球座標を用いた変数分離 のバックアップソース(No.2)

更新

[[量子力学Ⅰ]]

#contents
&mathjax();

* 極座標 [#x544b994]

&math(
\begin{cases}
x=r\sin\theta\cos\phi\\
y=r\sin\theta\sin\phi\\
z=r\cos\theta
\end{cases}
);


** 微分の変換 [#xba4e8e0]

 &math(df=\frac{\PD f}{\PD r}dr+\frac{\PD f}{\PD \theta}d\theta+\frac{\PD f}{\PD \phi}d\phi);

に、&math(dr=\frac{\PD r}{\PD x}dx,\ d\theta=\frac{\PD \theta}{\PD x}dx,\ d\phi=\frac{\PD \phi}{\PD x}dx); を代入すれば、

 &math(df=\frac{\PD f}{\PD r}\frac{\PD r}{\PD x}dx+\frac{\PD f}{\PD \theta}\frac{\PD \theta}{\PD x}dx+\frac{\PD f}{\PD \phi}\frac{\PD \phi}{\PD x}dx);

変形して、

 &math(\frac{\PD }{\PD x}f=\Big(\frac{\PD r}{\PD x}\frac{\PD}{\PD r}+\frac{\PD \theta}{\PD x}\frac{\PD}{\PD \theta}+\frac{\PD \phi}{\PD x}\frac{\PD}{\PD \phi}\Big)f);

したがって、

 &math(\frac{\PD }{\PD x}=\frac{\PD r}{\PD x}\frac{\PD}{\PD r}+\frac{\PD \theta}{\PD x}\frac{\PD}{\PD \theta}+\frac{\PD \phi}{\PD x}\frac{\PD}{\PD \phi}\\
);

同様にして、

 &math(
\begin{cases}
\displaystyle\frac{\PD }{\PD x}=\frac{\PD r}{\PD x}\frac{\PD}{\PD r}+\frac{\PD \theta}{\PD x}\frac{\PD}{\PD \theta}+\frac{\PD \phi}{\PD x}\frac{\PD}{\PD \phi}\\[4mm]
\displaystyle\frac{\PD }{\PD y}=\frac{\PD r}{\PD x}\frac{\PD}{\PD r}+\frac{\PD \theta}{\PD x}\frac{\PD}{\PD \theta}+\frac{\PD \phi}{\PD x}\frac{\PD}{\PD \phi}\\[4mm]
\displaystyle\frac{\PD }{\PD z}=\frac{\PD r}{\PD x}\frac{\PD}{\PD r}+\frac{\PD \theta}{\PD x}\frac{\PD}{\PD \theta}+\frac{\PD \phi}{\PD x}\frac{\PD}{\PD \phi}\\
\end{cases}
);

のように変換される。

** 具体的に計算する [#t892cda9]

全微分の時と異なり、&math(\frac{\PD r}{\PD x}\ne\Big(\frac{\PD x}{\PD r}\Big)^{-1}); であることに注意せよ。

&math(r^2=x^2+y^2+z^2); より、&math(2r\frac{\PD r}{\PD x}=2x); などが得られて、

 &math(
\begin{cases}
\displaystyle\frac{\PD r}{\PD x}=\frac{x}{r}=\sin\theta\cos\phi\\[4mm]
\displaystyle\frac{\PD r}{\PD y}=\frac{y}{r}=\sin\theta\sin\phi\\[4mm]
\displaystyle\frac{\PD r}{\PD z}=\frac{z}{r}=\cos\theta\\
\end{cases}
);

&math(\tan^2\theta=\frac{x^2+y^2}{z^2}); より、
&math(\frac{1}{\cos^2\theta}\frac{\PD \theta}{\PD y}=\frac{2x}{z^2});

&math(\frac{\not\!2\tan\theta}{\cos^2\theta}\frac{\PD \theta}{\PD x}=\frac{\not\! 2x}{z^2});、
&math(\frac{\not\!2\tan\theta}{\cos^2\theta}\frac{\PD \theta}{\PD y}=\frac{\not\! 2y}{z^2});、
&math(\frac{\not\!2\tan\theta}{\cos^2\theta}\frac{\PD \theta}{\PD z}=-\not\!2\frac{x^2+y^2}{z^3});、

