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シュレーディンガー方程式を解く6〜ラプラシアン4〜

ついつい寝ダメしちゃった日曜日,みなさまいかがお過ごしでしょうか.
(・∀・)

うち,テレビないんで,YouTubeで日本のドキュメンタリィ見るとか,あまり有意義でなく過ごせて非常に満足です(;ω;)


さて.シュレーディンガー方程式の演算子部分を球座標に変換するために四苦八苦しましただな.
これ,39歳で初めて量子論習ってから4年来,ずっとやりたかったんですだ.

もうちょっとなんでがんばりますんだ.
(^▽^)

さて.
演算子の中身はそれぞれ,


{\begin{align}

\displaystyle\frac{\partial^2}{\partial x^2}


&=\sin^2{\theta}\cos^2{\phi}\frac{\partial^2}{\partial r^2}-\frac{2}{r^2}\sin{\theta}\cos{\theta}\cos^2{\phi}\frac{\partial}{\partial\theta}+\frac{2}{r}\sin{\theta}\cos{\theta}\cos^2{\phi}\frac{\partial^2}{\partial r\partial\theta}\\
&\ \ \ \ +\frac{1}{r^2}\sin{\phi}\cos{\phi}\frac{\partial}{\partial\phi}-\frac{2}{r}\sin{\phi}\cos{\phi}\frac{\partial^2}{\partial r\partial\theta}+\frac{1}{r}\cos^2{\theta}\cos^2{\phi}\frac{\partial}{\partial r}\\

&\ \ \ \ +\frac{1}{r^2}\cos^2{\theta}\cos^2{\phi}\frac{\partial^2}{\partial\theta^2}+\frac{1}{r^2}\frac{\cos^2{\theta}\sin{\phi}\cos{\phi}}{\sin^2{\theta}}\frac{\partial}{\partial\phi}\\

&\ \ \ \ -\frac{2}{r^2}\frac{\cos{\theta}\sin{\phi}\cos{\phi}}{\sin{\theta}}\frac{\partial^2}{\partial\theta\partial\phi}+\frac{\sin^2{\phi}}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\sin^2{\phi}\cos{\theta}}{\sin{\theta}}\frac{\partial}{\partial\theta}\\

&\ \ \ \ +\frac{1}{r^2}\frac{\sin{\phi}\cos{\phi}}{\sin^2{\theta}}\frac{\partial}{\partial\phi}+\frac{1}{r^2}\frac{\sin^2{\phi}}{\sin^2{\theta}}\frac{\partial^2}{\partial\phi^2}
\end{align}
}



{\begin{align}

\displaystyle\frac{\partial^2}{\partial y^2}

&=\sin^2{\theta}\sin^2{\phi}\frac{\partial^2}{\partial r^2}-\frac{2}{r^2}\sin{\theta}\cos{\theta}\sin^2{\phi}\frac{\partial}{\partial\theta}+\frac{2}{r}\sin{\theta}\cos{\theta}\sin^2{\phi}\frac{\partial^2}{\partial r\partial\theta}\\

&\ \ \ \ -\frac{1}{r^2}\sin{\phi}\cos{\phi}\frac{\partial}{\partial\phi}+\frac{2}{r}\sin{\phi}\cos{\phi}\frac{\partial^2}{\partial r\partial\phi}+\frac{1}{r}\cos^2{\theta}\sin^2{\phi}\frac{\partial}{\partial r}\\

&\ \ \ \ +\frac{1}{r^2}\cos^2{\theta}\sin^2{\phi}\frac{\partial^2}{\partial\theta^2}-\frac{1}{r^2}\frac{\cos^2{\theta}\sin{\phi}\cos{\phi}}{\sin^2{\theta}}\frac{\partial}{\partial\phi}\\

&\ \ \ \ +\frac{2}{r^2}\frac{\cos{\theta}\sin{\phi}\cos{\phi}}{\sin{\theta}}\frac{\partial^2}{\partial\theta\partial\phi}+\frac{1}{r}\cos^2{\phi}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\cos{\theta}\cos^2{\phi}}{\sin{\theta}}\frac{\partial}{\partial\theta}\\

