\(\)
Problem 7-6
Show, for a particle moving in three-dimentional space \(x,y,z,\)
\begin{equation}
\Big\langle \big(x_{k+1}-x_k\big)^{2} \Big\rangle=\Big\langle \big(y_{k+1}-y_k\big)^{2} \Big\rangle
=\Big\langle \big(z_{k+1}-z_k\big)^{2} \Big\rangle=-\frac{\hbar\varepsilon}{im}\langle 1 \rangle
\tag{7-50}
\end{equation}
\Big\langle \big(x_{k+1}-x_k\big)^{2} \Big\rangle=\Big\langle \big(y_{k+1}-y_k\big)^{2} \Big\rangle
=\Big\langle \big(z_{k+1}-z_k\big)^{2} \Big\rangle=-\frac{\hbar\varepsilon}{im}\langle 1 \rangle
\tag{7-50}
\end{equation}
\begin{align}
\Big\langle \big(x_{k+1}-x_k\big)\big(y_{k+1}-y_k\big) \Big\rangle
&=\Big\langle \big(z_{k+1}-z_k\big)\big(x_{k+1}-x_k\big)\Big\rangle\\
&=\Big\langle \big(y_{k+1}-y_k\big)\big(z_{k+1}-z_k\big)\Big\rangle=0
\tag{7-51}
\end{align}
\Big\langle \big(x_{k+1}-x_k\big)\big(y_{k+1}-y_k\big) \Big\rangle
&=\Big\langle \big(z_{k+1}-z_k\big)\big(x_{k+1}-x_k\big)\Big\rangle\\
&=\Big\langle \big(y_{k+1}-y_k\big)\big(z_{k+1}-z_k\big)\Big\rangle=0
\tag{7-51}
\end{align}
( 解答 ) 3次元空間に於ける式 (7-35) \(\sim\) 式 (7-38) は次となる：
\begin{align}
\def\mb#1{\mathbf{#1}}
&S[\mb{x}(t)]=\int_{t_1}^{t_2}dt\,\left\{\frac{m}{2}\dot{\mb{x}}^{2}-V(\mb{x})\right\},\quad
\delta S=-\int_{t_1}^{t_2}dt\,\left\{m\ddot{\mb{x}}+\frac{d V(\mb{x})}{dt}\right\}\delta \mb{x}(t),\\
&S=\sum_i \left[\frac{m}{2\varepsilon}\left\{(x_{i+1}-x_i)^{2}+(y_{i+1}-y_i)^{2}+(z_{i+1}-z_i)^{2}\right\}
-V(\mb{x}_i)\varepsilon\right]
\tag{1}
\end{align}
\def\mb#1{\mathbf{#1}}
&S[\mb{x}(t)]=\int_{t_1}^{t_2}dt\,\left\{\frac{m}{2}\dot{\mb{x}}^{2}-V(\mb{x})\right\},\quad
\delta S=-\int_{t_1}^{t_2}dt\,\left\{m\ddot{\mb{x}}+\frac{d V(\mb{x})}{dt}\right\}\delta \mb{x}(t),\\
&S=\sum_i \left[\frac{m}{2\varepsilon}\left\{(x_{i+1}-x_i)^{2}+(y_{i+1}-y_i)^{2}+(z_{i+1}-z_i)^{2}\right\}
-V(\mb{x}_i)\varepsilon\right]
\tag{1}
\end{align}
このとき \(\partial S/\partial x_k\)， \(\partial S/\partial y_k\)， \(\partial S/\partial z_k\) は次となる：
\begin{align}
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
\ppdiff{S}{x_k}&=-\varepsilon\left\{m\frac{x_{k+1}-2x_k+x_{k-1}}{\varepsilon^{2}}
-\ppdiff{V(\mb{x}_k)}{x_k}\right\},\\
\ppdiff{S}{y_k}&=-\varepsilon\left\{m\frac{y_{k+1}-2y_k+y_{k-1}}{\varepsilon^{2}}
-\ppdiff{V(\mb{x}_k)}{y_k}\right\},\\
\ppdiff{S}{z_k}&=-\varepsilon\left\{m\frac{z_{k+1}-2z_k+z_{k-1}}{\varepsilon^{2}}
-\ppdiff{V(\mb{x}_k)}{z_k}\right\}
\tag{2}
\end{align}
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
\ppdiff{S}{x_k}&=-\varepsilon\left\{m\frac{x_{k+1}-2x_k+x_{k-1}}{\varepsilon^{2}}
