Zurück zu Harmonischer Oszillator | Hoch zum Quantenmechanik Anker | Vor zu Drehimpulsquantelung

Drehimpuls-Operator:

${\displaystyle {\hat {L}}:={\hat {x}}\times {\hat {p}}={\frac {\hbar }{i}}x\times \nabla }$

${\displaystyle {\begin{matrix}{\hat {L}}_{1}&=&{\frac {\hbar }{i}}x_{2}{\frac {\partial }{\partial x_{3}}}-{\frac {\hbar }{i}}x_{3}{\frac {\partial }{\partial x_{2}}}\\{\hat {L}}_{2}&=&{\frac {\hbar }{i}}x_{3}{\frac {\partial }{\partial x_{1}}}-{\frac {\hbar }{i}}x_{1}{\frac {\partial }{\partial x_{3}}}\\{\hat {L}}_{3}&=&{\frac {\hbar }{i}}x_{1}{\frac {\partial }{\partial x_{2}}}-{\frac {\hbar }{i}}x_{2}{\frac {\partial }{\partial x_{1}}}\end{matrix}}}$

Kommutator und Antikommutator

Der Kommutator zweier Operatoren ist definiert durch: ${\displaystyle \left[{\hat {A}},{\hat {B}}\right]:={\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}}=-\left({\hat {B}}{\hat {A}}-{\hat {A}}{\hat {B}}\right)=-\left[{\hat {B}},{\hat {A}}\right]}$ Entsprechend definiert man den Antikommutator ${\displaystyle \left\{{\hat {A}},{\hat {B}}\right\}:={\hat {A}}{\hat {B}}+{\hat {B}}{\hat {A}}={\hat {B}}{\hat {A}}+{\hat {A}}{\hat {B}}=\left\{{\hat {B}},{\hat {A}}\right\}}$

${\displaystyle {\begin{matrix}{\hat {L}}_{1}{\hat {L}}_{2}&=&\left({\frac {\hbar }{i}}x_{2}{\frac {\partial }{\partial x_{3}}}-{\frac {\hbar }{i}}x_{3}{\frac {\partial }{\partial x_{2}}}\right)\left({\frac {\hbar }{i}}x_{3}{\frac {\partial }{\partial x_{1}}}-{\frac {\hbar }{i}}x_{1}{\frac {\partial }{\partial x_{3}}}\right)\\&=&-\hbar ^{2}\left(x_{2}x_{3}{\frac {\partial ^{2}}{\partial x_{1}\partial x_{3}}}+x_{2}{\frac {\partial }{\partial x_{1}}}-x_{1}x_{2}{\frac {\partial ^{2}}{\partial x_{3}^{2}}}-x_{3}^{2}{\frac {\partial ^{2}}{\partial x_{1}\partial x_{2}}}+x_{1}x_{3}{\frac {\partial ^{2}}{\partial x_{2}\partial x_{3}}}\right)\end{matrix}}}$

${\displaystyle {\begin{matrix}{\hat {L}}_{2}{\hat {L}}_{1}&=&\left({\frac {\hbar }{i}}x_{3}{\frac {\partial }{\partial x_{1}}}-{\frac {\hbar }{i}}x_{1}{\frac {\partial }{\partial x_{3}}}\right)\left({\frac {\hbar }{i}}x_{2}{\frac {\partial }{\partial x_{3}}}-{\frac {\hbar }{i}}x_{3}{\frac {\partial }{\partial x_{2}}}\right)\\&=&-\hbar ^{2}\left(x_{2}x_{3}{\frac {\partial ^{2}}{\partial x_{1}\partial x_{3}}}-x_{3}^{2}{\frac {\partial ^{2}}{\partial x_{1}\partial x_{2}}}-x_{1}x_{2}{\frac {\partial ^{2}}{\partial x_{3}^{2}}}+x_{1}x_{3}{\frac {\partial ^{2}}{\partial x_{2}\partial x_{3}}}+x_{1}{\frac {\partial }{\partial x_{2}}}\right)\end{matrix}}}$

