Two Examples of Localized Wave Packets

Lets now try two examples of a wave packet localized in \bgroup\color{black}$k$\egroup and properly normalized at \bgroup\color{black}$t=0$\egroup.
  1. A ``square'' packet: $A(k)={1\over\sqrt{a}}$ for $k_0-{a\over 2}<k<k_0+{a\over 2}$ and 0 elsewhere.
  2. A Gaussian packet: $A(k)=\left({2\alpha\over\pi}\right)^{1/4}e^{-\alpha (k-k_0)^2}$.

These are both localized in momentum about \bgroup\color{black}$p=\hbar k_0$\egroup.

Check the normalization of (1).

\begin{displaymath}\bgroup\color{black} \int\limits_{-\infty}^\infty \vert A(k)\...
...imits_{k_0-{a\over 2}}^{k_0+{a\over 2}}dk={1\over a}a=1 \egroup\end{displaymath}

Check the normalization of (2) using the result for a definite integral of a Gaussian \bgroup\color{black}$\int\limits_{-\infty}^{\infty}\; dx\; e^{-ax^2} = \sqrt{\pi\over a}.$\egroup

\begin{displaymath}\bgroup\color{black} \int\limits_{-\infty}^\infty \vert A(k)\...
=\sqrt{2\alpha\over\pi}\sqrt{\pi\over 2\alpha}=1 \egroup\end{displaymath}

So now we take the Fourier Transform of (1) right here.

f(x)&=&{1\over \sqrt{2\pi}}\int\limits_{-\infty}^\infty A(k) e...
...a\over 2\pi}e^{ik_0x}{2\sin\left({ax\over 2}\right)\over ax} \\

Note that \bgroup\color{black}${2\sin\left({ax\over 2}\right)\over ax}$\egroup is equal to 1 at \bgroup\color{black}$x=0$\egroup and that it decreases from there. If you square this, it should remind you of a single slit diffraction pattern! In fact, the single slit gives us a square localization in position space and the F.T. is this \bgroup\color{black}${\sin(x)\over x}$\egroup function.

The Fourier Transform of a Gaussian wave packet \bgroup\color{black}$A(k)=\left({2\alpha\over\pi}\right)^{1/4}e^{-\alpha (k-k_0)^2}$\egroup is

\begin{displaymath}\bgroup\color{black}f(x)=\left({1\over 2\pi\alpha}\right)^{1/4} e^{ik_0x} e^{-{x^2\over 4\alpha}}\egroup\end{displaymath}

also a Gaussian. We will show later that a Gaussian is the best one can do to localize a particle in position and momentum at the same time.

In both of these cases of \bgroup\color{black}$f(x)$\egroup (transformed from a normalized \bgroup\color{black}$A(k)$\egroup localized in momentum space) we see

We have achieved our goal of finding states that represent one free particle. We see that we can have states which are localized both in position space and momentum space. We achieved this by making wave packets which are superpositions of states with definite momentum. The wave packets, while localized, have some width in \bgroup\color{black}$x$\egroup and in \bgroup\color{black}$p$\egroup.

Jim Branson 2013-04-22