球座標における微分演算子/メモ

(215d) 更新

目次

演習:偏微分の計算

解答

(1)

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

  \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}

(2)

\tan^2\theta=\frac{x^2+y^2}{z^2} より、 \frac{1}{\cos^2\theta}\frac{\PD \theta}{\PD y}=\frac{2x}{z^2}

\frac{\not\!2\tan\theta}{\cos^2\theta}\frac{\PD \theta}{\PD x}=\frac{\not\! 2x}{z^2} \frac{\not\!2\tan\theta}{\cos^2\theta}\frac{\PD \theta}{\PD y}=\frac{\not\! 2y}{z^2} \frac{\not\!2\tan\theta}{\cos^2\theta}\frac{\PD \theta}{\PD z}=-\not\!2\frac{x^2+y^2}{z^3}

  \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}

(3)

\tan\phi=\frac{y}{x} より、

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

  \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}

球座標のラプラシアン

\frac{\PD^2}{\PD x^2} &=\Big(\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}\Big)^2\\ &=\sin\theta\cos\phi \frac{\PD}{\PD r}\Big(\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}\Big)\\ &\ \ \ +\frac{1}{r}\cos\theta\cos\phi \frac{\PD}{\PD \theta} \Big(\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}\Big)\\ &\ \ \ -\frac{\sin\phi}{r\sin\theta} \frac{\PD}{\PD \phi} \Big(\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}\Big)\\ &=\sin^2\theta\cos^2\phi \frac{\PD^2}{\PD r^2} -\frac{\sin\theta\cos\theta\cos^2\phi}{r^2} \frac{\PD}{\PD \theta} +\frac{\sin\theta\cos\theta\cos^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta} +\frac{\sin\phi\cos\phi}{r^2} \frac{\PD}{\PD \phi} -\frac{\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}\\ &\ \ \ +\frac{\cos^2\theta\cos^2\phi}{r} \frac{\PD}{\PD r} +\frac{\sin\theta\cos\theta\cos^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta} +\frac{-\sin\theta\cos\theta\cos^2\phi}{r^2}\frac{\PD}{\PD \theta} +\frac{\cos^2\theta\cos^2\phi}{r^2} \frac{\PD^2}{\PD \theta^2}\\ &\hspace{9cm}+\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi} -\frac{\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi} \\ &\ \ \ +\frac{\sin^2\phi}{r} \frac{\PD}{\PD r} -\frac{\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi} +\frac{\cos\theta\sin^2\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta} -\frac{\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi} +\frac{\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi} +\frac{\sin^2\phi}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2}\\ &=\sin^2\theta\cos^2\phi \frac{\PD^2}{\PD r^2} -\frac{\sin\theta\cos\theta\cos^2\phi}{r^2} \frac{\PD}{\PD \theta} +\frac{2\sin\theta\cos\theta\cos^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta} +\frac{\sin\phi\cos\phi}{r^2} \frac{\PD}{\PD \phi} -\frac{2\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}\\ &\ \ \ +\frac{\cos^2\theta\cos^2\phi}{r} \frac{\PD}{\PD r} +\frac{-\sin\theta\cos\theta\cos^2\phi}{r^2}\frac{\PD}{\PD \theta} +\frac{\cos^2\theta\cos^2\phi}{r^2} \frac{\PD^2}{\PD \theta^2} +\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}\\ &\ \ \ -\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi} +\frac{\sin^2\phi}{r} \frac{\PD}{\PD r} +\frac{\cos\theta\sin^2\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta} +\frac{\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi} +\frac{\sin^2\phi}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2}\\

