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Finite square well

  1. Mar 3, 2005 #1

    I have a problem with the finite square well. I have to analyze the odd bound states of the finite square well,

    [tex] V(x)=
    -V_0 & \text{for } -a<x<a\\
    0 & \text{otherwise}

    Specifically, I have to examine the limiting cases (wide, deep well and narrow, shallow well) and find out, if there is always at least one odd bound state.

    When I try to determine the energies of these odd states, I find that E=0 is always a solution. Is E=0 a bound state, a scattering state or what?

    Also, what exactly are scattering states?
  2. jcsd
  3. Mar 3, 2005 #2


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    Okay,did you find the "normal" /general sollution...?Odd states refer to the behavior of wavefunctions under parity...In your case,depending on the E:V,it could be only [itex] \sin [/itex] or [itex] \sinh [/tex].

    Bound states are normalizable states,physical states according to I-st postulate...In your case,which would be those...??
    Scattering states would correspond to nonnormalizable states...

  4. Mar 3, 2005 #3


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    Scattering states are those that satisfy the Somerfeld Radiation Condition, which is obeyed if a state behaves like a plane wave at infinity. Drop a rock in a water wave, say from a speed boat. The wave pattern will settle down and stabilize-- the original wave will still be going, as will "scattered waves" generated by the rock -- and these waves will behave like free waves, once they've gone out a bit from the rock's splash.
    Reilly Atkinson
  5. Mar 3, 2005 #4
    My question is really, does E=0 correspond to a bound state? I hope not, because this implies that there is always one odd, bound state (and there isn't according to the book).
  6. Mar 3, 2005 #5


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    No,the wave function is a "sine",which is not a bound state...It's not normalizable...

  7. Mar 3, 2005 #6
    Ok, thanks!
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