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Potential well solutions question

  1. Mar 18, 2012 #1
    Hey,

    I've been trying to work out how, for a finite well of high Vo and width L, the interior solution has the form L Sin(kx + d),

    I see that if d=0 then the solution resembles an infinite well, so that implies d depends inversely on the wells potential. But I can't work out what d comes from, and why the constant at the front is the length of the well (why is the amplitude of the wave the length of the well)

    d is just the phase of the wave, so does it represent an adjustment to assure all the continuity requirements are satisfied?

    Ive also been trying to work out what the k is, when I go through the process of finding the wave function inside I end up with (s/k)Asinkx + Acoskx after applying the boundary conditions.

    where k2=2mE/[itex]\hbar^{2}[/itex]

    and s2=2m(Vo-E)/[itex]\hbar^{2}[/itex]

    i've been trying to work this out because ive also read that the below equation should also satisfy the internal solutions

    ka = n[itex]\pi - 2Sin^{-1}(\frac{k\hbar}{\sqrt{2mV_{o}}})[/itex]

    this makes it seem like d resembles the inverse sin term, which makes sense because if Vo ->infinity you would get the solution to an infinite potential well

    Anyone have any ideas?
     
  2. jcsd
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