量子力学Ⅰ/球面調和関数/メモ の履歴(No.2)
更新解答:$m$ に関する漸化式†
(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)\\ );