問題 3-5 の解答例

Problem 3-5

Use the results of Prob. 3-2 and Eq. (3.42) to show that the wave function of a free particle satisfies the equation $ \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{equation} -\frac{\hbar}{i}\frac{\partial \psi}{\partial t}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2} \psi}{\partial x^{2}} \tag{3.43} \end{equation}

which is the Schrödinger equation for a free particle.

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(解答) 問題 3-2 の式3.18)と式(3.42)を用い，$t_b\to t$，$x_b\to x$，$x_3\to x^{'}$，$t_3\to t^{'}$とおくと次が得られる：

\begin{align} &-\frac{\hbar}{i} \frac{\partial}{\partial t} K=-\frac{\hbar^{2}}{2m} \frac{\partial^{2}}{\partial x^{2}} K,\tag{1}\\ &\psi(x,t)=\int_{-\infty}^{\infty}dx^{'}\,K(x,t;x^{'},t^{'})\,\psi(x^{'},t^{'})\tag{2} \end{align}

式(2)に $\displaystyle{-\frac{\hbar}{i}\pdiff{t}}$ を作用させる．右辺では, $\psi(x',t')$ に時間変数 $t$ は含まないので, これは$K$のみに作用することに注意する．その後, 式(1)を利用すると,

\begin{align} -\frac{\hbar}{i}\pdiff{t}\psi(x,t) &=\int_{-\infty}^{\infty} dx^{'}\,\left\{-\frac{\hbar}{i}\pdiff{t} K(x,t ; x^{'},t^{'})\right\} \psi(x^{'},t^{'})\notag\\ &=\int_{-\infty}^{\infty} dx^{'}\,\left\{-\frac{\hbar^{2}}{2m}\Bpdiff{x} K(x,t ; x^{'},t^{'})\right\}\,\psi(x^{'},t^{'})\notag\\ &=-\frac{\hbar^{2}}{2m} \Bpdiff{x} \int_{-\infty}^{\infty} dx^{'}\,K(x,t ; x^{'},t^{'})\,\psi(x^{'},t^{'})\notag\\ &=-\frac{\hbar^{2}}{2m} \Bpdiff{x}\psi(x,t) \tag{3} \end{align}

よって式(3.43)が言える：

\begin{equation} -\frac{\hbar}{i}\pdiff{t}\psi(x,t)=-\frac{\hbar^{2}}{2m}\Bpdiff{x}\psi(x,t) \tag{4} \end{equation}