Editing Das Wasserstoffatom (relativistsich)

<noinclude>{{Scripthinweis|Quantenmechanik|7|5}}</noinclude>

In einem rotationssymmetrischen Potenzial haben wir als Dirac- Hamiltonian:

:<math>\begin{align}

& H=\left( c\bar{\alpha }\bar{p}+{{m}_{0}}{{c}^{2}}\beta +V(r) \right) \\

& {{p}_{r}}:=\frac{1}{r}\left( \bar{r}\bar{p}-i\hbar  \right) \\

& {{\alpha }_{r}}:=\frac{1}{r}\bar{\alpha }\bar{r} \\

& \hbar Q:=\beta \left( \tilde{\bar{\sigma }}\bar{L}+\hbar  \right) \\

\end{align}</math>

Dabei sind <math>{{p}_{r}},{{\alpha }_{r}},\hbar Q</math>

hermitesche Operatoren

Man kann den Hamilton- Operator schreiben als:

:<math>H=\left( c{{\alpha }_{r}}{{p}_{r}}+\frac{ic}{r}{{\alpha }_{r}}\beta \hbar Q+{{m}_{0}}{{c}^{2}}\beta +V(r) \right)</math>

Beweis:

:<math>\begin{align}

& {{\alpha }_{r}}{{p}_{r}}+\frac{i}{r}{{\alpha }_{r}}\beta \hbar Q={{\alpha }_{r}}\left[ \frac{1}{r}\left( \bar{r}\bar{p}-i\hbar  \right)+\frac{i}{r}{{\beta }^{2}}\left( \tilde{\bar{\sigma }}\bar{L}+\hbar  \right) \right] \\

& {{\beta }^{2}}=1 \\

& =\frac{{{\alpha }_{r}}}{r}\left( \bar{r}\bar{p}+i\tilde{\bar{\sigma }}\bar{L} \right)=\frac{1}{{{r}^{2}}}\left[ \left( \bar{\alpha }\bar{r} \right)\left( \bar{r}\bar{p} \right)+i\left( \bar{\alpha }\bar{r} \right)\left( \tilde{\bar{\sigma }}\bar{L} \right) \right] \\

& i\left( \bar{\alpha }\bar{r} \right)\left( \tilde{\bar{\sigma }}\bar{L} \right)=i\left( \bar{\alpha }\bar{r} \right)\left( \bar{r}\bar{p} \right)-i{{r}^{2}}\left( \bar{\alpha }\bar{p} \right) \\

& \Rightarrow {{\alpha }_{r}}{{p}_{r}}+\frac{i}{r}{{\alpha }_{r}}\beta \hbar Q=\frac{1}{{{r}^{2}}}\left[ \left( \bar{\alpha }\bar{r} \right)\left( \bar{r}\bar{p} \right)+i\left( \bar{\alpha }\bar{r} \right)\left( \tilde{\bar{\sigma }}\bar{L} \right) \right]=\bar{\alpha }\bar{p} \\

\end{align}</math>

Es gilt weiter:

:<math>\left[ \hbar Q,H \right]=0</math>
.
 Somit existieren gemeinsame Eigenzustände zu <math>H</math>

und <math>\hbar Q</math>

'''Eigenwerte von '''<math>\hbar Q</math>

:

:<math>\begin{align}

& {{\left( \hbar Q \right)}^{2}}=\beta \left( \tilde{\bar{\sigma }}\bar{L}+\hbar  \right)\beta \left( \tilde{\bar{\sigma }}\bar{L}+\hbar  \right)={{\beta }^{2}}{{\left( \tilde{\bar{\sigma }}\bar{L}+\hbar  \right)}^{2}} \\

& \left[ \beta ,\tilde{\bar{\sigma }} \right]=0=\left( \begin{matrix}

\left[ 1,\tilde{\bar{\sigma }} \right] & {}  \\

{} & -\left[ 1,\tilde{\bar{\sigma }} \right]  \\

\end{matrix} \right) \\

& {{\beta }^{2}}=1 \\

& \Rightarrow {{\left( \hbar Q \right)}^{2}}=\left( \tilde{\bar{\sigma }}\bar{L} \right)\left( \tilde{\bar{\sigma }}\bar{L} \right)+2\hbar \left( \tilde{\bar{\sigma }}\bar{L} \right)+{{\hbar }^{2}} \\

