Quantenmechanik/ Drehimpuls-Algebra

Drehimpuls-Operator:

$\hat{L}:=\hat{x}\times \hat{p}=\frac{\hslash }{i}x\times \mathrm{\nabla }$

$\begin{array}{ccc}{\hat{L}}_1& =& \frac{\hslash }{i}{x}_2\frac{\partial }{\partial x_3}-\frac{\hslash }{i}{x}_3\frac{\partial }{\partial x_2}\\ {\hat{L}}_2& =& \frac{\hslash }{i}{x}_3\frac{\partial }{\partial x_1}-\frac{\hslash }{i}{x}_1\frac{\partial }{\partial x_3}\\ {\hat{L}}_3& =& \frac{\hslash }{i}{x}_1\frac{\partial }{\partial x_2}-\frac{\hslash }{i}{x}_2\frac{\partial }{\partial x_1}\end{array}$

Kommutator und Antikommutator

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

$\begin{array}{ccc}{\hat{L}}_1{\hat{L}}_2& =& (\frac{\hslash }{i}{x}_2\frac{\partial }{\partial x_3}-\frac{\hslash }{i}{x}_3\frac{\partial }{\partial x_2})(\frac{\hslash }{i}{x}_3\frac{\partial }{\partial x_1}-\frac{\hslash }{i}{x}_1\frac{\partial }{\partial x_3})\\ & =& -{\hslash }^2(x_2{x}_3\frac{{\partial }^2}{\partial x_1\partial x_3}+x_2\frac{\partial }{\partial x_1}-x_1{x}_2\frac{{\partial }^2}{\partial x_3^2}-x_3^2\frac{{\partial }^2}{\partial x_1\partial x_2}+x_1{x}_3\frac{{\partial }^2}{\partial x_2\partial x_3})\end{array}$

$\begin{array}{ccc}{\hat{L}}_2{\hat{L}}_1& =& (\frac{\hslash }{i}{x}_3\frac{\partial }{\partial x_1}-\frac{\hslash }{i}{x}_1\frac{\partial }{\partial x_3})(\frac{\hslash }{i}{x}_2\frac{\partial }{\partial x_3}-\frac{\hslash }{i}{x}_3\frac{\partial }{\partial x_2})\\ & =& -{\hslash }^2(x_2{x}_3\frac{{\partial }^2}{\partial x_1\partial x_3}-x_3^2\frac{{\partial }^2}{\partial x_1\partial x_2}-x_1{x}_2\frac{{\partial }^2}{\partial x_3^2}+x_1{x}_3\frac{{\partial }^2}{\partial x_2\partial x_3}+x_1\frac{\partial }{\partial x_2})\end{array}$

$[{\hat{L}}_1,{\hat{L}}_2]={\hat{L}}_1{\hat{L}}_2-{\hat{L}}_2{\hat{L}}_1=-{\hslash }^2(x_2\frac{\partial }{\partial x_1}-x_1\frac{\partial }{\partial x_2})=-i\hslash (\frac{\hslash }{i}{x}_2\frac{\partial }{\partial x_1}-\frac{\hslash }{i}{x}_1\frac{\partial }{\partial x_2})=i\hslash {\hat{L}}_3$

Analog erhällt man $[{\hat{L}}_2,{\hat{L}}_3]=i\hslash {\hat{L}}_1$ und $[{\hat{L}}_3,{\hat{L}}_1]=i\hslash {\hat{L}}_2$

mit den Regeln

$[A,B]=-[B,A]$ und $[A,A]=0$

kennt man somit alle Kommutatoren der Drehimpulsoperatoren.

Gesamtdrehimpuls

${\hat{L}}^2:={\hat{L}}_1^2+{\hat{L}}_2^2+{\hat{L}}_3^2$

${\hat{L}}_1{\hat{L}}_2^2={\hat{L}}_2{\hat{L}}_1{\hat{L}}_2+[{\hat{L}}_1,{\hat{L}}_2]{\hat{L}}_2={\hat{L}}_2^2{\hat{L}}_1+{\hat{L}}_2[{\hat{L}}_1,{\hat{L}}_2]+[{\hat{L}}_1,{\hat{L}}_2]{\hat{L}}_2={\hat{L}}_2^2{\hat{L}}_1+i\hslash ({\hat{L}}_2{\hat{L}}_3+{\hat{L}}_3{\hat{L}}_2)$