&\ \ \ \ -\frac{1}{r^2}\frac{\sin{\phi}\cos{\phi}}{\sin^2{\theta}}\frac{\partial}{\partial\phi}+\frac{1}{r^2}\frac{\cos^2{\phi}}{\sin^2{\theta}}\frac{\partial^2}{\partial\phi^2}\\

\end{align}\\
}



{
\begin{align}
\displaystyle\frac{\partial^2}{\partial z^2}

&=\displaystyle\cos^2{\theta}\frac{\partial^2}{\partial r^2}+\frac{2}{r^2}\sin{\theta}\cos{\theta}\frac{\partial}{\partial\theta}-\frac{2}{r}\sin{\theta}\cos{\theta}\frac{\partial^2}{\partial r\partial\theta}\\

&\ \ \ \ +\frac{1}{r}\sin^2{\theta}\frac{\partial}{\partial r}+\frac{1}{r^2}\sin^2{\theta}\frac{\partial^2}{\partial\theta^2}
\end{align}
}



こうなりましただ.
これらは,


{\Delta=\nabla\cdot\nabla=\displaystyle\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}}


と,あたかもベクトルの内積のように計算されますじゃ.
まあ実際に同じ性質なんでしょうよ.

そして項の一番右端の演算子にあわせてまとめていきますんじゃ.


例えば,各第1項は,


{\displaystyle
\left(\sin^2{\theta}\cos^2{\phi}+\sin^2{\theta}\sin^2{\phi}+\cos^2{\theta}\right)\frac{\partial^2}{\partial r^2}
}


これは,{\sin^2{\alpha}+\cos^2{\alpha}=1}が存分に使えまして,三角関数が残らず1になって


{
\displaystyle\left(\sin^2{\theta}\cos^2{\phi}+\sin^2{\theta}\sin^2{\phi}+\cos^2{\theta}\right)\frac{\partial^2}{\partial r^2}=\frac{\partial^2}{\partial r^2}
}


となりますだ.

...この調子で残らずまとめていくと,かなり気持ちよく各項消えていき,残ったのは以下ですじゃ.


{\begin{align}
\Delta=
\displaystyle\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}\frac{1}{\sin^2{\theta}}\frac{\partial^2}{\partial\phi^2}
\end{align}
}


(;ω;)感涙….

自力でがんばれたのはここまでで,このあとにもう一段階,見やすくした式の形があるようですだな.


{\begin{align}
\Delta=
\displaystyle\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}
\end{align}
}


これにより,シュレーディンガー方程式は,ポテンシャルを{V}とすると,


{\left[-\displaystyle\frac{\hbar^2}{2m}\left(\displaystyle\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)+V(\textbf{x})\right]\Psi(\mathbf{x})=E\Psi(\mathbf{x})}


だったのが,


{\begin{align}
\left[-\displaystyle\frac{\hbar^2}{2m}\left\{\displaystyle\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}\right\}+V(\mathbf{r})\right]\Psi(\mathbf{r})=E\Psi(\mathbf{r})
\end{align}
}


と,このように,球座標での表現に変換できたわけですじゃ!

(;ω;)大感涙.


〜〜〜〜〜〜〜〜〜〜〜〜〜〜〜〜〜〜
お礼コーナー.
今回までのパートは,

・筑波大武内修先生のWebSite
・EMANさま
obeliskさまのWebSite「三次元極座標におけるラプラシアン」のコーナー

を勉強し,たどり着くことができました.

世間の教科書の大半では「数学等の教科書にゆずる」との記載で結果のみが示されており,そのまま過ごしてしまって,こんにちに至りましただ.

このように,少なくとも自分はWebで公開されている教材や資料にたのむところは大であります.


(ー人ー)感謝.

シュレーディンガー方程式を解く4〜ラプラシアン2〜

前口上です(・∀・)

ラプラシアン極座標変換ということであれば,それを示している媒体は天下にゴマンとございますじゃ.

このブログはお勉強ノートでして,ゆえに,わちきらは作業過程を示さねばならん.
お勉強ノートだから...

