問題 8-5 の解答例

Feynman-Hibbs cover
\(\)
Problem 8-5
A transition element which employs the same wave function as both the initial and final states is called an expectation value. [1]Complare this definition of expectation value with the definition of the expected value of an operator given in Sec. 5-3, particularly in Eq. (5-46).
Thus the expectation value of \(F\) for the ground state \(\Phi_0\) of Eq. (8-83) is

\begin{equation}
\def\reverse#1{\frac{1}{#1}}
\def\half{\frac{1}{2}}
\def\BraKet#1#2#3{\left\langle #1\left| #2 \right| #3 \right\rangle}
\BraKet{\Phi_0}{F}{\Phi_0}=\int\!\dotsb \iint \Phi_{0}^{*}\, F \,\Phi_0\,dQ_0\,dQ_1\dotsb dQ_{N-1}
\tag{8-84}
\end{equation}

( The integral over complex variables is defined as equal to the corresponding integral over real normal coordinates \(Q_{\alpha}^{\,C}\) and \(Q_{\alpha}^{\,S}\). )
Show that the following expectation values are correct (for \(\alpha\neq 0\)): [2]【 注意】 この問題は, J.J.Sakuraiの § 2.3 (p.126 \(\sim\) ) に記してある次のことに相当している.すなわち, 調和振動子の基底状態での位置についての … Continue reading
\begin{align}
&\BraKet{\Phi_0}{Q_{\alpha}}{\Phi_0}=\BraKet{\Phi_0}{Q^{*}_{\alpha}}{\Phi_0}=0,\notag\\
&\BraKet{\Phi_0}{Q_{\alpha}^{2}}{\Phi_0}=\BraKet{\Phi_0}{Q^{* 2}_{\alpha}}{\Phi_0}=0,\notag\\
&\BraKet{\Phi_0}{Q^{*}_{\alpha}Q_{\alpha}}{\Phi_0}=\frac{\hbar}{2\omega_{\alpha}}\BraKet{\Phi_0}{1}{\Phi_0}, \tag{8-85}\\
&\BraKet{\Phi_0}{Q^{*}_{\alpha}Q_{\beta}}{\Phi_0}=0\qquad \mathrm{if}\quad \alpha\neq \beta\notag
\end{align}


( 解答 )  前の問題 8-4 から, \(\Phi_0\) と \(\Phi_0^{\,*}\) は次のように書ける:

\begin{align}
\Phi_0&=\prod_{\alpha=0}^{\half(N-1)}A_{\alpha}^{\,C}\exp\left\{-\frac{1}{2\hbar}\omega_{\alpha}\Big(Q_{\alpha}^{\,C}\Big)^{2}\right\}
\prod_{\alpha=1}^{\half(N-1)}A_{\alpha}^{\,S}\exp\left\{-\frac{1}{2\hbar}\omega_{\alpha}\Big(Q_{\alpha}^{\,S}\Big)^{2}\right\}\notag\\
&=A\prod_{\alpha=0}^{\half(N-1)} \exp\left[ -\frac{\omega_{\alpha}}{2\hbar}
\left\{\Big(Q_{\alpha}^{\,C}\Big)^{2}+\Big(Q_{\alpha}^{\,S}\Big)^{2}\right\}\right],\notag\\
\Phi_0^{\,*} &=A^{*}\prod_{\beta=0}^{\half(N-1)} \exp\left[ -\frac{\omega_{\beta}}{2\hbar}
\left\{\Big(Q_{\beta}^{\,C}\Big)^{2}+\Big(Q_{\beta}^{\,S}\Big)^{2}\right\}\right]
\tag{1}
\end{align}

このときの \(Q_\alpha^{\,C}\) と \(Q_\alpha^{\,S}\) が一緒に合わさって, この系の \(N\) 個の基準座標 \(Q_i\) を形成するのであった.
\begin{equation*}
Q_i=\left[\,Q_0^{\,C},\,Q_1^{\,C},\dotsb,\,Q_{\reverse{2}(N-1)}^{\,C},\,Q_1^{\,S},\,Q_2^{\,S},\dotsb,
\,Q_{\reverse{2}(N-1)}^{\,S}\,\right]
\end{equation*}