\frac{\PD^2}{\PD y^2} &=\Big( \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} \Big)^2\\ &= \sin^2\theta\sin^2\phi \frac{\PD^2}{\PD r^2} -\frac{\sin\theta\cos\theta\sin^2\phi}{r^2} \frac{\PD}{\PD \theta} +\frac{2\sin\theta\cos\theta\sin^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta} -\frac{\sin\phi\cos\phi}{r^2} \frac{\PD}{\PD \phi} +\frac{2\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}\\ &\ \ \ +\frac{\cos^2\theta\sin^2\phi}{r} \frac{\PD}{\PD r} +\frac{-\sin\theta\cos\theta\sin^2\phi}{r^2} \frac{\PD}{\PD \theta} +\frac{\cos^2\theta\sin^2\phi}{r^2} \frac{\PD^2}{\PD \theta^2} -\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi} +\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi}\\ &\ \ \ +\frac{\cos^2\phi}{r} \frac{\PD}{\PD r} +\frac{\cos\theta\cos^2\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta} +\frac{-\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi} +\frac{\cos^2\phi}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2}

\frac{\PD^2}{\PD z^2} &=\Big( \cos\theta \frac{\PD}{\PD r} -\frac{1}{r}\sin\theta \frac{\PD}{\PD \theta} \Big)^2\\ &= \cos^2\theta \frac{\PD^2}{\PD r^2} +\frac{\sin\theta\cos\theta}{r^2} \frac{\PD}{\PD \theta} -\frac{2\sin\theta\cos\theta}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta} \\ &\ \ \ +\frac{\sin^2\theta}{r} \frac{\PD}{\PD r} +\frac{\sin\theta\cos\theta}{r^2} \frac{\PD}{\PD \theta} +\frac{\sin^2\theta}{r^2} \frac{\PD^2}{\PD \theta^2}

