Das Wasserstoffatom (relativistsich)

Quantenmechanikvorlesung von Prof. Dr. E. Schöll, PhD

Der Artikel Das Wasserstoffatom (relativistsich) basiert auf der Vorlesungsmitschrift von Franz- Josef Schmitt des 7.Kapitels (Abschnitt 5) der Quantenmechanikvorlesung von Prof. Dr. E. Schöll, PhD.

Das Wasserstoffatom (relativistsich)	Relativistische Quntenmechanik
	Kovariante Schreibweise der Relativitätstheorie Klein- Gordon- Gleichung Dirac- Gleichung für Elektronen Der nichtrelativistische Grenzfall Das Wasserstoffatom (relativistsich)	Schrödingsche Wellenmechanik Formalisierung der Quantenmechanik Drehimpuls Spin und Systeme identischer Teilchen Näherungsmethoden Streutheorie Relativistische Quntenmechanik

In einem rotationssymmetrischen Potenzial haben wir als Dirac- Hamiltonian:

${\begin{aligned}&H=\left(c{\bar {\alpha }}{\bar {p}}+{{m}_{0}}{{c}^{2}}\beta +V(r)\right)\\&{{p}_{r}}:={\frac {1}{r}}\left({\bar {r}}{\bar {p}}-i\hbar \right)\\&{{\alpha }_{r}}:={\frac {1}{r}}{\bar {\alpha }}{\bar {r}}\\&\hbar Q:=\beta \left({\tilde {\bar {\sigma }}}{\bar {L}}+\hbar \right)\\\end{aligned}}$

Dabei sind ${{p}_{r}},{{\alpha }_{r}},\hbar Q$

hermitesche Operatoren

Man kann den Hamilton- Operator schreiben als:

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

Beweis:

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

Es gilt weiter:

$\left[\hbar Q,H\right]=0$

. Somit existieren gemeinsame Eigenzustände zu $H$

und $\hbar Q$

Eigenwerte von $\hbar Q$

${\begin{aligned}&{{\left(\hbar Q\right)}^{2}}=\beta \left({\tilde {\bar {\sigma }}}{\bar {L}}+\hbar \right)\beta \left({\tilde {\bar {\sigma }}}{\bar {L}}+\hbar \right)={{\beta }^{2}}{{\left({\tilde {\bar {\sigma }}}{\bar {L}}+\hbar \right)}^{2}}\\&\left[\beta ,{\tilde {\bar {\sigma }}}\right]=0=\left({\begin{matrix}\left[1,{\tilde {\bar {\sigma }}}\right]&{}\\{}&-\left[1,{\tilde {\bar {\sigma }}}\right]\\\end{matrix}}\right)\\&{{\beta }^{2}}=1\\&\Rightarrow {{\left(\hbar Q\right)}^{2}}=\left({\tilde {\bar {\sigma }}}{\bar {L}}\right)\left({\tilde {\bar {\sigma }}}{\bar {L}}\right)+2\hbar \left({\tilde {\bar {\sigma }}}{\bar {L}}\right)+{{\hbar }^{2}}\\&\left({\tilde {\bar {\sigma }}}{\bar {L}}\right)\left({\tilde {\bar {\sigma }}}{\bar {L}}\right)={{L}^{2}}+i{\tilde {\bar {\sigma }}}\left({\bar {L}}\times {\bar {L}}\right)\\&\left({\bar {L}}\times {\bar {L}}\right)=i\hbar {\bar {L}}\\&\Rightarrow \left({\tilde {\bar {\sigma }}}{\bar {L}}\right)\left({\tilde {\bar {\sigma }}}{\bar {L}}\right)={{L}^{2}}-\hbar {\tilde {\bar {\sigma }}}({\bar {L}})\\\end{aligned}}$

Somit:

${\begin{aligned}&{{\left(\hbar Q\right)}^{2}}={{L}^{2}}+\hbar {\tilde {\bar {\sigma }}}{\bar {L}}+{{\hbar }^{2}}={{\left({\bar {L}}+{\frac {\hbar }{2}}{\tilde {\bar {\sigma }}}\right)}^{2}}+{\frac {{\hbar }^{2}}{4}}\\&mit\ {{\left({\bar {L}}+{\frac {\hbar }{2}}{\tilde {\bar {\sigma }}}\right)}^{2}}={{L}^{2}}+\hbar {\tilde {\bar {\sigma }}}{\bar {L}}+{\frac {{\hbar }^{2}}{4}}{{\tilde {\bar {\sigma }}}^{2}}\\&{{\tilde {\bar {\sigma }}}^{2}}=3\\&\left({\bar {L}}+{\frac {\hbar }{2}}{\tilde {\bar {\sigma }}}\right)={\bar {J}}\\\end{aligned}}$

