\(\)
Problem 7-9
Use this result Eq. (7-68) to show that if \(S\) corresponds to a harmonic oscillator
\begin{equation}
S=\frac{m}{2}\int_{t_1}^{t_2} \big[ \dot{x}^{2} -\omega^{2}x^{2} \big]\,dt
\tag{1}
\end{equation}
S=\frac{m}{2}\int_{t_1}^{t_2} \big[ \dot{x}^{2} -\omega^{2}x^{2} \big]\,dt
\tag{1}
\end{equation}
then
\begin{align}
&\left\langle \exp\left\{ \frac{i}{\hbar}\int_{t_1}^{t_2} f(t)x(t)\,dt\right\}\right\rangle\\
&\quad =\big\langle 1 \big\rangle \exp\bigg\{\frac{i}{\hbar}\frac{m\omega}{2\sin\omega(t_2-t_1)}
\bigg[ \frac{2x_2}{mw}\int_{t_1}^{t_2} f(t)\sin\omega(t-t_1)\,dt \\
&\qquad\qquad +\frac{2x_1}{m\omega}\int_{t_1}^{t_2} f(t)\sin\omega(t_2-t)\,dt\\
&\qquad\qquad -\frac{2}{m^{2}\omega^{2}}\int_{t_1}^{t_2}dt\int_{t_1}^{t} f(t)f(s)\sin\omega(t_2-t)\sin\omega(s-t_1)\,ds \bigg]\bigg\}
\tag{2}
\end{align}
&\left\langle \exp\left\{ \frac{i}{\hbar}\int_{t_1}^{t_2} f(t)x(t)\,dt\right\}\right\rangle\\
&\quad =\big\langle 1 \big\rangle \exp\bigg\{\frac{i}{\hbar}\frac{m\omega}{2\sin\omega(t_2-t_1)}
\bigg[ \frac{2x_2}{mw}\int_{t_1}^{t_2} f(t)\sin\omega(t-t_1)\,dt \\
&\qquad\qquad +\frac{2x_1}{m\omega}\int_{t_1}^{t_2} f(t)\sin\omega(t_2-t)\,dt\\
&\qquad\qquad -\frac{2}{m^{2}\omega^{2}}\int_{t_1}^{t_2}dt\int_{t_1}^{t} f(t)f(s)\sin\omega(t_2-t)\sin\omega(s-t_1)\,ds \bigg]\bigg\}
\tag{2}
\end{align}
where \(x_1\) and \(x_2\) are the initial and final coordinates of the oscillator.
( 解答 ) 調和振動子の古典的経路に対する作用 \(S_{cl}\) は, 問題 2-2 中の式 (2-9) で与えれている：
\begin{equation}
S_{cl}=\frac{m\omega}{2\sin\omega T}\left[(x_2^{2}+x_1^{2})\cos\omega T-2x_2x_1\right]
\tag{3}
\end{equation}
S_{cl}=\frac{m\omega}{2\sin\omega T}\left[(x_2^{2}+x_1^{2})\cos\omega T-2x_2x_1\right]
\tag{3}
\end{equation}
ただし \(T=t_2-t_1\) である． また, 外力が働いている調和振動子の古典的経路に対する作用 \(S’_{cl}\) は, 問題 3-11 中の式 (3-66) で与えれている：
\begin{align}
S’_{cl}=\frac{m\omega}{2\sin\omega T}&\bigg[(x_2^{2}+x_1^{2})\cos\omega T-2x_2x_1\\
&+\frac{2x_2}{m\omega}\int_{t_1}^{t_2}dt\,f(t)\sin\omega(t-t_1)\\
&+\frac{2x_1}{m\omega}\int_{t_1}^{t_2}dt\,f(t)\sin\omega(t_2-t)\\
&-\frac{2}{m^{2}\omega^{2}}\int_{t_1}^{t_2}dt\int_{t_1}^{t}ds\,f(t)f(s)\sin\omega(t_2-t)\sin\omega(s-t_1)\bigg]
\tag{4}
\end{align}
S’_{cl}=\frac{m\omega}{2\sin\omega T}&\bigg[(x_2^{2}+x_1^{2})\cos\omega T-2x_2x_1\\
&+\frac{2x_2}{m\omega}\int_{t_1}^{t_2}dt\,f(t)\sin\omega(t-t_1)\\