$[{\hat{L}}_1,{\hat{L}}_2^2]=i\hslash ({\hat{L}}_2{\hat{L}}_3+{\hat{L}}_3{\hat{L}}_2)=i\hslash \{{\hat{L}}_2,{\hat{L}}_3\}$

$[{\hat{L}}_3,{\hat{L}}_2^2]={\hat{L}}_2[{\hat{L}}_3,{\hat{L}}_2]+[{\hat{L}}_3,{\hat{L}}_2]{\hat{L}}_2=-i\hslash \{{\hat{L}}_2,{\hat{L}}_1\}$

Entsprechend errechnet man:

$[{\hat{L}}_2,{\hat{L}}_3^2]=i\hslash \{{\hat{L}}_3,{\hat{L}}_1\}$

$[{\hat{L}}_1,{\hat{L}}_3^2]=-i\hslash \{{\hat{L}}_3,{\hat{L}}_2\}$

Somit gelangt man schliesslich zu:

$[{\hat{L}}_1,{\hat{L}}^2]=[{\hat{L}}_1,{\hat{L}}_1^2]+[{\hat{L}}_1,{\hat{L}}_2^2]+[{\hat{L}}_1,{\hat{L}}_3^2]=i\hslash \{{\hat{L}}_2,{\hat{L}}_3\}-i\hslash \{{\hat{L}}_3,{\hat{L}}_2\}=0$

Analog gilt:

$[{\hat{L}}_2,{\hat{L}}^2]=0$

und

$[{\hat{L}}_3,{\hat{L}}^2]=0$

Der Gesamtdrehimpuls vertauscht also mit jeder Komponente des Drehimpulses.

Aufsteiger und Absteiger

Aufsteiger ${\hat{L}}_{+}:={\hat{L}}_1+i{\hat{L}}_2$

Absteiger ${\hat{L}}_{-}:={\hat{L}}_1-i{\hat{L}}_2$

$[{\hat{L}}^2,L_{\pm }]=[{\hat{L}}^2,L_1]\pm i[{\hat{L}}^2,L_2]=0\pm i\cdot 0=0$

$[{\hat{L}}_3,L_{\pm }]=[{\hat{L}}_3,L_1]\pm i[{\hat{L}}_3,L_2]=i\hslash {\hat{L}}_2\pm \hslash {\hat{L}}_1=\pm \hslash ({\hat{L}}_1\pm i{L}_2)=\pm \hslash L_{\pm }$

Offenbar gilt:

${\hat{L}}_1=\frac{1}{2}({\hat{L}}_{+}+{\hat{L}}_{-})$

sowie

${\hat{L}}_2=\frac{1}{2i}({\hat{L}}_{+}-{\hat{L}}_{-})$

Und somit

$\begin{array}{ccc}{\hat{L}}_1^2+{\hat{L}}_2^2& =& \frac{1}{4}({\hat{L}}_{+}^2+{\hat{L}}_{+}{\hat{L}}_{-}+{\hat{L}}_{-}{\hat{L}}_{+}+{\hat{L}}_{-}^2)\\ & & -\frac{1}{4}({\hat{L}}_{+}^2-{\hat{L}}_{+}{\hat{L}}_{-}-{\hat{L}}_{-}{\hat{L}}_{+}+{\hat{L}}_{-}^2)\\ & =& \frac{1}{2}({\hat{L}}_{+}{\hat{L}}_{-}+{\hat{L}}_{-}{\hat{L}}_{+})\end{array}$

Weiterhin ist

$[{\hat{L}}_{+},{\hat{L}}_{-}]=[{\hat{L}}_1+i{\hat{L}}_2,{\hat{L}}_1-i{\hat{L}}_2]=-i[{\hat{L}}_1,{\hat{L}}_2]+i[{\hat{L}}_2,{\hat{L}}_1]=2i[{\hat{L}}_2,{\hat{L}}_1]=2\hslash {\hat{L}}_3$

Damit

${\hat{L}}_1^2+{\hat{L}}_2^2=\frac{1}{2}({\hat{L}}_{+}{\hat{L}}_{-}+{\hat{L}}_{-}{\hat{L}}_{+})={\hat{L}}_{+}{\hat{L}}_{-}-\hslash {\hat{L}}_3$

und schließlich

${\hat{L}}^2={\hat{L}}_1^2+{\hat{L}}_2^2+{\hat{L}}_3^2={\hat{L}}_{+}{\hat{L}}_{-}-\hslash {\hat{L}}_3+{\hat{L}}_3^2$