問題 3-3 の解答例
Problem 3-3
By squaring the amplitude given in Eq. (3.20) and then integrating over , show that the probability of passage through the original sharp-edged slit is
\begin{equation}
P(\text{going through})=\frac{m}{2\pi\hbar T} 2b
\tag{3.35}
\end{equation}
In the course of this problem the integral
\begin{equation}
\int_{-\infty}^{\infty} e^{i ax}\,dx=2\pi\,\delta(a)
\tag{3.36}
\end{equation}
will appear. This is the integral representation of the Dirac delta function of $a$.
Thus the quantum-mechanical results agree with the idea that the probability that a particle goes through a slit is equal to the probability that the particle arrives at the slit.
(解答) 式(3.20) は $\displaystyle{\alpha=\frac{m}{2\hbar\theta}}$, $\displaystyle{\beta=\frac{m}{2\hbar T}}$ とすると次に書ける： $ \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{${\displaystyle\strut #1}$}} $
\begin{align}
\psi(x)&=\int_{b}^{b} d y\,K(x+x_{0}; x_{0}+y)\,K(x_{0}+y; 0)\tag{1}\\
&=\int_{-b}^{b} d y\,\left(\frac{m}{2\pi i\hbar\theta}\right)^{1/2}\exp\left\{\frac{i m (x-y)^{2}}{2\hbar\theta}\right\}
\left(\frac{m}{2\pi i \hbar T}\right)^{1/2}\exp\left\{\frac{i m(x_{0}+y)^{2}}{2\hbar T}\right\}\notag\\
&=\frac{m}{2\pi i\hbar\sqrt{\theta T}}\int_{-b}^{b}\,e^{i \alpha (y-x)^{2}}\,e^{i \beta (y+x_0)^{2}}\,d y\notag\\
&=\frac{m}{2\pi i \hbar \sqrt{\theta T}}\,e^{i(\alpha x^{2}+\beta x^{2})} \int_{-b}^{b}\, e^{i\{(\alpha+\beta)y^{2}+2\beta x_{0} y-2\alpha x y\}}\,dy
\tag{2}
\end{align}
の絶対値の2乗を考えると, 位相部分は $|e^{i\gamma}|^{2}=1$なので除ける．よって次となる：
\begin{align}
\left|\psi(x)\right|^{2}&=\psi^{*}(x)\psi(x)]\notag\\
&=\left|\frac{m}{2\pi i\hbar\sqrt{\theta T}}\right|^{2}\int_{-b}^{b}
e^{-i\{(\alpha+\beta)y'^{2}+2\beta x_{0}y'-2\alpha x y' \} }\, dy' \int_{-b}^{b}\,e^{ i \{(\alpha+\beta)y^{2}+2\beta x_{0} y-2\alpha x y\} }\,dy\notag\\
&=\frac{m^{2}}{4\pi^{2}\hbar^{2}\theta T} \int_{-b}^{b} dy' \int_{-b}^{b} dy\ e^{-i \{(\alpha+\beta)(y'^{2}-y^{2})
+2\beta x_{0} (y'-y)\} }\, e^{i 2\alpha (y'-y)x}
\tag{3}
\end{align}
よってスリットを通り抜ける確率 $P$ は, 式(3.36)を利用し, また, デルタ関数の性質 $$ \delta(ax)=\frac{1}{a}\delta(x) $$ に注意して次となる：
\begin{align}
P &=\int_{-\infty}^{\infty} |\psi(x)|^{2}\,dx=\frac{m^{2}}{4\pi^{2}\hbar^{2}\theta T}\int_{-b}^{b} dy' \int_{-b}^{b} dy\,
e^{-i\{ (\alpha+\beta)(y'^{2}-y^{2})+2\beta x_{0}(y'-y)\} } \int_{-\infty}^{\infty}\, e^{i2\alpha (y'-y)x}\,dx
\notag\\
&=\frac{m^{2}}{4\pi^{2}\hbar^{2}\theta T} \int_{-b}^{b} dy \int_{-b}^{b} dy'\,e^{-i\{ (\alpha+\beta)(y'^{2}-y^{2})
+2\beta x_{0} (y'-y)\} }\cdot\frac{1}{2\alpha}\cdot 2\pi\,\delta(y'-y)
\notag\\
&=\frac{m^{2}}{4\pi^{2}\hbar^{2}\theta T}\cdot\frac{\pi}{\alpha}\int_{-b}^{b}dy\int_{-b}^{b}dy'\,e^{-i\{(\alpha+\beta)(y'^{2}-y^{2})+2\beta x_{0}(y'-y)\} }\,\delta(y'-y)
\tag{4}
\end{align}
$dy'$ 積分を実行すると, デルタ関数の性質から $y'=y$ のときだけが残る．従って指数関数部分は $1$ となる．また $\displaystyle{\alpha\theta=\frac{m}{2\hbar}}$ である．以上から, 確率 $P$ として次が得られる：
\begin{equation}
P=\frac{m^{2}}{4\pi\hbar^{2}T}\cdot\frac{1}{\alpha\theta} \int_{-b}^{b}\,dy=\frac{m^{2}}{4\pi\hbar^{2}T}\cdot
\frac{2\hbar}{m}\cdot 2b=\frac{m}{2\pi\hbar T}\cdot 2b
\tag{5}
\end{equation}
式(1)の被積分関数内の核 $K(x_0+y;0)$ の絶対値の2乗は「粒子がスリットの中の点 $x_0+y$ に到達する確率分布関数」と見做すことが出来る．それを表現したのが本文中の式(3.34)である：
\begin{equation}
P(x_0+y)\,dy=|K(x_0+y;0)|^{2}\,dy=\frac{m}{2\pi \hbar T}\,dy
\tag{3.34}
\end{equation}
これにスリット幅 $2b$ を掛け合わせたものは「スリット内に粒子が到達する全確率」である：
\begin{equation}
\int_{-b}^{b}P(x_0+y)\,dy=\int_{-b}^{b}\frac{m}{2\pi \hbar T}\,dy=\frac{m}{2\pi \hbar T}\,2b
\tag{6}
\end{equation}
この結果が, ちょうどこの問題の解答(5)すなわち「粒子が元のスリットを通り抜ける確率」に一致していることに注意しよう!!．