よって, 式 (8-84) は次のように表わすことが出来る:
\begin{align}
&\BraKet{\Phi_0}{F}{\Phi_0}=\int dQ_0\dotsb \int dQ_{N-1}\ \Phi_0^{\,*}\,F\,\Phi_0\notag\\
&=|A|^{2}\idotsint dQ_0^{\,C}\dotsb dQ_{\reverse{2}(N-1)}^{\,C}\, dQ_1^{\,S}\dotsb dQ_{\reverse{2}(N-1)}^{\,S}\notag\\
&\times \prod_{\nu=0}^{\half(N-1)} \exp\left[ -\frac{\omega_{\nu}}{2\hbar}
\left\{\Big(Q_{\nu}^{\,C}\Big)^{2}+\Big(Q_{\nu}^{\,S}\Big)^{2}\right\}\right]
\,F\,\prod_{\mu=0}^{\half(N-1)} \exp\left[ -\frac{\omega_{\mu}}{2\hbar}
\left\{\Big(Q_{\mu}^{\,C}\Big)^{2}+\Big(Q_{\mu}^{\,S}\Big)^{2}\right\}\right]
\tag{2}
\end{align}

また \(Q_\alpha^{\,2}\) , \(Q_\alpha^{\,*\,2}\) , \(Q_\alpha^{\,*}Q_\alpha\) , \(Q_\alpha^{\,*}Q_\beta\) は次となる:
\begin{align}
&Q_\alpha^{\,2}=\reverse{2}\left(Q_\alpha^{\,C}-iQ_\alpha^{\,S}\right)^{2}
=\reverse{2}\left\{\left(Q_\alpha^{\,C}\right)^{2}-i2Q_\alpha^{\,C}Q_\alpha^{\,S}-\left(Q_\alpha^{\,S}\right)^{2}
\right\}\notag\\
&Q_\alpha^{\,*\,2}=\reverse{2}\left(Q_\alpha^{\,C}+iQ_\alpha^{\,S}\right)^{2}
=\reverse{2}\left\{\left(Q_\alpha^{\,C}\right)^{2}+i2Q_\alpha^{\,C}Q_\alpha^{\,S}-\left(Q_\alpha^{\,S}\right)^{2}
\right\}\tag{3}\\
&Q_\alpha^{\,*}Q_\alpha=\reverse{2}\left(Q_\alpha^{\,C}+iQ_\alpha^{\,S}\right)\left(Q_\alpha^{\,C}-iQ_\alpha^{\,S}
\right)=\reverse{2}\left\{\left(Q_\alpha^{\,C}\right)^{2}+\left(Q_\alpha^{\,S}\right)^{2}\right\}\notag\\
&Q_\alpha^{\,*}Q_\beta=\reverse{2}\left(Q_\alpha^{\,C}+iQ_\alpha^{\,S}\right)\left(Q_\beta^{\,C}-iQ_\beta^{\,S}
\right)=\reverse{2}\left(Q_\alpha^{\,C}Q_\beta^{\,C}+Q_\alpha^{\,S}Q_\beta^{\,S}\right)+\frac{i}{2}
\left(Q_\alpha^{\,S}Q_\beta^{\,C}+Q_\alpha^{\,C}Q_\beta^{\,S}\right)\notag
\end{align}

さらに, 変数 \(Q_\alpha\) の経路積分を変数 \(Q_\alpha^{\,C}\) 及び \(Q_\alpha^{\,S}\) の経路積分へ変数変換することは, 複素平面に於ける単なる座標回転に相当している.従って, 変数変換によって余分な因子は付かずまた積分範囲も変わらない.以上の事をもとに, 各々の期待値を式 (8-84) により求めて行こう.

まず \(\BraKet{\Phi_0}{Q_\alpha}{\Phi_0}\) 及び \(\BraKet{\Phi_0}{Q_\alpha^{\,*}}{\Phi_0}\) を考えると,

