Wavepackets [k-space to z-space]

1. Jan 10, 2007

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

1. The problem statement, all variables and given/known data
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

2. Relevant equations

again, as attached

3. 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

Attached Files:

• Equation.JPG
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2. Jan 11, 2007

dextercioby

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

Daniel.

3. Jan 19, 2007

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 lenght 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).

Last edited: Jan 19, 2007