\(\)
Problem 7-15
Show that the rule works for two successive momenta, that is,
\begin{align}
\Bigg\langle \chi \Bigg| m\frac{x_{k+1} – x_k}{\varepsilon}m\frac{x_k – x_{k-1}}{\varepsilon}
\Bigg| \psi \Bigg\rangle
&=\int_{-\infty}^{\infty} \chi^{*}(x,t)\,p\,p\,\psi(x,t)\,dx\notag\\
&=-\hbar^{2}\int_{-\infty}^{\infty} \chi^{*}(x,t)\frac{\partial^{2}}{\partial x^{2}} \psi(x,t)\,dx
\tag{7-97}
\end{align}
\Bigg\langle \chi \Bigg| m\frac{x_{k+1} – x_k}{\varepsilon}m\frac{x_k – x_{k-1}}{\varepsilon}
\Bigg| \psi \Bigg\rangle
&=\int_{-\infty}^{\infty} \chi^{*}(x,t)\,p\,p\,\psi(x,t)\,dx\notag\\
&=-\hbar^{2}\int_{-\infty}^{\infty} \chi^{*}(x,t)\frac{\partial^{2}}{\partial x^{2}} \psi(x,t)\,dx
\tag{7-97}
\end{align}
(注意) 原書では, 式 (7-97) の表現が次のようになっている：
\begin{align*}
\def\pdiff#1{\frac{\partial}{\partial #1}}
\def\ppdiff#1#2{\frac{\partial #1}{\partial #2}}
\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\frac{x_{k+1}-x_k}{\varepsilon}m
\frac{x_k-x_{k-1}}{\varepsilon}\,\right|\psi\,\right\rangle
&=\iint_{-\infty}^{\infty} \chi^{*}(y,t)\,p\,p\, \psi(x,t)\,dx\,dy\\
&=-\hbar^{2}\iint_{-\infty}^{\infty}\chi^{*}(y,t)\frac{\partial^{2}}{\partial x^{2}}\psi(x,t)\,dx\,dy
\end{align*}
しかし,「連続する２つの量の場合には訳書のような表現式の方が良いのではないか」と思われたため,そちらに従って書き換えたので注意すべし．
( 解答 ) 本文の式 (7-96) 及びその導出手順を利用して求めて行こう．
速やかに続けて測定される２つの量を含む式 \(G\) を考える． \(p_k\equiv m(x_{k+1}-x_k)/\varepsilon\) として,
\begin{equation}
G=\frac{m(x_{k+1} – x_k)}{\varepsilon}\cdot \frac{m(x_k – x_{k-1})}{\varepsilon}
=p_k \cdot \frac{m(x_k – x_{k-1})}{\varepsilon}
=\frac{m}{\varepsilon}p_k (x_k-x_{k-1})
\tag{1}
\end{equation}
G=\frac{m(x_{k+1} – x_k)}{\varepsilon}\cdot \frac{m(x_k – x_{k-1})}{\varepsilon}
=p_k \cdot \frac{m(x_k – x_{k-1})}{\varepsilon}
=\frac{m}{\varepsilon}p_k (x_k-x_{k-1})
\tag{1}
\end{equation}
とする．このとき式 (7-96) は次のように表わせることになる：
\begin{equation}
\left\langle\,\chi\left|\,m\frac{x_{k+1}-x_k}{\varepsilon}x_k\,\right|\psi\,\right\rangle
=\BraKet{\chi}{p_k x_k}{\psi}=\int^{\infty}_{-\infty}dx\,\chi^{*}(x,t)\,px\,\psi(x,t)
\tag{2}
\end{equation}
\left\langle\,\chi\left|\,m\frac{x_{k+1}-x_k}{\varepsilon}x_k\,\right|\psi\,\right\rangle
=\BraKet{\chi}{p_k x_k}{\psi}=\int^{\infty}_{-\infty}dx\,\chi^{*}(x,t)\,px\,\psi(x,t)
\tag{2}
\end{equation}
この式 (2) を用いると \(G\) の遷移要素要素は次となる：
\begin{align}