\begin{align}
&\BraKet{\Phi_0}{Q_\alpha}{\Phi_0}\notag\\
&=|A|^{2}\idotsint dQ_0^{\,C}\dotsb dQ_{\reverse{2}(N-1)}^{\,C}\, dQ_1^{\,S}\dotsb dQ_{\reverse{2}(N-1)}^{\,S}\notag\\
&\times \prod_{\nu=0}^{\half(N-1)} \exp\left[ -\frac{\omega_{\nu}}{2\hbar}
\left\{\Big(Q_{\nu}^{\,C}\Big)^{2}+\Big(Q_{\nu}^{\,S}\Big)^{2}\right\}\right]
\,\reverse{\sqrt{2}}\left\{Q_\alpha^{\,C}-iQ_\alpha^{\,S}\right\}\notag\\
&\times \prod_{\mu=0}^{\half(N-1)} \exp\left[ -\frac{\omega_{\mu}}{2\hbar}
\left\{\Big(Q_{\mu}^{\,C}\Big)^{2}+\Big(Q_{\mu}^{\,S}\Big)^{2}\right\}\right]\notag\\
&=\frac{|A|^{2}}{\sqrt{2}}\left[\prod_{\mu\neq\alpha}\int dQ_\mu^{\,C}\,
\exp\left\{-\frac{\omega_\mu}{\hbar}\left(Q_\mu^{\,C}\right)^{2}\right\}\int dQ_{\alpha}^{\,C}\,Q_{\alpha}^{\,C}
\exp\left\{-\frac{\omega_{\alpha}}{\hbar}\left(Q_{\alpha}^{\,C}\right)^{2}\right\}\prod_{\nu=1}\int dQ_\nu^{\,S}\,
\exp\left\{-\frac{\omega_\nu}{\hbar}\left(Q_\nu^{\,S}\right)^{2}\right\}\right.\notag\\
&\quad\left. -i\prod_{\mu=0}\int dQ_\mu^{\,C}\,\exp\left\{-\frac{\omega_\mu}{\hbar}\left(Q_\mu^{\,C}\right)^{2}\right\}
\prod_{\nu\ne\alpha}\int dQ_\nu^{\,S}\,\exp\left\{-\frac{\omega_\nu}{\hbar}\left(Q_\nu^{\,S}\right)^{2}\right\}
\int dQ_{\alpha}^{\,S}\,Q_{\alpha}^{\,S}\exp\left\{-\frac{\omega_{\alpha}}{\hbar}\left(Q_{\alpha}^{\,S}\right)^{2}\right\}\right]
\tag{4}
\end{align}

このとき, 第1項と第2項の両方に次の形の積分が因子として含まれている:
\begin{equation}
\int_{-\infty}^{\infty} xe^{-ax^{2}}\,dx=0
\tag{5}
\end{equation}

この積分は, 被積分関数が奇関数であるためにゼロである.従って, 全体もゼロとなる. \(Q_\alpha^{\,*}\) の場合では, 第2項目の符号がプラスとなるだけで同様な議論となる.よって, 次が言える:
\begin{equation}
\BraKet{\Phi_0}{Q_\alpha}{\Phi_0}=0,\quad \BraKet{\Phi_0}{Q_\alpha^{\,*}}{\Phi_0}=0
\tag{6}
\end{equation}