では作業報告書状態だろうと,れっつらGo!
(・◇・)ゞ


{
\begin{align}
\left(\displaystyle\frac{\partial}{\partial x}\cdot\displaystyle\frac{\partial}{\partial x}\right)&=\left(\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}\right)\left(\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}\right)\\
&=\sin{\theta}\cos{\phi}\frac{\partial}{\partial r}\sin{\theta}\cos{\phi}\frac{\partial}{\partial r}+\sin{\theta}\cos{\phi}\frac{\partial}{\partial r}\frac{1}{r}\cos{\theta}\cos{\phi}\frac{\partial}{\partial\theta}+\sin{\theta}\cos{\phi}\frac{\partial}{\partial r}\left(-\frac{\sin{\phi}}{r\sin{\theta}}\frac{\partial}{\partial\phi}\right)\\
&\ \ \ \ +\frac{1}{r}\cos{\theta}\cos{\phi}\frac{\partial}{\partial\theta}\sin{\theta}\cos{\phi}\frac{\partial}{\partial r}+\frac{1}{r}\cos{\theta}\cos{\phi}\frac{\partial}{\partial\theta}\frac{1}{r}\cos{\theta}\cos{\phi}\frac{\partial}{\partial\theta}+\frac{1}{r}\cos{\theta}\cos{\phi}\frac{\partial}{\partial\theta}\left(-\frac{\sin{\phi}}{r\sin{\theta}}\frac{\partial}{\partial\phi}\right)\\
&\ \ \ \ +\left(-\frac{\sin{\phi}}{r\sin{\theta}}\frac{\partial}{\partial\phi}\right)\sin{\theta}\cos{\phi}\frac{\partial}{\partial r}+\left(-\frac{\sin{\phi}}{r\sin{\theta}}\frac{\partial}{\partial\phi}\right)\frac{1}{r}\cos{\theta}\cos{\phi}\frac{\partial}{\partial\theta}+\left(-\frac{\sin{\phi}}{r\sin{\theta}}\frac{\partial}{\partial\phi}\right)\left(-\frac{\sin{\phi}}{r\sin{\theta}}\frac{\partial}{\partial\phi}\right)\\
\\
&=\sin^2{\theta}\cos^2{\phi}\frac{\partial^2}{\partial r^2}+\sin{\theta}\cos{\theta}\cos^2{\phi}\left(\frac{\partial}{\partial r}\frac{1}{r}\frac{\partial}{\partial\theta}+\frac{1}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial\theta}\right)-\sin{\phi}\cos{\phi}\left(\frac{\partial}{\partial r}\frac{1}{r}\frac{\partial}{\partial\phi}+\frac{1}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial\phi}\right)\\&\ \ \ \ +\frac{1}{r}\cos{\theta}\cos^2{\phi}\left(\frac{\partial}{\partial\theta}\sin{\theta}\frac{\partial}{\partial r}+\sin{\theta}\frac{\partial}{\partial\theta}\frac{\partial}{\partial r}\right)+\frac{1}{r^2}\cos{\theta}\cos^2{\phi}\left(\frac{\partial}{\partial\theta}\cos{\theta}\frac{\partial}{\partial\theta}+\cos{\theta}\frac{\partial}{\partial\theta}\frac{\partial}{\partial\theta}\right)-\frac{1}{r^2}\cos{\theta}\sin{\phi}\cos{\phi}\left(\frac{\partial}{\partial\theta}\frac{1}{\sin{\theta}}\frac{\partial}{\partial\phi}+\frac{1}{\sin{\theta}}\frac{\partial}{\partial\theta}\frac{\partial}{\partial\phi}\right)\\
&\ \ \ \ -\frac{1}{r}\sin{\phi}\left(\frac{\partial}{\partial\phi}\cos{\phi}\frac{\partial}{\partial r}+\cos{\phi}\frac{\partial}{\partial\phi}\frac{\partial}{\partial r}\right)-\frac{\sin{\phi}\cos{\theta}}{r^2\sin{\theta}}\left(\frac{\partial}{\partial\phi}\cos{\phi}\frac{\partial}{\partial\theta}+\cos{\phi}\frac{\partial}{\partial\phi}\frac{\partial}{\partial\theta}\right)+\frac{\sin{\phi}}{r^2\sin^2{\theta}}\left(\frac{\partial}{\partial\phi}\sin{\phi}\frac{\partial}{\partial\phi}+\sin{\phi}\frac{\partial}{\partial\phi}\frac{\partial}{\partial\phi}\right)\\
