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[[File:Fabry Perot Etalon Rings Fringes.png|right|thumb|200px|[[Interference (wave propagation)|Interference fringes]], showing fine structure (splitting) of a cooled [[deuterium]] source, viewed through a [[Fabry-Pérot étalon]].]]
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In [[atomic physics]], the '''fine structure''' describes the splitting of the [[spectral line]]s 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 [[Degenerate energy level|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'')<sup>2</sup>, where ''Z'' is the [[atomic number]] and ''α'' is the [[fine-structure constant]], a [[dimensionless quantity|dimensionless number]] equal to approximately {{val|7.297|e=-3}}.
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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 (quantum mechanics)|Hamiltonian]] is given by
: <math>H=H_{0}+H_{\mathrm{kinetic}}+H_{\mathrm{so}}+H_{\mathrm{Darwinian}}.\!</math>
 
==Kinetic energy relativistic correction==
Classically, the kinetic energy term of the [[Hamiltonian Mechanics|Hamiltonian]] is
 
: <math>T=\frac{p^{2}}{2m}.</math>
 
However, when considering a more accurate theory of Nature viz. [[special relativity]], we must use a relativistic form of the kinetic energy,
 
: <math>T=\sqrt{p^{2}c^{2}+m^{2}c^{4}}-mc^{2},</math>
 
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]] ( specifically a [[Binomial series]] ), we find
 
: <math>T=\frac{p^{2}}{2m}-\frac{p^{4}}{8m^{3}c^{2}}+\cdots.</math>
 
Then, the first order correction to the Hamiltonian is
 
: <math>H_{\mathrm{kinetic}}=-\frac{p^{4}}{8m^{3}c^{2}}.</math>
 
Using this as a [[Perturbation theory (quantum mechanics)|perturbation]], we can calculate the first order energy corrections due to relativistic effects.
 
: <math>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</math>
 
where <math>\psi^{0}</math> is the unperturbed wave function.  Recalling the unperturbed Hamiltonian, we see
 
: <math>H^{0}\vert\psi^{0}\rangle=E_{n}\vert\psi^{0}\rangle </math>
 
: <math>\left(\frac{p^{2}}{2m}+V\right)\vert\psi^{0}\rangle=E_{n}\vert\psi^{0}\rangle </math>
 
: <math>p^{2}\vert\psi^{0}\rangle=2m(E_{n}-V)\vert\psi^{0}\rangle</math>
 
We can use this result to further calculate the relativistic correction:
 
: <math>E_{n}^{(1)}=-\frac{1}{8m^{3}c^{2}}\langle\psi^{0}\vert p^{2}p^{2}\vert\psi^{0}\rangle</math>
 
: <math>E_{n}^{(1)}=-\frac{1}{8m^{3}c^{2}}\langle\psi^{0}\vert (2m)^{2}(E_{n}-V)^{2}\vert\psi^{0}\rangle </math>
 
: <math>E_{n}^{(1)}=-\frac{1}{2mc^{2}}(E_{n}^{2}-2E_{n}\langle V\rangle +\langle V^{2}\rangle )</math>
 
For the hydrogen atom, <math>V=\frac{e^{2}}{r}</math>, <math>\langle V\rangle=\frac{-e^{2}}{a_{0}n^{2}}</math>, and <math>\langle V^{2}\rangle=\frac{e^{4}}{(l+1/2)n^{3}a_{0}^{2}}</math> where <math>a_{0}</math> is the [[Bohr Radius]], <math>n</math> is the [[principal quantum number]] and <math>l</math> is the [[azimuthal quantum number]]. Therefore the relativistic correction for the hydrogen atom is
 
: <math>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)</math>
 
where we have used:
 
:<math> E_n = - \frac{e^2}{2 a_0 n^2} </math>
 
On final calculation, the order of magnitude for the relativistic correction to the ground state is <math> -9.056 \times 10^{-4}\ \text{eV}</math>.
 