& \left( \tilde{\bar{\sigma }}\bar{L} \right)\left( \tilde{\bar{\sigma }}\bar{L} \right)={{L}^{2}}+i\tilde{\bar{\sigma }}\left( \bar{L}\times \bar{L} \right) \\

& \left( \bar{L}\times \bar{L} \right)=i\hbar \bar{L} \\

& \Rightarrow \left( \tilde{\bar{\sigma }}\bar{L} \right)\left( \tilde{\bar{\sigma }}\bar{L} \right)={{L}^{2}}-\hbar \tilde{\bar{\sigma }}(\bar{L}) \\

\end{align}</math>

Somit:

:<math>\begin{align}

& {{\left( \hbar Q \right)}^{2}}={{L}^{2}}+\hbar \tilde{\bar{\sigma }}\bar{L}+{{\hbar }^{2}}={{\left( \bar{L}+\frac{\hbar }{2}\tilde{\bar{\sigma }} \right)}^{2}}+\frac{{{\hbar }^{2}}}{4} \\

& mit\ {{\left( \bar{L}+\frac{\hbar }{2}\tilde{\bar{\sigma }} \right)}^{2}}={{L}^{2}}+\hbar \tilde{\bar{\sigma }}\bar{L}+\frac{{{\hbar }^{2}}}{4}{{{\tilde{\bar{\sigma }}}}^{2}} \\

& {{{\tilde{\bar{\sigma }}}}^{2}}=3 \\

& \left( \bar{L}+\frac{\hbar }{2}\tilde{\bar{\sigma }} \right)=\bar{J} \\

\end{align}</math>

Schließlich also

:<math>{{\left( \hbar Q \right)}^{2}}={{\bar{J}}^{2}}+\frac{{{\hbar }^{2}}}{4}</math>

Die Eigenwerte von <math>{{\bar{J}}^{2}}</math>

sind jedoch bekannt, nämlich <math>\hbar j\left( j+1 \right)</math>

mit <math>j=l\pm s=\frac{1}{2},\frac{3}{2},...</math>

:<math>\begin{align}

& {{\left( \hbar Q \right)}^{2}}\left| j \right\rangle =\left( {{\hbar }^{2}}j(j+1)+\frac{{{\hbar }^{2}}}{4} \right)\left| j \right\rangle ={{\hbar }^{2}}{{(j+\frac{1}{2})}^{2}}\left| j \right\rangle  \\

& {{(j+\frac{1}{2})}^{2}}:={{q}^{2}} \\

\end{align}</math>

Somit:

:<math>\begin{align}

& \left( \hbar Q \right)\left| j \right\rangle =\left( \hbar q \right)\left| j \right\rangle  \\

& q=\pm 1,\pm 2,... \\

\end{align}</math>

Es bleibt das radiale Eigenwertproblem für

:<math>H=\left( c{{\alpha }_{r}}{{p}_{r}}+\frac{ic}{r}{{\alpha }_{r}}\beta \hbar Q+{{m}_{0}}{{c}^{2}}\beta +V(r) \right)</math>

'''Geeignete Darstellung für '''<math>{{\alpha }_{r}}</math>

:

:<math>\begin{align}

& {{\left( {{\alpha }_{r}} \right)}^{2}}=\frac{1}{{{r}^{2}}}\left( \bar{\alpha }\bar{r} \right)\left( \bar{\alpha }\bar{r} \right)=\frac{1}{{{r}^{2}}}{{\alpha }^{\mu }}{{\alpha }^{\nu }}{{x}^{\mu }}{{x}^{\nu }}=\frac{1}{2{{r}^{2}}}\left( {{\alpha }^{\mu }}{{\alpha }^{\nu }}+{{\alpha }^{\nu }}{{\alpha }^{\mu }} \right){{x}^{\mu }}{{x}^{\nu }} \\

