量子力学Ⅰ/水素原子/メモ のバックアップ(No.1)

解く †

　

と置けば、

　&math( \frac{\PD^2}{\PD\rho^2}X(\rho)e^{-\rho/n} &=\frac{\PD}{\PD\rho}X'(\rho)e^{-\rho/n}-\frac{1}{n}\frac{\PD}{\PD\rho}X(\rho)e^{-\rho/n}\\ &=X''(\rho)e^{-\rho/n}-\frac{2}{n}X'(\rho)e^{-\rho/n}+\frac{n}{n^2}X(\rho)e^{-\rho/n}\\ );

を使って、

　&math( X''(\rho)-\frac{2}{n}X'(\rho)+\cancel{\frac{1}{n^2}X(\rho)}

1. \left\{\frac{2}{\rho}-\frac{l(l+1)}{\rho^2}\right\}X(\rho)-\cancel{\frac{1}{n^2} X(\rho)}=0 );

　&math( X''(\rho)-\frac{2}{n}X'(\rho)+\left\{\frac{2}{\rho}-\frac{l(l+1)}{\rho^2}\right\}X(\rho)=0 );

と置けば、

　&math( \sum_{i=0}^\infty k(k-1)c_k\rho^{k-2}-\frac{2}{n}\sum_{i=0}^\infty kc_k\rho^{k-1}+\left\{\frac{2}{\rho}-\frac{l(l+1)}{\rho^2}\right\}\sum_{i=0}^\infty c_k\rho^k=0 );

　&math( \sum_{i=0}^\infty (k+2)(k+1)c_{k+2}\rho^k-\frac{2}{n}\sum_{i=0}^\infty (k+1)c_{k+1}\rho^k+\sum_{i=0}^\infty 2c_k\rho^{k-2}-\sum_{i=0}^\infty l(l+1)c_k\rho^{k-1}=0 );

　&math( \sum_{i=0}^\infty (k+2)(k+1)c_{k+2}\rho^k-\frac{2}{n}\sum_{i=0}^\infty (k+1)c_{k+1}\rho^k+\sum_{i=-2}^\infty 2c_{k+2}\rho^k-\sum_{i=-1}^\infty l(l+1)c_{k+1}\rho^{k}=0 );

　&math( &2c_0\rho^{-2}+\{2c_{1}-l(l+1)c_0\}\rho^{-1}+\\ &\sum_{i=0}^\infty \big[\big\{(k+2)(k+1)+2\big\}c_{k+2}-\big\{2(k+1)/n+l(l+1)\big\}c_{k+1}\big]\rho^k=0\\ );

すなわち、

　&math( &c_0=0\\ &c_1=\frac{l(l+1)}{2}c_0\\ &\big\{(k+2)(k+1)+2\big\}c_{k+2}-\big\{2(k+1)/n+l(l+1)\big\}c_{k+1}=0 );