Schrodinger equation Eigenfunction problem

[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?

 

[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.

 

[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.

 

[joex444]

I don't know what "$$p\hat \psi(x) = p \psi(x)$$" means.

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.

 

[dextercioby]

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.

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

Daniel.

 

[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.

 

[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??)...

 

[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 canceled by the x dependence in U(x). Then E will be specified by the requirement that U(x) go to zero at infinity.

 

[SCDL]

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