次に \(\BraKet{\Phi_0}{Q_\alpha^{\,2}}{\Phi_0}\) 及び \(\BraKet{\Phi_0}{Q_\alpha^{\,*\,2}}{\Phi_0}\) を考える.式 (2) の \(F\) に, 式 (3) の \(\Phi_0^{\,2}\) または \(\Phi_0^{\,*\,2}\) を代入すると,
\begin{align}
&\BraKet{\Phi_0}{Q_\alpha^{\,2}}{\Phi_0}\notag\\
&=|A|^{2}\idotsint dQ_0^{\,C}\dotsb dQ_{\reverse{2}(N-1)}^{\,C}\, dQ_1^{\,S}\dotsb dQ_{\reverse{2}(N-1)}^{\,S}
\prod_{\nu=0}^{\half(N-1)} \exp\left[ -\frac{\omega_{\nu}}{2\hbar}
\left\{\Big(Q_{\nu}^{\,C}\Big)^{2}+\Big(Q_{\nu}^{\,S}\Big)^{2}\right\}\right]\notag\\
&\times \,\reverse{2}\left\{\left(Q_\alpha^{\,C}\right)^{2}-\left(Q_\alpha^{\,S}\right)^{2}-2i Q_\alpha^{\,C}Q_\alpha^{\,S}\right\}\,
\prod_{\mu=0}^{\half(N-1)}\exp\left[ -\frac{\omega_{\mu}}{2\hbar}\left\{\Big(Q_{\mu}^{\,C}\Big)^{2}+\Big(Q_{\mu}^{\,S}\Big)^{2}\right\}
\right]\notag\\
&=\frac{|A|^{2}}{2}\left[\prod_{\mu\neq\alpha}\int dQ_\mu^{\,C}\,
\exp\left\{-\frac{\omega_\mu}{\hbar}\left(Q_\mu^{\,C}\right)^{2}\right\}\int dQ_{\alpha}^{\,C}\,\left(Q_{\alpha}^{\,C}\right)^{2}
\exp\left\{-\frac{\omega_{\alpha}}{\hbar}\left(Q_{\alpha}^{\,C}\right)^{2}\right\}
\prod_{\nu=1}\int dQ_\nu^{\,S}\,\exp\left\{-\frac{\omega_\nu}{\hbar}\left(Q_\nu^{\,S}\right)^{2}\right\}\right.\notag\\
&\quad-\prod_{\mu\ne\alpha}\int dQ_\mu^{\,C}\,\exp\left\{-\frac{\omega_\mu}{\hbar}\left(Q_\mu^{\,C}\right)^{2}\right\}
\prod_{\nu\ne\alpha}\int dQ_\nu^{\,S}\,\exp\left\{-\frac{\omega_\nu}{\hbar}\left(Q_\nu^{\,S}\right)^{2}\right\}
\int dQ_{\alpha}^{\,S}\,\left(Q_{\alpha}^{\,S}\right)^{2}
\exp\left\{-\frac{\omega_{\alpha}}{\hbar}\left(Q_{\alpha}^{\,S}\right)^{2}\right\}\notag\\
&\quad\left. -2i \prod_{\mu\ne\alpha}\int dQ_\mu^{\,C}\,\exp\left\{-\frac{\omega_\mu}{\hbar}\left(Q_\mu^{\,C}\right)^{2}\right\}
\prod_{\nu\ne\alpha}\int dQ_\nu^{\,S}\,\exp\left\{-\frac{\omega_\nu}{\hbar}\left(Q_\nu^{\,S}\right)^{2}\right\}
\int dQ_{\alpha}^{\,C}\,Q_{\alpha}^{\,C}\exp\left\{-\frac{\omega_{\alpha}}{\hbar}\left(Q_{\alpha}^{\,C}\right)^{2}\right\}
\int dQ_{\alpha}^{\,S}\,Q_{\alpha}^{\,S}\exp\left\{-\frac{\omega_{\alpha}}{\hbar}\left(Q_{\alpha}^{\,S}\right)^{2}\right\}\right]
\tag{7}
\end{align}

このとき 第 1 項と第 2 項の積分は, 変数的には等価な2重積分であるから互いに打ち消し合う.また, 第 3 項は前述の式(4)と同じ理由からゼロである.よって, 式 (7) 全体もゼロとなることは明らかである.

\(Q_{\alpha}^{*}\) の場合は, 第 3 項の符号がプラスになるだけであるから, やはりゼロとなる.以上から, 次の結論となる:

\begin{equation}
\BraKet{\Phi_0}{Q_\alpha^{\,2}}{\Phi_0}=0,\quad \BraKet{\Phi_0}{Q_\alpha^{\,*\,2}}{\Phi_0}=0
\tag{8}
\end{equation}