Schließlich also

${{\left(\hbar Q\right)}^{2}}={{\bar {J}}^{2}}+{\frac {{\hbar }^{2}}{4}}$

Die Eigenwerte von ${{\bar {J}}^{2}}$

sind jedoch bekannt, nämlich $\hbar j\left(j+1\right)$

mit $j=l\pm s={\frac {1}{2}},{\frac {3}{2}},...$

${\begin{aligned}&{{\left(\hbar Q\right)}^{2}}\left|j\right\rangle =\left({{\hbar }^{2}}j(j+1)+{\frac {{\hbar }^{2}}{4}}\right)\left|j\right\rangle ={{\hbar }^{2}}{{(j+{\frac {1}{2}})}^{2}}\left|j\right\rangle \\&{{(j+{\frac {1}{2}})}^{2}}:={{q}^{2}}\\\end{aligned}}$

Somit:

${\begin{aligned}&\left(\hbar Q\right)\left|j\right\rangle =\left(\hbar q\right)\left|j\right\rangle \\&q=\pm 1,\pm 2,...\\\end{aligned}}$

Es bleibt das radiale Eigenwertproblem für

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

Geeignete Darstellung für ${{\alpha }_{r}}$

${\begin{aligned}&{{\left({{\alpha }_{r}}\right)}^{2}}={\frac {1}{{r}^{2}}}\left({\bar {\alpha }}{\bar {r}}\right)\left({\bar {\alpha }}{\bar {r}}\right)={\frac {1}{{r}^{2}}}{{\alpha }^{\mu }}{{\alpha }^{\nu }}{{x}^{\mu }}{{x}^{\nu }}={\frac {1}{2{{r}^{2}}}}\left({{\alpha }^{\mu }}{{\alpha }^{\nu }}+{{\alpha }^{\nu }}{{\alpha }^{\mu }}\right){{x}^{\mu }}{{x}^{\nu }}\\&\left({{\alpha }^{\mu }}{{\alpha }^{\nu }}+{{\alpha }^{\nu }}{{\alpha }^{\mu }}\right)=2{{\delta }^{\mu \nu }}\\&{\frac {1}{2{{r}^{2}}}}2{{x}^{\mu }}{{x}^{\mu }}={\frac {{r}^{2}}{{r}^{2}}}=1\\&{{\alpha }_{r}}\beta +\beta {{\alpha }_{r}}={\frac {1}{r}}\left({\bar {\alpha }}\beta +\beta {\bar {\alpha }}\right){\bar {r}}\\&\left({\bar {\alpha }}\beta +\beta {\bar {\alpha }}\right)=0\Rightarrow {\frac {1}{r}}\left({\bar {\alpha }}\beta +\beta {\bar {\alpha }}\right){\bar {r}}=0\\\end{aligned}}$

Für

$\beta =\left({\begin{matrix}1&0\\0&-1\\\end{matrix}}\right)$

kann dies durch die Darstellung ${{\alpha }_{r}}=\left({\begin{matrix}0&-i\\i&0\\\end{matrix}}\right)$

mit ${{\alpha }_{r}}={{\alpha }_{r}}^{+}$

erfüllt werden:

${\begin{aligned}&{{\alpha }_{r}}\beta =\left({\begin{matrix}0&i\\i&0\\\end{matrix}}\right)\\&\beta {{\alpha }_{r}}=\left({\begin{matrix}0&-i\\-i&0\\\end{matrix}}\right)\\\end{aligned}}$

Es gilt:

${\begin{aligned}&{{p}_{r}}={\frac {1}{r}}\left({\bar {r}}{\bar {p}}-i\hbar \right)\\&{\bar {r}}{\bar {p}}={\frac {\hbar }{i}}r{\frac {\partial }{\partial r}}\\&{{p}_{r}}={\frac {1}{r}}\left({\frac {\hbar }{i}}r{\frac {\partial }{\partial r}}-i\hbar \right)=-i\hbar \left({\frac {\partial }{\partial r}}+{\frac {1}{r}}\right)\\\end{aligned}}$