\BraKet{\chi}{G}{\psi}&=\frac{m}{\varepsilon}\BraKet{\chi}{p_k x_k}{\psi}
-\frac{m}{\varepsilon}\BraKet{\chi}{p_k x_{k-1}}{\psi}\notag\\
&=\frac{m}{\varepsilon}\int^{\infty}_{-\infty}dx\,\chi^{*}(x,t)\,px\,\psi(x,t)
-\frac{m}{\varepsilon}\BraKet{\chi}{p_k x_{k-1}}{\psi}
\tag{3}
\end{align}
\BraKet{\chi}{G}{\psi}&=\frac{m}{\varepsilon}\BraKet{\chi}{p_k x_k}{\psi}
-\frac{m}{\varepsilon}\BraKet{\chi}{p_k x_{k-1}}{\psi}\notag\\
&=\frac{m}{\varepsilon}\int^{\infty}_{-\infty}dx\,\chi^{*}(x,t)\,px\,\psi(x,t)
-\frac{m}{\varepsilon}\BraKet{\chi}{p_k x_{k-1}}{\psi}
\tag{3}
\end{align}
この第2項中の遷移要素 \(\BraKet{\chi}{p_k x_{k-1}}{\psi}\) を考える． \(x=x_k,y=x_{k-1},t=t_k\) とすると,
\begin{align}
\BraKet{\chi}{p_k x_{k-1}}{\psi}&=\bra{\chi}p_k\left(\int dx_k\ket{x_k}\bra{x_k}\right)
\left(\int dx_{k-1}\ket{x_{k-1}}\bra{x_{k-1}}\right)\,x_{k-1}\,\ket{\psi}\notag\\
&=\int dx_{k}\int dx_{k-1}\BK{\chi}{x_k}\,p_k\,\BK{x_k}{x_{k-1}}\,x_{k-1}\,\BK{x_{k-1}}{\psi}\notag\\
&=\int dx\int dy\,\chi^{*}(x,t)\,p(t)\,K(x,t;y,t-\varepsilon)\,y\,\psi(y,t-\varepsilon)\\
&=\int dx\,\chi^{*}(x,t)\,p(t)\int dy\,K(x,t;y,t-\varepsilon)\,y\,\psi(y,t-\varepsilon)
\tag{4}
\end{align}
\BraKet{\chi}{p_k x_{k-1}}{\psi}&=\bra{\chi}p_k\left(\int dx_k\ket{x_k}\bra{x_k}\right)
\left(\int dx_{k-1}\ket{x_{k-1}}\bra{x_{k-1}}\right)\,x_{k-1}\,\ket{\psi}\notag\\
&=\int dx_{k}\int dx_{k-1}\BK{\chi}{x_k}\,p_k\,\BK{x_k}{x_{k-1}}\,x_{k-1}\,\BK{x_{k-1}}{\psi}\notag\\
&=\int dx\int dy\,\chi^{*}(x,t)\,p(t)\,K(x,t;y,t-\varepsilon)\,y\,\psi(y,t-\varepsilon)\\
&=\int dx\,\chi^{*}(x,t)\,p(t)\int dy\,K(x,t;y,t-\varepsilon)\,y\,\psi(y,t-\varepsilon)
\tag{4}
\end{align}
そこで, 上式中にある \(dy\) 積分を考えてみる．経路積分ゼミナールを参照すると次が成立するようだ：
\begin{equation}
\int dy\,K(x,t;y,t-\varepsilon)\,y\,\psi(y,t-\varepsilon)
=x\psi(x,t)+\frac{i\varepsilon\hbar}{m}\pdiff{x}\psi(x,t)
\tag{5}
\end{equation}
\int dy\,K(x,t;y,t-\varepsilon)\,y\,\psi(y,t-\varepsilon)
=x\psi(x,t)+\frac{i\varepsilon\hbar}{m}\pdiff{x}\psi(x,t)
\tag{5}
\end{equation}
この式が成立することは, \(\tau=t-\varepsilon\) そして \(f(y,\tau)=y\psi(y,\tau)\) とし, シュレディンガー方程式から言える演算子関係：\(\displaystyle{\pdiff{t}=\frac{H}{i\hbar}}\) 及び式 (4-23)：\(\displaystyle{Hx-xH=-\frac{\hbar^{2}}{m}\pdiff{x}}\) を用いることで以下のように示すことが出来る：
\begin{align}
\int dy\,K(x,t;y,t-\varepsilon)&y\,\psi(y,t-\varepsilon)
