\(\)
Problem 7-4
Consider the path integral
\begin{equation}
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
\int \mathscr{D}x(t)\,\ppdiff{F}{x_k}\,e^{iS[x(t)]/\hbar}
\tag{7-31}
\end{equation}
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
\int \mathscr{D}x(t)\,\ppdiff{F}{x_k}\,e^{iS[x(t)]/\hbar}
\tag{7-31}
\end{equation}
Where \(t_k\) is some intermediate time not at either end point. The path integral is simply an integral over all the points \(x_i\). So we integrate by parts to get
\begin{equation}
\int \mathscr{D}x(t)\,\ppdiff{F}{x_k}\,e^{iS[x(t)]/\hbar}=-\frac{i}{\hbar}\int \mathscr{D}x(t)\,F\ppdiff{S}{x_k}e^{iS[x(t)]/\hbar}
\tag{7-32}
\end{equation}
\int \mathscr{D}x(t)\,\ppdiff{F}{x_k}\,e^{iS[x(t)]/\hbar}=-\frac{i}{\hbar}\int \mathscr{D}x(t)\,F\ppdiff{S}{x_k}e^{iS[x(t)]/\hbar}
\tag{7-32}
\end{equation}
dropping the integrated part.
Discuss why the integrated part vanishes.
( 解答 ) 式 (7-13) の導出を参考にすると, 遷移要素 \(\displaystyle{\left\langle\ppdiff{F}{x_k}\right\rangle_{S}}\) は次のように書くことが出来る：
\begin{align}
\def\pdiff#1{\frac{\partial}{\partial#1}}
\def\mb#1{\mathbf{#1}}
\left\langle\ppdiff{F}{x_k}\right\rangle_{S}&\equiv\left\langle\chi\left|\ppdiff{F}{x_k}\right|\phi\right\rangle_{S}
=\int dx_b\int dx_a\int_{a}^{b}\mathscr{D}x(t)\,\chi^{*}(x_b)\ppdiff{F}{x_k}e^{iS/\hbar}\psi(x_a)\\
&=\int dx_b\int dx_a\int_{-\infty}^{\infty} dx_k\,\chi^{*}(x_b)K_0(b,k)\ppdiff{F}{x_k}K_0(k,a)\psi(x_a)\\
&=\int_{-\infty}^{\infty} dx_k\,\chi^{*}(k)\ppdiff{F}{x_k}\psi(x_k)
\tag{1}
\end{align}
\def\pdiff#1{\frac{\partial}{\partial#1}}
\def\mb#1{\mathbf{#1}}
\left\langle\ppdiff{F}{x_k}\right\rangle_{S}&\equiv\left\langle\chi\left|\ppdiff{F}{x_k}\right|\phi\right\rangle_{S}
=\int dx_b\int dx_a\int_{a}^{b}\mathscr{D}x(t)\,\chi^{*}(x_b)\ppdiff{F}{x_k}e^{iS/\hbar}\psi(x_a)\\
&=\int dx_b\int dx_a\int_{-\infty}^{\infty} dx_k\,\chi^{*}(x_b)K_0(b,k)\ppdiff{F}{x_k}K_0(k,a)\psi(x_a)\\
&=\int_{-\infty}^{\infty} dx_k\,\chi^{*}(k)\ppdiff{F}{x_k}\psi(x_k)
\tag{1}
\end{align}
ただし,
\begin{equation}
\chi^{*}(k)=\int_{-\infty}^{\infty} dx_b\,\chi^{*}(x_b)K_0(b,k),\quad
\psi(k)=\int_{-\infty}^{\infty} dx_a\,K_0(k,a)\,\psi(x_a)
\tag{2}
\end{equation}
\chi^{*}(k)=\int_{-\infty}^{\infty} dx_b\,\chi^{*}(x_b)K_0(b,k),\quad
\psi(k)=\int_{-\infty}^{\infty} dx_a\,K_0(k,a)\,\psi(x_a)
\tag{2}
\end{equation}
そこで, 式 (1) に於いて \(x_k\) についての部分積分を行うと,
\begin{align}
\left\langle\ppdiff{F}{x_k}\right\rangle_{S}&=\int_{-\infty}^{\infty} dx_k\,\chi^{*}(k)\ppdiff{F}{x_k}\psi(x_k)\\
&=\Bigg[\chi^{*}(x_k)\,F\,\psi(x_k)\Bigg|_{-\infty}^{\infty} -\int_{-\infty}^{\infty} dx_k\,\chi^{*}(x_k)\,F\pdiff{x_k}\psi(x_k)
\tag{3}
\end{align}
\left\langle\ppdiff{F}{x_k}\right\rangle_{S}&=\int_{-\infty}^{\infty} dx_k\,\chi^{*}(k)\ppdiff{F}{x_k}\psi(x_k)\\
&=\Bigg[\chi^{*}(x_k)\,F\,\psi(x_k)\Bigg|_{-\infty}^{\infty} -\int_{-\infty}^{\infty} dx_k\,\chi^{*}(x_k)\,F\pdiff{x_k}\psi(x_k)
\tag{3}
\end{align}
一般に「無限遠で波動関数はゼロと考える」から, \(\chi^{*}(x_k)=0,\psi(x_k)=0, \mathrm{when}\ x_k\to\pm\infty\) とする．従って, 上式 (3) 第1項の積分部分はゼロであるとして無視してよい！．
また, 上式 (3) の第2項は, 本文の式 (7-32) のようになり次式が言える：
\begin{align}
&-\int dx_k\,\chi^{*}(x_k)\,F\,\pdiff{x_k}\psi(x_k)=-\int dx_b\int dx_a\int \mathscr{D}x(t)\,\chi^{*}(x_b)\,F\,\pdiff{x_k}\,
e^{iS[x_k]/\hbar}\psi(x_a)\\
&\quad=-\int dx_b\int dx_a\int \mathscr{D}x(t)\,\chi^{*}(x_b)\,F\,\frac{i}{\hbar}\ppdiff{S}{x_k}\,e^{iS[x_k]/\hbar}\psi(x_a)\\
&\quad=-\frac{i}{\hbar}\left\langle \chi \left| F\ppdiff{S}{x_k}\right|\psi\right\rangle_{S}
=-\frac{i}{\hbar}\left\langle F\ppdiff{S}{x_k}\right\rangle_{S}
\tag{4}
\end{align}
&-\int dx_k\,\chi^{*}(x_k)\,F\,\pdiff{x_k}\psi(x_k)=-\int dx_b\int dx_a\int \mathscr{D}x(t)\,\chi^{*}(x_b)\,F\,\pdiff{x_k}\,
e^{iS[x_k]/\hbar}\psi(x_a)\\
&\quad=-\int dx_b\int dx_a\int \mathscr{D}x(t)\,\chi^{*}(x_b)\,F\,\frac{i}{\hbar}\ppdiff{S}{x_k}\,e^{iS[x_k]/\hbar}\psi(x_a)\\
&\quad=-\frac{i}{\hbar}\left\langle \chi \left| F\ppdiff{S}{x_k}\right|\psi\right\rangle_{S}
=-\frac{i}{\hbar}\left\langle F\ppdiff{S}{x_k}\right\rangle_{S}
\tag{4}
\end{align}
