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  応援いただけましたら励みになります(・ω・)ノシ       にほんブログ村 科学ブログ 物理学へ

シュレーディンガー方程式を解く2

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!