シュレーディンガー方程式を解く2 - 広島文理科大学波動幾何学研究所

第2回ですじゃ．

前回はラプラシアンの中身を作るところの出だしまででしただ．

${\displaystyle\frac{\partial r}{\partial x}=\frac{x}{r}=\sin{\theta}\cos{\phi}}$

この調子だと，他の変数についてもいけそうだなや！

${\displaystyle\frac{\partial r}{\partial y}=\frac{y}{r}=\sin{\theta}\sin{\phi}}$

${\displaystyle\frac{\partial r}{\partial z}=\frac{z}{r}=\cos{\theta}}$

よしよし．

それでは次は ${\theta}$ ．

${\displaystyle\frac{\partial}{\partial x}=\frac{\partial\theta}{\partial x}\frac{\partial}{\partial \theta}}$

前回の(3)でarctanにしちゃってるけど，

${\tan^{2}{\theta}=\displaystyle\frac{x^2+y^2}{z^2}}$

こいつを使いましょうぞ．

${\displaystyle\frac{\partial}{\partial x}\displaystyle\frac{x^2+y^2}{z^2}=\frac{\partial\theta}{\partial x}\frac{\partial}{\partial \theta}\tan^{2}{\theta}}$

${\displaystyle\frac{\partial}{\partial x}\displaystyle\frac{1}{z^2}\cdot(x^2+y^2)=\frac{\partial\theta}{\partial x}\frac{2\tan{\theta}}{\cos^2{\theta}}}$

${\displaystyle\frac{2x}{z^2}=\frac{\partial\theta}{\partial x}\frac{2\tan{\theta}}{\cos^2{\theta}}}$

関数の積の微分，前々回の結果など使いました．

wave-geometry.hatenablog.com

式を整理します．

${\displaystyle\frac{\partial\theta}{\partial x}=\frac{x}{z^2}\frac{\cos^2{\theta}}{\tan{\theta}}}$

${\displaystyle\frac{1}{\tan{\theta}}=\displaystyle\frac{\cos{\theta}}{\sin{\theta}}}$ だし，

${\cos{\theta}=\displaystyle\frac{z}{r}}$ だし，

${x=r\sin{\theta}\cos{\phi}}$ だしで，

${\displaystyle\frac{\partial\theta}{\partial x}=\frac{r\sin{\theta}\cos{\phi}}{z^2}\frac{z^2}{r^2}\frac{\cos{\theta}}{\sin{\theta}}}$

${\displaystyle\frac{\partial\theta}{\partial x}=\frac{1}{r}\cos{\theta}\cos{\phi}}$

次はy，

${\displaystyle\frac{2y}{z^2}=\frac{\partial\theta}{\partial y}\frac{2\tan{\theta}}{\cos^2{\theta}}}$

${\displaystyle\frac{\partial\theta}{\partial y}=\frac{y}{z^2}\frac{\cos^2{\theta}}{\tan{\theta}}}$

${\displaystyle\frac{\partial\theta}{\partial y}=\frac{r\sin{\theta}\sin{\phi}}{z^2}\frac{z^2}{r^2}\frac{\cos{\theta}}{\sin{\theta}}}$

${\displaystyle\frac{\partial\theta}{\partial y}=\frac{1}{r}\cos{\theta}\sin{\phi}}$

${\theta}$ の最後はz，これはちょっとだけ違う．

${\displaystyle\frac{\partial}{\partial z}\left\{z^{-2}(x^2+y^2)\right\}=\frac{\partial\theta}{\partial z}\frac{2\tan{\theta}}{\cos^2{\theta}}}$

${\displaystyle\frac{-2(x^2+y^2)}{z^3}=\frac{\partial\theta}{\partial z}\frac{2\tan{\theta}}{\cos^2{\theta}}}$

${\displaystyle\frac{\partial\theta}{\partial z}=\frac{-(x^2+y^2)}{z^3}\frac{\cos^2{\theta}}{\tan{\theta}}}$

${\displaystyle=-\frac{r^2\sin^2{\theta}}{r^3\cos^{3}{\theta}}\frac{\cos^3{\theta}}{\sin{\theta}}}$

${\displaystyle=-\frac{1}{r}\sin{\theta}}$

これで ${\theta}$ の段は出揃いましただな．

${\displaystyle\frac{\partial\theta}{\partial x}=\frac{1}{r}\cos{\theta}\cos{\phi}}$

${\displaystyle\frac{\partial\theta}{\partial y}=\frac{1}{r}\cos{\theta}\sin{\phi}}$