同様にして \(\BraKet{\Phi_0}{Q_\alpha^{\,*}Q_\alpha}{\Phi_0}\) の場合は,
\begin{align}
&\BraKet{\Phi_0}{Q_\alpha^{\,*}Q_{\alpha}}{\Phi_0}\notag\\
&=|A|^{2}\idotsint dQ_0^{\,C}\dotsb dQ_{\reverse{2}(N-1)}^{\,C}\, dQ_1^{\,S}\dotsb dQ_{\reverse{2}(N-1)}^{\,S}
\prod_{\nu=0}^{\half(N-1)} \exp\left[ -\frac{\omega_{\nu}}{2\hbar}
\left\{\Big(Q_{\nu}^{\,C}\Big)^{2}+\Big(Q_{\nu}^{\,S}\Big)^{2}\right\}\right]\notag\\
&\times \,\reverse{2}\left\{\left(Q_\alpha^{\,C}\right)^{2}+\left(Q_\alpha^{\,S}\right)^{2}\right\}\,\prod_{\mu=0}^{\half(N-1)}
\exp\left[ -\frac{\omega_{\mu}}{2\hbar}\left\{\Big(Q_{\mu}^{\,C}\Big)^{2}+\Big(Q_{\mu}^{\,S}\Big)^{2}\right\}\right]\notag\\
&=\frac{|A|^{2}}{2}\left[\prod_{\mu\neq\alpha}\int dQ_\mu^{\,C}\,
\exp\left\{-\frac{\omega_\mu}{\hbar}\left(Q_\mu^{\,C}\right)^{2}\right\}\int dQ_{\alpha}^{\,C}\,\left(Q_{\alpha}^{\,C}\right)^{2}
\exp\left\{-\frac{\omega_{\alpha}}{\hbar}\left(Q_{\alpha}^{\,C}\right)^{2}\right\}
\prod_{\nu=1}\int dQ_\nu^{\,S}\,\exp\left\{-\frac{\omega_\nu}{\hbar}\left(Q_\nu^{\,S}\right)^{2}\right\}\right.\notag\\
&\quad\left.+\prod_{\mu=0}\int dQ_\mu^{\,C}\,\exp\left\{-\frac{\omega_\mu}{\hbar}\left(Q_\mu^{\,C}\right)^{2}\right\}
\prod_{\nu\ne\alpha}\int dQ_\nu^{\,S}\,\exp\left\{-\frac{\omega_\nu}{\hbar}\left(Q_\nu^{\,S}\right)^{2}\right\}
\int dQ_{\alpha}^{\,S}\,\left(Q_{\alpha}^{\,S}\right)^{2}
\exp\left\{-\frac{\omega_{\alpha}}{\hbar}\left(Q_{\alpha}^{\,S}\right)^{2}\right\}\right]
\tag{9}
\end{align}

ここで, 巻末の積分公式を利用する:
\begin{align}
I_A(a)&=\int_{-\infty}^{\infty}e^{-ax^{2}}\,dx=\sqrt{\frac{\pi}{a}} \quad\rightarrow\quad
I_{A}\left(\frac{\omega_\alpha}{\hbar}\right)=\sqrt{\frac{\pi\hbar}{\omega_{\alpha}}}\ ,\notag\\
I_B(a)&=\int_{-\infty}^{\infty}x^{2}e^{-ax^{2}}\,dx=\reverse{2a}\sqrt{\frac{\pi}{a}} \quad\rightarrow\quad
I_B\left(\frac{\omega_\alpha}{\hbar}\right)=\frac{\hbar}{2\omega_{\alpha}}\sqrt{\frac{\pi\hbar}{\omega_{\alpha}}}
\tag{10}
\end{align}

すると, 上式 (9) は
\begin{align}
&\BraKet{\Phi_0}{Q_\alpha^{\,*}Q_{\alpha}}{\Phi_0}\notag\\
&=\frac{|A|^{2}}{2}\left\{\,\prod_{\mu\neq\alpha}I_{A}\left(\frac{\omega_\mu}{\hbar}\right)I_B\left(\frac{\omega_\alpha}{\hbar}\right)
\prod_{\nu=1} I_{A}\left(\frac{\omega_{\nu}}{\hbar}\right)
+ \prod_{\mu=0} I_{A}\left(\frac{\omega_\mu}{\hbar}\right)
\prod_{\nu\ne\alpha} I_{A}\left(\frac{\omega_\nu}{\hbar}\right) I_B\left(\frac{\omega_\alpha}{\hbar}\right)\,\right\}\notag\\
&=\frac{|A|^{2}}{2}\left\{\,\prod_{\mu\neq\alpha}\sqrt{\frac{\pi\hbar}{\omega_{\mu}}}\cdot
\frac{\hbar}{2\omega_{\alpha}}\cdot \sqrt{\frac{\pi\hbar}{\omega_{\alpha}}}
\prod_{\nu=1}\sqrt{\frac{\pi\hbar}{\omega_{\nu}}}
+\prod_{\mu=0}\sqrt{\frac{\pi\hbar}{\omega_{\mu}}}
\prod_{\nu\ne\alpha}\sqrt{\frac{\pi\hbar}{\omega_{\nu}}}\cdot \frac{\hbar}{2\omega_{\alpha}}\cdot
\sqrt{\frac{\pi\hbar}{\omega_{\alpha}}}\ \right\}\notag\\
&=\frac{\hbar}{2\omega_{\alpha}}\cdot |A|^{2} \prod_{\mu=0}^{\half(N-1)}\sqrt{\frac{\pi\hbar}{\omega_{\mu}}}\,
\prod_{\nu=1}^{\half(N-1)}\sqrt{\frac{\pi\hbar}{\omega_{\nu}}}\notag\\
&=\frac{\hbar}{2\omega_{\alpha}}\cdot |A|^{2}\sqrt{\frac{\pi\hbar}{\omega_0}}
\prod_{\mu=1}^{\half(N-1)}\frac{\pi\hbar}{\omega_{\mu}}
\tag{11}
\end{align}

