Fine structure

The fine structure of the line Spectral line of a Atome corresponds to its separation in several components of Fréquence S very close, detectable by a Spectroscope of good resolution.

This structure is explained within the framework of the Quantum physics. It is due to the lifting partial of the degeneration of a Energy level of the Modèle of Bohr because of three corrections:

coupling of the magnetic Moment of spin of the electron with the Magnetic field generated by its movement (orbital magnetic moment);
the taking into account of the relativistic movement of the electron;
the effect of the Zitterbewegung of the electron, which makes that this one feels the average nuclear electric field on an area, and not in a specific way.

The discovery of the fine structure of the atomic Hydrogène was worth the Nobel Prize of physics to Willis Eugene Lamb in 1955.

Relativistic correction

In the case slightly relativistic, the Hamiltonien is written

H = \ frac {p^2} {2m} - \ frac {p^4} {8m^3c^2}

On the basis of the Hamiltonian of the solution not-relativist H of clean states $\ psi_ {nlm_l}$ of energy E ,

H = H_0 - \ frac {1} {2mc^2} (H_0-V) ^2

where V represents the potential, the Théorie of the disturbances makes it possible to write:

\ Delta E^ \ mathrm {rel} _ {nlm_l} = - \ frac {1} {8mc^2} \ left \ langle \ psi_ {nlm_l} | (H_0-V) ^2 | \psi_{nlm_l} \right\rangle

As follows:

\ Delta E^ \ mathrm {rel} _ {nlm_l} = - \ frac {1} {8mc^2} \ left (E_ {N} ^2 - 2E_ {N} \ langle \ psi_ {nlm_l}|V|\ psi_ {nlm_l} \ rangle + \ langle \ psi_ {nlm_l}|V^2|\ psi_ {nlm_l} \ rangle \ right)

In the case of a Hydrogénoïde, the potential is Coulomb and the nondisturbed clean states are harmonic spherical. The expression above becomes:

\ Delta E^ \ mathrm {rel} _ {nlm_l} = - \ frac {(Z \ alpha) ^2} {N} \ left (\ frac {1} {l+1/2} - \ frac {3} {4n} \ right) |E_ {N}|

Zitterbewegung

Spin-orbit coupling

Origin of the perturbatif term

The Mécanique quantum relativist reveals, amongst other things, the fact that the electrons have a spin. This one generates a magnetic Moment of spin

$\ vec {M_s} = \ frac {Q} {m_e} \ vec {S}$

As the electron moves in an environment where reign the Electric field created by the loads of the core and others electrons, according to the restricted Relativité, the electron, in its reference frame, perceives a Magnetic field called field motionnel

$\ vec {B'} = - \ frac {\ vec {v} \ wedge \ vec {E}} {c^2}$

Energy associated with this interaction is thus

$W_ {so} = - \ vec {M_s} \ cdot \ vec {B'}$

As the reference frame of the electron is in rotation and not galiléen, the calculation of the field motionnel requires to make two changes of reference frames (one in translation and one in rotation). The calculation made by Thomas gives

$W_ {so} = \ frac {1} {2m_e^2c^2} \ frac {1} {R} \ frac \ vec {L} \ cdot \ vec {S}$

with $\ vec {L}$ the kinetic Moment of the electron around the core and $\ vec {S}$ the kinetic Moment of Spin of the electron.

It is common to note this term

$W_ {so} = \ xi (R) \ vec {L} \ cdot \ vec {S} {\ textrm ~~avec~~} \ xi (R) = \ frac {1} {2m_e^2c^2} \ frac {1} {R} \ frac$

what makes it possible to emphasize the purely radial term.

Calculation in disturbance

On the assumption that this term contributes a weak share to the energy in front of the principal term

H_0

, one can treat it in disturbance. But before, it is advisable to notice that the term

\ vec {L} \ cdot \ vec {S}

does not commutate not with

\ vec {L}

and

\ vec {S}

. It is thus essential to find a new Whole Complete of Observable which Commutent (ECOC). With this intention, the kinetic Moment total

$\ vec {J} ~ \ stackrel {def} {=} \ sum \ vec {L} ~~ \ Leftrightarrow ~~ \ vec {J} = \ vec {L} + \ vec {S}$

is used instead of each kinetic moments and the new ECOC becomes $H, L^2, S^2, J^2, J_z$ . The base of the common clean vectors becomes then $\ left| \ psi_ {nlsjm_j} \ right \ rangle$ with $m_j = m_l + m_s$ . It results from it

$J^2 = L^2 + S^2 + 2 \ vec {L} \ cdot \ vec {S} ~~ \ Leftrightarrow ~~ \ vec {L} \ cdot \ vec {S} = \ frac {1} {2} \ left (J^2 - L^2 - S^2 \ right)$

from where

$W_ {so} = \ frac {1} {2} \ xi (R) \ left (J^2 - L^2 - S^2 \ right)$

The Théorie of the disturbances makes it possible to write:

$\ Delta E_ {nlsj} ^ {so} = \ frac {1} {2} \ left \ langle \ psi_ {nlsjm_j} \ left| \ xi (R) \ left (J^2 - L^2 - S^2 \ right) \ right| \psi_{nlsjm_j} \right\rangle$

By posing

\ frac {A_ {nl}} {\ hbar^2} = \ int_0^ \ infty \ left| R_{nl} \right|^2 \ xi (R) r^2 {\ mathrm D} r

the result is:

\ Delta E_ {nlsj} ^ {so} = \ frac {A_ {nl}} {2} \ left J (j+1) - L (l+1) - S (s+1) \ right

Example with the alkaline ones

Here

s = 1/2

thus

s (s+1) = 3/4

Is $l = 0$ , then $j = s$ from where $\ Delta E_ {nlsj} ^ {so} = 0$
Is $l \ neq 0$ , then:

$j = L + \ frac {1} {2}$ thus $\ Delta E_ {nlsj} ^ {so} = \ frac {A_ {nl}} {2} \ times l$
$j = L - \ frac {1} {2}$ thus $\ Delta E_ {nlsj} ^ {so} = - \ frac {A_ {nl}} {2} \ times (l+1)$

Except for the layers S, there is a lifting partial of the degeneration of the energy levels. That results in an unfolding of these levels (Example of the Sodium which has an unfolding of the yellow Emission line in two lines respectively with 589,0nm and 589,6nm)

The barycentre of the level is not moved.

See too

Random links: