問題 5-2 の解答例
Problem 5-2
If we transform only the time and not the spatial variables, defining
\begin{equation}
k(x_{2},E_{2}; x_{1},E_{1})=\int d t_{1} \int d t_{2}\, e^{+ i E_{2} t_{2}/\hbar} K(x_{2},t_{2}; x_{1},t_{1}) e^{- i E_{1} t_{1}/\hbar}
\tag{5.20}
\end{equation}
show that for a system with a time-independent hamiltonian $H$
\begin{equation}
k(x_2,E_2; x_1,E_1)=2\pi i \hbar^{2}\delta(E_2-E_1)\sum_{m}\frac{\phi_{m}(x_{b})\phi^{*}_{m}(x_{1})}{E_{1}-E_{m}+i\varepsilon}
\tag{5.21}
\end{equation}
where and $\phi_m(x)$ are the eigenvalues and eigenfunctions of $H$.
( 解答 ) 式 4.59) より $K(x_2,t_2; x_1,t_1)$ は次である： $ \def\ket#1{|#1\rangle} \def\bra#1{\langle#1|} \def\BK#1#2{\langle #1|#2\rangle} \def\BraKet#1#2#3{\langle#1|#2|#3\rangle} \def\ppdiff#1#2{\frac{\partial #1}{\partial #2}} \def\odiff#1{\frac{d}{d #1}} \def\pdiff#1{\frac{\partial}{\partial #1}} \def\Bppdiff#1#2{\frac{\partial^{2}#1}{\partial #2^{2}}} \def\Bpdiff#1{\frac{\partial^{2}}{\partial #1^{2}}} \def\mb#1{\mathbf{#1}} \def\ds#1{\mbox{${\displaystyle\strut #1}$}} \def\mfrac#1#2{\frac{#1}{#2}} $
\begin{equation}
K(x_2,t_2; x_1,t_1)=\sum_{n=0}^{\infty} \phi_{n}(x_2)\phi_{n}^{*}(x_1)\,e^{i E_{n}(t_{2}-t_{1})/\hbar} \quad \text{when}
\quad t_{2} > t_{1},\quad \text{otherwise}\quad 0
\label{eq1}
\end{equation}
これを式 (5.20) に代入すると,
\begin{align}
k(x_2,E_2;x_1,E_1)&=\int_{-\infty}^{\infty}dt_1\int_{-\infty}^{\infty}dt_{2}\,e^{i E_{2}t_{2}/\hbar} K(x_2,t_2; x_1,t_1)\,
e^{-i E_{1}t_{1}/\hbar}\notag\\
&=\int_{-\infty}^{\infty} dt_1 \int_{t_{1}}^{\infty}\, dt_{2}\,e^{i E_{2} t_{2}/\hbar}
\left(\sum_{m=1}^{\infty} \phi_{m}(x_{2})\phi_{m}^{*}(x_{1})\,e^{-i E_{m}(t_{2}-t_{1})/\hbar}\right) e^{-i E_{1}t_{1}/\hbar}
\label{eq2}
\end{align}
本文の式 (5.14) を導出したときと同様に, $t_2=t_1+\theta$ と置くと $dt_2=d\theta$ で積分範囲は $[0,\infty]$ となるから,
\begin{align}
k(x_2,E_2;x_1,E_1)&=\int_{-\infty}^{\infty} dt_1 \int_{t_{1}}^{\infty}\,dt_2\,e^{i E_{2} t_{2}/\hbar}\,e^{-i E_{1} t_{1}/\hbar}
\sum_{m=1}^{\infty} \phi_{m}(x_2)\phi_{m}^{*}(x_{1})\,e^{-i E_{m}(t_{2}-t_{1})/\hbar}\notag\\
&=\int_{-\infty}^{\infty} dt_1\int_{0}^{\infty} d\theta\ e^{i E_{2}(t_{1}+\theta)/\hbar}\,e^{-i E_{1} t_{1}/\hbar}
\sum_{m=1}^{\infty}\phi_{m}(x_{2})\phi_{m}^{*}(x_{1})\,e^{-i E_{m}\theta/\hbar}\notag\\
&=\int_{-\infty}^{\infty} dt_1\,e^{i(E_{2}-E_{1})t_{1}/\hbar}\sum_{m=1}^{\infty}\phi_{m}(x_{2})\phi_{m}^{*}(x_{1})
\int_{0}^{\infty}\,d\theta\,e^{i (E_{2}-E_{m})\theta/\hbar}
\label{eq3}
\end{align}
\begin{equation}
\delta(ax)=\frac{1}{a}\delta(x)=\frac{1}{2\pi a}\int_{-\infty}^{\infty} e^{ikx}\,dk\quad\rightarrow\quad
\int_{-\infty}^{\infty} dt_1\,e^{i(E_{2}-E_{1})t_{1}/\hbar}=2\pi\hbar\,\delta(E_{2}-E_{1})
\label{eq4}
\end{equation}
また最後の積分は式 (5.15) $\sim$ 式 (5.17) から次に書ける：
\begin{equation}
\int_{0}^{\infty} d\theta\,e^{i \omega \theta}=\frac{i}{\omega+i\varepsilon} \quad\rightarrow\quad
\int_{0}^{\infty} d\theta\,e^{i(E_{2}-E_{m})\theta/\hbar}=\frac{i}{(E_{2}-E_{m})/\hbar +i\varepsilon}
\label{eq5}
\end{equation}
従って, 式\eqref{eq3}は次となる：
\begin{align}
k(x_2,E_2;x_1,E_1)&=2\pi\hbar\,\delta(E_{2}-E_{1})\sum_{m=1}^{\infty}\phi_{m}(x_{2})\phi_{m}^{*}(x_{1})
\frac{i}{(E_{2}-E_{m})/\hbar+i\varepsilon}\notag\\
&=2\pi\hbar^{2}\,\delta(E_{2}-E_{1})\sum_{m=1}^{\infty} \phi_{m}(x_{2})\phi_{m}^{*}(x_{1})
\frac{i}{E_{2}-E_{m}+i\varepsilon^{'}}\notag\\
&=2\pi i \hbar^{2}\,\delta(E_{2}-E_{1})\,\sum_{m}
\frac{\phi_{m}(x_{2})\phi_{m}^{*}(x_{1})}{E_{2}-E_{m}+i\varepsilon^{'}},\quad \text{where}\quad\,\varepsilon^{'}=\hbar\varepsilon
\label{eq6}
\end{align}
ただし$\delta(E_2-E_1)$ があるので, 和中にある分母の $E_2$ は $E_1$ に置き換えてよい．よって次の結果式を得る：
\begin{equation}
k(x_2,E_2;x_1,E_1)=2\pi i \hbar^{2}\,\delta(E_{2}-E_{1})\,\sum_{m}
\frac{\phi_{m}(x_{2})\phi_{m}^{*}(x_{1})}{E_{1}-E_{m}+i\varepsilon^{'}}
\label{eq7}
\end{equation}
(注意) 原書では式 (5.21) の最初の因子が $2\pi\hbar$ となっているが, これは明らかにミスプリントである．校訂版では修正されている．