-\ppdiff{V(\mb{x}_k)}{x_k}\right\},\\
\ppdiff{S}{y_k}&=-\varepsilon\left\{m\frac{y_{k+1}-2y_k+y_{k-1}}{\varepsilon^{2}}
-\ppdiff{V(\mb{x}_k)}{y_k}\right\},\\
\ppdiff{S}{z_k}&=-\varepsilon\left\{m\frac{z_{k+1}-2z_k+z_{k-1}}{\varepsilon^{2}}
-\ppdiff{V(\mb{x}_k)}{z_k}\right\}
\tag{2}
\end{align}
そして式 (7-40) に相当する式は次となる：
\begin{align}
&\left\langle\,\ppdiff{F}{x_k}\,\right\rangle_S=-\frac{i}{\hbar}\left\langle\,F\,\ppdiff{S}{x_k}
\,\right\rangle_S=\frac{i\varepsilon}{\hbar}\left\langle\,F\,\left\{\,m\left(\frac{x_{k+1}-2x_k+x_{k-1}}
{\varepsilon^{2}}\right)+\frac{d V(\mb{x}_k)}{d x_k}\,\right\}\,\right\rangle_S\tag{3-a}\\
&\left\langle\,\ppdiff{F}{y_k}\,\right\rangle_S=-\frac{i}{\hbar}\left\langle\,F\,\ppdiff{S}{y_k}
\,\right\rangle_S=\frac{i\varepsilon}{\hbar}\left\langle\,F\,\left\{\,m\left(\frac{y_{k+1}-2y_k+y_{k-1}}
{\varepsilon^{2}}\right)+\frac{d V(\mb{x}_k)}{dy_k}\,\right\}\,\right\rangle_S\tag{3-b}\\
&\left\langle\,\ppdiff{F}{z_k}\,\right\rangle_S=-\frac{i}{\hbar}\left\langle\,F\,\ppdiff{S}{z_k}
\,\right\rangle_S=\frac{i\varepsilon}{\hbar}\left\langle\,F\,\left\{\,m\left(\frac{z_{k+1}-2z_k+z_{k-1}}
{\varepsilon^{2}}\right)+\frac{d V(\mb{x}_k)}{d z_k}\,\right\}\,\right\rangle_S
\tag{3-c}
\end{align}
&\left\langle\,\ppdiff{F}{x_k}\,\right\rangle_S=-\frac{i}{\hbar}\left\langle\,F\,\ppdiff{S}{x_k}
\,\right\rangle_S=\frac{i\varepsilon}{\hbar}\left\langle\,F\,\left\{\,m\left(\frac{x_{k+1}-2x_k+x_{k-1}}
{\varepsilon^{2}}\right)+\frac{d V(\mb{x}_k)}{d x_k}\,\right\}\,\right\rangle_S\tag{3-a}\\
&\left\langle\,\ppdiff{F}{y_k}\,\right\rangle_S=-\frac{i}{\hbar}\left\langle\,F\,\ppdiff{S}{y_k}
\,\right\rangle_S=\frac{i\varepsilon}{\hbar}\left\langle\,F\,\left\{\,m\left(\frac{y_{k+1}-2y_k+y_{k-1}}
{\varepsilon^{2}}\right)+\frac{d V(\mb{x}_k)}{dy_k}\,\right\}\,\right\rangle_S\tag{3-b}\\
&\left\langle\,\ppdiff{F}{z_k}\,\right\rangle_S=-\frac{i}{\hbar}\left\langle\,F\,\ppdiff{S}{z_k}
\,\right\rangle_S=\frac{i\varepsilon}{\hbar}\left\langle\,F\,\left\{\,m\left(\frac{z_{k+1}-2z_k+z_{k-1}}
{\varepsilon^{2}}\right)+\frac{d V(\mb{x}_k)}{d z_k}\,\right\}\,\right\rangle_S
\tag{3-c}
\end{align}
すると例えば \(F=y_k\) の場合の上式 (3-b) は, 式 (7-44) \(\sim\) 式 (7-45) と同様にして,
\begin{align}
&\left\langle\,\ppdiff{y_k}{y_k}\,\right\rangle=-\frac{i\varepsilon}{\hbar}\left\langle\,y_k
\ppdiff{S}{y_k}\,\right\rangle_S\ \rightarrow\ \bigl\langle\,1\,\bigr\rangle
=\frac{i m}{\hbar}\left\langle\,y_k\frac{y_{k+1}-2y_k+y_{k-1}}{\varepsilon}-\varepsilon
\ppdiff{V(\mb{x}_k)}{y_k}\,\right\rangle\\
