問題 5-8 の解答例

Problem 5-8

Note that Eq. (5-44) implies $G_{A}^{*}(x,x^{'})=G_{A}(x^{'},x)$. With this in mind show that for any two wave functions $g(x)$ and $f(x)$, both of which approach $0$ as $x$ goes to $\pm\infty$,

\begin{equation} \int_{-\infty}^{\infty} dx\, g^{*}(x)\mathcal{A}f(x)=\int_{-\infty}^{\infty} dx\, \Big[\mathcal{A}g(x)\Big]^{*} f(x) \tag{5-47} \end{equation}

Any operator, such as $\mathcal{A}$, for which Eq. (5-47) holds is called hermitian [ cf. Eq. (4-30) ].

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( 解答 )　まず必要な式を示しておこう．物理量 $A$ の測定結果の平均値すなわち「期待値」を記号 $\langle A \rangle$ で表すならば, $ \def\bra#1{\langle#1|} \def\BK#1#2{\langle #1|#2\rangle} \def\BraKet#1#2#3{\langle#1|#2|#3\rangle} \def\ppdiff#1#2{\frac{\partial #1}{\partial #2}} \def\odiff#1{\frac{d}{d #1}} \def\pdiff#1{\frac{\partial}{\partial #1}} \def\Bppdiff#1#2{\frac{\partial^{2}#1}{\partial #2^{2}}} \def\Bpdiff#1{\frac{\partial^{2}}{\partial #1^{2}}} \def\mb#1{\mathbf{#1}} \def\ds#1{\mbox{${\displaystyle\strut #1}$}} \def\mfrac#1#2{\frac{#1}{#2}} $

\begin{align} \langle A \rangle &=\sum_{a}\sum_{b}\sum_{c}\dotsb \,a\,\left|F_{a, b, c, \dotsb}\right|^{2} \tag{5-40}\\ &=\sum_{a}\sum_{b}\sum_{c}\dotsb \,a\,\int dx\,\chi_{a, b, c,\dotsb}(x)f^{*}(x)\int dx^{'}\,\chi^{*}_{a, b, c, \dotsb}(x^{'})\,f(x^{'})\notag\\ &\equiv \int dx\,f^{*}(x)R(x), \tag{5-42} \end{align}

ただし,

\begin{align} R(x)&=\int dx^{'}\,G_{A}(x,x^{'})\,f(x^{'})\equiv \mathcal{A}f(x), \tag{5-43, 45}\\ G_{A}(x,x^{'})&=\sum_{a}\sum_{b}\sum_{c}\dotsb \,a\,\chi_{a,b,c,\dotsb}(x)\,\chi^{*}_{a, b, c,\dotsb}(x^{'}) \tag{5-44} \end{align}

最初に $G_{A}^{*}(x,x^{'})=G_{A}(x^{'},x)$ が成り立つことを確認して置こう．それは, 式 (5-44) の両辺の複素共役をとることで示すことが出来る．$a$ はオブザーバブル $A$ の固有値なので実数であることに注意すれば,

\begin{equation} G_{A}^{*}(x,x^{'})=\sum_{a}\sum_{b}\sum_{c}\dotsb \,a\,\chi_{a,b,c,\dotsb}(x^{'})\,\chi^{*}_{a, b, c,\dotsb}(x) =G_{A}(x^{'},x) \label{eq1} \end{equation}

さて, 式 (5-47) の左辺に式 (5-42)$\sim$ 式 (5-45) を用いて行くならば,

\begin{align} &\int_{-\infty}^{\infty} dx\, g^{*}(x)\mathcal{A}f(x)=\int dx\,g^{*}(x)\,R(x)\notag\\ &\quad=\int dx\,g^{*}(x)\int dx^{'}\,G_{A}(x,x^{'})\,f(x^{'})=\int dx\,g^{*}(x)\int dx^{'}\,G_{A}^{*}(x^{'},x)\,f(x^{'})\notag\\ &\quad=\int dx^{'}\,f(x^{'})\int dx\,g^{*}(x)\,G_{A}^{*}(x^{'},x)=\int dx^{'}\,f(x^{'})\,\left[ \mathcal{A}\,g(x^{'})\right]^{*} \label{eq2} \end{align}

この最後の結果式に於いて, 単なる変数文字の書き換え： $x^{'}\to x$ をするならば, 求めるべき式となる：

\begin{equation} \int_{-\infty}^{\infty} dx\, g^{*}(x)\mathcal{A}f(x)=\int_{-\infty}^{\infty} dx\,\left[ \mathcal{A}\,g(x)\right]^{*} f(x) \label{eq3} \end{equation}