hamiltonian, we see

$H^{0}\vert\psi^{0}\rangle=E_{n}\vert\psi^{0}\rangle$
$\left(\frac{p^{2}}{2m}+V\right)\vert\psi^{0}\rangle=E_{n}\vert\psi^{0}\rangle$
$p^{2}\vert\psi^{0}\rangle=2m(E_{n}-V)\vert\psi^{0}\rangle$

We can use this result to further calculate the relativistic correction:

$E_{n}^{(1)}=-\frac{1}{8m^{3}c^{2}}\langle\psi^{0}\vert p^{2}p^{2}\vert\psi^{0}\rangle$
$E_{n}^{(1)}=-\frac{1}{8m^{3}c^{2}}\langle\psi^{0}\vert (2m)^{2}(E_{n}-V)^{2}\vert\psi^{0}\rangle$
$E_{n}^{(1)}=-\frac{1}{2mc^{2}}(E_{n}^{2}-2E_{n}\langle V\rangle +\langle V^{2}\rangle )$

For the hydrogen atom, $V=\frac{e^{2}}{r}$, $\langle V\rangle=\frac{-e^{2}}{a_{0}n^{2}}$, and $\langle V^{2}\rangle=\frac{e^{4}}{(l+1/2)n^{3}a_{0}^{2}}$ where $a_{0}$ is the Bohr Radius, $n$ is the principal quantum number and $l$ is the azimuthal quantum number. Therefore the relativistic correction for the hydrogen atom is

$E_{n}^{(1)}=-\frac{1}{2mc^{2}}\left(E_{n}^{2}+2E_{n}\frac{e^{2}}{a_{0}n^{2}} +\frac{e^{4}}{(l+1/2)n^{3}a_{0}^{2}}\right)=-\frac{E_{n}^{2}}{2mc^{2}}\left(\frac{4n}{l+1/2}-3\right)$

where we have used:

$E_n = - \frac{e^2}{2 a_0 n^2}$

On final calculation, the order of magnitude for the relativistic correction to the ground state is $-9.056 \times 10^{-4}\ \text{eV}$.

## Spin-orbit coupling

For a hydrogen-like atom with $Z$ protons, orbital momentum $\vec L$ and electron spin $\vec S$, the spin-orbit term is given by:

$H_{so}=\frac{1}{2} \left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)\left(\frac{g_s}{2m_{e}^{2}c^{2}}\right)\frac{\vec L\cdot\vec S}{r^{3}}$

$m_e$ is the electron mass, $\epsilon_0$ is the vacuum permittivity and $g_s$ is the spin g-factor. $r$ is the distance of the electron from the nucleus.

The spin-orbit correction arises when we shift from the standard frame of reference (where the electron orbits the nucleus) into one where the electron is stationary and the nucleus instead orbits it. In this case the orbiting nucleus functions as an effective current loop, which in turn will generate a magnetic field. However, the electron itself has a magnetic moment due to its intrinsic angular momentum. The two magnetic vectors, $\vec B$ and $\vec\mu_s$ couple together so that there is a certain energy cost depending on their relative orientation. This gives rise to the energy correction of the form

$\Delta E_{SO} = \xi (r)\vec L \cdot \vec S$

Notice that there is a factor of 2, called the Thomas precession, which comes from the relativistic calculation that changes back to the electron's frame from the nucleus frame.

Since

$\left\langle \frac {1}{r^3} \right\rangle = \frac {Z^3}{n^3 a_0^3} \frac {1} {l (l+\frac{1}{2}) (l + 1)}$
$\left\langle \vec L \cdot \vec S \right\rangle = \frac {\hbar^2} {2} ( j(j+1) - l(l+1) - s(s+1) )$

the expectation value for the Hamiltonian is:

$\left\langle H_{SO} \right\rangle = \frac{E_n{}^2}{m_e c^2} \left( n \frac{j(j+1)-l(l+1)-\frac{3}{4}}{l \left( l+\frac{1}{2}\right) (l+1) } \right)$

Thus the order of magnitude for the spin-orbital coupling is $\frac{Z^4}{n^3(j+1/2)} 10^{-5}\text{ eV}$.

