Squaring $e^{-ax^2}$ graph

Question

When studying wavefunctions in Chemistry, we modelled an oscillator in the form $e^{-ax^2}$.

When I squared this wave function (taking $a = 1$) and plotted the graph, I found that the line decayed faster than my original function.

This is shown below.

I was trying to work out mathematically why this was the case but I couldn't understand for as long as I tried.

Why is this the case?

Answer 1

Because, it is :

$$\left(e^{-x^2}\right)^2 = e^{-2x^2} < e^{-x^2}$$

The exponent of the squared expression is bigger (or equal for some distinct cases), thus with the minus sign, you have a decaying exponential function. Of course, the decay becomes bigger if the exponent grows bigger.

Answer 2

Simply for $\alpha\in (0,1)$: $$(1-\alpha)^2=1-(2\alpha-\alpha^2)<1-\alpha$$