${\displaystyle\frac{\partial\theta}{\partial z}=-\frac{1}{r}\sin{\theta}}$

次は ${\phi}$ の段．
（・〜・）がんばりますだ．

${\tan{\phi}=\displaystyle\frac{y}{x}\cdots(1)}$ を使いますのじゃ．

まずはx，

${\displaystyle\frac{\partial}{\partial x}\left(\frac{1}{x}\cdot y\right)=\frac{\partial\phi}{\partial x}\frac{1}{\cos^2{\phi}}}$

${-\displaystyle\frac{y}{x^2}=\frac{\partial\phi}{\partial x}\frac{1}{\cos^2{\phi}}}$

${\displaystyle\frac{\partial\phi}{\partial x}=-\cos^2{\phi}\frac{r\sin{\theta}\sin{\phi}}{r^2\sin^2{\theta}\cos^2{\phi}}}$

${\displaystyle=-\frac{\sin{\phi}}{r\sin{\theta}}}$

次はy，

${\displaystyle\frac{\partial}{\partial y}\left(\frac{1}{x}\cdot y\right)=\frac{\partial\phi}{\partial y}\frac{1}{\cos^2{\phi}}}$

${\displaystyle\frac{1}{x}=\frac{\partial\phi}{\partial x}\frac{1}{\cos^2{\phi}}}$

${\displaystyle\frac{\partial\phi}{\partial y}=\frac{\cos^2{\phi}}{r\sin{\theta}\cos{\phi}}}$

${\displaystyle=\frac{\cos{\phi}}{r\sin{\theta}}}$

最後に，zは(1)の右辺がゼロになるんで，どうあってもゼロですだなぁ．

${\displaystyle\frac{\partial}{\partial z}\left(\frac{1}{x}\cdot y\right)=\frac{\partial\phi}{\partial z}\frac{1}{\cos^2{\phi}}}$

${\displaystyle\frac{\partial\phi}{\partial z}=0}$

（・∀・）ヨシヨシ．

まとめますと，

${\displaystyle\frac{\partial r}{\partial x}=\sin{\theta}\cos{\phi}}$

${\displaystyle\frac{\partial r}{\partial y}=\sin{\theta}\sin{\phi}}$

${\displaystyle\frac{\partial r}{\partial z}=\cos{\theta}}$

${\displaystyle\frac{\partial\theta}{\partial x}=\frac{1}{r}\cos{\theta}\cos{\phi}}$

${\displaystyle\frac{\partial\theta}{\partial y}=\frac{1}{r}\cos{\theta}\sin{\phi}}$

${\displaystyle\frac{\partial\theta}{\partial z}=-\frac{1}{r}\sin{\theta}}$

${\displaystyle\frac{\partial\phi}{\partial x}=-\frac{\sin{\phi}}{r\sin{\theta}}}$

${\displaystyle\frac{\partial\phi}{\partial y}=\frac{\cos{\phi}}{r\sin{\theta}}}$

${\displaystyle\frac{\partial\phi}{\partial z}=0}$

得られたこれらを，以下の式に代入すればよいのですじゃ．

${\displaystyle\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}+\frac{\partial\phi}{\partial x}\frac{\partial}{\partial \phi}}$

${\displaystyle\frac{\partial}{\partial y}=\frac{\partial r}{\partial y}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial y}\frac{\partial}{\partial \theta}+\frac{\partial\phi}{\partial y}\frac{\partial}{\partial \phi}}$

${\displaystyle\frac{\partial}{\partial z}=\frac{\partial r}{\partial z}\frac{\partial}{\partial r}+\frac{\partial \theta}{\partial z}\frac{\partial}{\partial \theta}+\frac{\partial\phi}{\partial z}\frac{\partial}{\partial \phi}}$

それっ，

${\displaystyle\frac{\partial}{\partial x}=\sin{\theta}\cos{\phi}\frac{\partial}{\partial r}+\frac{1}{r}\cos{\theta}\cos{\phi}\frac{\partial}{\partial \theta}-\frac{\sin{\phi}}{r\sin{\theta}}\frac{\partial}{\partial \phi}}$

${\displaystyle\frac{\partial}{\partial y}=\sin{\theta}\sin{\phi}\frac{\partial}{\partial r}+\frac{1}{r}\cos{\theta}\sin{\phi}\frac{\partial}{\partial \theta}+\frac{\cos{\phi}}{r\sin{\theta}}\frac{\partial}{\partial \phi}}$

${\displaystyle\frac{\partial}{\partial z}=\cos{\theta}\frac{\partial}{\partial r}-\frac{1}{r}\sin{\theta}\frac{\partial}{\partial \theta}}$

（；ω；）感涙
今回はここまで！