第5章 中心力場のSchrödinger方程式②

5.2 極座標による3次元のSchrödinger方程式(つづき)

(5.14)が正しいSchrödinger方程式と同等であることを確かめる．
古典論では$\mathbf{L}=\mathbf{r}\times\mathbf{p}$なので，量子論でも

\begin{align}
\hat{L}_x
&=
\hat{y}\hat{p}_z-\hat{z}\hat{p}_y
=
\frac{\hbar}{i}
\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)
\\
\hat{L}_y
&=
\hat{z}\hat{p}_x-\hat{x}\hat{p}_z
=
\frac{\hbar}{i}
\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)
\\
\hat{L}_z
&=
\hat{x}\hat{p}_y-\hat{y}\hat{p}_x
=
\frac{\hbar}{i}
\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)
\end{align}

\tag{5.15}

とする．(5.15)は(5.8)を使って極座標に変換できる．
(5.8)より，

\frac{\partial x}{\partial r}=\sin\theta\cos\varphi, \quad
\frac{\partial x}{\partial \theta}=r\cos\theta\cos\varphi, \quad
\frac{\partial x}{\partial \varphi}=-r\sin\theta\sin\varphi
\tag{5.15.1}

\frac{\partial y}{\partial r}=\sin\theta\sin\varphi, \quad
\frac{\partial y}{\partial \theta}=r\cos\theta\sin\varphi, \quad
\frac{\partial y}{\partial \varphi}=r\sin\theta\cos\varphi
\tag{5.15.2}

\frac{\partial z}{\partial r}=\cos\theta, \quad
\frac{\partial z}{\partial \theta}=-r\sin\theta, \quad
\frac{\partial z}{\partial \varphi}=0
\tag{5.15.3}

一階の偏微分演算子は，

\frac{\partial}{\partial r}
=\frac{\partial x}{\partial r}\frac{\partial}{\partial x}
+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}
+\frac{\partial z}{\partial r}\frac{\partial}{\partial z}
=\sin\theta\cos\varphi\frac{\partial}{\partial x}
+\sin\theta\sin\varphi\frac{\partial}{\partial y}
+\cos\theta\frac{\partial}{\partial z}

\tag{5.15.4}

\frac{\partial}{\partial \theta}
=\frac{\partial x}{\partial \theta}\frac{\partial}{\partial x}
+\frac{\partial y}{\partial \theta}\frac{\partial}{\partial y}
+\frac{\partial z}{\partial \theta}\frac{\partial}{\partial z}
=r\cos\theta\cos\varphi\frac{\partial}{\partial x}
+r\cos\theta\sin\varphi\frac{\partial}{\partial y}
-r\sin\theta\frac{\partial}{\partial z}

\tag{5.15.5}

\frac{\partial}{\partial \varphi}
=\frac{\partial x}{\partial \varphi}\frac{\partial}{\partial x}
+\frac{\partial y}{\partial \varphi}\frac{\partial}{\partial y}
+\frac{\partial z}{\partial \varphi}\frac{\partial}{\partial z}
=-r\sin\theta\sin\varphi\frac{\partial}{\partial x}
+r\sin\theta\cos\varphi\frac{\partial}{\partial y}

\tag{5.15.6}

(5.15.4)×$r\sin\theta$より

r\sin\theta\frac{\partial}{\partial r}
=r\sin^2\theta\cos\varphi\frac{\partial}{\partial x}
+r\sin^2\theta\sin\varphi\frac{\partial}{\partial y}
+r\sin\theta\cos\theta\frac{\partial}{\partial z}

(5.15.5)×$\cos\theta$より

\cos\theta\frac{\partial}{\partial\theta}
=r\cos^2\theta\cos\varphi\frac{\partial}{\partial x}
+r\cos^2\theta\sin\varphi\frac{\partial}{\partial y}
-r\sin\theta\cos\theta\frac{\partial}{\partial z}

両式を加えると，

r\sin\theta\frac{\partial}{\partial r}+\cos\theta\frac{\partial}{\partial\theta}
=r\cos\varphi\frac{\partial}{\partial x}+r\sin\varphi\frac{\partial}{\partial y}
\tag{5.15.7}

(5.15.6)×$\sin\varphi$より

\sin\varphi\frac{\partial}{\partial\varphi}
=-r\sin\theta\sin^2\varphi\frac{\partial}{\partial x}
+r\sin\theta\sin\varphi\cos\varphi\frac{\partial}{\partial y}

(5.15.7)×$\sin\theta\cos\varphi$より

r\sin^2\theta\cos\varphi\frac{\partial}{\partial r}
+\sin\theta\cos\theta\cos\varphi\frac{\partial}{\partial \theta}
=r\sin\theta\cos^2\varphi\frac{\partial}{\partial x}
+r\sin\theta\sin\varphi\cos\varphi\frac{\partial}{\partial y}