 &math(
\begin{cases}
\displaystyle\frac{\PD \theta}{\PD x}=\frac{r\sin\theta\cos\phi}{r^2\cos^2\theta}\frac{\cos^2\theta}{\tan\theta}=\frac{1}{r}\cos\theta\cos\phi\\[4mm]
\displaystyle\frac{\PD \theta}{\PD y}=\frac{r\sin\theta\sin\phi}{r^2\cos^2\theta}\frac{\cos^2\theta}{\tan\theta}=\frac{1}{r}\cos\theta\sin\phi\\[4mm]
\displaystyle\frac{\PD \theta}{\PD z}=-\frac{r^2\sin^2\theta}{r^3\cos^3\theta}\frac{\cos^2\theta}{\tan\theta}=-\frac{1}{r}\sin\theta
\end{cases}
);

&math(\tan\phi=\frac{y}{x}); より、

&math(\frac{1}{\cos^2\phi}\frac{\PD\phi}{\PD x}=-\frac{y}{x^2});、
&math(\frac{1}{\cos^2\phi}\frac{\PD\phi}{\PD y}=\frac{1}{x});、
&math(\frac{1}{\cos^2\phi}\frac{\PD\phi}{\PD z}=0); であるから、

 &math(
\begin{cases}
\displaystyle\frac{\PD \phi}{\PD x}=-\frac{r\sin\theta\sin\phi}{r^2\sin^2\theta\cos^2\phi}\cos^2\phi=-\frac{\sin\phi}{r\sin\theta}\\[4mm]
\displaystyle\frac{\PD \phi}{\PD y}=\frac{1}{r\sin\theta\cos\phi}\cos^2\phi=\frac{\cos\phi}{r\sin\theta}\\[4mm]
\displaystyle\frac{\PD \phi}{\PD z}=0
\end{cases}
);

これらを代入して、

 &math(
\begin{cases}

\displaystyle\frac{\PD}{\PD x}=
\sin\theta\cos\phi \frac{\PD}{\PD r}
+\frac{1}{r}\cos\theta\cos\phi \frac{\PD}{\PD \theta}
-\frac{\sin\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\\[4mm]

\displaystyle\frac{\PD}{\PD y}=
\sin\theta\sin\phi \frac{\PD}{\PD r}
+\frac{1}{r}\cos\theta\sin\phi \frac{\PD}{\PD \theta}
+\frac{\cos\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\\[4mm]

\displaystyle \frac{\PD}{\PD z}=
\cos\theta \frac{\PD}{\PD r}
-\frac{1}{r}\sin\theta \frac{\PD}{\PD \theta}\\[4mm]

\end{cases}
);

** 球座標表示のラプラシアン [#w0c305d4]

&math(\triangle=\nabla^2=\frac{\PD^2}{\PD x^2}+\frac{\PD^2}{\PD y^2}+\frac{\PD^2}{\PD z^2});

に上記を代入するだけ! ・・・ 実際やってみるとえらい大変。→ [[計算の詳細>量子力学Ⅰ/中心力場内の粒子の運動/メモ#o621468a]]

結果だけまとめると、

&math(\nabla^2=\frac{\PD^2}{\PD r^2}+\frac{2}{r} \frac{\PD}{\PD r}+\frac{1}{r^2}\Lambda);

ただし、

&math(\Lambda=\frac{1}{\sin\theta} \frac{\PD}{\PD \theta}
\Big(\sin\theta\frac{\PD}{\PD \theta}\Big)+\frac{1}{\sin^2\theta} \frac{\PD^2}{\PD \phi^2});

* 極座標で表わしたシュレーディンガー方程式 [#mfb957f5]

* 球関数 $Y^m_l(\theta,\phi)$:角運動量の固有関数 [#s564caed]

* 動径方向の固有関数 [#y54f7421]

** 球形の箱形ポテンシャル [#u7cec361]

** 3次元調和振動子 [#tf639724]

** 水素原子(静電ポテンシャル) [#c5480582]

Counter: 34042 (from 2010/06/03), today: 1, yesterday: 0