足せばいい(本気?)。

&\frac{\PD^2}{\PD x^2}+\frac{\PD^2}{\PD y^2}+\frac{\PD^2}{\PD z^2}\\ &= \sin^2\theta\cos^2\phi \frac{\PD^2}{\PD r^2} -\frac{\sin\theta\cos\theta\cos^2\phi}{r^2} \frac{\PD}{\PD \theta} +\frac{2\sin\theta\cos\theta\cos^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta} +\frac{\sin\phi\cos\phi}{r^2} \frac{\PD}{\PD \phi} -\frac{2\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}\\ &\ \ \ +\frac{\cos^2\theta\cos^2\phi}{r} \frac{\PD}{\PD r} +\frac{-\sin\theta\cos\theta\cos^2\phi}{r^2}\frac{\PD}{\PD \theta} +\frac{\cos^2\theta\cos^2\phi}{r^2} \frac{\PD^2}{\PD \theta^2} +\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi}\\ &\ \ \ -\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi} +\frac{\sin^2\phi}{r} \frac{\PD}{\PD r} +\frac{\cos\theta\sin^2\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta} +\frac{\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi} +\frac{\sin^2\phi}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2}\\ &\ \ \ +\sin^2\theta\sin^2\phi \frac{\PD^2}{\PD r^2} -\frac{\sin\theta\cos\theta\sin^2\phi}{r^2} \frac{\PD}{\PD \theta} +\frac{2\sin\theta\cos\theta\sin^2\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta} -\frac{\sin\phi\cos\phi}{r^2} \frac{\PD}{\PD \phi} +\frac{2\sin\phi\cos\phi}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \phi}\\ &\ \ \ +\frac{\cos^2\theta\sin^2\phi}{r} \frac{\PD}{\PD r} +\frac{-\sin\theta\cos\theta\sin^2\phi}{r^2} \frac{\PD}{\PD \theta} +\frac{\cos^2\theta\sin^2\phi}{r^2} \frac{\PD^2}{\PD \theta^2} -\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi} +\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi}\\ &\ \ \ +\frac{\cos^2\phi}{r} \frac{\PD}{\PD r} +\frac{\cos\theta\cos^2\phi}{r^2\sin\theta} \frac{\PD}{\PD \theta} +\frac{-\sin\phi\cos\phi}{r^2\sin^2\theta} \frac{\PD}{\PD \phi} +\frac{\cos^2\phi}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2}\\ &\ \ \ +\cos^2\theta \frac{\PD^2}{\PD r^2} +\frac{\sin\theta\cos\theta}{r^2} \frac{\PD}{\PD \theta} -\frac{2\sin\theta\cos\theta}{r} \frac{\PD}{\PD r}\frac{\PD}{\PD \theta} \\ &\ \ \ -\frac{\sin^2\theta}{r} \frac{\PD}{\PD r}- -\frac{\sin\theta\cos\theta}{r^2} \frac{\PD}{\PD \theta} -\frac{\sin^2\theta}{r^2} \frac{\PD^2}{\PD \theta^2}\\ &= (\sin^2\theta\cos^2\phi+\sin^2\theta\sin^2\phi+\cos^2\theta)\frac{\PD^2}{\PD r^2} \\&\ \ \ +\Big(\frac{\cos^2\theta\cos^2\phi}{r}+\frac{\sin^2\phi}{r}+\frac{\cos^2\theta\sin^2\phi}{r}+\frac{\cos^2\phi}{r} +\frac{\sin^2\theta}{r}\Big) \frac{\PD}{\PD r} \\&\ \ \ +\Big(-\cancel{\frac{\sin\theta\cos\theta\cos^2\phi}{r^2}}+\cancel{\frac{-\sin\theta\cos\theta\cos^2\phi}{r^2}}+\frac{\cos\theta\sin^2\phi}{r^2\sin\theta}-\cancel{\frac{\sin\theta\cos\theta\sin^2\phi}{r^2}}\\ &\hspace{4cm}+\cancel{\frac{-\sin\theta\cos\theta\sin^2\phi}{r^2}}+\frac{\cos\theta\cos^2\phi}{r^2\sin\theta} +\cancel{\frac{\sin\theta\cos\theta}{r^2}}+\cancel{\frac{\sin\theta\cos\theta}{r^2}}\Big) \frac{\PD}{\PD \theta} \\&\ \ \ +\Big(\frac{\cos^2\theta\cos^2\phi}{r^2}+\frac{\cos^2\theta\sin^2\phi}{r^2}+\frac{\sin^2\theta}{r^2}\Big) \frac{\PD^2}{\PD \theta^2} \\&\ \ \ +\Big(\cancel{\frac{\sin\phi\cos\phi}{r^2}}+\cancel{\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta}}+\cancel{\frac{\sin\phi\cos\phi}{r^2\sin^2\theta}}-\cancel{\frac{\sin\phi\cos\phi}{r^2}}-\cancel{\frac{\cos^2\theta\sin\phi\cos\phi}{r^2\sin^2\theta}}-\cancel{\frac{\sin\phi\cos\phi}{r^2\sin^2\theta}}\Big) \frac{\PD}{\PD \phi} \\&\ \ \ +\Big(\frac{\sin^2\phi}{r^2\sin^2\theta}+\frac{\cos^2\phi}{r^2\sin^2\theta}\Big) \frac{\PD^2}{\PD \phi^2} \\&\ \ \ +\Big(\cancel{\frac{2\sin\theta\cos\theta\cos^2\phi}{r}}+\cancel{\frac{2\sin\theta\cos\theta\sin^2\phi}{r}}-\cancel{\frac{2\sin\theta\cos\theta}{r}}\Big) \frac{\PD}{\PD r}\frac{\PD}{\PD \theta} \\&\ \ \ +\Big(-\cancel{\frac{2\sin\phi\cos\phi}{r}}+\cancel{\frac{2\sin\phi\cos\phi}{r}}\Big) \frac{\PD}{\PD r}\frac{\PD}{\PD \phi} \\&\ \ \ +\Big(-\cancel{\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta}}+\cancel{\frac{2\cos\theta\sin\phi\cos\phi}{r^2\sin\theta}}\Big)\frac{\PD}{\PD \theta}\frac{\PD}{\PD \phi}\\ &= \frac{\PD^2}{\PD r^2}+\frac{2}{r} \frac{\PD}{\PD r}+\frac{\cos\theta}{r^2\sin\theta} \frac{\PD}{\PD \theta} +\frac{1}{r^2}\frac{\PD^2}{\PD \theta^2}+\frac{1}{r^2\sin^2\theta} \frac{\PD^2}{\PD \phi^2} \\ &= \frac{\PD^2}{\PD r^2}+\frac{2}{r} \frac{\PD}{\PD r} +\frac{1}{r^2}\underbrace{\bigg[\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}\bigg]}_{=\,\Lambda}