Also

$H=\hbar c\left({\begin{matrix}0&-1\\1&0\\\end{matrix}}\right)\left({\frac {\partial }{\partial r}}+{\frac {1}{r}}\right)-{\frac {c\hbar q}{r}}\left({\begin{matrix}0&1\\1&0\\\end{matrix}}\right)+{{m}_{0}}{{c}^{2}}\left({\begin{matrix}1&0\\0&-1\\\end{matrix}}\right)+V\left({\begin{matrix}1&0\\0&1\\\end{matrix}}\right)$

Ansatz für den Radialanteil

$\left({\begin{matrix}{{\phi }_{a}}\\{{\phi }_{b}}\\\end{matrix}}\right){\tilde {\ }}{\frac {1}{r}}\left({\begin{matrix}F(r)\\G(r)\\\end{matrix}}\right)$

Eingesetzt in die Eigenwertgleichung für H:

$\left({\begin{matrix}{{\phi }_{a}}\\{{\phi }_{b}}\\\end{matrix}}\right){\tilde {\ }}{\frac {1}{r}}\left({\begin{matrix}F(r)\\G(r)\\\end{matrix}}\right)$

folgt:

${\begin{aligned}&-{\frac {\hbar c}{r}}{\frac {dG}{dr}}-{\frac {c\hbar q}{{r}^{2}}}G+{\frac {{{m}_{0}}{{c}^{2}}}{r}}F+{\frac {V}{r}}F=E{\frac {F}{r}}\\&{\frac {\hbar c}{r}}{\frac {dF}{dr}}-{\frac {c\hbar q}{{r}^{2}}}F-{\frac {{{m}_{0}}{{c}^{2}}}{r}}G+{\frac {V}{r}}G=E{\frac {G}{r}}\\&V=-{\frac {{e}^{2}}{4\pi {{\varepsilon }_{0}}}}{\frac {1}{r}}\\\end{aligned}}$

Also:

${\begin{aligned}&\left(E-{{m}_{0}}{{c}^{2}}-V\right)F+\hbar c{\frac {dG}{dr}}+{\frac {c\hbar q}{r}}G=0\\&\left(E+{{m}_{0}}{{c}^{2}}-V\right)G-\hbar c{\frac {dF}{dr}}+{\frac {c\hbar q}{r}}F=0\\&V=-{\frac {{e}^{2}}{4\pi {{\varepsilon }_{0}}}}{\frac {1}{r}}\\\end{aligned}}$

Skalentransformation:

${\begin{aligned}&{{a}_{1}}={\frac {{{m}_{0}}{{c}^{2}}+E}{\hbar c}}\\&{{a}_{2}}={\frac {{{m}_{0}}{{c}^{2}}-E}{\hbar c}}\\&a={\sqrt {{{a}_{1}}{{a}_{2}}}}={\frac {\sqrt {{{m}_{0}}^{2}{{c}^{4}}-{{E}^{2}}}}{\hbar c}}\\\end{aligned}}$

Führt man des weiteren ein:

${\begin{aligned}&\rho :=ar\\&\gamma :={\frac {{e}^{2}}{4\pi {{\varepsilon }_{0}}\hbar c}}\approx {\frac {1}{137}}\\\end{aligned}}$

Also einen skalierten Radius und die Feinstrukturkonstante,

wodurch sich auch das Potenzial vereinfacht zu:

${\frac {V}{\hbar ca}}=-{\frac {\gamma }{\rho }}$

${\begin{aligned}&\left({\frac {d}{d\rho }}+{\frac {q}{\rho }}\right)G-\left({\frac {{a}_{2}}{a}}-{\frac {\gamma }{\rho }}\right)F=0\\&\left({\frac {d}{d\rho }}-{\frac {q}{\rho }}\right)F-\left({\frac {{a}_{1}}{a}}+{\frac {\gamma }{\rho }}\right)G=0\\\end{aligned}}$

Randbedingung:

$F(\rho ),G(\rho )$

regulär bei $\rho \to 0$

$F(\rho ),G(\rho )\to 0$

für $\rho \to \infty$

Betrachte ${\begin{aligned}&\left|E\right|<{{m}_{0}}{{c}^{2}}\Rightarrow {{a}_{1}},{{a}_{2}}>0\\&a\in R\\\end{aligned}}$

also gebundene Zustände

Asymptotisches Verhalten:

${\begin{aligned}&\rho \to \infty \\&\Rightarrow G{\acute {\ }}={\frac {{a}_{2}}{a}}F\quad F{\acute {\ }}={\frac {{a}_{1}}{a}}G\\&\Rightarrow G{\acute {\ }}{\acute {\ }}=G,\quad F{\acute {\ }}{\acute {\ }}=F\\&\Rightarrow G={{e}^{-\rho }}=F=G={{e}^{-\rho }}\\\end{aligned}}$

Weil ${{e}^{+\rho }}$

divergiert !

${\begin{aligned}&\rho \to 0\\&\Rightarrow G{\acute {\ }}+{\frac {q}{\rho }}G+{\frac {\gamma }{\rho }}F=0\\&F{\acute {\ }}-{\frac {q}{\rho }}F-{\frac {\gamma }{\rho }}G=0\\\end{aligned}}$

Ansatz:

${\begin{aligned}&F(\rho )={{f}_{0}}{{\rho }^{\lambda }}\\&G(\rho )={{g}_{0}}{{\rho }^{\lambda }}\\&\Rightarrow \left(\lambda +q\right){{g}_{0}}+\gamma {{f}_{0}}=0\\&\left(\lambda -q\right){{f}_{0}}-\gamma {{g}_{0}}=0\\\end{aligned}}$

Es existieren nichttriviale Lösungen ${{f}_{0}},{{g}_{0}}$

, falls $\left(\lambda +q\right)\left(\lambda -q\right)+{{\gamma }^{2}}={{\lambda }^{2}}-{{q}^{2}}+{{\gamma }^{2}}=0$

Also $\lambda =\pm {\sqrt {{{q}^{2}}-{{\gamma }^{2}}}}>0$

und regulär bei $\rho \to 0$

Ansatz:

${\begin{aligned}&F(\rho )={{\rho }^{\lambda }}{{e}^{-\rho }}f\left(\rho \right)\\&G(\rho )={{\rho }^{\lambda }}{{e}^{-\rho }}g\left(\rho \right)\\&\Rightarrow g{\acute {\ }}-g+{\frac {\lambda +q}{\rho }}g-\left({\frac {{a}_{2}}{a}}-{\frac {\gamma }{\rho }}\right)f=0\\&f{\acute {\ }}-f+{\frac {\lambda -q}{\rho }}f-\left({\frac {{a}_{1}}{a}}+{\frac {\gamma }{\rho }}\right)g=0\\\end{aligned}}$

Die Lösung erfolgt über einen Potenzreihenansatz:

${\begin{aligned}&f(\rho )=\sum \limits _{k=0}^{\infty }{{{f}_{k}}{{\rho }^{k}}}\Rightarrow f{\acute {\ }}(\rho )=\sum \limits _{k=1}^{\infty }{k{{f}_{k}}{{\rho }^{k-1}}}=\sum \limits _{k=0}^{\infty }{(k+1){{f}_{k+1}}{{\rho }^{k}}}\\&g(\rho )=\sum \limits _{k=0}^{\infty }{{{g}_{k}}{{\rho }^{k}}}\Rightarrow g{\acute {\ }}(\rho )=\sum \limits _{k=1}^{\infty }{k{{g}_{k}}{{\rho }^{k-1}}}\\&{\frac {f(\rho )}{\rho }}=\sum \limits _{k=0}^{\infty }{{{f}_{k}}{{\rho }^{k-1}}=}{\frac {{f}_{0}}{\rho }}+\sum \limits _{k=0}^{\infty }{{{f}_{k+1}}{{\rho }^{k}}}\\\end{aligned}}$

usw... wird dies ebenfalls für $g{\acute {\ }}(\rho ),{\frac {g(\rho )}{\rho }}$

aufgestellt

Koeffizientenvergleich liefert:

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

bis auf Normfaktor

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

k=0,1,2,.... Rekursionsformel !!