&\text{therefore}\\
&\left\langle y_k\frac{y_{k+1}-y_k}{\varepsilon}\right\rangle-\left\langle y_k
\frac{y_k-y_{k-1}}{\varepsilon}\right\rangle\simeq \frac{\hbar}{im}\bigl\langle\,1\,\bigr\rangle\ \rightarrow\
\left\langle y_k\frac{y_{k+1}-y_k}{\varepsilon}\right\rangle-\left\langle y_{k+1}
\frac{y_{k+1}-y_k}{\varepsilon}\right\rangle\simeq \frac{\hbar}{im}\bigl\langle\,1\,\bigr\rangle
\tag{4}
\end{align}
&\left\langle\,\ppdiff{y_k}{y_k}\,\right\rangle=-\frac{i\varepsilon}{\hbar}\left\langle\,y_k
\ppdiff{S}{y_k}\,\right\rangle_S\ \rightarrow\ \bigl\langle\,1\,\bigr\rangle
=\frac{i m}{\hbar}\left\langle\,y_k\frac{y_{k+1}-2y_k+y_{k-1}}{\varepsilon}-\varepsilon
\ppdiff{V(\mb{x}_k)}{y_k}\,\right\rangle\\
&\text{therefore}\\
&\left\langle y_k\frac{y_{k+1}-y_k}{\varepsilon}\right\rangle-\left\langle y_k
\frac{y_k-y_{k-1}}{\varepsilon}\right\rangle\simeq \frac{\hbar}{im}\bigl\langle\,1\,\bigr\rangle\ \rightarrow\
\left\langle y_k\frac{y_{k+1}-y_k}{\varepsilon}\right\rangle-\left\langle y_{k+1}
\frac{y_{k+1}-y_k}{\varepsilon}\right\rangle\simeq \frac{\hbar}{im}\bigl\langle\,1\,\bigr\rangle
\tag{4}
\end{align}
よって, 式 (7-49) に相当する式として次式が得られる：
\begin{equation}
\left\langle\,\left(\frac{y_{k+1}-y_k}{\varepsilon}\right)^{2}\,\right\rangle=-\frac{\hbar}{im\varepsilon}
\bigl\langle\,1\,\bigr\rangle,\quad\rightarrow\
\bigl\langle\,(y_{k+1}-y_k)^{2}\,\bigr\rangle=-\frac{\hbar\varepsilon}{im}\bigl\langle\,1\,\bigr\rangle
\tag{5}
\end{equation}
\left\langle\,\left(\frac{y_{k+1}-y_k}{\varepsilon}\right)^{2}\,\right\rangle=-\frac{\hbar}{im\varepsilon}
\bigl\langle\,1\,\bigr\rangle,\quad\rightarrow\
\bigl\langle\,(y_{k+1}-y_k)^{2}\,\bigr\rangle=-\frac{\hbar\varepsilon}{im}\bigl\langle\,1\,\bigr\rangle
\tag{5}
\end{equation}
同様にして \(F=x_k\)， \(F=z_k\) の場合を考えれば, 明らかに
\begin{equation}
\bigl\langle\,(x_{k+1}-x_k)^{2}\,\bigr\rangle=-\frac{\hbar\varepsilon}{im}\bigl\langle\,1\,\bigr\rangle,\quad
\bigl\langle\,(z_{k+1}-z_k)^{2}\,\bigr\rangle=-\frac{\hbar\varepsilon}{im}\bigl\langle\,1\,\bigr\rangle
\tag{6}
\end{equation}
\bigl\langle\,(x_{k+1}-x_k)^{2}\,\bigr\rangle=-\frac{\hbar\varepsilon}{im}\bigl\langle\,1\,\bigr\rangle,\quad
\bigl\langle\,(z_{k+1}-z_k)^{2}\,\bigr\rangle=-\frac{\hbar\varepsilon}{im}\bigl\langle\,1\,\bigr\rangle
\tag{6}
\end{equation}
よって, 式 (7-50) が言える．
また, 例えば \(F=y_k\) とした場合の上式 (3-c) を考えると,
\begin{align}
\left\langle\,\ppdiff{y_k}{z_k}\,\right\rangle_S=0 &=\frac{i\varepsilon}{\hbar}\left\langle\,y_k\left\{m
\left(\frac{z_{k+1}-2z_k+z_{k-1}}{\varepsilon^{2}}\right)+\ppdiff{V(\mb{x}_k)}{z_k}\right\}\,\right\rangle_S\\
