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Potential well solutions

  1. Mar 16, 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 the reflection of any particles which moved past the well?

    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 the question also says that the solution inside the well should also satisfy

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

    which makes me think the d resembles that arcsin the above equation

    Anyone have any ideas?
     
  2. jcsd
  3. Mar 18, 2012 #2

    Redbelly98

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    It looks like you're trying to solve the time-independent Schrodinger equation, which should have been listed as a "Relevant Equation" in the homework template. It also puts this in the Advanced Physics category, so I'll move the thread there.
    I don't think the constant is simply L -- the units are wrong -- but at any rate it would be found by normalizing the wavefunction.
    It represents the fact that the potential is not zero at the well boundary x=0. This is related to the fact that the potential is nonzero beyond the boundary as well, i.e. the particle's wavefunction penetrates beyond the boundary. I wouldn't really call that reflection though.
    where a = ???
    It's hard to tell exactly what you are stuck on. Is it just in trying to relate the two forms of solution you have given,
    [tex](s/k)A\sin kx + A\cos kx \ \small \text{ and } \normalsize \ B \sin(kx + d) \text{,}[/tex]
    to each other?
     
  4. Mar 18, 2012 #3
    Yea that's what I'm stuck on,

    Sorry that ka = ... formula I wrote was from a set of notes where the the width of the well is a,

    so a = L,

    I've tried substituting kl into the interior solution, ψ(L)
    where
    ψ(x)=(s/k)Asinkx + Acoskx
    so ψ(L)=(s/k)AsinkL + AcoskL
    and ended up with
    ψ(L)=-A cos(n[itex]\pi[/itex])
    assuming that the k in that formula is the same as the k i posted before
     
  5. Mar 19, 2012 #4

    Redbelly98

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    Try using trig angle-sum identities on sin(kx+d). I'm pretty sure that will work out to a form that you can more easily equate with
    (s/k)Asinkx+Acoskx​

    Okay.
    I'm not following you here, mainly because I can't do this in my head and don't have time right now to work it out on paper to verify what you are saying.

    But, if you are satisfied with the "(s/k)Asinkx+Acoskx" form of the solution, we should probably just concentrate on seeing how "Sin(kx + d)" is equivalent. Will you also be needing to normalize the wavefunctions, and finding the energy eigenvalues, or are you good with how to do that?
     
  6. Mar 22, 2012 #5
    Hey,

    I worked it out, the d in the sign was just to compensate for the lack of cos in the solution,

    I found the d was something like arcTan(k/s) and it was consistent with the boundaries and stuff,

    Thanks for your help!
     
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