\(\)
Problem 7-14
Show that the transition element of \(\displaystyle{m\left(\frac{x_{k+1}-x_{k}}{\varepsilon}\right)f(x_{k+1})}\) is equivalent to that of \((f\cdot p)\).
( 注意 ) 原書の問題文では「transition amplitude」となっている．しかし,ここは「transition element」であろうと思うので修正した．
( 解答 ) \((m/\varepsilon)(x_{k+1}-x_k)\) は, 式 (7-72) より時刻 \(t_k\) に於ける運動量 \(p_k\) である．また \(f(x_{k+1})\) は, 時刻 \(t_{k+1}\) に於けるオブザーバブルと考えられる．すると \((m/\varepsilon)(x_{k+1}-x_k)f(x_{k+1})\) は時間が連続した一連の量 \(p_k f(x_{k+1})\) と見做せる．従って, 問題文の前に書かれている本文の規則から,「この量の遷移要素の積分的な定義は, それに相当する演算子 \(\hat{p}\) 及び \(\hat{f}\) を時間順序に従って右から左へ書いて行けば良い」．よって \(\langle \hat{f}\hat{p}\rangle\) となる：
\begin{equation}
\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}
\left\langle \chi \left| m\frac{x_{k+1}-x_k}{\varepsilon}f(x_{k+1})\right|\psi\right\rangle
= \int \chi^{*}(x,t)\,\hat{f}\hat{p}\,\psi(x,t)\,dx=\langle \chi|\,\hat{f}\hat{p}\,|\psi\rangle
\tag{1}
\end{equation}
\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}
\left\langle \chi \left| m\frac{x_{k+1}-x_k}{\varepsilon}f(x_{k+1})\right|\psi\right\rangle
= \int \chi^{*}(x,t)\,\hat{f}\hat{p}\,\psi(x,t)\,dx=\langle \chi|\,\hat{f}\hat{p}\,|\psi\rangle
\tag{1}
\end{equation}
この結論の式 (1) を, 式 (7-96) を導いた本文の記述と同じ手順でも求めて見よう．
式 (7-91) の \(F\) の代わりに \(F=(m/\varepsilon)(x_{k+1}-x_k)f(x_{k+1})\) を用いる．この \(F\) の遷移要素は,
\begin{align}
\left\langle \chi \Big| m\left(\frac{x_{k+1}-x_k}{\varepsilon}\right)f(x_{k+1})\Big|\psi\right\rangle
&=\frac{m}{\varepsilon}\big\langle\chi \big| x_{k+1}f(x_{k+1})\big|\psi\big\rangle
-\frac{m}{\varepsilon}\big\langle\chi \big| x_{k}f(x_{k+1})\big|\psi\big\rangle\notag\\
&=\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k+1}\big|\psi\big\rangle
-\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k}\big|\psi\big\rangle
\tag{1}
\end{align}
\left\langle \chi \Big| m\left(\frac{x_{k+1}-x_k}{\varepsilon}\right)f(x_{k+1})\Big|\psi\right\rangle
&=\frac{m}{\varepsilon}\big\langle\chi \big| x_{k+1}f(x_{k+1})\big|\psi\big\rangle
-\frac{m}{\varepsilon}\big\langle\chi \big| x_{k}f(x_{k+1})\big|\psi\big\rangle\notag\\
&=\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k+1}\big|\psi\big\rangle
-\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k}\big|\psi\big\rangle
\tag{1}
\end{align}
ただし \(x_{k+1}\) と \(f(x_{k+1})\) は交換可能, また, \(x_{k}\) と \(f(x_{k+1})\) も交換可能であると見做せることを利用して,「時間順に並び替えておく」ことに注意する．
式 (1) の右辺の各項を展開して行く．まず第 1 項は, \(x=x_{k+1}\) , \(y=x_k\) , \(t=t_k\) とすると,
\begin{align}
\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k+1}\big|\psi\big\rangle
&=\frac{m}{\varepsilon}\bra{\chi}\left(\int dx_{k+1}\,\ket{x_{k+1}}\bra{x_{k+1}}\right)
f(x_{k+1})\,x_{k+1}\left(\int dx_k\,\ket{x_k}\bra{x_k}\right)\ket{\psi}\notag\\
&=\frac{m}{\varepsilon}\int dx_{k+1}\int dx_k\,\BK{\chi}{x_{k+1}}f(x_{k+1})\,x_{k+1}\BK{x_{k+1}}{x_k}