&=\frac{im}{\hbar}\left\langle\,y_k\left(\frac{z_{k+1}-z_k}{\varepsilon}-\frac{z_k-z_{k-1}}{\varepsilon}
\right)+\varepsilon\frac{y_k}{m}\ppdiff{V}{z_k}\,\right\rangle_S\\
\rightarrow\quad&\left\langle\,y_k\left(\frac{z_{k+1}-z_k}{\varepsilon}\right)-y_k\left(\frac{z_k-z_{k-1}}{\varepsilon}\right)\,\right\rangle_S\simeq 0
\tag{7}
\end{align}
\left\langle\,\ppdiff{y_k}{z_k}\,\right\rangle_S=0 &=\frac{i\varepsilon}{\hbar}\left\langle\,y_k\left\{m
\left(\frac{z_{k+1}-2z_k+z_{k-1}}{\varepsilon^{2}}\right)+\ppdiff{V(\mb{x}_k)}{z_k}\right\}\,\right\rangle_S\\
&=\frac{im}{\hbar}\left\langle\,y_k\left(\frac{z_{k+1}-z_k}{\varepsilon}-\frac{z_k-z_{k-1}}{\varepsilon}
\right)+\varepsilon\frac{y_k}{m}\ppdiff{V}{z_k}\,\right\rangle_S\\
\rightarrow\quad&\left\langle\,y_k\left(\frac{z_{k+1}-z_k}{\varepsilon}\right)-y_k\left(\frac{z_k-z_{k-1}}{\varepsilon}\right)\,\right\rangle_S\simeq 0
\tag{7}
\end{align}
このとき, 式 (7-46) と式 (7-47) の同等性と同じ論理を用いるならば,
\begin{align}
&\left\langle\,y_k\left(\frac{z_{k+1}-z_k}{\varepsilon}\right)-y_k\left(\frac{z_k-z_{k-1}}{\varepsilon}
\right)\,\right\rangle
=\left\langle\,y_k\left(\frac{z_{k+1}-z_k}{\varepsilon}\right)-y_{k+1}\left(\frac{z_{k+1}-z_k}{\varepsilon}
\right)\,\right\rangle=0\\
&\rightarrow\ \left\langle\,(y_k-y_{k+1})\left(\frac{z_{k+1}-z_k}{\varepsilon}\right)\,\right\rangle
=-\frac{1}{\varepsilon}\bigl\langle\,(y_{k+1}-y_k)(z_{k+1}-z_k)\,\bigr\rangle=0
\tag{8}
\end{align}
&\left\langle\,y_k\left(\frac{z_{k+1}-z_k}{\varepsilon}\right)-y_k\left(\frac{z_k-z_{k-1}}{\varepsilon}
\right)\,\right\rangle
=\left\langle\,y_k\left(\frac{z_{k+1}-z_k}{\varepsilon}\right)-y_{k+1}\left(\frac{z_{k+1}-z_k}{\varepsilon}
\right)\,\right\rangle=0\\
&\rightarrow\ \left\langle\,(y_k-y_{k+1})\left(\frac{z_{k+1}-z_k}{\varepsilon}\right)\,\right\rangle
=-\frac{1}{\varepsilon}\bigl\langle\,(y_{k+1}-y_k)(z_{k+1}-z_k)\,\bigr\rangle=0
\tag{8}
\end{align}
よって,
\begin{equation}
\bigl\langle\,(y_{k+1}-y_k)(z_{k+1}-z_k)\,\bigr\rangle=0
\tag{9}
\end{equation}
\bigl\langle\,(y_{k+1}-y_k)(z_{k+1}-z_k)\,\bigr\rangle=0
\tag{9}
\end{equation}
同様に \(F=z_k\) とした場合の式(3-1)や， \(F=x_k\) とした場合の上式 (3-b) を考えることで類似的な結果が求まる：
\begin{equation}
\bigl\langle\,(z_{k+1}-z_k)(x_{k+1}-x_k)\,\bigr\rangle=0,\quad
\bigl\langle\,(x_{k+1}-x_k)(y_{k+1}-y_k)\,\bigr\rangle=0
\tag{10}
\end{equation}
\bigl\langle\,(z_{k+1}-z_k)(x_{k+1}-x_k)\,\bigr\rangle=0,\quad
\bigl\langle\,(x_{k+1}-x_k)(y_{k+1}-y_k)\,\bigr\rangle=0
\tag{10}
\end{equation}
よって, 式 (7-51) が言える．