\\
&=\sin^2{\theta}\cos^2{\phi}\frac{\partial^2}{\partial r^2}-\frac{1}{r^2}\sin{\theta}\cos{\theta}\cos^2{\phi}\frac{\partial}{\partial\theta}+\frac{1}{r}\sin{\theta}\cos{\theta}\cos^2{\phi}\frac{\partial^2}{\partial r\partial\theta}+\frac{1}{r^2}\sin{\phi}\cos{\phi}\frac{\partial}{\partial\phi}-\frac{1}{r}\sin{\phi}\cos{\phi}\frac{\partial^2}{\partial r\partial\theta}\\
&\ \ \ \ +\frac{1}{r}\cos^2{\theta}\cos^2{\phi}\frac{\partial}{\partial r}+\frac{1}{r}\sin{\theta}\cos{\theta}\cos^2{\phi}\frac{\partial^2}{\partial r\partial\theta}-\frac{1}{r^2}\sin{\theta}\cos{\theta}\cos^2{\phi}\frac{\partial}{\partial\theta}+\frac{1}{r^2}\cos^2{\theta}\cos^2{\phi}\frac{\partial^2}{\partial\theta^2}+\frac{1}{r^2}\frac{\cos^2{\theta}\sin{\phi}\cos{\phi}}{\sin^2{\theta}}\frac{\partial}{\partial\phi}\\
&\ \ \ \ -\frac{1}{r^2}\frac{\cos{\theta}\sin{\phi}\cos^2{\phi}}{\sin{\theta}}\frac{\partial^2}{\partial\theta\partial\phi}+\frac{\sin^2{\phi}}{r}\frac{\partial}{\partial r}-\frac{1}{r}\sin{\phi}\cos{\phi}\frac{\partial^2}{\partial r\partial\phi}+\frac{1}{r^2}\frac{\sin^2{\phi}\cos{\theta}}{\sin{\theta}}\frac{\partial}{\partial\theta}-\frac{1}{r^2}\frac{\sin{\phi}\cos{\phi}\cos{\theta}}{\sin{\theta}}\frac{\partial^2}{\partial\theta\partial\phi}\\
&\ \ \ \ +\frac{1}{r^2}\frac{\sin{\phi}\cos{\phi}}{\sin^2{\theta}}\frac{\partial}{\partial\phi}+\frac{1}{r^2}\frac{\sin^2{\phi}}{\sin^2{\theta}}\frac{\partial^2}{\partial\phi^2}\\
\\
&=\sin^2{\theta}\cos^2{\phi}\frac{\partial^2}{\partial r^2}-\frac{2}{r^2}\sin{\theta}\cos{\theta}\cos^2{\phi}\frac{\partial}{\partial\theta}+\frac{2}{r}\sin{\theta}\cos{\theta}\cos^2{\phi}\frac{\partial^2}{\partial r\partial\theta}+\frac{1}{r^2}\sin{\phi}\cos{\phi}\frac{\partial}{\partial\phi}\\
&\ \ \ \ -\frac{2}{r}\sin{\phi}\cos{\phi}\frac{\partial^2}{\partial r\partial\theta}+\frac{1}{r}\cos^2{\theta}\cos^2{\phi}\frac{\partial}{\partial r}+\frac{1}{r^2}\cos^2{\theta}\cos^2{\phi}\frac{\partial^2}{\partial\theta^2}+\frac{1}{r^2}\frac{\cos^2{\theta}\sin{\phi}\cos{\phi}}{\sin^2{\theta}}\frac{\partial}{\partial\phi}-\frac{2}{r^2}\frac{\cos{\theta}\sin{\phi}\cos{\phi}}{\sin{\theta}}\frac{\partial^2}{\partial\theta\partial\phi}\\
&\ \ \ \ +\frac{\sin^2{\phi}}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\sin^2{\phi}\cos{\theta}}{\sin{\theta}}\frac{\partial}{\partial\theta}+\frac{1}{r^2}\frac{\sin{\phi}\cos{\phi}}{\sin^2{\theta}}\frac{\partial}{\partial\phi}+\frac{1}{r^2}\frac{\sin^2{\phi}}{\sin^2{\theta}}\frac{\partial^2}{\partial\phi^2}
\end{align}
}


(;ω;)
ほロほろ…暑さもあってちょっと気分悪くなりましただ.