恐らくもっと簡単に求める方法もあるはず。

別解

2次元の極座標表示を利用すると少し楽らしい。
https://twitter.com/Paul_Painleve/status/995843040244711424

2次元の極座標:

\begin{cases} x&=r\cos\theta\\ y&=r\sin\theta \end{cases}     \begin{cases} r^2=x^2+y^2\\ \tan^2\theta=y^2/x^2 \end{cases}

より、

2r\frac{\partial r}{\partial x}=2x=2r\cos\theta,\ \ 2r\frac{\partial r}{\partial y}=2y=2r\sin\theta

\frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial x}=-\frac{y}{x^2}=-\frac{r\sin\theta}{r^2\cos^2\theta},\ \ \frac{1}{\cos^2\theta}\frac{\partial\theta}{\partial y}=\frac{1}{x}=\frac{1}{r\cos\theta}

などを使って、

  \frac{\partial}{\partial x} &=\left(\frac{\partial r}{\partial x}\right)\frac{\partial}{\partial r} +\left(\frac{\partial \theta}{\partial x}\right)\frac{\partial}{\partial \theta} =\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\\
  \frac{\partial}{\partial y} &=\left(\frac{\partial r}{\partial y}\right)\frac{\partial}{\partial r} +\left(\frac{\partial \theta}{\partial y}\right)\frac{\partial}{\partial \theta} =\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\\

  \frac{\partial^2}{\partial x^2}f &=\left(\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right) \left(\cos\theta\frac{\partial f}{\partial r}-\frac{\sin\theta}{r}\frac{\partial f}{\partial \theta}\right)\\ &=\cos^2\theta\frac{\partial^2 f}{\partial r^2} -\cos\theta\sin\theta\left(-\frac{1}{r^2}\frac{\partial f}{\partial \theta}+\frac{1}{r}\frac{\partial^2f}{\partial r\partial\theta}\right)\\ &\hspace{1cm}-\frac{\sin\theta}{r}\left(-\sin\theta\frac{\partial f}{\partial r}+\cos\theta\frac{\partial^2 f}{\partial\theta\partial r}\right) +\frac{\sin\theta}{r^2}\left(\cos\theta\frac{\partial f}{\partial \theta}+\sin\theta\frac{\partial^2f}{\partial\theta^2}\right)\\ &=\left[\cos^2\theta\frac{\partial^2 }{\partial r^2}+\frac{\sin^2\theta}{r}\frac{\partial}{\partial r} -2r\cos\theta\sin\theta\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial}{\partial\theta}\right) +\frac{\sin^2\theta}{r^2}\frac{\partial^2}{\partial\theta^2} \right]f
  \frac{\partial^2}{\partial y^2}f &=\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right) \left(\sin\theta\frac{\partial f}{\partial r}+\frac{\cos\theta}{r}\frac{\partial f}{\partial \theta}\right)\\ &=\sin^2\theta\frac{\partial^2 f}{\partial r^2} +\sin\theta\cos\theta\left(-\frac{1}{r^2}\frac{\partial f}{\partial \theta}+\frac{1}{r}\frac{\partial^2f}{\partial r\partial\theta}\right)\\ &\hspace{1cm}+\frac{\cos\theta}{r}\left(\cos\theta\frac{\partial f}{\partial r}-\sin\theta\frac{\partial^2 f}{\partial\theta\partial r}\right) +\frac{\cos\theta}{r^2}\left(-\sin\theta\frac{\partial f}{\partial \theta}+\cos\theta\frac{\partial^2f}{\partial\theta^2}\right)\\ &=\left[\sin^2\theta\frac{\partial^2 }{\partial r^2}+\frac{\cos^2\theta}{r}\frac{\partial}{\partial r} +2r\cos\theta\sin\theta\frac{\partial}{\partial r}\left(\frac{1}{r}\frac{\partial}{\partial\theta}\right) +\frac{\cos^2\theta}{r^2}\frac{\partial^2}{\partial\theta^2} \right]f\\