\BK{x_k}{\psi}\notag\\
&=\frac{m}{\varepsilon}\int dx\int dy\,\chi^{*}(x,t+\varepsilon)\,f(x)\,x\,
K(x,t+\varepsilon;y,t)\,\psi(y,t)
\tag{2}
\end{align}
\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k+1}\big|\psi\big\rangle
&=\frac{m}{\varepsilon}\bra{\chi}\left(\int dx_{k+1}\,\ket{x_{k+1}}\bra{x_{k+1}}\right)
f(x_{k+1})\,x_{k+1}\left(\int dx_k\,\ket{x_k}\bra{x_k}\right)\ket{\psi}\notag\\
&=\frac{m}{\varepsilon}\int dx_{k+1}\int dx_k\,\BK{\chi}{x_{k+1}}f(x_{k+1})\,x_{k+1}\BK{x_{k+1}}{x_k}
\BK{x_k}{\psi}\notag\\
&=\frac{m}{\varepsilon}\int dx\int dy\,\chi^{*}(x,t+\varepsilon)\,f(x)\,x\,
K(x,t+\varepsilon;y,t)\,\psi(y,t)
\tag{2}
\end{align}
ここで, 式 (3-42) を利用すると,
\begin{equation}
\int dy\,K(x,t+\varepsilon; y,t)\,\psi(y,t)=\psi(x,t+\varepsilon)
\tag{3}
\end{equation}
\int dy\,K(x,t+\varepsilon; y,t)\,\psi(y,t)=\psi(x,t+\varepsilon)
\tag{3}
\end{equation}
また,
\begin{equation}
\chi^{*}(x,t+\varepsilon)\approx \chi^{*}(x,t)\left(1+\frac{i\varepsilon}{\hbar}H\right),\quad
\psi(x,t+\varepsilon)\approx \left(1-\frac{i\varepsilon}{\hbar}H\right)\psi(x,t)
\tag{4}
\end{equation}
\chi^{*}(x,t+\varepsilon)\approx \chi^{*}(x,t)\left(1+\frac{i\varepsilon}{\hbar}H\right),\quad
\psi(x,t+\varepsilon)\approx \left(1-\frac{i\varepsilon}{\hbar}H\right)\psi(x,t)
\tag{4}
\end{equation}
と出来る．従って, 式 (2) は次となる：
\begin{align}
\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k+1}\big|\psi\big\rangle
&=\frac{m}{\varepsilon}\int dx\int dy\,\chi^{*}(x,t+\varepsilon)\,f(x)\,x\,
K(x,t+\varepsilon;y,t)\,\psi(y,t)\notag\\
&=\frac{m}{\varepsilon}\int dx\,\chi^{*}(x,t+\varepsilon)\,f(x)\,x\,\psi(x,t+\varepsilon)\\
&=\frac{m}{\varepsilon}\int dx\,\chi^{*}(x,t)\left(1＋\frac{i\varepsilon}{\hbar}H\right)
\,f(x)\,x\,\left(1-\frac{i\varepsilon}{\hbar}H\right)\,\psi(x,t)
\tag{5}
\end{align}
\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k+1}\big|\psi\big\rangle
&=\frac{m}{\varepsilon}\int dx\int dy\,\chi^{*}(x,t+\varepsilon)\,f(x)\,x\,
K(x,t+\varepsilon;y,t)\,\psi(y,t)\notag\\
&=\frac{m}{\varepsilon}\int dx\,\chi^{*}(x,t+\varepsilon)\,f(x)\,x\,\psi(x,t+\varepsilon)\\
&=\frac{m}{\varepsilon}\int dx\,\chi^{*}(x,t)\left(1＋\frac{i\varepsilon}{\hbar}H\right)
\,f(x)\,x\,\left(1-\frac{i\varepsilon}{\hbar}H\right)\,\psi(x,t)
\tag{5}
\end{align}
次に第 2 項を展開すると,
\begin{align}
\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k}\big|\psi\big\rangle
&=\frac{m}{\varepsilon}\bra{\chi}\left(\int dx_{k+1}\,\ket{x_{k+1}}\bra{x_{k+1}}\right)
f(x_{k+1})\left(\int dx_k\,\ket{x_k}\bra{x_k}\right)\,x_{k}\,\ket{\psi}\notag\\
&=\frac{m}{\varepsilon}\int dx_{k+1}\int dx_k\,\BK{\chi}{x_{k+1}}
f(x_{k+1})\BK{x_{k+1}}{x_k}\,x_k\, \BK{x_k}{\psi}\notag\\
&=\frac{m}{\varepsilon}\int dx_{k+1}\int dx_k\,\chi^{*}(k+1)\,f(x_{k+1})\,K(k+1,k)\,x_k\,\psi(k)\notag\\
&=\frac{m}{\varepsilon}\int dx \int dy\,\chi^{*}(x,t+\varepsilon)\,f(x)\,K(x,t+\varepsilon; y,t)\,y\,\psi(y,t)
\tag{6}
\end{align}
\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k}\big|\psi\big\rangle
&=\frac{m}{\varepsilon}\bra{\chi}\left(\int dx_{k+1}\,\ket{x_{k+1}}\bra{x_{k+1}}\right)
f(x_{k+1})\left(\int dx_k\,\ket{x_k}\bra{x_k}\right)\,x_{k}\,\ket{\psi}\notag\\
&=\frac{m}{\varepsilon}\int dx_{k+1}\int dx_k\,\BK{\chi}{x_{k+1}}
f(x_{k+1})\BK{x_{k+1}}{x_k}\,x_k\, \BK{x_k}{\psi}\notag\\
&=\frac{m}{\varepsilon}\int dx_{k+1}\int dx_k\,\chi^{*}(k+1)\,f(x_{k+1})\,K(k+1,k)\,x_k\,\psi(k)\notag\\
&=\frac{m}{\varepsilon}\int dx \int dy\,\chi^{*}(x,t+\varepsilon)\,f(x)\,K(x,t+\varepsilon; y,t)\,y\,\psi(y,t)
\tag{6}
\end{align}
このとき \(g(y,t)\equiv y\,\psi(y,t)\) とすると, 式 (3-42) を利用して次とすることが肝要である：
\begin{align}
\int dy\,K(x,t+\varepsilon; y,t)\,y\,\psi(y,t)&=\int dy\,K(x,t+\varepsilon; y,t)g(y,t)
=g(x,t+\varepsilon)\notag\\
&=\left(1-\frac{i\varepsilon}{\hbar}H\right)\,x\,\psi(x,t)
\tag{7}
\end{align}
\int dy\,K(x,t+\varepsilon; y,t)\,y\,\psi(y,t)&=\int dy\,K(x,t+\varepsilon; y,t)g(y,t)
=g(x,t+\varepsilon)\notag\\
&=\left(1-\frac{i\varepsilon}{\hbar}H\right)\,x\,\psi(x,t)
\tag{7}
\end{align}
もしこれを次のように考えてしまうと, うまく結果を導くことが出来ないようなので注意する：
\begin{align}
\int dy\,K(x,t+\varepsilon; y,t)\,y\,\psi(y,t)&=\int dy\,K(x,t+\varepsilon; y,t)g(y,t)
=g(x,t+\varepsilon)\notag\\
&=x\psi(x,t+\varepsilon)=x\,\left(1-\frac{i\varepsilon}{\hbar}H\right)\psi(x,t)
\end{align}
\int dy\,K(x,t+\varepsilon; y,t)\,y\,\psi(y,t)&=\int dy\,K(x,t+\varepsilon; y,t)g(y,t)
=g(x,t+\varepsilon)\notag\\
&=x\psi(x,t+\varepsilon)=x\,\left(1-\frac{i\varepsilon}{\hbar}H\right)\psi(x,t)
\end{align}
式 (7) を式 (6) に用いると,
\begin{align}
\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k}\big|\psi\big\rangle
&＝\frac{m}{\varepsilon}\int dx\, \chi^{*}(x,t+\varepsilon)\,f(x)\int dy\,K(x,t+\varepsilon; y,t)\,
y\,\psi(y,t)\notag\\
&=\frac{m}{\varepsilon}\int dx\,\chi^{*}(x,t)\left(1+\frac{i\varepsilon}{\hbar}H\right)\,f(x)
\left(1-\frac{i\varepsilon}{\hbar}H\right)\,x\,\psi(x,t)
\tag{8}
\end{align}
\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k}\big|\psi\big\rangle
&＝\frac{m}{\varepsilon}\int dx\, \chi^{*}(x,t+\varepsilon)\,f(x)\int dy\,K(x,t+\varepsilon; y,t)\,
y\,\psi(y,t)\notag\\
&=\frac{m}{\varepsilon}\int dx\,\chi^{*}(x,t)\left(1+\frac{i\varepsilon}{\hbar}H\right)\,f(x)
\left(1-\frac{i\varepsilon}{\hbar}H\right)\,x\,\psi(x,t)
\tag{8}
\end{align}
\(\psi(x)=\psi(x,t)\) 及び \(\chi^{*}(x,t)=\chi^{*}(x)\) と略記する．すると, 式 (1) は式 (5) と式 (8) から次となる：
\begin{align}
&\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k+1}\big|\psi\big\rangle \notag\\
&=\frac{m}{\varepsilon}\left[ \int dx\,\chi^{*}(x)\left(1+\frac{i\varepsilon}{\hbar}H\right)