'''Note:''' In reality, <math>p^{4}</math> is not a [[Hermitian operator]] for hydrogen-like s-orbitals <math>(l = 0)</math>. 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 <math>l = 0</math>. 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 <math>l = 0</math>.
 
==Spin-orbit coupling==
: <math>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}}</math>
 
The [[Spin (physics)|spin]]-orbit correction arises when we shift from the standard [[frame of reference]] (where the [[electron]] orbits the [[Atomic nucleus|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, <math>\vec B</math> and <math>\vec\mu_s</math> 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
 
:<math> \Delta E_{SO} = \xi (r)\vec L \cdot \vec S</math>
 
Notice that there is a factor of 2, which comes from the relativistic calculation that changes back to the electron's frame from nucleus frame by [[Llewellyn Thomas]]. This factor also called the Thomas factor.
 
since
 
: <math> \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)}</math>
 
: <math> \left\langle \vec L \cdot \vec S \right\rangle = \frac {\hbar^2} {2} ( j(j+1) - l(l+1) - s(s+1) )</math>
 
the expectation value for the Hamiltonian is:
 
: <math> \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)</math>
 
Thus the order of magnitude for the spin-orbital coupling is <math> \frac{Z^4}{n^3(j+1/2)} 10^{-5}\text{ eV}</math>.
 
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 ==
: <math> 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)</math>
 
: <math> H_{\mathrm{Darwinian}}=\frac{\hbar^{2}}{8m_{e}^{2}c^{2}}\,4\pi\left(\frac{Ze^2}{4\pi \epsilon_{0}}\right)| \psi(0)^2|</math>
 
:: <math> \psi (0) = 0 \text{ for } l > 0 </math>
:: <math> \psi (0) = \frac{1}{\sqrt{4\pi}}\,2 \left( \frac {Z}{n a_0} \right)^\frac {3}{2} \text{ for } l = 0 </math>
 
: <math> H_{\mathrm{Darwinian}}=\frac{2n}{m_e c^{2}}\,E_n^2</math>
 
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 {{val|9.057|e=-5|u=eV}}.
 
The [[Charles Galton Darwin|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 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:<ref>{{Cite book
| publisher = Butterworth-Heinemann
| isbn = 978-0-7506-3371-0
| last = Berestetskii
| first = V. B.
| coauthors = E. M. Lifshitz, L. P. Pitaevskii
| title = Quantum electrodynamics
| year = 1982
}}</ref>
 
: <math>\Delta E = \frac{E_{n}(Z\alpha)^{2}}{n}\left( \frac{1}{j + 1/2} - \frac{3}{4n} \right)\,,</math>
where <math>j</math> is the [[total angular momentum]] (<math>j = 1/2</math> if <math>l = 0</math> and <math>j = l \pm 1/2</math> otherwise). It is worth noting that this expression was first obtained by [[Arnold Sommerfeld|A. Sommerfeld]] based on the [[Old quantum theory|old Bohr theory]], i.e., before the modern [[quantum mechanics]] was formulated.
 
[[File:New Microsoft Office PowerPoint Presentation.png|frame|center]]
 
==See also==
* [[Spin-orbit interaction]]
* [[Angular momentum coupling]]
* [[Hyperfine structure]]
 
==References==
{{Reflist}}
*{{cite book | author=Griffiths, David J.|title=Introduction to Quantum Mechanics (2nd ed.) | publisher=Prentice Hall |year=2004 |isbn=0-13-805326-X}}
*{{cite book | author=[[Liboff, Richard L.]] | title=Introductory Quantum Mechanics | publisher=Addison-Wesley | year=2002 | isbn=0-8053-8714-5}}
 
==External links==
*[http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/hydfin.html#c1 Hyperphysics: Fine Structure]
*[http://farside.ph.utexas.edu/teaching/qmech/lectures/node107.html University of Texas: The fine structure of hydrogen]
 
[[Category:Atomic physics]]

Latest revision as of 20:19, 1 January 2015

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