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Solving Schrod. Eqn. for Potential Barrier with E > V

  1. Sep 29, 2013 #1
    I have a question that's driving me insane and I'm sure there's a simple answer that I'm missing for some reason, but I'm not getting my a-ha moment.

    Consider 2 cases from intro QM:

    Infinite square well

    Potential barrier with E > V0

    For the infinite square well, the Schrodinger eqn gives

    d2ψ/dx2 + k2ψ = 0

    Since k2 > 0, this gives oscillating solutions (some combination of sines and cosines). Correct?

    For the potential barrier with E > V0, the Schrodinger eqn gives (with the potential jumping from 0 to V0 at x = 0)

    d2ψ/dx2 + k2IE = 0 (where kI = √2mE/(hbar2)

    prior to the potential step (i.e. for x < 0) and

    d2ψ/dx2 + k2IIE = 0 (where kII = √2m(E-V0)/(hbar2)

    after the potential step (for x > 0)

    My problem is this: As far as I can tell, both kI and kII are positive numbers (since E > V0 > 0). Why, then, do they give non-oscillating solutions? Whereas, for the infinite square well, a positive k gave oscillating solutions?

    Am I making a sign error here or just forgetting something from differential equations? What is different about these equations that one of them gives oscillating, and the other non-oscillating, solutions?

    Thanks for any help!
     
  2. jcsd
  3. Sep 29, 2013 #2

    stevendaryl

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    Why do you think the solution is non-oscillating? What do you think the solution is?
     
  4. Sep 29, 2013 #3
    Well, I think it all hinges on the sign of k2.

    If I have a differential equation

    d2f(x)/dx2 = αf(x),

    then if α < 0, the equation is satisfied by f(x) = sin(x√α) (or f(x) = cos(x√α) or combination of the two) (oscillating). Whereas if α > 0, the equation is satisfied by f(x) = exp(x√α) (non-oscillating). Which I verify by actually differentiating the functions.
     
  5. Sep 29, 2013 #4
    As far as what is leading me to go against my own reasoning, every text I've consulted, both physical and online, tells me that in the region x > 0 where V0 > 0 (but E > V0), the solutions to the appropriate Schr. Eqn are of the form

    Cexp(ikx) + Dexp(-ikx)

    where k is the wave number appropriate to the region (kII in the original post).
     
  6. Sep 29, 2013 #5
    Ok, I seriously feel like a heel. I see my mistake. The complex exponentials DO describe an oscillating function, according to Euler's equation.

    Ugh. Sorry about that :(
     
  7. Sep 29, 2013 #6

    stevendaryl

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    Yes, and that's an oscillating solution. Do you know the equation: exp(ikx) = cos(kx) + i sin(kx)?
     
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