\,f(x)\,x\,\left(1-\frac{i\varepsilon}{\hbar}H\right)\,\psi(x)
-\int dx\,\chi^{*}(x)\left(1+\frac{i\varepsilon}{\hbar}H\right)\,f(x)
\left(1-\frac{i\varepsilon}{\hbar}H\right)\,x\,\psi(x)\right] \notag\\
&\approx \frac{m}{\varepsilon}\left[\int dx\,\chi^{*}(x)f(x)x\psi(x)+\int dx\,\chi^{*}(x)f(x)x\left(
-\frac{i\varepsilon}{\hbar}H\right)\psi(x)
+\int dx\,\chi^{*}(x)\left(\frac{i\varepsilon}{\hbar}H\right)f(x)x\psi(x)\right. \notag\\
&\qquad\left. -\int dx\,\chi^{*}(x)f(x)x\psi(x)-\int dx\,\chi^{*}(x)f(x)
\left(-\frac{i\varepsilon}{\hbar}H\right)x\psi(x)
-\int dx\,\chi^{*}(x)\left(\frac{i\varepsilon}{\hbar}H\right)f(x)x\psi(x)\,\right]\notag\\
&=\frac{m}{\varepsilon}\cdot \left(\frac{-i\varepsilon}{\hbar}\right)\left[
\int dx\,\chi^{*}f(x)xH\psi(x)-\int dx\,\chi^{*}(x)f(x)Hx\psi(x)\right]\notag\\
&=\int dx\,\chi^{*}(x)f(x)\left[\frac{im}{\hbar}\big(Hx-xH\big)\right]\psi(x)
\tag{9}
\end{align}
&\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k+1}\big|\psi\big\rangle \notag\\
&=\frac{m}{\varepsilon}\left[ \int dx\,\chi^{*}(x)\left(1+\frac{i\varepsilon}{\hbar}H\right)
\,f(x)\,x\,\left(1-\frac{i\varepsilon}{\hbar}H\right)\,\psi(x)
-\int dx\,\chi^{*}(x)\left(1+\frac{i\varepsilon}{\hbar}H\right)\,f(x)
\left(1-\frac{i\varepsilon}{\hbar}H\right)\,x\,\psi(x)\right] \notag\\
&\approx \frac{m}{\varepsilon}\left[\int dx\,\chi^{*}(x)f(x)x\psi(x)+\int dx\,\chi^{*}(x)f(x)x\left(
-\frac{i\varepsilon}{\hbar}H\right)\psi(x)
+\int dx\,\chi^{*}(x)\left(\frac{i\varepsilon}{\hbar}H\right)f(x)x\psi(x)\right. \notag\\
&\qquad\left. -\int dx\,\chi^{*}(x)f(x)x\psi(x)-\int dx\,\chi^{*}(x)f(x)
\left(-\frac{i\varepsilon}{\hbar}H\right)x\psi(x)
-\int dx\,\chi^{*}(x)\left(\frac{i\varepsilon}{\hbar}H\right)f(x)x\psi(x)\,\right]\notag\\
&=\frac{m}{\varepsilon}\cdot \left(\frac{-i\varepsilon}{\hbar}\right)\left[
\int dx\,\chi^{*}f(x)xH\psi(x)-\int dx\,\chi^{*}(x)f(x)Hx\psi(x)\right]\notag\\
&=\int dx\,\chi^{*}(x)f(x)\left[\frac{im}{\hbar}\big(Hx-xH\big)\right]\psi(x)
\tag{9}
\end{align}
ここで, 式 (4-23) より \(\displaystyle{Hx-xH=-\frac{\hbar^{2}}{m}\pdiff{x}}\), また, \(\displaystyle{\frac{\hbar}{i}\pdiff{x}}\) は運動量演算子 \(\hat{p}\) であった．よって, 式 (9) は
\begin{equation}
\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k+1}\big|\psi\big\rangle
=\int dx\,\chi^{*}(x)f(x)\frac{\hbar}{i}\pdiff{x}\psi(x)\\
= \Big\langle \chi \Big| \hat{f}(x)\hat{p}\Big|\psi\Big\rangle
\tag{10}
\end{equation}
\frac{m}{\varepsilon}\big\langle \chi \big| f(x_{k+1})\,x_{k+1}\big|\psi\big\rangle
=\int dx\,\chi^{*}(x)f(x)\frac{\hbar}{i}\pdiff{x}\psi(x)\\
= \Big\langle \chi \Big| \hat{f}(x)\hat{p}\Big|\psi\Big\rangle
\tag{10}
\end{equation}
すなわち \((m/\varepsilon)(x_{k+1}-x_k)f(x_{k+1})\) の遷移要素は \(\hat{f}(x)\hat{p}\) の遷移要素に等価であることが示された．
