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 first order relativistic corrections.

The gross structure of line spectra is the line spectra predicted by 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 (Zα)², where Z is the atomic number and α is the fine-structure constant, a dimensionless number equal to approximately 7.297×10⁻³.

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

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

Kinetic energy relativistic correction

Classically, the kinetic energy term of the Hamiltonian is

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

However, when considering 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. Expanding this in a Taylor 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 ψ⁰ 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 a₀ 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 $-9.056 \times 10^{-4}\ \text{eV}$.

Note: In reality, p⁴ is not a Hermitian operator for hydrogen-like s-orbitals (l = 0). The use of first order quantum perturbation theory requires that the perturbing Hamiltonian be Hermitian. Thus, the proof shown above is not entirely rigorous when l = 0. Despite this shortcoming, comparison with the exact answer (derived from the Dirac equation) shows that the result shown above is correct to the first order, even when (l = 0).

Spin-orbit coupling

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

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, which is come from the relativistic calculation of change back to electron frame from nucleus frame by Llewellyn Thomas. This factor also called the Thomas factor.

since

$\left\langle \frac {1}{r^3} \right\rangle = \frac {1}{n^3 a_0^3} \frac {1} {l (l+\frac{1}{2}) (l + 1)}$

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

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 $\frac{Z}{n^3} 10^{-5}\text{ eV}$.

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

$E_{\mathrm{Darwin}}=\frac{\hbar^{2}}{8m_{e}^{2}c^{2}}\,4\pi\left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)\delta^{3}\left(\vec r\right)$

$E_{\mathrm{Darwin}}=\frac{\hbar^{2}}{8m_{e}^{2}c^{2}}\,4\pi\left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)| \psi(0)^2|$

ψ(0) = 0 for l > 0

$\psi (0) = \frac{1}{\sqrt{4\pi}}\,2 \left( \frac {Z}{n a_0} \right)^\frac {3}{2} \text{ for } l = 0$

$E_{\mathrm{Darwin}}=\frac{2n}{m_e c^{2}}\,E_n^2$

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.

Another mechanism that affects only the s-state is the Lamb shift. The reader should not mix up the Darwin term and 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 :

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

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.

See also

• Spin-orbit interaction
• Angular momentum coupling
• Hyperfine structure
• Lamb shift

References

• Griffiths, David J. (2004). Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. ISBN 0-13-805326-X. 
• Liboff, Richard L. (2002). Introductory Quantum Mechanics. Addison-Wesley. ISBN 0-8053-8714-5. 

External links

• Hyperphysics: Fine Structure
• University of Texas: The fine structure of hydrogen