両式を加えると，

r\sin^2\theta\cos\varphi\frac{\partial}{\partial r}
+\sin\theta\cos\theta\cos\varphi\frac{\partial}{\partial \theta}
-\sin\varphi\frac{\partial}{\partial\varphi}
=r\sin\theta\frac{\partial}{\partial x}

すなわち

\frac{\partial}{\partial x}=
\sin\theta\cos\varphi\frac{\partial}{\partial r}
+\frac{\cos\theta\cos\varphi}{r}\frac{\partial}{\partial\theta}
-\frac{\sin\varphi}{r\sin\theta}\frac{\partial}{\partial\varphi}
\tag{5.15.8}

また(5.15.6)×$\cos\varphi$より

\cos\varphi\frac{\partial}{\partial \varphi}
=-r\sin\theta\sin\varphi\cos\varphi\frac{\partial}{\partial x}
+r\sin\theta\cos^2\varphi\frac{\partial}{\partial y}

(5.15.7)×$\sin\theta\sin\varphi$より

r\sin^2\sin\varphi\frac{\partial}{\partial r}
+\sin\theta\cos\theta\sin\varphi\frac{\partial}{\partial \theta}
=r\sin\theta\sin\varphi\cos\varphi\frac{\partial}{\partial x}
+r\sin\theta\sin^2\varphi\frac{\partial}{\partial y}

両式を加えると，

r\sin^2\sin\varphi\frac{\partial}{\partial r}
+\sin\theta\cos\theta\sin\varphi\frac{\partial}{\partial \theta}
+\cos\varphi\frac{\partial}{\partial \varphi}
=
r\sin\theta\frac{\partial}{\partial y}

すなわち

\frac{\partial}{\partial y}
=
\sin\theta\sin\varphi\frac{\partial}{\partial r}
+\frac{\cos\theta\sin\varphi}{r}\frac{\partial}{\partial \theta}
+\frac{\cos\varphi}{r\sin\theta}\frac{\partial}{\partial \varphi}
\tag{5.15.9}

(5.15.8)，(5.15.9)を(5.15.4)に代入すると，

\frac{\partial}{\partial r}
=
\sin^2\theta\cos^2\varphi\frac{\partial}{\partial r}
+\frac{\sin\theta\cos\theta\cos^2\varphi}{r}\frac{\partial}{\partial \theta}
-\frac{\sin\varphi\cos\varphi}{r}\frac{\partial}{\partial \varphi}
+\sin^2\theta\sin^2\varphi\frac{\partial}{\partial r}
+\frac{\sin\theta\cos\theta\sin^2\varphi}{r}\frac{\partial}{\partial \theta}
+\frac{\sin\varphi\cos\varphi}{r}\frac{\partial}{\partial \varphi}
+\cos\theta\frac{\partial}{\partial z},

\cos^2\theta\frac{\partial}{\partial r}=
\frac{\sin\theta\cos\theta}{r}\frac{\partial}{\partial \theta}
+\cos\theta\frac{\partial}{\partial z}

すなわち

\frac{\partial}{\partial z}
=
\cos\theta\frac{\partial}{\partial r}
-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}
\tag{5.15.10}

(5.8),(5.15.9),(5.15.10)より,

\begin{align}
y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}
&=
r\sin\theta\cos\theta\sin\varphi\frac{\partial}{\partial r}
-\sin^2\theta\sin\varphi\frac{\partial}{\partial \theta}
-r\sin\theta\cos\theta\sin\varphi\frac{\partial}{\partial r}
-\cos^2\theta\sin\varphi\frac{\partial}{\partial \theta}
-\cot\theta\cos\varphi\frac{\partial}{\partial \varphi},
\\
&=
-\left(
\sin\varphi\frac{\partial}{\partial \theta}
+\cot\theta\cos\varphi\frac{\partial}{\partial \varphi}
\right)
\end{align}

(5.8),(5.15.8),(5.15.10)より,

\begin{align}
z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}
&=
r\sin\theta\cos\theta\cos\varphi\frac{\partial}{\partial r}
+\cos^2\theta\cos\varphi\frac{\partial}{\partial \theta}
-\cot\theta\sin\varphi\frac{\partial}{\partial \varphi}
-r\sin\theta\cos\theta\cos\varphi\frac{\partial}{\partial r}
+\sin^2\theta\cos\varphi\frac{\partial}{\partial \theta}
\\
&=
\cos\varphi\frac{\partial}{\partial \theta}
-\cot\theta\sin\varphi\frac{\partial}{\partial \varphi}
\end{align}

