# Finding the potential function from the wavefunction

Kaguro
Homework Statement:
A particle is confined in one dimension by a potential V(x).
Given psi=[(x/a)^n]*exp(-x/a), find the potential.
Relevant Equations:
H (psi)=E(psi)
H ={ -(h_bar)^2/2m +V}
I would differentiate this twice and plug it into the S.E, but for that I'll need E. Which I don't have. Please provide me some direction.

Gold Member
$$H\psi(x)=[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)]\psi(x)$$
$$\psi(x)=(\frac{x}{a})^n e^{-x/a}$$?

• BvU
Kaguro
Sorry, I forgot typing the derivative.
Anyways, I can find the derivative and write the whole LHS, but for RHS, E(psi) , I need E. That's the main problem. What do I do with my lack of E?

Gold Member
What is the result of double differential of ##\psi?##

Kaguro
##\frac{d^2 \psi}{dx^2} = e^{-x/a}*[n(n-1)(x/a)^{n-2} -(2n+1)(x/a)^n + (x/a)^{n+2}]##

Gold Member
$$= \{n(n-1) (x/a)^{-2}+(x/a)^2-(2n+1)\} \psi(x) = \frac{2m}{\hbar^2}(V(x)-E)\psi(x)$$
So you find V(x) and E if your calculation is right. I am afraid not.

Last edited:
• Kaguro
Kaguro
V(x) = ## \frac{\hbar ^2}{2m}((n)(n-1)(x/a)^{-2}+(x/a)^2)##
E = ## \frac{\hbar ^2(2n+1)}{2m}##

Kaguro
Thank You very much!

I was too worked up over not having E...

Gold Member
Check your math and do the same way.
$$\frac{d}{dx}e^{-x/a}=-\frac{1}{a} e^{-x/a}$$

• Kaguro
Kaguro
Oh man! I forgot the chain rule! How can I forget the chain rule in final year undergrad?!

So, finally V(x) = ##\frac{\hbar^2}{2m}(\frac{n(n-1)}{x^2} -\frac{2n}{ax})##

Gold Member
You see ##|\psi|## increases to infinity for ##x→-\infty## though ##V \rightarrow 0##. Your teacher may explain physical meaning of the problem.