& \left( {{\alpha }^{\mu }}{{\alpha }^{\nu }}+{{\alpha }^{\nu }}{{\alpha }^{\mu }} \right)=2{{\delta }^{\mu \nu }} \\

& \frac{1}{2{{r}^{2}}}2{{x}^{\mu }}{{x}^{\mu }}=\frac{{{r}^{2}}}{{{r}^{2}}}=1 \\

& {{\alpha }_{r}}\beta +\beta {{\alpha }_{r}}=\frac{1}{r}\left( \bar{\alpha }\beta +\beta \bar{\alpha } \right)\bar{r} \\

& \left( \bar{\alpha }\beta +\beta \bar{\alpha } \right)=0\Rightarrow \frac{1}{r}\left( \bar{\alpha }\beta +\beta \bar{\alpha } \right)\bar{r}=0 \\

\end{align}</math>

Für

:<math>\beta =\left( \begin{matrix}

1 & 0  \\

0 & -1  \\

\end{matrix} \right)</math>

kann dies durch die Darstellung <math>{{\alpha }_{r}}=\left( \begin{matrix}

0 & -i  \\

i & 0  \\

\end{matrix} \right)</math>

mit <math>{{\alpha }_{r}}={{\alpha }_{r}}^{+}</math>

erfüllt werden:

:<math>\begin{align}

& {{\alpha }_{r}}\beta =\left( \begin{matrix}

0 & i  \\

i & 0  \\

\end{matrix} \right) \\

& \beta {{\alpha }_{r}}=\left( \begin{matrix}

0 & -i  \\

-i & 0  \\

\end{matrix} \right) \\

\end{align}</math>

Es gilt:

:<math>\begin{align}

& {{p}_{r}}=\frac{1}{r}\left( \bar{r}\bar{p}-i\hbar  \right) \\

& \bar{r}\bar{p}=\frac{\hbar }{i}r\frac{\partial }{\partial r} \\

& {{p}_{r}}=\frac{1}{r}\left( \frac{\hbar }{i}r\frac{\partial }{\partial r}-i\hbar  \right)=-i\hbar \left( \frac{\partial }{\partial r}+\frac{1}{r} \right) \\

\end{align}</math>

Also

:<math>H=\hbar c\left( \begin{matrix}

0 & -1  \\

1 & 0  \\

\end{matrix} \right)\left( \frac{\partial }{\partial r}+\frac{1}{r} \right)-\frac{c\hbar q}{r}\left( \begin{matrix}

0 & 1  \\

1 & 0  \\

\end{matrix} \right)+{{m}_{0}}{{c}^{2}}\left( \begin{matrix}

1 & 0  \\

0 & -1  \\

\end{matrix} \right)+V\left( \begin{matrix}

1 & 0  \\

0 & 1  \\

\end{matrix} \right)</math>

Ansatz für den Radialanteil

:<math>\left( \begin{matrix}

{{\phi }_{a}}  \\

{{\phi }_{b}}  \\

\end{matrix} \right)\tilde{\ }\frac{1}{r}\left( \begin{matrix}

F(r)  \\

G(r)  \\

\end{matrix} \right)</math>

Eingesetzt in die Eigenwertgleichung für H:

:<math>\left( \begin{matrix}

{{\phi }_{a}}  \\

{{\phi }_{b}}  \\

\end{matrix} \right)\tilde{\ }\frac{1}{r}\left( \begin{matrix}

F(r)  \\

G(r)  \\

\end{matrix} \right)</math>

folgt:

:<math>\begin{align}

& -\frac{\hbar c}{r}\frac{dG}{dr}-\frac{c\hbar q}{{{r}^{2}}}G+\frac{{{m}_{0}}{{c}^{2}}}{r}F+\frac{V}{r}F=E\frac{F}{r} \\

& \frac{\hbar c}{r}\frac{dF}{dr}-\frac{c\hbar q}{{{r}^{2}}}F-\frac{{{m}_{0}}{{c}^{2}}}{r}G+\frac{V}{r}G=E\frac{G}{r} \\

& V=-\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}\frac{1}{r} \\

\end{align}</math>

Also:

:<math>\begin{align}

& \left( E-{{m}_{0}}{{c}^{2}}-V \right)F+\hbar c\frac{dG}{dr}+\frac{c\hbar q}{r}G=0 \\

& \left( E+{{m}_{0}}{{c}^{2}}-V \right)G-\hbar c\frac{dF}{dr}+\frac{c\hbar q}{r}F=0 \\

& V=-\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}\frac{1}{r} \\

\end{align}</math>

====Skalentransformation:====
:<math>\begin{align}

& {{a}_{1}}=\frac{{{m}_{0}}{{c}^{2}}+E}{\hbar c} \\

& {{a}_{2}}=\frac{{{m}_{0}}{{c}^{2}}-E}{\hbar c} \\

& a=\sqrt{{{a}_{1}}{{a}_{2}}}=\frac{\sqrt{{{m}_{0}}^{2}{{c}^{4}}-{{E}^{2}}}}{\hbar c} \\

\end{align}</math>

Führt man des weiteren ein:

:<math>\begin{align}

& \rho :=ar \\

& \gamma :=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}\hbar c}\approx \frac{1}{137} \\

\end{align}</math>

Also einen skalierten Radius und die Feinstrukturkonstante,

wodurch sich auch das Potenzial vereinfacht zu:

:<math>\frac{V}{\hbar ca}=-\frac{\gamma }{\rho }</math>

:

:<math>\begin{align}

& \left( \frac{d}{d\rho }+\frac{q}{\rho } \right)G-\left( \frac{{{a}_{2}}}{a}-\frac{\gamma }{\rho } \right)F=0 \\

& \left( \frac{d}{d\rho }-\frac{q}{\rho } \right)F-\left( \frac{{{a}_{1}}}{a}+\frac{\gamma }{\rho } \right)G=0 \\

\end{align}</math>

<u>'''Randbedingung:'''</u>

:<math>F(\rho ),G(\rho )</math>

regulär bei <math>\rho \to 0</math>

:<math>F(\rho ),G(\rho )\to 0</math>

für <math>\rho \to \infty </math>

<u>'''Betrachte '''</u><math>\begin{align}

& \left| E \right|<{{m}_{0}}{{c}^{2}}\Rightarrow {{a}_{1}},{{a}_{2}}>0 \\

& a\in R \\

\end{align}</math>

also gebundene Zustände

'''Asymptotisches Verhalten:'''

:<math>\begin{align}

& \rho \to \infty  \\

& \Rightarrow G\acute{\ }=\frac{{{a}_{2}}}{a}F\quad F\acute{\ }=\frac{{{a}_{1}}}{a}G \\

& \Rightarrow G\acute{\ }\acute{\ }=G,\quad F\acute{\ }\acute{\ }=F \\

& \Rightarrow G={{e}^{-\rho }}=F=G={{e}^{-\rho }} \\

\end{align}</math>

Weil <math>{{e}^{+\rho }}</math>

divergiert!

:<math>\begin{align}

& \rho \to 0 \\

& \Rightarrow G\acute{\ }+\frac{q}{\rho }G+\frac{\gamma }{\rho }F=0 \\

& F\acute{\ }-\frac{q}{\rho }F-\frac{\gamma }{\rho }G=0 \\

\end{align}</math>

Ansatz:

:<math>\begin{align}

& F(\rho )={{f}_{0}}{{\rho }^{\lambda }} \\

& G(\rho )={{g}_{0}}{{\rho }^{\lambda }} \\

& \Rightarrow \left( \lambda +q \right){{g}_{0}}+\gamma {{f}_{0}}=0 \\

& \left( \lambda -q \right){{f}_{0}}-\gamma {{g}_{0}}=0 \\

\end{align}</math>

Es existieren nichttriviale Lösungen <math>{{f}_{0}},{{g}_{0}}</math>
,
 falls <math>\left( \lambda +q \right)\left( \lambda -q \right)+{{\gamma }^{2}}={{\lambda }^{2}}-{{q}^{2}}+{{\gamma }^{2}}=0</math>

Also <math>\lambda =\pm \sqrt{{{q}^{2}}-{{\gamma }^{2}}}>0</math>

und regulär bei <math>\rho \to 0</math>

Ansatz:

:<math>\begin{align}

& F(\rho )={{\rho }^{\lambda }}{{e}^{-\rho }}f\left( \rho  \right) \\

& G(\rho )={{\rho }^{\lambda }}{{e}^{-\rho }}g\left( \rho  \right) \\

& \Rightarrow g\acute{\ }-g+\frac{\lambda +q}{\rho }g-\left( \frac{{{a}_{2}}}{a}-\frac{\gamma }{\rho } \right)f=0 \\

& f\acute{\ }-f+\frac{\lambda -q}{\rho }f-\left( \frac{{{a}_{1}}}{a}+\frac{\gamma }{\rho } \right)g=0 \\

\end{align}</math>

Die Lösung erfolgt über einen Potenzreihenansatz:

:<math>\begin{align}

& f(\rho )=\sum\limits_{k=0}^{\infty }{{{f}_{k}}{{\rho }^{k}}}\Rightarrow f\acute{\ }(\rho )=\sum\limits_{k=1}^{\infty }{k{{f}_{k}}{{\rho }^{k-1}}}=\sum\limits_{k=0}^{\infty }{(k+1){{f}_{k+1}}{{\rho }^{k}}} \\

& g(\rho )=\sum\limits_{k=0}^{\infty }{{{g}_{k}}{{\rho }^{k}}}\Rightarrow g\acute{\ }(\rho )=\sum\limits_{k=1}^{\infty }{k{{g}_{k}}{{\rho }^{k-1}}} \\

& \frac{f(\rho )}{\rho }=\sum\limits_{k=0}^{\infty }{{{f}_{k}}{{\rho }^{k-1}}=}\frac{{{f}_{0}}}{\rho }+\sum\limits_{k=0}^{\infty }{{{f}_{k+1}}{{\rho }^{k}}} \\

\end{align}</math>

usw... wird dies ebenfalls für <math>g\acute{\ }(\rho ),\frac{g(\rho )}{\rho }</math>

aufgestellt

Koeffizientenvergleich liefert:

:<math>\begin{align}

& O\left( \frac{1}{\rho } \right):\left( \lambda +q \right){{g}_{0}}+\gamma {{f}_{0}}=0\quad \quad \left( \lambda -q \right){{f}_{0}}-\gamma {{g}_{0}}=0 \\

& \Rightarrow {{f}_{0}},{{g}_{0}} \\

\end{align}</math>

bis auf Normfaktor

:<math>\begin{align}

& O\left( {{\rho }^{k}} \right):\left( \lambda +q+k+1 \right){{g}_{k+1}}-{{g}_{k}}+\gamma {{f}_{k+1}}-\frac{{{a}_{2}}}{a}{{f}_{k}}=0 \\

& \left( \lambda -q+k+1 \right){{f}_{k+1}}-{{f}_{k}}+\gamma {{g}_{k+1}}-\frac{{{a}_{1}}}{a}{{g}_{k}}=0 \\

\end{align}</math>

k=0,1,2,....  Rekursionsformel!!