(5.8),(5.15.8),(5.15.9)より,

\begin{align}
x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}
&=
r\sin^2\theta\sin\varphi\cos\varphi\frac{\partial}{\partial r}
+\sin\theta\cos\theta\sin\varphi\cos\varphi\frac{\partial}{\partial \theta}
+\cos^2\varphi\frac{\partial}{\partial \varphi}
-r\sin^2\theta\sin\varphi\cos\varphi\frac{\partial}{\partial r}
-\sin\theta\cos\theta\sin\varphi\cos\varphi\frac{\partial}{\partial \theta}
+\sin^2\varphi\frac{\partial}{\partial \varphi}
\\
&=
\frac{\partial}{\partial \varphi}
\end{align}

これら3式を(5.15)に代入すると,

\begin{align}
\hat{L}_x&=-\frac{\hbar}{i}\left(
\sin\varphi\frac{\partial}{\partial \theta}
+\cot\theta\cos\varphi\frac{\partial}{\partial \varphi}
\right)
\\
\hat{L}_y&=\frac{\hbar}{i}\left(
\cos\varphi\frac{\partial}{\partial \theta}
-\cot\theta\sin\varphi\frac{\partial}{\partial \varphi}
\right)
\\
\hat{L}_z&=\frac{\hbar}{i}
\frac{\partial}{\partial \varphi}
\end{align}
\tag{5.16}

となる.

さらに,

\begin{align}
\left(
\sin\varphi\frac{\partial}{\partial \theta}
+\cot\theta\cos\varphi\frac{\partial}{\partial \varphi}
\right)^2
&=
\sin\varphi\frac{\partial}{\partial \theta}
\left(
\sin\varphi\frac{\partial}{\partial \theta}
+\cot\theta\cos\varphi\frac{\partial}{\partial \varphi}
\right)
+
\cot\theta\cos\varphi\frac{\partial}{\partial \varphi}
\left(
\sin\varphi\frac{\partial}{\partial \theta}
+\cot\theta\cos\varphi\frac{\partial}{\partial \varphi}
\right)
\\
&=
\sin^2\varphi\frac{\partial^2}{\partial \theta^2}
+\sin\varphi\cos\varphi\frac{\partial}{\partial \theta}\left(\cot\theta\right)\frac{\partial}{\partial \varphi}
+\cot\theta\cos\varphi\frac{\partial}{\partial \theta}\frac{\partial}{\partial \varphi}\left(\sin\varphi\right)
+\cot^2\theta\cos\varphi\frac{\partial}{\partial \varphi}\left(\cos\varphi\frac{\partial}{\partial \varphi}\right)
\\
&=
\sin^2\varphi\frac{\partial^2}{\partial \theta^2}
+\sin\varphi\cos\varphi
\left(
-\frac{1}{\sin^2\theta}
+\cot\theta\frac{\partial}{\partial \theta}
\right)\frac{\partial}{\partial \varphi}
+\cot\theta\cos\varphi\frac{\partial}{\partial \theta}
\left(
\cos\varphi+\sin\varphi\frac{\partial}{\partial \varphi}
\right)
+\cot^2\theta\cos\varphi
\left(
-\sin\varphi\frac{\partial}{\partial \varphi}
+\cos\varphi\frac{\partial^2}{\partial \varphi^2}
\right)
\end{align}

\begin{align}
\left(
\cos\varphi\frac{\partial}{\partial \theta}
-\cot\theta\sin\varphi\frac{\partial}{\partial \varphi}
\right)^2
&=
\cos\varphi\frac{\partial}{\partial \theta}
\left(
\cos\varphi\frac{\partial}{\partial \theta}
-\cot\theta\sin\varphi\frac{\partial}{\partial \varphi}
\right)
-\cot\theta\sin\varphi\frac{\partial}{\partial \varphi}
\left(
\cos\varphi\frac{\partial}{\partial \theta}
-\cot\theta\sin\varphi\frac{\partial}{\partial \varphi}
\right)
\\
&=
\cos^2\varphi\frac{\partial^2}{\partial \theta^2}
-\sin\varphi\cos\varphi\frac{\partial}{\partial \theta}\left(\cot\theta\right)\frac{\partial}{\partial \varphi}
-\cot\theta\sin\varphi\frac{\partial}{\partial \theta}\frac{\partial}{\partial \varphi}\left(\cos\varphi\right)
+\cot^2\theta\sin\varphi\frac{\partial}{\partial \varphi}\left(\sin\varphi\frac{\partial}{\partial \varphi}\right)
\\
&=
\cos^2\varphi\frac{\partial^2}{\partial \theta^2}
-\sin\varphi\cos\varphi
\left(
-\frac{1}{\sin^2\theta}+\cot\theta\frac{\partial}{\partial \theta}
\right)\frac{\partial}{\partial \varphi}
-\cot\theta\sin\varphi\frac{\partial}{\partial \theta}
\left(
-\sin\varphi+\cos\varphi\frac{\partial}{\partial \varphi}
\right)
+\cot^2\theta\sin\varphi
\left(
\cos\varphi\frac{\partial}{\partial \varphi}
+\sin\varphi\frac{\partial^2}{\partial \varphi^2}
\right)
\end{align}

