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.

 

[anuttarasammyak]

H\psi(x)=[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)]\psi(x)
How about calculating RHS with
\psi(x)=(\frac{x}{a})^n e^{-x/a}?

 

[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?

 

[anuttarasammyak]

What is the result of double differential of $\psi?$

 

[Kaguro]

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

 

[anuttarasammyak]

= \{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.

 

[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...

 

[anuttarasammyak]

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

 

[Kaguro]

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

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

 

[anuttarasammyak]

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