:<math>\begin{align}

& a\left[ \left( \lambda +q+k+1 \right){{g}_{k+1}}-{{g}_{k}}+\gamma {{f}_{k+1}}-\frac{{{a}_{2}}}{a}{{f}_{k}} \right]-{{a}_{2}}\left[ \left( \lambda -q+k+1 \right){{f}_{k+1}}-{{f}_{k}}+\gamma {{g}_{k+1}}-\frac{{{a}_{1}}}{a}{{g}_{k}} \right]=0 \\

& \Rightarrow \left[ a\left( \lambda +q+k+1 \right)+{{a}_{2}}\gamma  \right]{{g}_{k+1}}=\left[ {{a}_{2}}\left( \lambda -q+k+1 \right)-a\gamma  \right]{{f}_{k+1}} \\

\end{align}</math>

'''Verhalten für große k:'''

:<math>ak{{g}_{k+1}}\approx {{a}_{2}}k{{f}_{k+1}}\Rightarrow {{f}_{k}}\approx \frac{a}{{{a}_{2}}}{{g}_{k}}</math>

Dies kann man einsetzen in

:<math>\left( \lambda +q+k+1 \right){{g}_{k+1}}-{{g}_{k}}+\gamma {{f}_{k+1}}-\frac{{{a}_{2}}}{a}{{f}_{k}}=0</math>

und es folgt:

:<math>\begin{align}

& \left( k+1 \right){{g}_{k+1}}\approx 2{{g}_{k}} \\

& \Rightarrow \frac{{{g}_{k+1}}}{{{g}_{k}}}\approx \frac{2}{k+1}\Rightarrow {{g}_{k+1}}\approx \frac{{{2}^{k+1}}}{\left( k+1 \right)!}{{g}_{0}} \\

& \Rightarrow g(\rho )\tilde{\ }{{e}^{2\rho }} \\

& \Rightarrow f(\rho )\tilde{\ }{{e}^{2\rho }} \\

\end{align}</math>

Falls die Potenzreihen

:<math>f(\rho )=\sum\limits_{k=0}^{\infty }{{{f}_{k}}{{\rho }^{k}}},g(\rho )=\sum\limits_{k=0}^{\infty }{{{g}_{k}}{{\rho }^{k}}}</math>

nicht abbrechen, so divergiert <math>\begin{align}

& F(\rho )={{\rho }^{\lambda }}{{e}^{-\rho }}f\left( \rho  \right) \\

& G(\rho )={{\rho }^{\lambda }}{{e}^{-\rho }}g\left( \rho  \right) \\

\end{align}</math>

exponentiell für <math>\rho \to \infty \Rightarrow F(\rho ),G(\rho )\tilde{\ }{{e}^{\rho }}</math>

Dies ist jedoch ein Widerspruch zu den gesetzten Randbedingungen!

Also muss es einen Abbruch bei <math>k=n\acute{\ }\in N</math>

geben:

:<math>{{f}_{n\acute{\ }+1}}={{g}_{n\acute{\ }+1}}=0</math>

Setzt man dies in die Rekursionsformel ein, so folgt:

:<math>\begin{align}

& -{{g}_{n\acute{\ }}}-\frac{{{a}_{2}}}{a}{{f}_{n\acute{\ }}}=0\Rightarrow {{a}_{2}}{{f}_{n\acute{\ }}}=-a{{g}_{n\acute{\ }}} \\

& -{{f}_{n\acute{\ }}}-\frac{{{a}_{1}}}{a}{{g}_{n\acute{\ }}}=0\Rightarrow a{{f}_{n\acute{\ }}}=-{{a}_{1}}{{g}_{n\acute{\ }}} \\

\end{align}</math>

Diese beiden Gleichungen stimmen jedoch für alle f,g überein, da

:<math>\frac{{{a}_{2}}}{a}=\frac{a}{{{a}_{1}}}</math>

Setzt man <math>{{a}_{2}}{{f}_{n\acute{\ }}}=-a{{g}_{n\acute{\ }}}</math>

in <math>\left[ a\left( \lambda +q+k+1 \right)+{{a}_{2}}\gamma  \right]{{g}_{k+1}}=\left[ {{a}_{2}}\left( \lambda -q+k+1 \right)-a\gamma  \right]{{f}_{k+1}}</math>

ein, so folgt mit <math>k+1=n\acute{\ }</math>

:

