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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

    Vo>0
    E<Vo

    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

    psi=A*exp(ikx)+B*exp(-ikx).

    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
    psi=A(exp(ikx)-exp(-ikx))

    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
    I think this may help you http://scienceworld.wolfram.com/physics/Half-InfiniteSquarePotentialWell.html

    I had a similar problem and it did.

    Your psi functions don't look right for the two regions x>0. Also, E<Vo implies a bound particle, it will be in the well, but may leak into the region x>a somewhat.
     
  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.
     
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