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

Verhalten für große k:

$ak{{g}_{k+1}}\approx {{a}_{2}}k{{f}_{k+1}}\Rightarrow {{f}_{k}}\approx {\frac {a}{{a}_{2}}}{{g}_{k}}$

Dies kann man einsetzen in

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

und es folgt:

${\begin{aligned}&\left(k+1\right){{g}_{k+1}}\approx 2{{g}_{k}}\\&\Rightarrow {\frac {{g}_{k+1}}{{g}_{k}}}\approx {\frac {2}{k+1}}\Rightarrow {{g}_{k+1}}\approx {\frac {{2}^{k+1}}{\left(k+1\right)!}}{{g}_{0}}\\&\Rightarrow g(\rho ){\tilde {\ }}{{e}^{2\rho }}\\&\Rightarrow f(\rho ){\tilde {\ }}{{e}^{2\rho }}\\\end{aligned}}$

Falls die Potenzreihen

$f(\rho )=\sum \limits _{k=0}^{\infty }{{{f}_{k}}{{\rho }^{k}}},g(\rho )=\sum \limits _{k=0}^{\infty }{{{g}_{k}}{{\rho }^{k}}}$

nicht abbrechen, so divergiert ${\begin{aligned}&F(\rho )={{\rho }^{\lambda }}{{e}^{-\rho }}f\left(\rho \right)\\&G(\rho )={{\rho }^{\lambda }}{{e}^{-\rho }}g\left(\rho \right)\\\end{aligned}}$

exponentiell für $\rho \to \infty \Rightarrow F(\rho ),G(\rho ){\tilde {\ }}{{e}^{\rho }}$

Dies ist jedoch ein Widerspruch zu den gesetzten Randbedingungen !

Also muss es einen Abbruch bei $k=n{\acute {\ }}\in N$

geben:

${{f}_{n{\acute {\ }}+1}}={{g}_{n{\acute {\ }}+1}}=0$

Setzt man dies in die Rekursionsformel ein, so folgt:

${\begin{aligned}&-{{g}_{n{\acute {\ }}}}-{\frac {{a}_{2}}{a}}{{f}_{n{\acute {\ }}}}=0\Rightarrow {{a}_{2}}{{f}_{n{\acute {\ }}}}=-a{{g}_{n{\acute {\ }}}}\\&-{{f}_{n{\acute {\ }}}}-{\frac {{a}_{1}}{a}}{{g}_{n{\acute {\ }}}}=0\Rightarrow a{{f}_{n{\acute {\ }}}}=-{{a}_{1}}{{g}_{n{\acute {\ }}}}\\\end{aligned}}$

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

${\frac {{a}_{2}}{a}}={\frac {a}{{a}_{1}}}$

Setzt man ${{a}_{2}}{{f}_{n{\acute {\ }}}}=-a{{g}_{n{\acute {\ }}}}$

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

ein, so folgt mit $k+1=n{\acute {\ }}$

${\begin{aligned}&{\frac {a\left(\lambda +q+n{\acute {\ }}\right)+{{a}_{2}}\gamma }{a}}=-{\frac {\left[{{a}_{2}}\left(\lambda -q+n{\acute {\ }}\right)-a\gamma \right]}{{a}_{2}}}\\&\lambda +q+n{\acute {\ }}+{\frac {{a}_{2}}{a}}\gamma +\lambda -q+n{\acute {\ }}+{\frac {a}{{a}_{2}}}\gamma =0\\&2a\left(\lambda +n{\acute {\ }}\right)=\left({\frac {{a}^{2}}{{a}_{2}}}-{{a}_{2}}\right)\gamma ={\frac {2E}{\hbar c}}\gamma \\&{\frac {{a}^{2}}{{a}_{2}}}={{a}_{1}}\\&{{a}^{2}}{{\left(\lambda +n{\acute {\ }}\right)}^{2}}={\frac {{E}^{2}}{{{\hbar }^{2}}{{c}^{2}}}}{{\gamma }^{2}}\\\end{aligned}}$

Weiter gilt:

${\begin{aligned}&{{a}^{2}}={\frac {{{m}_{0}}^{2}{{c}^{4}}-{{E}^{2}}}{{{\hbar }^{2}}{{c}^{2}}}}\\&\Rightarrow \left({{m}_{0}}^{2}{{c}^{4}}-{{E}^{2}}\right){{\left(\lambda +n{\acute {\ }}\right)}^{2}}={{E}^{2}}{{\gamma }^{2}}\\\end{aligned}}$

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

$E={\frac {{{m}_{0}}{{c}^{2}}}{\sqrt {1+{{\left({\frac {\gamma }{\lambda +n{\acute {\ }}}}\right)}^{2}}}}}$