他方, 遷移振幅は次である:
\begin{equation}
\BraKet{\Phi_0}{1}{\Phi_0}=\idotsint dQ_0^{\,C}\dotsb dQ_{\reverse{2}(N-1)}^{\,C}\dotsb dQ_0^{\,S}\dotsb
dQ_{\reverse{2}(N-1)}^{\,S}\ \Phi_0^{\,*}\,\Phi_0
\tag{12}
\end{equation}

これは式 (4) に於いて \(F=1\) とすればよいから,
\begin{align}
&\BraKet{\Phi_0}{1}{\Phi_0}\notag\\
&=|A|^{2}\idotsint dQ_0^{\,C}\dotsb dQ_{\reverse{2}(N-1)}^{\,C}\, dQ_1^{\,S}\dotsb dQ_{\reverse{2}(N-1)}^{\,S}\notag\\
&\times \prod_{\nu=0}^{\half(N-1)} \exp\left[ -\frac{\omega_{\nu}}{2\hbar}
\left\{\Big(Q_{\nu}^{\,C}\Big)^{2}+\Big(Q_{\nu}^{\,S}\Big)^{2}\right\}\right]
\prod_{\mu=0}^{\half(N-1)} \exp\left[ -\frac{\omega_{\mu}}{2\hbar}
\left\{\Big(Q_{\mu}^{\,C}\Big)^{2}+\Big(Q_{\mu}^{\,S}\Big)^{2}\right\}\right]\notag\\
&=|A|^{2}\,\prod_{\mu=0}^{\half(N-1)}\int dQ_{\mu}^{\,C}\,\exp\left\{-\frac{\omega_{\mu}}{\hbar}
\left(Q_{\mu}^{\,C}\right)^{2}\right\}\,\prod_{\nu=1}^{\reverse{2}(N-1)}
\int dQ_{\mu}^{\,C}\,\exp\left\{-\frac{\omega_{\nu}}{\hbar}\left(Q_{\nu}^{\,C}\right)^{2}\right\}\notag\\
&=|A|^{2}\,\prod_{\mu=0}^{\half(N-1)} I_A\left(\frac{\omega_{\mu}}{\hbar}\right)\,\prod_{\nu=1}^{\reverse{2}(N-1)}
I_A\left(\frac{\omega_{\nu}}{\hbar}\right)
=|A|^{2}\,\prod_{\mu=0}^{\half(N-1)}\sqrt{\frac{\pi\hbar}{\omega_{\mu}}}\,\prod_{\nu=1}^{\reverse{2}(N-1)}
\sqrt{\frac{\pi\hbar}{\omega_{\nu}}}\notag\\
&=|A|^{2}\sqrt{\frac{\pi\hbar}{\omega_0}} \prod_{\mu=1}^{\half(N-1)}\frac{\pi\hbar}{\omega_{\mu}}
\tag{13}
\end{align}

この結果は, ちょうど式 (11) の因子の一部になっていることが分かる.従って, 式 (11) は次のように表わせることになる:
\begin{align}
\BraKet{\Phi_0}{Q_\alpha^{\,*}Q_{\alpha}}{\Phi_0}
&=\frac{\hbar}{2\omega_{\alpha}}\cdot |A|^{2}\sqrt{\frac{\pi\hbar}{\omega_0}}
\prod_{\mu=1}^{\half(N-1)}\frac{\pi\hbar}{\omega_{\mu}}\notag\\
&=\frac{\hbar}{2\omega_{\alpha}}\BraKet{\Phi_0}{1}{\Phi_0}
\tag{14}
\end{align}

