A particle has a wave function given by Ae^(-x^2/2L^2) and Energy h(bar)/2mL^2

A particle has a wave function given by Ae^(-x^2/2L^2) and Energy h(bar)/2mL^2, where L is ome length. (a) Find potential energy as a function of x.

So, I know I have to use the Schrodinger equation, time independent, because that gives me

(-h(bar)^2/2m)(d^2si(x)/dx^2) + V(x)si(x) = Esi(x)

Just not sure how to set this up and how to figure out the second derivative with respect to x of Ae^(-x^2/2L^2)

(BTW, you have a typo in your question: it should be E = ħ²/2mL², not ħ/2mL²).

Schrödinger's equation (time-independent, one-dimensional) is:

Eψ = (-ħ²/2m)d²ψ/dx² + Vψ

To keep this answer easy to read, I'm going to move the goal-posts slightly and set ħ²/2m = L = 1. You shouldn't find it too difficult to adapt this to the full problem. So, with my odd choice of units we've got:

(1). Eψ = -d²ψ/dx² + Vψ
(2). ψ = A.exp(-x²/2)
(3). E = 1.

From (1) and (3),

(4). Vψ = ψ + d²ψ/dx²

So let's plug in (2). Note that the "normalisation" factor A doesn't matter, since we can divide it out on both sides of (4), and so I'll just ignore it by setting

ψ = exp(-x²/2)

The first derivative of ψ is:

dψ/dx = -x.exp(-x²/2)

by using the chain rule. Applying the product rule and chain rule, the second derivative is:

d²ψ/dx² = -exp(-x²/2) + x²exp(-x²/2) = (x² - 1)exp(-x²/2)

The factor of exp(-x²/2) can be divided out on each side of (4), and we're left with|:

V = 1 + (x² - 1) = x².

which is a harmonic oscillator potential. When you put your factors of ħ, m and L back in to do the full problem you should find:

V = ħ²x²/2mL⁴.