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.