=\int dy\,K(x,\tau+\varepsilon;y,\tau)\,y\,\psi(y,\tau)=f(x,\tau+\varepsilon),\notag\\
f(x,\tau+\varepsilon)&\approx f(x,\tau)+\varepsilon\pdiff{\tau}f(x,\tau)
=f(x,\tau)+\varepsilon\frac{H}{i\hbar}f(x,\tau)\notag\\
&=\left(1+\frac{\varepsilon H}{i\hbar}\right)f(x,\tau)
=\left(1-\frac{i\varepsilon}{\hbar}H\right)\,x\,\psi(x,\tau)\notag\\
&=x\psi(x,t-\varepsilon)-\frac{i\varepsilon}{\hbar}Hx\psi(x,t-\varepsilon)\notag\\
&=x\left\{\psi(x,t)-\varepsilon\ppdiff{\psi}{t}\right\}
-\frac{i\varepsilon}{\hbar}Hx\left\{\psi(x,t)-\varepsilon\ppdiff{\psi}{t}\right\}\notag\\
&=x\psi(x,t)-\frac{i\varepsilon}{\hbar}Hx\psi(x,t)+\frac{i\varepsilon}{\hbar}xH\psi(x,t)\notag\\
&=x\psi(x,t)-\frac{i\varepsilon}{\hbar}\big(Hx-xH\big)\psi(x,t)\notag\\
&=x\psi(x,t)-\frac{i\varepsilon}{\hbar}\left(-\frac{\hbar^{2}}{m}\pdiff{x}\right)\psi(x,t)\notag\\
&=x\psi(x,t)+\frac{i\varepsilon\hbar}{m}\pdiff{x}\psi(x,t)
\tag{5′}
\end{align}
\int dy\,K(x,t;y,t-\varepsilon)&y\,\psi(y,t-\varepsilon)
=\int dy\,K(x,\tau+\varepsilon;y,\tau)\,y\,\psi(y,\tau)=f(x,\tau+\varepsilon),\notag\\
f(x,\tau+\varepsilon)&\approx f(x,\tau)+\varepsilon\pdiff{\tau}f(x,\tau)
=f(x,\tau)+\varepsilon\frac{H}{i\hbar}f(x,\tau)\notag\\
&=\left(1+\frac{\varepsilon H}{i\hbar}\right)f(x,\tau)
=\left(1-\frac{i\varepsilon}{\hbar}H\right)\,x\,\psi(x,\tau)\notag\\
&=x\psi(x,t-\varepsilon)-\frac{i\varepsilon}{\hbar}Hx\psi(x,t-\varepsilon)\notag\\
&=x\left\{\psi(x,t)-\varepsilon\ppdiff{\psi}{t}\right\}
-\frac{i\varepsilon}{\hbar}Hx\left\{\psi(x,t)-\varepsilon\ppdiff{\psi}{t}\right\}\notag\\
&=x\psi(x,t)-\frac{i\varepsilon}{\hbar}Hx\psi(x,t)+\frac{i\varepsilon}{\hbar}xH\psi(x,t)\notag\\
&=x\psi(x,t)-\frac{i\varepsilon}{\hbar}\big(Hx-xH\big)\psi(x,t)\notag\\
&=x\psi(x,t)-\frac{i\varepsilon}{\hbar}\left(-\frac{\hbar^{2}}{m}\pdiff{x}\right)\psi(x,t)\notag\\
&=x\psi(x,t)+\frac{i\varepsilon\hbar}{m}\pdiff{x}\psi(x,t)
\tag{5′}
\end{align}
従って式 (4) は, この式 (5) を用いて次となる：
\begin{align}
\BraKet{\chi}{p_k x_{k-1}}{\psi}
&=\int dx\,\chi^{*}(x,t)\,p(t)\int dy\,K(x,t;y,t-\varepsilon)\,y\,\psi(y,t-\varepsilon)\notag\\
&=\int dx\,\chi^{*}(x,t)\,p(t)\left\{x\psi(x,t)+\frac{i\varepsilon\hbar}{m}\pdiff{x}\psi(x,t)\right\}\notag\\
&=\int dx\,\chi^{*}(x,t)\,p\,x\,\psi(x,t)+\frac{i\varepsilon\hbar}{m}
\int dx\,\chi^{*}(x,t)\,p\,\pdiff{x}\psi(x,t)