${\displaystyle \left[{\hat {L}}_{1},{\hat {L}}_{2}\right]={\hat {L}}_{1}{\hat {L}}_{2}-{\hat {L}}_{2}{\hat {L}}_{1}=-\hbar ^{2}\left(x_{2}{\frac {\partial }{\partial x_{1}}}-x_{1}{\frac {\partial }{\partial x_{2}}}\right)=-i\hbar \left({\frac {\hbar }{i}}x_{2}{\frac {\partial }{\partial x_{1}}}-{\frac {\hbar }{i}}x_{1}{\frac {\partial }{\partial x_{2}}}\right)=i\hbar {\hat {L}}_{3}}$

Analog erhällt man ${\displaystyle \left[{\hat {L}}_{2},{\hat {L}}_{3}\right]=i\hbar {\hat {L}}_{1}}$ und ${\displaystyle \left[{\hat {L}}_{3},{\hat {L}}_{1}\right]=i\hbar {\hat {L}}_{2}}$

mit den Regeln

${\displaystyle \left[A,B\right]=-\left[B,A\right]}$ und ${\displaystyle \left[A,A\right]=0}$

kennt man somit alle Kommutatoren der Drehimpulsoperatoren.

Gesamtdrehimpuls

${\displaystyle {\hat {L}}^{2}:={\hat {L}}_{1}^{2}+{\hat {L}}_{2}^{2}+{\hat {L}}_{3}^{2}}$

${\displaystyle {\hat {L}}_{1}{\hat {L}}_{2}^{2}={\hat {L}}_{2}{\hat {L}}_{1}{\hat {L}}_{2}+\left[{\hat {L}}_{1},{\hat {L}}_{2}\right]{\hat {L}}_{2}={\hat {L}}_{2}^{2}{\hat {L}}_{1}+{\hat {L}}_{2}\left[{\hat {L}}_{1},{\hat {L}}_{2}\right]+\left[{\hat {L}}_{1},{\hat {L}}_{2}\right]{\hat {L}}_{2}={\hat {L}}_{2}^{2}{\hat {L}}_{1}+i\hbar \left({\hat {L}}_{2}{\hat {L}}_{3}+{\hat {L}}_{3}{\hat {L}}_{2}\right)}$

${\displaystyle \left[{\hat {L}}_{1},{\hat {L}}_{2}^{2}\right]=i\hbar \left({\hat {L}}_{2}{\hat {L}}_{3}+{\hat {L}}_{3}{\hat {L}}_{2}\right)=i\hbar \left\{{\hat {L}}_{2},{\hat {L}}_{3}\right\}}$

${\displaystyle \left[{\hat {L}}_{3},{\hat {L}}_{2}^{2}\right]={\hat {L}}_{2}\left[{\hat {L}}_{3},{\hat {L}}_{2}\right]+\left[{\hat {L}}_{3},{\hat {L}}_{2}\right]{\hat {L}}_{2}=-i\hbar \left\{{\hat {L}}_{2},{\hat {L}}_{1}\right\}}$

Entsprechend errechnet man:

${\displaystyle \left[{\hat {L}}_{2},{\hat {L}}_{3}^{2}\right]=i\hbar \left\{{\hat {L}}_{3},{\hat {L}}_{1}\right\}}$

${\displaystyle \left[{\hat {L}}_{1},{\hat {L}}_{3}^{2}\right]=-i\hbar \left\{{\hat {L}}_{3},{\hat {L}}_{2}\right\}}$

Somit gelangt man schliesslich zu:

${\displaystyle \left[{\hat {L}}_{1},{\hat {L}}^{2}\right]=\left[{\hat {L}}_{1},{\hat {L}}_{1}^{2}\right]+\left[{\hat {L}}_{1},{\hat {L}}_{2}^{2}\right]+\left[{\hat {L}}_{1},{\hat {L}}_{3}^{2}\right]=i\hbar \left\{{\hat {L}}_{2},{\hat {L}}_{3}\right\}-i\hbar \left\{{\hat {L}}_{3},{\hat {L}}_{2}\right\}=0}$

Analog gilt:

${\displaystyle \left[{\hat {L}}_{2},{\hat {L}}^{2}\right]=0}$

und

${\displaystyle \left[{\hat {L}}_{3},{\hat {L}}^{2}\right]=0}$

Der Gesamtdrehimpuls vertauscht also mit jeder Komponente des Drehimpulses.

