- A ``square'' packet: for and 0 elsewhere.
- A Gaussian packet: .

These are both localized in momentum about .

Check the normalization of (1).

Check the normalization of (2) using the result for a
definite integral of a Gaussian

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

Note that is equal to 1 at 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 function.

The
Fourier Transform of a Gaussian
wave packet
is

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 (transformed from a normalized localized in momentum space) we see

- A coefficient which
**correctly normalizes**the state to 1, - - a
**wave corresponding to momentum**, - and a packet function which is
**localized**in .

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
and in
.

Jim Branson 2013-04-22