&+\frac{2x_1}{m\omega}\int_{t_1}^{t_2}dt\,f(t)\sin\omega(t_2-t)\\
&-\frac{2}{m^{2}\omega^{2}}\int_{t_1}^{t_2}dt\int_{t_1}^{t}ds\,f(t)f(s)\sin\omega(t_2-t)\sin\omega(s-t_1)\bigg]
\tag{4}
\end{align}
ただし表記を \(a\to1,\,b\to 2\) に変更した． 式 (4) から式 (3) を引き算することで \((i/\hbar)\{S’_{cl}-S_{cl}\}\) を求めると,
\begin{align}
\frac{i}{\hbar}\Big\{S’_{cl}-S_{cl}\Big\} &=\frac{i}{\hbar}\frac{m\omega}{2\sin\omega T}
\bigg[\frac{2x_2}{m\omega}\int_{t_1}^{t_2}dt\,f(t)\sin\omega(t-t_1)\\
&\qquad +\frac{2x_1}{m\omega}\int_{t_1}^{t_2}dt\,f(t)\sin\omega(t_2-t)\\
&\qquad -\frac{2}{m^{2}\omega^{2}}\int_{t_1}^{t_2}dt\int_{t_1}^{t}ds\,f(t)f(s)\sin\omega(t_2-t)\sin\omega(s-t_1)\bigg]
\tag{5}
\end{align}
\frac{i}{\hbar}\Big\{S’_{cl}-S_{cl}\Big\} &=\frac{i}{\hbar}\frac{m\omega}{2\sin\omega T}
\bigg[\frac{2x_2}{m\omega}\int_{t_1}^{t_2}dt\,f(t)\sin\omega(t-t_1)\\
&\qquad +\frac{2x_1}{m\omega}\int_{t_1}^{t_2}dt\,f(t)\sin\omega(t_2-t)\\
&\qquad -\frac{2}{m^{2}\omega^{2}}\int_{t_1}^{t_2}dt\int_{t_1}^{t}ds\,f(t)f(s)\sin\omega(t_2-t)\sin\omega(s-t_1)\bigg]
\tag{5}
\end{align}
この結果式 (5) を 式 (7-68) に代入すれば, 題意の式 (2) が得られる：
\begin{align}
&\left\langle\,\exp\left\{\frac{i}{\hbar}\int f(t)x(t)\,dt\right\}\,\right\rangle
=\big\langle\,1\,\big\rangle\exp\left\{\frac{i}{\hbar}(S^{‘}_{cl}-S_{cl})\right\}\tag{7-68}\\
&\quad =\big\langle 1 \big\rangle \exp\bigg\{\frac{i}{\hbar}\frac{m\omega}{2\sin\omega(t_2-t_1)}
\bigg[ \frac{2x_2}{mw}\int_{t_1}^{t_2} f(t)\sin\omega(t-t_1)\,dt \\
&\qquad\qquad +\frac{2x_1}{m\omega}\int_{t_1}^{t_2} f(t)\sin\omega(t_2-t)\,dt\\
&\qquad\qquad -\frac{2}{m^{2}\omega^{2}}\int_{t_1}^{t_2}dt\int_{t_1}^{t} f(t)f(s)\sin\omega(t_2-t)\sin\omega(s-t_1)\,ds \bigg]\bigg\}
\tag{6}
\end{align}
&\left\langle\,\exp\left\{\frac{i}{\hbar}\int f(t)x(t)\,dt\right\}\,\right\rangle
=\big\langle\,1\,\big\rangle\exp\left\{\frac{i}{\hbar}(S^{‘}_{cl}-S_{cl})\right\}\tag{7-68}\\
&\quad =\big\langle 1 \big\rangle \exp\bigg\{\frac{i}{\hbar}\frac{m\omega}{2\sin\omega(t_2-t_1)}
\bigg[ \frac{2x_2}{mw}\int_{t_1}^{t_2} f(t)\sin\omega(t-t_1)\,dt \\
&\qquad\qquad +\frac{2x_1}{m\omega}\int_{t_1}^{t_2} f(t)\sin\omega(t_2-t)\,dt\\
&\qquad\qquad -\frac{2}{m^{2}\omega^{2}}\int_{t_1}^{t_2}dt\int_{t_1}^{t} f(t)f(s)\sin\omega(t_2-t)\sin\omega(s-t_1)\,ds \bigg]\bigg\}
\tag{6}
\end{align}
ただし \(T=t_2-t_1\) である．