\tag{6}
\end{align}
\BraKet{\chi}{p_k x_{k-1}}{\psi}
&=\int dx\,\chi^{*}(x,t)\,p(t)\int dy\,K(x,t;y,t-\varepsilon)\,y\,\psi(y,t-\varepsilon)\notag\\
&=\int dx\,\chi^{*}(x,t)\,p(t)\left\{x\psi(x,t)+\frac{i\varepsilon\hbar}{m}\pdiff{x}\psi(x,t)\right\}\notag\\
&=\int dx\,\chi^{*}(x,t)\,p\,x\,\psi(x,t)+\frac{i\varepsilon\hbar}{m}
\int dx\,\chi^{*}(x,t)\,p\,\pdiff{x}\psi(x,t)
\tag{6}
\end{align}
この結果式 (6) を式 (3) に用いると, 求めるべき式 (7-97) が得られる：
\begin{align}
&\BraKet{\chi}{G}{\psi}
=\left\langle\,\chi\left|\,m\frac{x_{k+1}-x_k}{\varepsilon}x_k\,\right|\psi\,\right\rangle\notag\\
&\quad =\frac{m}{\varepsilon}\int^{\infty}_{-\infty}dx\,\chi^{*}(x,t)\,px\,\psi(x,t)
-\frac{m}{\varepsilon}\BraKet{\chi}{p_k x_{k-1}}{\psi}\notag\\
&\quad =\frac{m}{\varepsilon}\int^{\infty}_{-\infty}dx\,\chi^{*}(x,t)\,px\,\psi(x,t)\notag\\
&\qquad -\frac{m}{\varepsilon}\left\{\int dx\,\chi^{*}(x,t)\,p\,x\,\psi(x,t)+\frac{i\varepsilon\hbar}{m}
\int dx\,\chi^{*}(x,t)\,p\,\pdiff{x}\psi(x,t)\right\}\notag\\
&\quad =\int dx\,\chi^{*}(x,t)\,p\,\frac{\hbar}{i}\pdiff{x}\psi(x,t)\notag\\
&\quad = \int dx\,\chi^{*}(x,t)\,p\,p\,\psi(x,t)\notag\\
&\quad = \int dx\,\chi^{*}(x,t)\,\left(\frac{\hbar}{i}\pdiff{x}\right)
\left(\frac{\hbar}{i}\pdiff{x}\right)\psi(x,t)\notag\\
&\quad = -\hbar^{2}\int dx\,\chi^{*}(x,t)\,\frac{\partial^{2}}{\partial x^{2}}\psi(x,t)
\tag{7-97}
\end{align}
&\BraKet{\chi}{G}{\psi}
=\left\langle\,\chi\left|\,m\frac{x_{k+1}-x_k}{\varepsilon}x_k\,\right|\psi\,\right\rangle\notag\\
&\quad =\frac{m}{\varepsilon}\int^{\infty}_{-\infty}dx\,\chi^{*}(x,t)\,px\,\psi(x,t)
-\frac{m}{\varepsilon}\BraKet{\chi}{p_k x_{k-1}}{\psi}\notag\\
&\quad =\frac{m}{\varepsilon}\int^{\infty}_{-\infty}dx\,\chi^{*}(x,t)\,px\,\psi(x,t)\notag\\
&\qquad -\frac{m}{\varepsilon}\left\{\int dx\,\chi^{*}(x,t)\,p\,x\,\psi(x,t)+\frac{i\varepsilon\hbar}{m}
\int dx\,\chi^{*}(x,t)\,p\,\pdiff{x}\psi(x,t)\right\}\notag\\
&\quad =\int dx\,\chi^{*}(x,t)\,p\,\frac{\hbar}{i}\pdiff{x}\psi(x,t)\notag\\
&\quad = \int dx\,\chi^{*}(x,t)\,p\,p\,\psi(x,t)\notag\\
&\quad = \int dx\,\chi^{*}(x,t)\,\left(\frac{\hbar}{i}\pdiff{x}\right)
\left(\frac{\hbar}{i}\pdiff{x}\right)\psi(x,t)\notag\\
&\quad = -\hbar^{2}\int dx\,\chi^{*}(x,t)\,\frac{\partial^{2}}{\partial x^{2}}\psi(x,t)
\tag{7-97}
\end{align}
