The Square Wave Packet

Given the following one dimensional probability amplitude in the position variable x, compute the probability distribution in momentum space. Show that the uncertainty principle is roughly satisfied.

$$\bgroup\color{black}\psi (x) = {1\over\sqrt{2a}}\egroup$$

for $-a<x<a$, otherwise $\psi (x)=0$.

Its normalized.

$$\bgroup\color{black}\int\limits_{-\infty}^\infty\; \psi^*\psi dx=\int\limits_{-a}^a\; {1\over 2a}dx = 1\egroup$$

Take the Fourier Transform.

$$\begin{eqnarray*} \phi(k)&=&{1\over\sqrt{2\pi}}\int\limits_{-a}^a {1\over\sqrt{2... ...{4\pi a}}[-2i\sin{ka}] = -\sqrt{1\over\pi a}{\sin(ka)\over k}\\ \end{eqnarray*}$$

Now estimate the width of the two probability distributions.

$$\begin{eqnarray*} \Delta x=2a\\ \vert\phi(k)\vert^2={1\over\pi a}{\sin^2(ka)\over k^2}\\ \Delta k={2\pi\over a}\\ \Delta x\Delta k=4\pi\\ \end{eqnarray*}$$