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

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SUMMARY

The discussion focuses on determining the width of a Gaussian wavepacket in real space over time, specifically in the context of quantum mechanics. The wavefunction is expressed as |\psi|^2, which follows the form of a Gaussian curve, where the width is derived from the equation exp{- (z - vt)^2/A(t)} = 1/2. The width of the wavepacket increases over time due to the dispersive nature of the Schrödinger equation, which causes different Fourier components to propagate at varying velocities. Key insights include the relationship between the width and the uncertainty principle, represented by the equation ΔzΔp = ħ/2.

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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
 

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

Daniel.
 
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).
 
Last edited:

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