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Step barrier

  1. May 6, 2004 #1
    Consider a potential step located at x=0, such that u=0 to the left of the step (for x<0), and u = U to the right of the step (x>0).

    A particle with total energy E > U traveling from left to right has:

    [tex]\psi_1(x) = Ae^{ik_1x} + Be^{-ik_1x} \;\textrm{for}\; x\le0 \;\textrm{and} \;k_1 =\sqrt\left(\frac{2mE}{\hbar^2}\right)[/tex]

    and

    [tex]\psi_2(x) = Ce^{ik_2x} + De^{-ik_2x} \;\textrm{for}\; x\ge0 \;\textrm{and} \;k_2 =\sqrt\left(\frac{2m(E-U)}{\hbar^2}\right)[/tex]

    I would say that [tex]Ae^{ik_1x}[/tex] portrays a wave propagating to the right, with kinetic energy equal to E and [tex]Be^{-ik_1x}[/tex] portrays a wave propagating to the left, and represents the probability of the particle being reflected by the barrier despite the fact that the particle has sufficient energy to pass over the barrier.

    Then I would say that [tex]Ce^{ik_2x}[/tex] portrays the wave continuing to propagate towards the right, with reduced kinetic energy E-U, having surpassed the barrier, and that D = 0 because in the region x>0 there is nothing to cause a wave to propagate towards the left, and therefore [tex]\psi_2[/tex], instead of what I wrote above, is actually just

    [tex]\psi_2(x) = Ce^{ik_2x} \;\textrm{for}\; x\ge0 \;\textrm{and} \;k_2 =\sqrt\left(\frac{2m(E-U)}{\hbar^2}\right)[/tex]

    Then, of course, there are matching conditions to be met at x=0, etc...

    The book's solution gives the same two equations that I wrote up at the top (without explanation), and states that " D = 0 at x = 0 ".

    Please tell me if my explanations are correct (or how to correct them if necessary) and, is D=0 only at x=0, or is D=0, period.
     
  2. jcsd
  3. May 6, 2004 #2

    Doc Al

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    Staff: Mentor

    I like your explanations and I agree with you that D = 0, period. The problem is that of a particle moving from left to right, so the only left-moving component is the reflected wave in region 1. (We ignore particles coming from x = +∞ moving towards x = 0.)
     
  4. May 6, 2004 #3
    Thank you.

    At least I feel that I'm beginning to see the QM pattern, even if I can't exactly say that it makes sense. :rolleyes:
     
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