Solution to 1992 Problem 27

We will use the fact that
\begin{align}\int_{\infty}^{\infty} e^{- (x - a)^2} dx = \sqrt{\pi} \label{eqn:1}\end{align}
where a is any complex number.
We assume that f(k) is a Gaussian and that \Delta k is its standard deviation. Then
\begin{align*}f(k) = \frac{1}{\sqrt{2 \pi \Delta k ^2}} e^{- (k - k_0)^2/2\Delta k^2}\end{align*}
where the prefactor of 1/\sqrt{2 \pi \Delta k ^2} is for normalization. Then,
\begin{align*}\psi(x,t) &= \int_{-\infty}^{\infty} e^{i (k x - \omega t)} \frac{1}{\sqrt{2 \pi \Delta k ^2}} e^{- (k - k_...
where equation (1) was used to evaluate the integral.
This implies that
\begin{align*}\boxed{\Delta k = \frac{1}{\Delta x_0}}\end{align*}
Therefore, answer (B) is correct.

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