最後に \(\BraKet{\Phi_0}{Q_\alpha^{\,*}Q_\beta}{\Phi_0}\) の場合 (ただし \(\alpha\neq \beta\) ) を考えると,

\begin{align}
&\BraKet{\Phi_0}{Q_\alpha^{\,*}Q_{\beta}}{\Phi_0}\notag\\
&=|A|^{2}\idotsint dQ_0^{\,C}\dotsb dQ_{\reverse{2}(N-1)}^{\,C}\, dQ_1^{\,S}\dotsb dQ_{\reverse{2}(N-1)}^{\,S}
\prod_{\nu=0}^{\half(N-1)} \exp\left[ -\frac{\omega_{\nu}}{2\hbar}
\left\{\Big(Q_{\nu}^{\,C}\Big)^{2}+\Big(Q_{\nu}^{\,S}\Big)^{2}\right\}\right]\notag\\
&\times \,\reverse{2}\left\{Q_\alpha^{\,C}Q_\beta^{\,C}+Q_\alpha^{\,S}Q_\beta^{\,S}
+i\left(Q_\alpha^{\,S}Q_\beta^{\,C}+Q_\alpha^{\,C}Q_\beta^{\,S}\right)\right\}\,\prod_{\mu=0}^{\half(N-1)}
\exp\left[ -\frac{\omega_{\mu}}{2\hbar}\left\{\Big(Q_{\mu}^{\,C}\Big)^{2}+\Big(Q_{\mu}^{\,S}\Big)^{2}\right\}\right]
\tag{15}
\end{align}

従って,
\begin{align}
&\BraKet{\Phi_0}{Q_\alpha^{\,*}Q_{\beta}}{\Phi_0}\notag\\
&=\frac{|A|^{2}}{2}\prod_{\mu\neq\alpha,\,\beta}\int dQ_\mu^{\,C}\,
\exp\left\{-\frac{\omega_\mu}{\hbar}\left(Q_\mu^{\,C}\right)^{2}\right\}\prod_{\nu\neq\alpha,\,\beta}
\int dQ_\nu^{\,S}\,\exp\left\{-\frac{\omega_\nu}{\hbar}\left(Q_\nu^{\,S}\right)^{2}\right\}\notag\\
&\quad\times\left[\sum_{k=C,S}\int dQ_\alpha^{\,k}\,Q_\alpha^{\,k}
\exp\left\{-\frac{\omega_\alpha}{\hbar}\left(Q_\alpha^{\,k}\right)^{2}\right\}
\int dQ_\beta^{\,k}\,Q_\beta^{\,k}\exp\left\{-\frac{\omega_\beta}{\hbar}\left(Q_\beta^{\,k}\right)^{2}\right\}\right.\notag\\
&\qquad+i\int dQ_\alpha^{\,S}\,Q_\alpha^{\,S}
\exp\left\{-\frac{\omega_\alpha}{\hbar}\left(Q_\alpha^{\,S}\right)^{2}\right\}
\int dQ_\beta^{\,C}\,Q_\beta^{\,C}\exp\left\{-\frac{\omega_\beta}{\hbar}\left(Q_\beta^{\,C}\right)^{2}\right\}\notag\\
&\qquad+\left. i\int dQ_\alpha^{\,C}\,Q_\alpha^{\,C}
\exp\left\{-\frac{\omega_\alpha}{\hbar}\left(Q_\alpha^{\,C}\right)^{2}\right\}
\int dQ_\beta^{\,S}\,Q_\beta^{\,S}\exp\left\{-\frac{\omega_\beta}{\hbar}\left(Q_\beta^{\,S}\right)^{2}\right\}\right]
\tag{16}
\end{align}

このとき \([\ ]\) 中の積分項の全てが, 式(5): \(\int_{-\infty}^{\infty} x e^{-ax^{2}}\,dx=0\) の形になっている.従って全体も明らかにゼロとなるので,
\begin{equation}
\BraKet{\Phi_0}{Q_\alpha^{\,*}Q_\beta}{\Phi_0}=0
\tag{14}
\end{equation}

References

References
1 Complare this definition of expectation value with the definition of the expected value of an operator given in Sec. 5-3, particularly in Eq. (5-46).
2 【 注意】 この問題は, J.J.Sakuraiの § 2.3 (p.126 \(\sim\) ) に記してある次のことに相当している.すなわち, 調和振動子の基底状態での位置についての \(x\) , \(x^{2}\) の期待値(平均値) や運動量についての \(p\) , \(p^{2}\) の期待値(平均値) は次である:
\begin{equation}
\langle\,x\,\rangle =0,\quad \langle \,x^{2}\,\rangle =\frac{\hbar}{2m\omega},\quad
\langle \,p\,\rangle =0,\quad \langle \,p^{2}\,\rangle =\frac{\hbar m\omega}{2}
\tag{15}
\end{equation}