:<math>\begin{align}

& \frac{a\left( \lambda +q+n\acute{\ } \right)+{{a}_{2}}\gamma }{a}=-\frac{\left[ {{a}_{2}}\left( \lambda -q+n\acute{\ } \right)-a\gamma  \right]}{{{a}_{2}}} \\

& \lambda +q+n\acute{\ }+\frac{{{a}_{2}}}{a}\gamma +\lambda -q+n\acute{\ }+\frac{a}{{{a}_{2}}}\gamma =0 \\

& 2a\left( \lambda +n\acute{\ } \right)=\left( \frac{{{a}^{2}}}{{{a}_{2}}}-{{a}_{2}} \right)\gamma =\frac{2E}{\hbar c}\gamma  \\

& \frac{{{a}^{2}}}{{{a}_{2}}}={{a}_{1}} \\

& {{a}^{2}}{{\left( \lambda +n\acute{\ } \right)}^{2}}=\frac{{{E}^{2}}}{{{\hbar }^{2}}{{c}^{2}}}{{\gamma }^{2}} \\

\end{align}</math>

Weiter gilt:

:<math>\begin{align}

& {{a}^{2}}=\frac{{{m}_{0}}^{2}{{c}^{4}}-{{E}^{2}}}{{{\hbar }^{2}}{{c}^{2}}} \\

& \Rightarrow \left( {{m}_{0}}^{2}{{c}^{4}}-{{E}^{2}} \right){{\left( \lambda +n\acute{\ } \right)}^{2}}={{E}^{2}}{{\gamma }^{2}} \\

\end{align}</math>

Löst man dies nach den exakten Energieeigenwerten, die sich damit ergeben, also nach E auf, so erhält man die Feinstrukturformel:

:<math>E=\frac{{{m}_{0}}{{c}^{2}}}{\sqrt{1+{{\left( \frac{\gamma }{\lambda +n\acute{\ }} \right)}^{2}}}}</math>

Mit der Feinstrukturkonstanten

:<math>\gamma \approx \frac{1}{137}</math>

:<math>\begin{align}

& \lambda =\sqrt{q} \\

& {{a}^{2}}=\frac{{{m}_{0}}^{2}{{c}^{4}}-{{E}^{2}}}{{{\hbar }^{2}}{{c}^{2}}} \\

& \Rightarrow \left( {{m}_{0}}^{2}{{c}^{4}}-{{E}^{2}} \right){{\left( \lambda +n\acute{\ } \right)}^{2}}={{E}^{2}}{{\gamma }^{2}} \\

\end{align}</math>

:<math>\begin{align}

& \lambda =\sqrt{{{q}^{2}}-{{\gamma }^{2}}}=\sqrt{{{\left( j+\frac{1}{2} \right)}^{2}}-{{\gamma }^{2}}} \\

& j=\frac{1}{2},\frac{3}{2},...,n\acute{\ }\in {{N}_{0}} \\

& j=l\pm s \\

\end{align}</math>

entwickelt man die Energieeigenwerte nach der Feinstrukturkonstanten bis <math>O\left( {{\gamma }^{4}} \right)</math>
,
 so folgt:

:<math>E={{m}_{0}}{{c}^{2}}\left[ 1-\frac{1}{2}{{\left( \frac{\gamma }{\lambda +n\acute{\ }} \right)}^{2}}+\frac{3}{8}{{\left( \frac{\gamma }{\lambda +n\acute{\ }} \right)}^{4}}+O\left( {{\gamma }^{6}} \right) \right]</math>

mit

:<math>\lambda \left( \gamma  \right)=|q|\sqrt{1-{{\left( \frac{\gamma }{q} \right)}^{2}}}=|q|\left[ 1-\frac{1}{2}{{\left( \frac{\gamma }{q} \right)}^{2}} \right]+O\left( {{\gamma }^{4}} \right)</math>