\left(\frac{\partial}{\partial \varphi}\right)^2
=
\frac{\partial^2}{\partial \varphi^2}

以上3式を,

\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}
\left(\sin\theta\frac{\partial}{\partial \theta}\right)
=
\frac{1}{\sin\theta}
\left(
\cos\theta\frac{\partial}{\partial \theta}+\sin\theta\frac{\partial^2}{\partial \theta^2}
\right)
=\cot\theta\frac{\partial}{\partial \theta}+\frac{\partial^2}{\partial \theta^2}

であることを用いて加えると，

\begin{align}
\left(
\sin\varphi\frac{\partial}{\partial \theta}
+\cot\theta\cos\varphi\frac{\partial}{\partial \varphi}
\right)^2
+
\left(
\cos\varphi\frac{\partial}{\partial \theta}
-\cot\theta\sin\varphi\frac{\partial}{\partial \varphi}
\right)^2
+
\left(\frac{\partial}{\partial \varphi}\right)^2
&=
\frac{\partial^2}{\partial \theta^2}
+\cot\theta\frac{\partial}{\partial \theta}
+
\left(\cot^2\theta+1\right)\frac{\partial^2}{\partial \varphi^2}
=
\frac{\partial^2}{\partial \theta^2}
+\cot\theta\frac{\partial}{\partial \theta}
+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \varphi^2}
\\
&=
\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}
\left(\sin\theta\frac{\partial}{\partial \theta}\right)
+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \varphi^2}
\end{align}

したがって,

\begin{align}
\left(\hat{L}\right)^2
&=\left(\hat{L}_x\right)^2+\left(\hat{L}_y\right)^2+\left(\hat{L}_z\right)^2
\\
&=
-\hbar^2\left[
\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}
\left(\sin\theta\frac{\partial}{\partial \theta}\right)
+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \varphi^2}
\right]
\end{align}
\tag{5.17}

また，

\hat{p}_r=\frac{\hbar}{i}\frac{1}{r}\frac{\partial}{\partial r}r

を2乗すると，

\begin{align}
\left(\hat{p}_r\right)^2
&=
-\hbar^2
\left(\frac{1}{r}\frac{\partial}{\partial r}r\right)
\left(\frac{1}{r}\frac{\partial}{\partial r}r\right)
\\
&=
-\hbar^2
\left(\frac{1}{r}\frac{\partial}{\partial r}r\right)
\left\{
\frac{1}{r}\left(1+r\frac{\partial}{\partial r}\right)
\right\}
\\
&=
-\hbar^2
\left(\frac{1}{r}\frac{\partial}{\partial r}r\right)
\left(\frac{1}{r}+\frac{\partial}{\partial r}\right)
\\
&=
-\hbar^2
\left(\frac{1}{r}\frac{\partial}{\partial r}\right)
\left(1+r\frac{\partial}{\partial r}\right)
\\
&=
-\hbar^2
\frac{1}{r}
\left(
\frac{\partial}{\partial r}+\frac{\partial}{\partial r}
+r\frac{\partial^2}{\partial r^2}
\right)
\\
&=
-\hbar^2
\left(
\frac{2}{r}\frac{\partial}{\partial r}
+\frac{\partial^2}{\partial r^2}
\right)
\end{align}

ここで，

\frac{1}{r}\frac{\partial^2}{\partial r^2}r
=
\frac{1}{r}\frac{\partial}{\partial r}\left(1+r\frac{\partial}{\partial r}\right)
=
\frac{1}{r}
\left(
\frac{\partial}{\partial r}+\frac{\partial}{\partial r}+r\frac{\partial^2}{\partial r^2}
\right)
=
\frac{2}{r}\frac{\partial}{\partial r}
+\frac{\partial^2}{\partial r^2}

より

\left(\hat{p}_r\right)^2=-\hbar^2\frac{1}{r}\frac{\partial^2}{\partial r^2}r
\tag{5.18}

なので，(5.9)，(5.17)，(5.18)より，

\frac{\hat{p}_r^2}{2m}
+\frac{\hat{L}_r^2}{2mr^2}
=
-\frac{\hbar^2}{2m}\Delta
\tag{5.19}

であることが確かめられた．したがって(5.14)が極座標のエルミート演算子のみで表された正しいハミルトニアンである．