なので、

  \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} &=\frac{\partial^2 }{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}\\ &=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}

を得る。これ自体有用な式なのだけれど、球座標系の計算にどう使うかというと、 \bm r から x\mathrm-y 平面に垂線を下ろした点と原点との距離を \rho=r\sin\theta と書き、

  (x,y,z)\to(\rho,\phi,z)\to(r,\theta,\phi)

の2段階の変数変換を考える。1段目は、

  \begin{cases} x=\rho\cos\phi\\ y=\rho\sin\phi\\ z=z \end{cases}

であるから、上記を参考に、

  \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} &=\frac{\partial^2 }{\partial \rho^2}+\frac{1}{\rho}\frac{\partial}{\partial \rho}+\frac{1}{\rho^2}\frac{\partial^2}{\partial\phi^2}\\

を得る。一方、2段目は、

  \begin{cases} \rho=r\sin\theta\\ \phi=\phi\\ z=r\cos\theta \end{cases}

であるから、上記を参考に、

  \frac{\partial^2}{\partial z^2}+\frac{\partial^2}{\partial \rho^2} &=\frac{\partial^2 }{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}\\

辺々加えると、 \partial^2/\partial\rho^2 の項は打消し合って、

  \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2} &=\frac{\partial^2 }{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r} +\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2} +\frac{1}{\rho}\frac{\partial}{\partial \rho} +\frac{1}{\rho^2}\frac{\partial^2}{\partial\phi^2}

2次元の \partial/\partial y の計算結果から、

  \frac{\partial}{\partial \rho} =\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\\

なので、

  \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2} &=\frac{\partial^2 }{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r} +\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2} +\frac{1}{r\sin\theta}\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}\right) +\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\\ &=\frac{\partial^2 }{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r} +\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2} +\frac{1}{r^2}\frac{\cos\theta}{\sin\theta}\frac{\partial}{\partial \theta} +\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\\ &=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) +\frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left( \sin\theta\frac{\partial}{\partial\theta}\right) +\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}\\

として、上で得たのと同じ結果が得られる。

これはこれで大変だけれど、完全に力ずくでやるより見通しが良い。

こちらもとても参考になる

http://irobutsu.a.la9.jp/PhysTips/Lap.html

「第1の方法:変分法を使え。」において

dr, d\theta, d\phi に対する \bm r の変化量はそれぞれ、 dr, r\,d\theta, r\sin\theta\,d\phi で、互いに直交するから、 これらの方向に x',y',z' 軸を取るのと比べれば、第一感では

  \nabla^2 &\overset{??}{=}\frac{\partial^2}{\partial r^2} +\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} +\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\\

などとなりそうなところ、実際には

  \nabla^2&=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) +\frac{1}{r^2}\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial \theta}\right) +\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\\ &\hspace{5mm}\uparrow\hspace{11mm}\uparrow\hspace{21mm}\uparrow\hspace{16mm}\uparrow

のように余計な因子が紛れ込むのだが、上記のリンク先ではラプラシアンが

  \nabla^2 &=\frac{1}{r^2\sin\theta}\frac{\partial}{\partial r}\left(r^2\sin\theta\frac{\partial}{\partial r}\right) +\frac{1}{r^2\sin\theta}\frac{1}{r}\frac{\partial}{\partial \theta}\left(r^2\sin\theta\frac{1}{r}\frac{\partial}{\partial \theta}\right) +\frac{1}{r^2\sin\theta}\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\left(r^2\sin\theta\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\right)\\