:<math>\begin{align}

& {{\left( \frac{1}{\lambda +n\acute{\ }} \right)}^{2}}=\frac{1}{{{\left[ n\acute{\ }+|q|-\frac{1}{2}\left( \frac{{{\gamma }^{2}}}{\left| q \right|} \right) \right]}^{2}}}+O\left( {{\gamma }^{4}} \right) \\
& n=n\acute{\ }+\left| q \right| \\
& n\acute{\ }=0,1,2,... \\
& \left| q \right|=j+\frac{1}{2}=1,2,.... \\
& {{\left( \frac{1}{\lambda +n\acute{\ }} \right)}^{2}}=\frac{1}{{{n}^{2}}}{{\left[ 1-\frac{1}{2}\left( \frac{{{\gamma }^{2}}}{\left| q \right|n} \right) \right]}^{-2}}+O\left( {{\gamma }^{4}} \right)=\frac{1}{{{n}^{2}}}\left[ 1+\left( \frac{{{\gamma }^{2}}}{\left| q \right|n} \right) \right]+O\left( {{\gamma }^{4}} \right)=\frac{1}{{{n}^{2}}}+\left( \frac{{{\gamma }^{2}}}{\left| q \right|{{n}^{3}}} \right)+O\left( {{\gamma }^{4}} \right) \\
& \left| q \right|=j+\frac{1}{2}=l\pm s+\frac{1}{2} \\
\end{align}</math>

Setzt man dies in die exakten Energieeigenwerte E ein, so folgt:
:<math>\begin{align}
& E={{m}_{0}}{{c}^{2}}\left[ 1-\left( \frac{{{\gamma }^{2}}}{2{{n}^{2}}} \right)-\left( \frac{{{\gamma }^{4}}}{2{{n}^{3}}} \right)\left( \frac{1}{j+\frac{1}{2}}-\frac{3}{4n} \right)+O\left( {{\gamma }^{6}} \right) \right] \\
& n=1,2,3 \\
& j=\frac{1}{2},\frac{3}{2},...,n-\frac{1}{2},wegen\ n=n\acute{\ }+j+\frac{1}{2} \\
& j=l\pm s \\
\end{align}</math>

<u>'''Diskussion'''</u>
:<math>O\left( {{\gamma }^{0}} \right):E={{m}_{0}}{{c}^{2}}</math>
Ruheenergie
:<math>O\left( {{\gamma }^{2}} \right):\Delta {{E}^{(2)}}=-{{m}_{0}}{{c}^{2}}\left( \frac{{{\gamma }^{2}}}{2{{n}^{2}}} \right)=-\frac{{{R}_{H}}}{{{n}^{2}}}</math>
nicht relativistisches, entartetes Energiespektrum
:<math>O\left( {{\gamma }^{4}} \right):\Delta {{E}^{(4)}}=-{{m}_{0}}{{c}^{2}}\left( \frac{{{\gamma }^{4}}}{2{{n}^{3}}} \right)\left( \frac{1}{j+\frac{1}{2}}-\frac{3}{4n} \right)</math>
Feinstruktur- Aufspaltung. Eine Aufhebung der j-Entartung durch Spin- Bahn- Kopplung.
Dabei bleibt die Freiheit der Ausrichtung der Achse des magnetischen Moments, also die <math>2(2j+1)</math>
- fache <math>{{m}_{j}}</math>
- Entartung+ Parität!
====Spektroskopische Beziehung der Feinstrukturterme: <math>n{{l}_{j}}</math>====
:<math>n=1:\quad j=\frac{1}{2}:\ 1{{s}_{\frac{1}{2}}}</math>
	
:<math>\begin{array}{*{35}{l}}
   {} & n=2:\quad j=\frac{1}{2}:\ 2{{s}_{\frac{1}{2}}}\quad 2{{p}_{\frac{1}{2}}}\quad \quad \quad \quad \quad \quad \quad n\overset{\acute{\ }}{\mathop{\ }}\,=1  \\
   {} & \quad \quad \quad \,j=\frac{3}{2}:\quad \quad \quad 2{{p}_{\frac{3}{2}}}\quad \quad \quad \quad \quad \quad \quad n\overset{\acute{\ }}{\mathop{\ }}\,=0  \\
\end{array}</math>

				n´=0.
.