Mit der Feinstrukturkonstanten

$\gamma \approx {\frac {1}{137}}$

${\begin{aligned}&\lambda ={\sqrt {q}}\\&{{a}^{2}}={\frac {{{m}_{0}}^{2}{{c}^{4}}-{{E}^{2}}}{{{\hbar }^{2}}{{c}^{2}}}}\\&\Rightarrow \left({{m}_{0}}^{2}{{c}^{4}}-{{E}^{2}}\right){{\left(\lambda +n{\acute {\ }}\right)}^{2}}={{E}^{2}}{{\gamma }^{2}}\\\end{aligned}}$

${\begin{aligned}&\lambda ={\sqrt {{{q}^{2}}-{{\gamma }^{2}}}}={\sqrt {{{\left(j+{\frac {1}{2}}\right)}^{2}}-{{\gamma }^{2}}}}\\&j={\frac {1}{2}},{\frac {3}{2}},...,n{\acute {\ }}\in {{N}_{0}}\\&j=l\pm s\\\end{aligned}}$

entwickelt man die Energieeigenwerte nach der Feinstrukturkonstanten bis $O\left({{\gamma }^{4}}\right)$

, so folgt:

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

mit

$\lambda \left(\gamma \right)=|q|{\sqrt {1-{{\left({\frac {\gamma }{q}}\right)}^{2}}}}=|q|\left[1-{\frac {1}{2}}{{\left({\frac {\gamma }{q}}\right)}^{2}}\right]+O\left({{\gamma }^{4}}\right)$

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

Setzt man dies in die exakten Energieeigenwerte E ein , so folgt:

${\begin{aligned}&E={{m}_{0}}{{c}^{2}}\left[1-\left({\frac {{\gamma }^{2}}{2{{n}^{2}}}}\right)-\left({\frac {{\gamma }^{4}}{2{{n}^{3}}}}\right)\left({\frac {1}{j+{\frac {1}{2}}}}-{\frac {3}{4n}}\right)+O\left({{\gamma }^{6}}\right)\right]\\&n=1,2,3\\&j={\frac {1}{2}},{\frac {3}{2}},...,n-{\frac {1}{2}},wegen\ n=n{\acute {\ }}+j+{\frac {1}{2}}\\&j=l\pm s\\\end{aligned}}$

Diskussion

$O\left({{\gamma }^{0}}\right):E={{m}_{0}}{{c}^{2}}$

Ruheenergie

$O\left({{\gamma }^{2}}\right):\Delta {{E}^{(2)}}=-{{m}_{0}}{{c}^{2}}\left({\frac {{\gamma }^{2}}{2{{n}^{2}}}}\right)=-{\frac {{R}_{H}}{{n}^{2}}}$

nicht relativistisches, entartetes Energiespektrum

$O\left({{\gamma }^{4}}\right):\Delta {{E}^{(4)}}=-{{m}_{0}}{{c}^{2}}\left({\frac {{\gamma }^{4}}{2{{n}^{3}}}}\right)\left({\frac {1}{j+{\frac {1}{2}}}}-{\frac {3}{4n}}\right)$

Feinstruktur- Aufspaltung. Eine Aufhebung der j-Entartung durch Spin- Bahn- Kopplung. Dabei bleibt die Freiheit der Ausrichtung der Achse des magnetischen Moments, also die $2(2j+1)$ - fache ${{m}_{j}}$ - Entartung+ Parität !

Spektroskopische Beziehung der Feinstrukturterme: $n{{l}_{j}}$

$n=1:\quad j={\frac {1}{2}}:\ 1{{s}_{\frac {1}{2}}}$

${\begin{array}{*{35}{l}}{}&n=2:\quad j={\frac {1}{2}}:\ 2{{s}_{\frac {1}{2}}}\quad 2{{p}_{\frac {1}{2}}}\quad \quad \quad \quad \quad \quad \quad n{\overset {\acute {\ }}{\mathop {\ } }}\,=1\\{}&\quad \quad \quad \,j={\frac {3}{2}}:\quad \quad \quad 2{{p}_{\frac {3}{2}}}\quad \quad \quad \quad \quad \quad \quad n{\overset {\acute {\ }}{\mathop {\ } }}\,=0\\\end{array}}$

n´=0 ..