# Schrodinger equation Eigenfunction problem

1. Nov 13, 2006

### joex444

So, here's the question:

$$\psi(x) = A*(\frac{x}{x_{0}})^n*e^(\frac{-x}{x_{0}})$$
Where A, n, and X0 are constants.

Using Schrodinger's equation, find the potential U(x) and energy E such that the given wave function is an eigenfunction (we can assume that at x = infinity there is 0 potential).

So, in my futile attempt at beginning this I tried to use $$\hat p\psi(x) = p \psi(x)$$ but my partial of psi with respect to x left me with a difference, due to the product rule which left me with $$-i*\hbar(\frac{1}{x_{0}} - \frac{n}{x}) = p$$ which isn't an eigenvalue for momentum. Using an E operator makes no sense, because there's no t term so the derivative goes to 0 and it just doesn't make sense.

Forget the right track, is this the right station?

Last edited: Nov 13, 2006
2. Nov 13, 2006

### quasar987

you're looking for values of U(x) and E such that

$$\frac{\hat{p}^2}{2m}\psi+U(x)\psi = E\psi$$

So I'd begin by expanding this equation (i.e. do the derivatives involved and simplify, etc) and then start looking for U(x) and E sucht that make the equality true.

3. Nov 13, 2006

### StatusX

I don't know what "$$p\hat \psi(x) = p \psi(x)$$" means. If $\hat \psi$ is the complex conjugate of $\psi$, it isn't true, and you're mistaking what the defintion of hermitian is.

You know that:

$$\frac{\hat p^2}{2m} \psi +U(x) \psi = E \psi$$

So solve for U. There is a solution for every E, but I'm guessing they don't all go to zero at infinity.

4. Nov 13, 2006

### joex444

Oh, sorry, my tex was backwards its fixed now.

OK, so using the p operator squared, that equation becomes the time independent schrodinger equation.

Solving for the second derivative of psi I obtained $$\frac{\partial^2\psi(x)}{\partial x^2} = \psi(x)(\frac{1}{x_{0}^2} - \frac{2n}{x} + \frac{(n-1)n}{x^2})$$. I thought it was kind of useful I could pull psi out, but the rest of it is fairly ugly. I've checked the derivation three times, so I'm fairly sure it's right. I also tried putting it all in a common denominator of $$x^2x_{0}^2$$ but its worse.

Last edited: Nov 14, 2006
5. Nov 14, 2006

### dextercioby

Not E, but the potential U(x) needs to go to zero at infinity. In this case, it does, since $n\geq 0$.

Daniel.

6. Nov 15, 2006

### joex444

So here's how I'm doing this derivative, correctly this time.
$$u = A(\frac{x}{x_{0}})^n$$
$$du = Ac^{-n}nx^{n-1}$$
$$v = e^{-x/x_{0}}$$
$$dv = -\frac{e^{-x/x_{0}}}{x_{0}}$$
$$\frac{\partial\psi(x)}{\partialx} = udv + vdu$$
$$\frac{\partial\psi(x)}{\partialx} = \psi(x)(\frac{-1}{x_{0}}+\frac{n}{x})$$
$$\frac{\partial^2\psi(x)}{\partialx^2} = (\frac{-1}{x_{0}}+\frac{n}{x})(\frac{\partial\psi(x)}{\partialx}) + \psi(x)\frac{\partial\psi(x)}{\partialx}(\frac{-1}{x_{0}}+\frac{n}{x})$$
$$\frac{\partial^2\psi(x)}{\partialx^2} = \psi(x)(\frac{n^2}{x^2}+\frac{1}{x_{0}^2}-\frac{2n}{xx_{0}}-\frac{n}{x^2})$$

Now, what do I do?

I haven't had differential equations so I'm pretty useless at that if it's needed.

7. Nov 16, 2006

### joex444

I have no idea what's going on here, because the way I see it I now have:

$$\frac{n^2}{x^2}+\frac{1}{x_{0}^2}-\frac{2n}{xx_{0}}-\frac{n}{x^2}+U(x)=E$$

So now the only restriction is that U(x) -> 0 as x -> infinity. Well, I'm not sure what I can do, it seems to me to be 2 unknowns and 1 equation so I get an infinite number of solutions.

If I let U(x) be 0, then E is found. If I let U(x) be e^-x then it satisfies the going to 0 thing and I have an E. If I let U(x) be $$e^{-x^2}$$ then I have yet another solution, and another E.

Perhaps I've hit upon something, that the solutions are of the form $$e^{-x^m}$$ where m is some integer (perhaps a real > 1??)...

8. Nov 16, 2006

### StatusX

E is a constant, so the LHS of the equation must also be a constant. Thus all the x dependence from the derivative terms must be cancelled by the x dependence in U(x). Then E will be specified by the requirement that U(x) go to zero at infinity.

9. Mar 7, 2010

### SCDL

I have the same problem to solve, but I don't know how to start, can someone help me?
Thank you.