Wavefunction and degree of localization

[argonsonic]

Homework Statement

Suppose that there is a wavefunction [itex]\Psi (x,0)[/itex] where 0 is referring to [itex]t[/itex]. Let us also say that [itex]a(k) = (C\alpha/\sqrt \pi )exp(-\alpha^2k^2)[/itex] is the spectral contents (spectral amplitudes) where [itex]k[/itex] is defined as wavenumber [itex]k[/itex]. [itex]\alpha[/itex] and [itex]C[/itex] is some constant

My question is, why do we calculate [itex]\Delta x[/itex] by looking at where the value of [itex]\Psi (x)[/itex] diminish by [itex]1/e[/itex] from the maximum possible value of [itex]\Psi (x)[/itex]?

Also, although the width of the [itex]\Psi (x)[/itex] packet is [itex]4\alpha[/itex], we define [itex]\Delta x[/itex] as [itex]\alpha[/itex]. Why is it like this?

Thanks.

Homework Equations

Fourier transform.

The Attempt at a Solution

 

[Nate810]

I don't know why we use the 1/e value to define \Delta x, but I do know why we use \alpha as \Delta x. The Fourier transform of a complex function \Psi (x) is calculated by looking at the spectral contents (a(k)), where k is defined as wavenumber k. Since the spectral amplitude a(k) is a Gaussian shape with a width of 4\alpha, this means that the Fourier transform of \Psi (x) has a width of \alpha. Thus, \Delta x is defined as \alpha.