の形に導かれる。

「1回目の微分」をした後に体積素 (r^2\sin\theta)dr\,d\theta\,d\phi の係数 r^2\sin\theta を掛け、「2回目の微分」をした後に同じ値で割る形になっている。

これは覚えやすい。

「第2の方法:ちゃんと基底ベクトルも微分しろ。」において

  \bm \nabla=\bm e_r\frac{\partial}{\partial r}+\bm e_\theta\frac{1}{r}\frac{\partial}{\partial \theta}+\bm e_\phi\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}

から

  \nabla^2&=\bm\nabla\cdot\bm\nabla\\ &=\bigg(\bm e_r\frac{\partial}{\partial r}+\bm e_\theta\frac{1}{r}\frac{\partial}{\partial \theta}+\bm e_\phi\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\bigg)\cdot \bigg(\bm e_r\frac{\partial}{\partial r}+\bm e_\theta\frac{1}{r}\frac{\partial}{\partial \theta}+\bm e_\phi\frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}\bigg)

のように計算する際、 \bm e_r,\bm e_\theta,\bm e_\phi は直交するため一見すると

  \nabla^2 &\overset{??}{=}\frac{\partial^2}{\partial r^2} +\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} +\frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\\

となりそうなところ、上記リンク先では

  \begin{pmatrix} \frac{\partial}{\partial r} \\ \frac{\partial}{\partial \theta} \\ \frac{\partial}{\partial \phi} \end{pmatrix} \begin{pmatrix} \bm e_r & \bm e_\theta & \bm e_\phi \end{pmatrix}=\begin{pmatrix} \bm 0 & \bm 0 & \bm 0 \\ \bm e_\theta & -\bm e_r & \bm 0 \\ (\sin\theta)\bm e_\phi & (\cos\theta)\bm e_\phi & -\bm e_r \end{pmatrix}

のように、 \bm e_r,\bm e_\theta,\bm e_\phi 自体が \theta,\phi の関数であることを考慮しなければならないことを指摘し、

  \bm e_\theta\frac{\partial\bm e_r}{\partial \theta} の項から \frac{1}{r}\frac{\partial}{\partial r}

  \bm e_\phi\frac{\partial\bm e_r}{\partial \theta} の項から \frac{1}{r}\frac{\partial}{\partial r} がもう1つ

  \bm e_\phi\frac{\partial\bm e_\theta}{\partial \theta} の項から \frac{\cos\theta}{r^2\sin\theta}\frac{\partial}{\partial\theta}

がそれぞれ出ることにより、正しいラプラシアンが得られることを示している。

球座標の角運動量演算子

\hat l_x&=-i\hbar\Big(y\frac{\PD}{\PD z}-z\frac{\PD}{\PD y}\Big)\\ &=-i\hbar\bigg[r\sin\theta\sin\phi\Big(\cancel{\cos\theta \frac{\PD}{\PD r}}-\frac{1}{r}\sin\theta \frac{\PD}{\PD \theta}\Big) -r\cos\theta\Big(\cancel{\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}\Big)\bigg]\\ &=i\hbar\Big(\sin\phi\frac{\PD}{\PD\theta}+\frac{\cos\phi}{\tan\theta}\frac{\PD}{\PD\phi}\Big)

\hat l_y&=-i\hbar\Big(z\frac{\PD}{\PD x}-x\frac{\PD}{\PD z}\Big)\\ &=-i\hbar\bigg[r\cos\theta\Big(\cancel{\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}\Big) -r\sin\theta\cos\phi\Big(\cancel{\cos\theta \frac{\PD}{\PD r}} -\frac{1}{r}\sin\theta \frac{\PD}{\PD \theta}\Big)\bigg]\\ &=i\hbar\Big(-\cos\phi\frac{\PD}{\PD\theta}+\frac{\sin\phi}{\tan\theta}\frac{\PD}{\PD\phi}\Big)