Aufsteiger und Absteiger

Aufsteiger ${\displaystyle {\hat {L}}_{+}:={\hat {L}}_{1}+i{\hat {L}}_{2}}$

Absteiger ${\displaystyle {\hat {L}}_{-}:={\hat {L}}_{1}-i{\hat {L}}_{2}}$

${\displaystyle \left[{\hat {L}}^{2},L_{\pm }\right]=\left[{\hat {L}}^{2},L_{1}\right]\pm i\left[{\hat {L}}^{2},L_{2}\right]=0\pm i\cdot 0=0}$

${\displaystyle \left[{\hat {L}}_{3},L_{\pm }\right]=\left[{\hat {L}}_{3},L_{1}\right]\pm i\left[{\hat {L}}_{3},L_{2}\right]=i\hbar {\hat {L}}_{2}\pm \hbar {\hat {L}}_{1}=\pm \hbar \left({\hat {L}}_{1}\pm iL_{2}\right)=\pm \hbar L_{\pm }}$

Offenbar gilt:

${\displaystyle {\hat {L}}_{1}={\frac {1}{2}}\left({\hat {L}}_{+}+{\hat {L}}_{-}\right)}$

sowie

${\displaystyle {\hat {L}}_{2}={\frac {1}{2i}}\left({\hat {L}}_{+}-{\hat {L}}_{-}\right)}$

Und somit

${\displaystyle {\begin{matrix}{\hat {L}}_{1}^{2}+{\hat {L}}_{2}^{2}&=&{\frac {1}{4}}\left({\hat {L}}_{+}^{2}+{\hat {L}}_{+}{\hat {L}}_{-}+{\hat {L}}_{-}{\hat {L}}_{+}+{\hat {L}}_{-}^{2}\right)\\&&-{\frac {1}{4}}\left({\hat {L}}_{+}^{2}-{\hat {L}}_{+}{\hat {L}}_{-}-{\hat {L}}_{-}{\hat {L}}_{+}+{\hat {L}}_{-}^{2}\right)\\&=&{\frac {1}{2}}\left({\hat {L}}_{+}{\hat {L}}_{-}+{\hat {L}}_{-}{\hat {L}}_{+}\right)\end{matrix}}}$

Weiterhin ist

${\displaystyle \left[{\hat {L}}_{+},{\hat {L}}_{-}\right]=\left[{\hat {L}}_{1}+i{\hat {L}}_{2},{\hat {L}}_{1}-i{\hat {L}}_{2}\right]=-i\left[{\hat {L}}_{1},{\hat {L}}_{2}\right]+i\left[{\hat {L}}_{2},{\hat {L}}_{1}\right]=2i\left[{\hat {L}}_{2},{\hat {L}}_{1}\right]=2\hbar {\hat {L}}_{3}}$

Damit

${\displaystyle {\hat {L}}_{1}^{2}+{\hat {L}}_{2}^{2}={\frac {1}{2}}\left({\hat {L}}_{+}{\hat {L}}_{-}+{\hat {L}}_{-}{\hat {L}}_{+}\right)={\hat {L}}_{+}{\hat {L}}_{-}-\hbar {\hat {L}}_{3}}$

und schließlich

${\displaystyle {\hat {L}}^{2}={\hat {L}}_{1}^{2}+{\hat {L}}_{2}^{2}+{\hat {L}}_{3}^{2}={\hat {L}}_{+}{\hat {L}}_{-}-\hbar {\hat {L}}_{3}+{\hat {L}}_{3}^{2}}$

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