# Determine the potential V(x)

1. Feb 25, 2010

### Denver Dang

1. The problem statement, all variables and given/known data

A particel is moving along the x-axis in a potential, V(x). The ground state of the wavefunction is:

$$$\psi \left( x \right)={{e}^{-\frac{1}{2}{{\left( \frac{x}{a} \right)}^{4}}}},$$$
where a is a constant of dimension length. We also know that V(a) = 0.

Determine V(x).

2. Relevant equations

The stationary Schrödinger equation:

$$$-\frac{{{\hbar }^{2}}}{2m}\frac{{{d}^{2}}\psi }{d{{x}^{2}}}+V\psi =E\psi$$$

3. The attempt at a solution

Well, first I derive the wavefunction two times and get:

\begin{align} & \frac{d}{dx}\psi \left( x \right)=-\frac{2}{{{a}^{4}}}{{x}^{3}}{{e}^{-\frac{1}{2}{{\left( \frac{x}{a} \right)}^{4}}}} \\ & \frac{d}{dx}\psi '\left( x \right)=\left( \frac{4{{x}^{6}}}{{{a}^{8}}}-\frac{6{{x}^{2}}}{{{a}^{4}}} \right){{e}^{-\frac{1}{2}{{\left( \frac{x}{a} \right)}^{4}}}} \\ \end{align}

Then I insert it in the Schrödinger equation.
And here is where I get confused. It's been a while since I've had QM, so I can't really remember what's next.
I figure that the potential is NOT just what I get when I isolate V in the Schrödinger equation. So what's is my next step ?

Regards

2. Feb 25, 2010

### gabbagabbahey

Why not?...Keep in mind that $E$, being an eigenvalue, must be a constant (no x-dependence).

3. Feb 25, 2010

### Denver Dang

So the potential V(x) is just:

$$$V\left( x \right)=\frac{2\hbar {{x}^{6}}-3{{a}^{4}}\hbar {{x}^{2}}+E{{a}^{8}}m}{{{a}^{8}}m}$$$ ?

4. Feb 25, 2010

### vela

Staff Emeritus
You still need to evaluate what E equals. E is an eigenvalue. Its value changes depending on the eigenstate of the system.

5. Feb 25, 2010

### Denver Dang

Ahhh yes...
But what is En for the stationary state ? Can't find it on the internet, and I haven't got my QM book in ages. If I recall correctly it's different for the infinite well, harmonic oscillator, finite well and free particle. So which one should I use ?

6. Feb 25, 2010

### gabbagabbahey

You tell us....you are told that V(a)=0, so what must E be for that to be true?

7. Feb 25, 2010

### Matterwave

You can use $$E=<\psi|H|\psi>$$ since your wave function is an eigenstate.

EDIT: Wait, disregard that...

8. Feb 26, 2010

### Denver Dang

I think I got it now...
V(a) = 0, which removes the potential part, replace the x's with a's and isolate E.
And then after that I use that E in the same Schrödinger equation and isolate V.

9. Feb 26, 2010

### gabbagabbahey

Another way to look at it is that you've already derived an expression for $V(x)$ that includes an unknown constant, $E$ (in post #3), and you are given an initial value, $V(a)=0$. So, you can determine the value of that constant by plugging $x=a$ into your expression, setting it equal to zero, and solving for $E$.