量子力学Ⅰ/球面調和関数/メモ のバックアップソース(No.2)

更新

#mathjax

* 解答:$m$ に関する漸化式 [#k309d54e]

(1)

 &math(
l_\pm Y_l{}^{\pm l}(\theta,\phi) 
&\propto \hbar e^{\pm i\phi}
\Big(\pm\frac{\PD}{\PD\theta}+\frac{i}{\tan\theta}\frac{\PD}{\PD\phi}\Big)
\sin^l\theta e^{\pm il\phi}\\
&=\hbar e^{\pm i(l+1)\phi}
\Big\{\pm(l\sin^{l-1}\theta\cos\theta)+\frac{i\cos\theta}{\sin\theta}\cdot(\pm il\sin^l\theta)\Big\}\\
&=\hbar e^{\pm i(l+1)\phi}\Big(\pm l\sin^{l-1}\theta\cos\theta\mp l\sin^{l-1}\theta\cos\theta\Big)\\
&= 0
);

(2)

 &math(
&\hat l_-\hat l_+Y_l{}^{m}(\theta,\phi)=\hat l_-\hbar\sqrt{(l-m)(l+m+1)}Y_l{}^{m+1}
=\hbar^2(l+m+1)(l-m)Y_l{}^m\\
&\hat l_+\hat l_-Y_l{}^{m}(\theta,\phi)=\hat l_+\hbar\sqrt{(l+m)(l-m+1)}Y_l{}^{m-1}
=\hbar^2(l+m)(l-m+1)Y_l{}^{m}\\
);

(3)

 &math(
\hat l_x^2+\hat l_y^2
&=\frac{1}{2}\left\{ (\hat l_x+i\hat l_y)(\hat l_x-i\hat l_y)+(\hat l_x-i\hat l_y)(\hat l_x+i\hat l_y)\right\}\\
&=\frac{1}{2}\left\{ \hat l_+\hat l_-+\hat l_-\hat l_+\right\}\\
);

(4)

 &math(
\langle \hat l_x^2+\hat l_y^2\rangle
&=\frac{\hbar^2}{2}\big\{(l+m+1)(l-m)+(l+m)(l-m+1)\big\}\\
&=\hbar^2(l^2-m^2+l)\\
);

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