Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: One-dimensional potential well problem

  1. Nov 14, 2006 #1
    My head's melting right now, because I've been stuck on this for the past 6 hours.
    There's a particle of mass moving in a potential well where
    V(x) = infinity at x<0
    V(X)=0, 0<x<a
    V(x)= Vo, x>a


    I'm assuming that the wavefunction at x<0 is 0, since there's an infinite potential there. The energy inside the potential well is just the kinetic energy, =(Hk)^2/2m, where H=h/2pi, so the wavefunction should be of the form
    psi= Aexp(-ikx )
    Now outside the well, at x>a, the energy should be E= Vo-Ek because the particle is bound in the well. We then get psi=Bexp(-Tx), where T is k with (Vo-E) instead of E.
    Am I wrong in assuming this? When I try to find the radius using
    X^2 + y^2 = R^2, where x=k=a*sqrt(2mE)/H and y=ai*(sqrt(2m(Vo-E))/H
    I get out a negative radius.
    Please help me see what I've done wrong, I'm sure I've got the energy value on the finite potential side wrong, but I can't see how.
  2. jcsd
  3. Nov 14, 2006 #2
    Haven't gone through your entire question to know if this is the only mistake, but I beleive you made a mistake determining psi in the well.

    When you solve the differential equation you should get a couple exponentials...something like


    Remember these are complex exponentials so they don't blow up at infinity and you can't eliminate terms like with real exponentials.

    Now your boundary conditions require that psi=0 for x</= 0, so plugging in 0 for x we get A+B=0, and hence A=-B, and psi is some sort of sine wave of the form

    I havent gone through the rest of your problem, and I'm not sure what radius you are referring to at the end.
  4. Nov 14, 2006 #3
  5. Nov 15, 2006 #4
    The radius comes in when we look at the wavefunctions across x=a, where they should be continuous, so psi1=psi2 and dpsi1/dx=dpsi2/dx. I have to divide one by the other to get a solution in the form -y=xcot(x), then rearrange the circle equation to get y, sub it in and the intersection gives the solution.
  6. Nov 15, 2006 #5
    Ok thanks, you've really helped me out, the website was especially helpful.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook