\(\)
Problem 6-3
For a free particle, Eq. (4-29) reduces to
\begin{equation}
\def\pdiff#1{\frac{\partial}{\partial #1}}
\def\Bpdiff#1{\frac{\partial^{2}}{\partial #1^{2}}}
-\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)+\frac{\hbar^{2}}{2m}\Bpdiff{x_b} K_0(b,a) =i\hbar\delta(t_b-t_a)\,\delta(x_b-x_a)
\tag{6-20}
\end{equation}
\def\pdiff#1{\frac{\partial}{\partial #1}}
\def\Bpdiff#1{\frac{\partial^{2}}{\partial #1^{2}}}
-\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)+\frac{\hbar^{2}}{2m}\Bpdiff{x_b} K_0(b,a) =i\hbar\delta(t_b-t_a)\,\delta(x_b-x_a)
\tag{6-20}
\end{equation}
Show, from this result and Eq. (6-19), that the kernel \(K_V\) satisfies the differential equation
\begin{equation}
-\frac{\hbar}{i}\pdiff{t_b}K_V(b,a)+\frac{\hbar^{2}}{2m}\Bpdiff{x_b} K_V(b,a)-V(b)K_V(b,a)
=i\hbar\delta(t_b-t_a)\,\delta(x_b-x_a)
\tag{6-21}
\end{equation}
-\frac{\hbar}{i}\pdiff{t_b}K_V(b,a)+\frac{\hbar^{2}}{2m}\Bpdiff{x_b} K_V(b,a)-V(b)K_V(b,a)
=i\hbar\delta(t_b-t_a)\,\delta(x_b-x_a)
\tag{6-21}
\end{equation}
(解答) 原書及び訳本では式 (6-21) の左辺第 3 項の符号が \(+\) となっているが, これは明らかにミスプリと思われるので修正しておく (校訂版では, 両辺を \(i\hbar\) で割った形に書き換えてある)．
\(K_V(b,a)\) は式 (6-19) から \(d\tau_c=dx_c dt_c\) として次であった：
\begin{equation}
K_V(b,a)=K_0(b,a)-\frac{i}{\hbar}\int d\tau_c\,K_0(b,c)\,V(c)\,K_V(c,a)
\tag{1}
\end{equation}
K_V(b,a)=K_0(b,a)-\frac{i}{\hbar}\int d\tau_c\,K_0(b,c)\,V(c)\,K_V(c,a)
\tag{1}
\end{equation}
この式 (1) の両辺に演算子 \((-\hbar/i)\partial/\partial t_b\) を作用させ, 問題文中の式 (6-20) 及び デルタ関数の性質を用いると,
\begin{align}
&-\frac{\hbar}{i}\pdiff{t_b}K_V(b,a)=-\frac{\hbar}{i}\pdiff{t_b} K_0(b,a)-\frac{i}{\hbar}
\int_{t_a}^{t_b}dt_c\int_{-\infty}^{\infty}dx_c\,\left\{-\frac{\hbar}{i}\pdiff{t_b}K_0(b,c)\right\}\,V(x_c,t_c)\,K_V(c,a)\\
&=-\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)\\
&\quad -\frac{i}{\hbar}\int_{t_a}^{t_b}dt_c\int_{-\infty}^{\infty}dx_c\,\left[-\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)
+i\hbar\,\delta(x_b-x_c)\,\delta(t_b-t_c)\right]\,V(c)\,K_V(c,a)\\
&=-\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)+\frac{i}{\hbar}\int_{t_a}^{t_b}dt_c\int_{-\infty}^{\infty}dx_c\,
\left\{\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)\right\}\,V(c)\,K_V(c,a)\\
&\quad+\int_{t_a}^{t_b}dt_c\int_{-\infty}^{\infty}dx_c\,V(x_c,t_c)\,K_V(c,a)\,\,\delta(t_b-t_c)\delta(x_b-x_c)\\
&=-\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)
+\frac{i}{\hbar}\int d\theta_c\,\left\{\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)\right\}\,V(c)\,K_V(c,a)+V(b)K_V(b,a)
\tag{2}
\end{align}
&-\frac{\hbar}{i}\pdiff{t_b}K_V(b,a)=-\frac{\hbar}{i}\pdiff{t_b} K_0(b,a)-\frac{i}{\hbar}
\int_{t_a}^{t_b}dt_c\int_{-\infty}^{\infty}dx_c\,\left\{-\frac{\hbar}{i}\pdiff{t_b}K_0(b,c)\right\}\,V(x_c,t_c)\,K_V(c,a)\\
&=-\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)\\
&\quad -\frac{i}{\hbar}\int_{t_a}^{t_b}dt_c\int_{-\infty}^{\infty}dx_c\,\left[-\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)
+i\hbar\,\delta(x_b-x_c)\,\delta(t_b-t_c)\right]\,V(c)\,K_V(c,a)\\
&=-\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)+\frac{i}{\hbar}\int_{t_a}^{t_b}dt_c\int_{-\infty}^{\infty}dx_c\,
\left\{\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)\right\}\,V(c)\,K_V(c,a)\\
&\quad+\int_{t_a}^{t_b}dt_c\int_{-\infty}^{\infty}dx_c\,V(x_c,t_c)\,K_V(c,a)\,\,\delta(t_b-t_c)\delta(x_b-x_c)\\
&=-\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)
+\frac{i}{\hbar}\int d\theta_c\,\left\{\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)\right\}\,V(c)\,K_V(c,a)+V(b)K_V(b,a)
\tag{2}
\end{align}
次に式 (1) の両辺に演算子 \((\hbar^{2}/2m)\partial^{2}/\partial x_b^{2}\) を作用させると,
\begin{equation}
\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_V(b,a)=\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,a)
-\frac{i}{\hbar}\int d\tau_c\,\left\{\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)\right\}\,V(c)\,K_V(c,a)
\tag{3}
\end{equation}
\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_V(b,a)=\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,a)
-\frac{i}{\hbar}\int d\tau_c\,\left\{\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)\right\}\,V(c)\,K_V(c,a)
\tag{3}
\end{equation}
以上の結果式 (2) と式 (3) とを足し合わせると,
\begin{align}
&-\frac{\hbar}{i}\pdiff{t_b}K_V(b,a)+\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_V(b,a)\\
&=-\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)\\
&\quad +\frac{i}{\hbar}\int d\theta_c\,\left\{\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)\right\}\,V(c)\,K_V(c,a)+V(b)K_V(b,a)\\
&\quad +\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,a)
-\frac{i}{\hbar}\int d\theta_c\,\left\{\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)\right\}\,V(c)\,K_V(c,a)\\
&= -\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)+\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,a)+V(b)K_V(b,a)\\
&=i\hbar\,\delta(t_b-t_a)\,\delta(x_b-x_a)+V(b)K_V(b,a)
\tag{4}
\end{align}
&-\frac{\hbar}{i}\pdiff{t_b}K_V(b,a)+\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_V(b,a)\\
&=-\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)\\
&\quad +\frac{i}{\hbar}\int d\theta_c\,\left\{\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)\right\}\,V(c)\,K_V(c,a)+V(b)K_V(b,a)\\
&\quad +\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,a)
-\frac{i}{\hbar}\int d\theta_c\,\left\{\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,c)\right\}\,V(c)\,K_V(c,a)\\
&= -\frac{\hbar}{i}\pdiff{t_b}K_0(b,a)+\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_0(b,a)+V(b)K_V(b,a)\\
&=i\hbar\,\delta(t_b-t_a)\,\delta(x_b-x_a)+V(b)K_V(b,a)
\tag{4}
\end{align}
ただし, 最後の式変形は問題文の式 (6-20) を利用している．以上から,
\begin{equation}
-\frac{\hbar}{i}\pdiff{t_b}K_V(b,a)+\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_V(b,a)
=i\hbar\,\delta(t_b-t_a)\,\delta(x_b-x_a)+V(b)K_V(b,a)
\tag{5}
\end{equation}
-\frac{\hbar}{i}\pdiff{t_b}K_V(b,a)+\frac{\hbar^{2}}{2m}\Bpdiff{x_b}K_V(b,a)
=i\hbar\,\delta(t_b-t_a)\,\delta(x_b-x_a)+V(b)K_V(b,a)
\tag{5}
\end{equation}
この結果式 (5) は, 問題で提示すべき式 (6-21) に等価である．
