\(\)
Problem 7-17
Prove Eq. (7-99) :
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
\def\pdiff#1{\frac{\partial}{\partial #1}}
\def\bra#1{\langle #1 |}
\def\ket#1{| #1 \rangle}
\def\BK#1#2{\langle #1 | #2 \rangle}
\def\BraKet#1#2#3{\langle #1 | #2 | #3 \rangle}
\left\langle\,\chi\left|\,m^{2}\frac{(x_{k+1}-x_k)^{2}}{\varepsilon^{2}}\,\right|\psi\,\right\rangle
=-\frac{m\hbar}{i\varepsilon}\BraKet{\chi}{1}{\psi}+\left\langle\,\chi
\left|\,m\frac{x_{k+1}-x_k}{\varepsilon}m\frac{x_k-x_{k-1}}{\varepsilon}\,\right|\psi\,\right\rangle
\tag{7-99}
\end{equation}
using Eq. (7-40) :
\left\langle\,\ppdiff{F}{x_k}\,\right\rangle
=\frac{i\varepsilon}{\hbar}\left\langle\,F\,\left[ m\left(\frac{x_{k+1}-2x_k+x_{k-1}}{\varepsilon^{2}}\right)+V'(x_k)\right]\,\right\rangle
\tag{7-40}
\end{equation}
with \(\displaystyle{F=m\frac{x_{k+1}-x_k}{\varepsilon}}\) ．
(注意) 原書では, 式 (7-99) 右辺 第1項のマイナス符号が抜けていると思われる．それは, 式 (7-99) と本文の前節 § 7-3 の式 (7-54) との同等性からも言えることである：
\left\langle\,\frac{m}{2}\left(\frac{x_{k+1}-x_k}{\varepsilon}\right)
\left(\frac{x_k-x_{k-1}}{\varepsilon}\right)\,\right\rangle
=\left\langle\,\frac{m}{2}\left(\frac{x_{k+1}-x_k}{\varepsilon}\right)^{2}\,\right\rangle
+\frac{\hbar}{2i\varepsilon}\Bigl\langle\,1\,\Bigr\rangle
\tag{7-54}
\end{equation}
ここでは, それを修正して表示しているので注意すべし． ( 経路積分ゼミナールでは, この式 (7-54) を根拠に回答している． )
( 解答 ) 式 (7-40) に於いてポテンシャル一定, 即ち \(\displaystyle{\frac{dV}{dx_k}=0}\) の場合を書くと次である：
\begin{align*}
\left\langle\,\ppdiff{F}{x_k}\,\right\rangle
&=\frac{i\varepsilon}{\hbar}\left\langle\,F\,m\frac{x_{k+1}-2x_k+x_{k-1}}{\varepsilon^{2}}\,\right\rangle
=\frac{i\varepsilon}{\hbar}\left\langle\,F\,m\frac{(x_{k+1}-x_k)-(x_k-x_{k-1})}{\varepsilon^{2}}\,\right\rangle\\
&=\frac{i}{\hbar}\left\langle\,F\,\left\{m\frac{x_{k+1}-x_k}{\varepsilon}
-m\frac{x_k-x_{k-1}}{\varepsilon}\right\}\,\right\rangle
\tag{1}
\end{align*}
この \(F\) として \(\displaystyle{F=m\frac{x_{k+1}-x_k}{\varepsilon}}\) を代入すると, その左辺及び右辺は,
\begin{align*}
\left\langle\,\ppdiff{F}{x_k}\,\right\rangle &=\left\langle\,\pdiff{x_k}
\left(m\frac{x_{k+1}-x_k}{\varepsilon}\right)\,\right\rangle=\left\langle\,-\frac{m}{\varepsilon}\,
\right\rangle=-\frac{m}{\varepsilon}\Bigl\langle\,1\,\Bigr\rangle, \tag{2}
\end{align*}
\begin{align*}
&\frac{i}{\hbar}\left\langle\,F\,\left\{m\frac{x_{k+1}-x_k}{\varepsilon}
-m\frac{x_k-x_{k-1}}{\varepsilon}\right\}\,\right\rangle\\
&\qquad =\frac{i}{\hbar}\left\langle\left(m\frac{x_{k+1}-x_k}{\varepsilon}\right)
\left\{m\frac{x_{k+1}-x_k}{\varepsilon}-m\frac{x_k-x_{k-1}}{\varepsilon}\right\}\,\right\rangle\\
&\qquad=\frac{i}{\hbar}\left\langle\,\left(m\frac{x_{k+1}-x_k}{\varepsilon}\right)^{2}\,\right\rangle
-\frac{i}{\hbar}\left\langle\,m\frac{x_{k+1}-x_k}{\varepsilon}\,m\frac{x_k-x_{k-1}}{\varepsilon}
\,\right\rangle \tag{3}
\end{align*}
従って, 式 (2) と式 (3) が等しいとして,
-\frac{m}{\varepsilon}\Bigl\langle\,1\,\Bigr\rangle=\frac{i}{\hbar}\left\langle\,
\left(m\frac{x_{k+1}-x_k}{\varepsilon}\right)^{2}\,\right\rangle-\frac{i}{\hbar}\left\langle\,m
\frac{x_{k+1}-x_k}{\varepsilon}\,m\frac{x_k-x_{k-1}}{\varepsilon}\,\right\rangle
\tag{4}
\end{equation*}
この両辺に \(\hbar/i\) を掛け合わせて適当に移項すれば,
\begin{align*}
&\left\langle\,m^{2}\left(\frac{x_{k+1}-x_k}{\varepsilon}\right)^{2}\,\right\rangle
=-\frac{m\hbar}{i\varepsilon}\Bigl\langle\,1\,\Bigr\rangle+\left\langle\,m\frac{x_{k+1}-x_k}{\varepsilon}
\,m\frac{x_k-x_{k-1}}{\varepsilon}\,\right\rangle
\tag{5}
\end{align*}
即ち 式 (7-99) が得られる：
\begin{align*}
\left\langle\,\chi\,\left|\,m^{2}\frac{(x_{k+1}-x_k)^{2}}{\varepsilon^{2}}\,\right|\,\psi\,
\right\rangle=-\frac{m\hbar}{i\varepsilon}\BraKet{\chi}{1}{\psi}++\left\langle\,m\frac{x_{k+1}-x_k}{\varepsilon}
\,m\frac{x_k-x_{k-1}}{\varepsilon}\,\right\rangle
\tag{7-99}
\end{align*}
