How Do You Determine the Width of a Gaussian Wavepacket in Real Space Over Time?

[Mazimillion]

Hi, this type of question has been confusing my slightly as of late, an a pointer in the right direction would be greatly appreciated

Homework Statement

The wavefunction associated with a Gaussian wavepacket propagating in free space can be shown to be [included as attachment - it's too complicated for here] where delta k is withe width of the wavepacket in k space and v is the velocity of the wavepacket.

Deduce an expression for the width of the wavepacket in real space (z-space)as a function of time

Homework Equations

again, as attached

The Attempt at a Solution

I'm suspecting it has something to do with Fourier Transforms, but I'm really stumped. it's probably straightforward, but I'm a bit blind to it at the moment

Thanks in advance

 

[dextercioby]

I have a hunch that $\Delta z\Delta p =\frac{\hbar}{2}$, since a gaussian wavepacket is minimizing the uncertainty relations.

Daniel.

 

[Jezuz]

To find the witdth of the wave packet you should consider the form of
$$|\psi|^2$$ .
This will have the form
$$\psi \propto \exp \left\{- \frac{(z - vt)^2}{A(t)} \right\}$$

This has the form of a Gaussian curve. The maximum occurs where $$z = vt$$ where the exponens takes on the value 1.
The width is given by the length between the points where the exponent is $$1/2$$. So the expression used to find the widht is
$$\exp \left\{ - \frac{(z-vt)^2}{A(t)} \right\} = \frac{1}{2}$$.
Solving this gives two solutions $$z_1(t)$$ and $$z_2 (t)$$ and the difference between these are the width of the wave packet.

You can expect that the width is increasing with time, since the Schrödinger equation has a dispersive term (a term that causes different Fourier components of the wave to propagate with different velocities).