きゅ,きゅうけい…

シュレーディンガー方程式を解く3〜ラプラシアン〜

wave-geometry.hatenablog.com


前回がんばってナブラの中身まで出しましただ.


{\nabla = \left(
\displaystyle\frac{\partial}{\partial x}\ 
\displaystyle\frac{\partial}{\partial y}\ 
\displaystyle\frac{\partial}{\partial z} 
\right)
}

{\left\{
\begin{array}
.\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}
\end{array}
\right.
}


よかよか(^▽^)

で,こんどはラプラシアン,すなわち


{\Delta=\nabla\cdot\nabla}


を求めるのですな.

交換関係の計算で勉強したように.演算子どうしの計算には,後ろに関数が有ると考えて計算するのが正しいのですな.

どういうことかというと,


{\displaystyle\frac{\partial}{\partial r}\sin{\theta}\frac{\partial}{\partial\theta}}


のようなときには,


{\displaystyle\frac{\partial}{\partial r}\sin{\theta}\frac{\partial\psi}{\partial\theta}}


この最後の肩の{\psi}みたいな,関数に作用した状態=ひとつの関数と見るんですな.

んで,


f:id:morio_roji1111:20180714114929p:plain


この赤と青の関数の積にかかった微分と見るんですわ.

関数の積の微分ですから,{(fg)'=f'g+fg'}を使って,


{\displaystyle\frac{\partial}{\partial r}\sin{\theta}\frac{\partial\psi}{\partial\theta}=\displaystyle\frac{\partial}{\partial r}\sin{\theta}\frac{\partial\psi}{\partial\theta}+\sin{\theta}\frac{\partial}{\partial r}\frac{\partial\psi}{\partial\theta}}


となるわけですじゃ.これ,全部ゼロで消えそうだな….

さて(・∀・).

まずは一番簡単な{\displaystyle\frac{\partial}{\partial z}}からいくのが常套手段のようですじゃ.
いきますだ.


{
\begin{align}
\left(\displaystyle\frac{\partial}{\partial z}\cdot\displaystyle\frac{\partial}{\partial z}\right)&=\left(\cos{\theta}\frac{\partial}{\partial r}-\frac{1}{r}\sin{\theta}\frac{\partial}{\partial\theta}\right)\left(\cos{\theta}\frac{\partial}{\partial r}-\frac{1}{r}\sin{\theta}\frac{\partial}{\partial\theta}\right)\\
&=\cos{\theta}\frac{\partial}{\partial r}\cos{\theta}\frac{\partial}{\partial r}-\cos{\theta}\frac{\partial}{\partial r}\frac{1}{r}\sin{\theta}\frac{\partial}{\partial\theta}\\
&\ \ \ \ \ -\frac{1}{r}\sin{\theta}\frac{\partial}{\partial\theta}\cos{\theta}\frac{\partial}{\partial r}-\frac{1}{r}\sin{\theta}\frac{\partial}{\partial\theta}\left(-\frac{1}{r}\sin{\theta}\frac{\partial}{\partial\theta}\right)
\end{align}
}


たとえば最終行第1項の{\displaystyle\cos{\theta}\frac{\partial}{\partial r}\cos{\theta}\frac{\partial}{\partial r}}

{\cos{\theta}}はただの数なので,演算子の前に集めちゃうんですな.

{\displaystyle\cos^2{\theta}\frac{\partial^2}{\partial r^2}}


こうなるわけか...(・〜・)この調子でいきますじゃ.


