Wavefunction and degree of localization

[argonsonic]

Homework Statement

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

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

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

Thanks.

Homework Equations

Fourier transform.

The Attempt at a Solution

 

[member 553083]

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.