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Schrodinger equation Eigenfunction problem

  1. Nov 13, 2006 #1
    So, here's the question:

    [tex]\psi(x) = A*(\frac{x}{x_{0}})^n*e^(\frac{-x}{x_{0}}) [/tex]
    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 [tex] \hat p\psi(x) = p \psi(x) [/tex] but my partial of psi with respect to x left me with a difference, due to the product rule which left me with [tex] -i*\hbar(\frac{1}{x_{0}} - \frac{n}{x}) = p [/tex] 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. jcsd
  3. Nov 13, 2006 #2

    quasar987

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

    [tex]\frac{\hat{p}^2}{2m}\psi+U(x)\psi = E\psi[/tex]

    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.
     
  4. Nov 13, 2006 #3

    StatusX

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

    You know that:

    [tex]\frac{\hat p^2}{2m} \psi +U(x) \psi = E \psi[/tex]

    So solve for U. There is a solution for every E, but I'm guessing they don't all go to zero at infinity.
     
  5. Nov 13, 2006 #4
    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 [tex]\frac{\partial^2\psi(x)}{\partial x^2} = \psi(x)(\frac{1}{x_{0}^2} - \frac{2n}{x} + \frac{(n-1)n}{x^2})[/tex]. 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 [tex]x^2x_{0}^2[/tex] but its worse.
     
    Last edited: Nov 14, 2006
  6. Nov 14, 2006 #5

    dextercioby

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

    Daniel.
     
  7. Nov 15, 2006 #6
    So here's how I'm doing this derivative, correctly this time.
    [tex] u = A(\frac{x}{x_{0}})^n [/tex]
    [tex]du = Ac^{-n}nx^{n-1} [/tex]
    [tex]v = e^{-x/x_{0}}[/tex]
    [tex]dv = -\frac{e^{-x/x_{0}}}{x_{0}}[/tex]
    [tex]\frac{\partial\psi(x)}{\partialx} = udv + vdu [/tex]
    [tex]\frac{\partial\psi(x)}{\partialx} = \psi(x)(\frac{-1}{x_{0}}+\frac{n}{x})[/tex]
    [tex]\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})[/tex]
    [tex]\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})[/tex]

    Now, what do I do?

    I haven't had differential equations so I'm pretty useless at that if it's needed.
     
  8. Nov 16, 2006 #7
    I have no idea what's going on here, because the way I see it I now have:

    [tex]\frac{n^2}{x^2}+\frac{1}{x_{0}^2}-\frac{2n}{xx_{0}}-\frac{n}{x^2}+U(x)=E[/tex]

    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 [tex]e^{-x^2}[/tex] then I have yet another solution, and another E.

    Perhaps I've hit upon something, that the solutions are of the form [tex]e^{-x^m}[/tex] where m is some integer (perhaps a real > 1??)...
     
  9. Nov 16, 2006 #8

    StatusX

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    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.
     
  10. Mar 7, 2010 #9
    I have the same problem to solve, but I don't know how to start, can someone help me?
    Thank you.
     
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