\hat l_z&=-i\hbar\Big(x\frac{\PD}{\PD y}-y\frac{\PD}{\PD x}\Big)\\ &=-i\hbar\bigg[r\sin\theta\cos\phi\Big(\cancel{\sin\theta\sin\phi \frac{\PD}{\PD r}} +\cancel{\frac{1}{r}\cos\theta\sin\phi \frac{\PD}{\PD \theta}} +\frac{\cos\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\Big)\\ &\hspace{1cm}-r\sin\theta\sin\phi\Big(\cancel{\sin\theta\cos\phi \frac{\PD}{\PD r}} +\cancel{\frac{1}{r}\cos\theta\cos\phi \frac{\PD}{\PD \theta}} -\frac{\sin\phi}{r\sin\theta} \frac{\PD}{\PD \phi}\Big)\bigg]\\ &=-i\hbar\frac{\PD}{\PD\phi}

\hat{\bm l}^2&=\hat l_x^2+\hat l_y^2+\hat l_z^2\\ &= -\hbar^2\Big(\sin\phi\frac{\PD}{\PD\theta}+\frac{\cos\phi}{\tan\theta}\frac{\PD}{\PD\phi}\Big)^2 -\hbar^2\Big(-\cos\phi\frac{\PD}{\PD\theta}+\frac{\sin\phi}{\tan\theta}\frac{\PD}{\PD\phi}\Big)^2 -\hbar^2\frac{\PD^2}{\PD\phi^2}\\ &= -\hbar^2\Big( \sin^2\phi\frac{\PD^2}{\PD\theta^2} -\cancel{\frac{\sin\phi\cos\phi}{\sin^2\theta}\frac{\PD}{\PD\phi}} +\cancel{\frac{2\sin\phi\cos\phi}{\tan\theta}\frac{\PD}{\PD\theta}\frac{\PD}{\PD\phi}} +\frac{\cos^2\phi}{\tan\theta}\frac{\PD}{\PD\theta} +\cancel{\frac{-\sin\phi\cos\phi}{\tan^2\theta}\frac{\PD}{\PD\phi}} +\frac{\cos^2\phi}{\tan^2\theta}\frac{\PD^2}{\PD\phi^2} \Big)\\ &\ \ \ -\hbar^2\Big( \cos^2\phi\frac{\PD^2}{\PD^2\theta} +\cancel{\frac{\sin\phi\cos\phi}{\sin^2\theta}\frac{\PD}{\PD\phi}} -\cancel{\frac{2\sin\phi\cos\phi}{\tan\theta}\frac{\PD}{\PD\theta}\frac{\PD}{\PD\phi}} +\frac{\sin^2\phi}{\tan\theta}\frac{\PD}{\PD\theta} +\cancel{\frac{\sin\phi\cos\phi}{\tan^2\theta}\frac{\PD}{\PD\phi}} +\frac{\sin^2\phi}{\tan^2\theta}\frac{\PD^2}{\PD\phi^2} \Big)\\ &\ \ \ -\hbar^2\frac{\PD^2}{\PD\phi^2}\\ &=-\hbar^2\Big[\frac{\PD^2}{\PD\theta^2}+\frac{1}{\tan\theta}\frac{\PD}{\PD\theta}+\Big(\frac{1}{\tan^2\theta}+1\Big)\frac{\PD^2}{\PD\phi^2}\Big]\\ &=-\hbar^2\Big(\frac{\PD^2}{\PD\theta^2}+\frac{\cos\theta}{\sin\theta}\frac{\PD}{\PD\theta}+\frac{1}{\sin^2\theta}\frac{\PD^2}{\PD\phi^2}\Big)\\ &=-\hbar^2\Big[\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}\Big]\\ &=-\hbar^2\hat\Lambda


Counter: 4677 (from 2010/06/03), today: 10, yesterday: 0