Remark: On the (n,l,s)=(n,0,1/2) and (n,l,s)=(n,1,-1/2) energy level, which the fine structure said their level are the same. If we take the g-factor to be 2.0031904622, then, the calculated energy level will be different by using 2 as g-factor. Only using 2 as the g-factor, we can match the energy level in the 1st order approximation of the relativistic correction. When using the higher order approximation for the relativistic term, the 2.0031904622 g-factor may agree with each other. However, if we use the g-factor as 2.0031904622, the result does not agree with the formula, which included every effect.

## Darwin term

One term in the non-relativistic expansion of the Dirac equation is given by:

$H_{\mathrm{Darwinian}}=\frac{\hbar^{2}}{8m_{e}^{2}c^{2}}\,4\pi\left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)\delta^{3}\left(\vec r\right)$
$\langle H_{\mathrm{Darwinian}} \rangle =\frac{\hbar^{2}}{8m_{e}^{2}c^{2}}\,4\pi\left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)| \psi(0)|^2$
$\psi (0) = 0 \text{ for } l > 0$
$\psi (0) = \frac{1}{\sqrt{4\pi}}\,2 \left( \frac {Z}{n a_0} \right)^\frac {3}{2} \text{ for } l = 0$
$H_{\mathrm{Darwinian}}=\frac{2n}{m_e c^{2}}\,E_n^2$

Thus, the Darwin term affects only the s-orbit. For example it gives the 2s-orbit the same energy as the 2p-orbit by raising the 2s-state by 9.057×10−5 eV.

The Darwin term changes the effective potential at the nucleus. It can be interpreted as a smearing out of the electrostatic interaction between the electron and nucleus due to zitterbewegung, or rapid quantum oscillations, of the electron. This can be motivated by a short calculation[1]

Quantum fluctuations allow for the creation of virtual electron-positron pairs with a lifetime estimated by the uncertainty principle $\Delta t \approx \hbar/\Delta E \approx \hbar/mc^2$. The distance the particles can move during this time is $\xi \approx c\Delta t \approx \hbar/mc = \lambda_c$, the Compton wavelength. The electrons of the atom interact with those pairs. This yields a fluctuating electron position $\vec r + \vec \xi$. Using a Taylor expansion, the effect on the potential $U$ can be estimated:

$U(\vec r + \vec\xi) \approx U(\vec r) + \xi\cdot\nabla U(\vec r) + \frac12 \sum_{ij} \xi_i\xi_j \partial_i\partial_j U(\vec r)$

Averaging over the fluctuations $\vec \xi$

$\overline\xi = 0, \quad \overline{\xi_i\xi_j} = \frac13 \overline{\vec\xi^2} \delta_{ij},$

gives the average potential

$\overline{U(\vec r + \vec\xi)} = U(\vec r) + \frac16 \overline{\vec\xi^2} \nabla^2 U(\vec r).$

Approximating $\overline{\vec\xi^2} \approx \lambda_c^2$, this yields the perturbation of the potential due to fluctuations:

$\delta U \approx \frac16 \lambda_c^2 \nabla^2 U = \frac{\hbar^2}{6m^2c^2}\nabla^2 U$

To compare with the expression above, plug in the Coulomb potential:

$\nabla^2 U = -\nabla^2 \frac{Z e^2}{4\pi\epsilon_0 r} = 4\pi \left(\frac{Z e^2}{4\pi\epsilon_0}\right) \delta(\vec r) \quad\Rightarrow\quad \delta U \approx \frac{\hbar^2}{6m^2c^2} 4\pi \left(\frac{Z e^2}{4\pi\epsilon_0}\right) \delta(\vec r)$

This is only slightly different.

Another mechanism that affects only the s-state is the Lamb shift. The reader should not confuse the Darwin term with the Lamb shift. The Darwin term makes the s-state and p-state the same energy, but the Lamb shift makes the s-state higher in energy than the p-state.

## Total effect

The total effect, obtained by summing the three components up, is given by the following expression:[2]

$\Delta E = \frac{E_{n}(Z\alpha)^{2}}{n}\left( \frac{1}{j + 1/2} - \frac{3}{4n} \right)\,,$

where $j$ is the total angular momentum ($j = 1/2$ if $l = 0$ and $j = l \pm 1/2$ otherwise). It is worth noting that this expression was first obtained by A. Sommerfeld based on the old Bohr theory, i.e., before the modern quantum mechanics was formulated.