{
\begin{align}
\left(\displaystyle\frac{\partial}{\partial z}\cdot\displaystyle\frac{\partial}{\partial z}\right)&=
\displaystyle\cos^2{\theta}\frac{\partial^2}{\partial r^2}-\cos{\theta}\sin{\theta}\left(\frac{\partial}{\partial r}\frac{1}{r}\frac{\partial}{\partial\theta}+\frac{1}{r}\frac{\partial}{\partial r}\frac{\partial}{\partial\theta}\right)\\
&\ \ \ \ -\frac{1}{r}\sin{\theta}\left(\frac{\partial}{\partial\theta}\cos{\theta}\frac{\partial}{\partial r}+\cos{\theta}\frac{\partial}{\partial\theta}\frac{\partial}{\partial r}\right)\\
&\ \ \ \ +\frac{1}{r^2}\sin{\theta}\left(\frac{\partial}{\partial\theta}\sin{\theta}\frac{\partial}{\partial\theta}+\sin{\theta}\frac{\partial}{\partial\theta}\frac{\partial}{\partial\theta}\right)
\end{align}
}


最後の項なんかも,


f:id:morio_roji1111:20180714123930p:plain


緑の{-\displaystyle\frac{1}{r}}は作用の対象じゃないんで前にでて,赤と青が積の微分の対象になったんですなァ.


結局,


f:id:morio_roji1111:20180714125756p:plain


赤青の項はめいめいまとまり,


{
\begin{align}
\left(\displaystyle\frac{\partial}{\partial z}\cdot\displaystyle\frac{\partial}{\partial z}\right)&=\displaystyle\cos^2{\theta}\frac{\partial^2}{\partial r^2}+\frac{2}{r^2}\sin{\theta}\cos{\theta}\frac{\partial}{\partial\theta}-\frac{2}{r}\sin{\theta}\cos{\theta}\frac{\partial^2}{\partial r\partial\theta}\\
&\ \ \ \ +\frac{1}{r}\sin^2{\theta}\frac{\partial}{\partial r}+\frac{1}{r^2}\sin^2{\theta}\frac{\partial^2}{\partial\theta^2}
\end{align}
}


(^▽^;)ふぅ...

一番項の少ない{\displaystyle\frac{\partial}{\partial z}}でも,結構大変(汗


ほかのやつもやらねばな...(;▽;)

ヤコビアン〜Change of Variables〜

前回極座標変換するのを頑張りましただ.

wave-geometry.hatenablog.com


ちょっと課題で地獄見たついでにメモ的に.
(;ω;)

方向微分っていうんですか?
これのことだったのか.


{\mathbf{J}=\left(
\begin{array}
.\displaystyle\frac{\partial x}{\partial r} && \displaystyle\frac{\partial x}{\partial\theta} && \displaystyle\frac{\partial x}{\partial\phi} \\
\displaystyle\frac{\partial y}{\partial r} && \displaystyle\frac{\partial y}{\partial\theta} && \displaystyle\frac{\partial y}{\partial\phi} \\
\displaystyle\frac{\partial z}{\partial r} && \displaystyle\frac{\partial z}{\partial\theta} && \displaystyle\frac{\partial z}{\partial\phi}
\end{array}
\right)
}


この{\mathbf{J}}こそがヤコビアンですだな.

これで変数変換も怖くない怖くない.


...う〜ん,前回だしたラプラシアンの変換と,分子分母が逆さまなんだな.
きっと役割がちがうのだろう.
ちなみにカルテジアンが分母,球座標が分子でやてみると,見事に{\displaystyle\frac{1}{r^2\sin{\theta}}}とサカサマに.


では,


{\left\{
\begin{array}
.x = r\sin{\theta}\cos{\phi}\\
y = r\sin{\theta}\sin{\phi}\\
z = r\cos{\theta}
\end{array}
\right.
}


これを使ってそのまま微分して中身作りますだな.

たとえば,


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


こんなふうに.

では,


{\mathbf{|J|}=\left|
\begin{array}
.\sin{\theta}\cos{\phi} && r\cos{\theta}\cos{\phi} && -r\sin{\theta}\sin{\phi} \\
\sin{\theta}\sin{\phi}  && r\cos{\theta}\sin{\phi} && r\sin{\theta}\cos{\phi} \\
\cos{\theta}  && -r\sin{\theta} && 0
\end{array}
\right|
\\
\\
=r^2\sin{\theta}
}


と,なんときっちり微小体積を球座標で積分する場合の差異がでてくるんですな.


{dxdydz \to |\mathbf{J}|dr d\theta d\phi}

{= r^2\sin{\theta}dr d\theta d\phi}


ほほう.
(・ω・)

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!