問題 4-4 の解答例

Problem 4-4

Show $$ \def\ppdiff#1#2{\frac{\partial #1}{\partial #2}} \def\pdiff#1{\frac{\partial}{\partial #1}} \def\mb#1{\mathbf{#1}} \def\reverse#1{\frac{1}{#1}} \def\mfrac#1#2{\frac{\,#1\,}{\,#2\,}} \def\Bpdiff#1{\frac{\partial^{2}}{\partial #1^{2}}} \Bpdiff{x} x = x\Bpdiff{x}+2\pdiff{x} \tag{4-22} $$ and therefore that, for the $H$ of Eq. (4-15), $$ Hx-xH=-\frac{\hbar^{2}}{m}\pdiff{x} \tag{4-23} $$ This operator notation is used a great deal in the conventional formulations of quantum mechanics.

(解答) 演算子はその右に作用を受ける関数 $f(x)$ が存在することで意味を持つことに留意すれば, 式(4-22)の左辺は次のような意味である：

\begin{align} \left(\Bpdiff{x}x\right)f(x)&=\left(\pdiff{x}\pdiff{x}\right)xf(x)=\pdiff{x}\left(\pdiff{x}xf(x)\right)\notag\\ &=\pdiff{x}\left(\ppdiff{x}{x}f(x)+x\ppdiff{f(x)}{x}\right)=\pdiff{x}\left(f(x)+x\ppdiff{f(x)}{x}\right)\notag\\ &=\ppdiff{f(x)}{x}+\pdiff{x}\left(x\ppdiff{f(x)}{x}\right)\notag\\ &=\ppdiff{f(x)}{x}+\ppdiff{x}{x}\ppdiff{f(x)}{x}+x\Bpdiff{x}f(x)=2\pdiff{x}f(x)+x\Bpdiff{x}f(x)\notag\\ &=\left(2\pdiff{x}+\Bpdiff{x}\right) f(x) \tag{1} \end{align}

これが任意の関数 $f(x)$ について成り立つべきである．従って, 式(4-22)が言える：

\begin{equation} \Bpdiff{x}x=2\pdiff{x}+x\Bpdiff{x} \tag{2} \end{equation}

また, 明らかに次が言える：

\begin{equation} V(x)x-xV(x)=0 \tag{3} \end{equation}

さらに, 式(4-15)から

\begin{equation} H=-\frac{\hbar^{2}}{2m}\Bpdiff{x}+V(x) \tag{4} \end{equation}

以上の式(2)$\sim$式(4)を用いれば, 式(4-23)が言える：

\begin{align} Hx-xH&=\left(-\mfrac{\hbar^{2}}{2m}\Bpdiff{x}+V(x)\right) x-x \left( -\mfrac{\hbar^{2}}{2m}\Bpdiff{x}+V(x)\right)\notag\\ &=-\mfrac{\hbar^{2}}{2m}\Bpdiff{x}x+V(x)x+x\mfrac{\hbar^{2}}{2m}\Bpdiff{x}-xV(x)\notag\\ &=-\mfrac{\hbar^{2}}{2m}\left(2\pdiff{x}+x\Bpdiff{x}\right)+x\mfrac{\hbar^{2}}{2m}\Bpdiff{x}+V(x)x-xV(x)\notag\\ &=-\mfrac{\hbar^{2}}{m}\pdiff{x}-\mfrac{\hbar^{2}}{2m}x\Bpdiff{x}+\mfrac{\hbar^{2}}{2m}x\Bpdiff{x}\notag\\ &=-\mfrac{\hbar^{2}}{m}\pdiff{x} \tag{5} \end{align}