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Wavepacket incident on a potential step

  1. May 19, 2013 #1
    Hello,
    I am writing a Fortran 95 program to model the scattering of a wavepacket by a potential step of height V0 at x=0. My wavepacket is formed by the superposition of numerous travellling waves of different k values. The wavepacket has the dispersion relation ω(k)=k2. I want the wavepacket to be in its undispersed state at t=0 at a start position x0. Therefore each component wave is composed of an incident wave, reflected wave, and transmitted wave.

    At x≤0 :
    ψ(x,t)=A(ei(k(x-x0)-ωt) +[itex]\frac{k-k'}{k+k'}[/itex]e-i(k(x+x0)-ωt))

    At x>0 :
    ψ(x,t)=A[itex]\frac{2k}{k+k'}[/itex]ei(k'(x-[itex]\frac{k}{k'}[/itex]x0)-ωt)

    The factor of [itex]\frac{k}{k'}[/itex] in the bottom equation was found analytically to ensure continuity in the wavefunctions at the boundary. Well, at least that's what I thought: It works as long as E>V0 otherwise there is discontinuity. Can anyone help me as to why? I dont think this is a Fortran programming problem, but more of a physics one...
     
  2. jcsd
  3. May 19, 2013 #2
    Exactly. Aren't you satisfied yet?
     
  4. May 19, 2013 #3
    Well not really: when I work through the maths, this factor should work fine in all situations, but in practice it only works when E>V0. If you have any idea why this might be then I would be very grateful. I'm fairly certain its not a problem with my code as I have applied it in many different ways all with the same result.
     
  5. May 19, 2013 #4
    Hi

    for E<V0, ψ for x>0 is no longer a plane wave but an exponentially dumping function as e^-Kx where k'=iK pure imaginary number.
     
    Last edited: May 19, 2013
  6. May 19, 2013 #5
    Hi,
    Thanks, but this is built into my program already: k' becomes complex reducing ψ=eik'x to e-αx the dumping function you describe.
     
  7. May 19, 2013 #6
    Hi.

    In both of the cases, e^ik'x = e^-Kx = 1 for x=0. So continuity conditions should be the same in the both cases.
    What is the discontinuity you are worrying about?
     
  8. May 19, 2013 #7
    Because if I do not offset by x0, then the wavepacket impacts the step at exactly t=0. Therefore to watch it scatter you must set the time range to be from -t to t. In the period -t to 0 the wavepacket starts off slightly dispersed, becoming less dispersed until it hits the barrier. This is unphysical so I have offset the position of the wavepacket at t=0 to x=x0. Although eikx and e-αx both equal 1 at x=0, eik(x-x0) and e-α(x-x0) are not equal there. So the factor [itex]\frac{k}{k'}[/itex] has to be introduced as above. This works fine on paper but not in practice.
     
  9. May 19, 2013 #8
    Hi.

    The dumping depends on the distance from the wall, x-0, not the distance from x0, x-x0.
    The wave function for x>0 is e-αx+ikx0.
     
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