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The Delta-Function Potential

  1. Oct 12, 2014 #1
    I'm reading through Griffiths Intro to QM 2nd Ed. and when it comes to bound/scattering states (2.5) they say:

    ##E<0 \implies## bound state

    ##E>0 \implies## scattering state

    Why doesn't this change depending on whether you have a positive or negative delta-function potential?
     
    Last edited: Oct 12, 2014
  2. jcsd
  3. Oct 12, 2014 #2

    Matterwave

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    It does change. If you have a positive delta-function potential, there are no bound states.
     
  4. Oct 12, 2014 #3
    Why is that, and why is it not the case for a negative delta-function potential?
     
  5. Oct 12, 2014 #4

    Matterwave

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    A positive delta function potential only has scattering states since it's just a scattering problem. The reason it has no bound states is the same reason that a positive finite potential well has no bound states. A positive delta-function potential has no way to "trap" a particle into a bound state.

    A negative delta-function potential has 1 (and only 1) bound state, but to figure that out, you actually have to go ahead and solve the problem.
     
  6. Oct 12, 2014 #5
    Ahh, I see! But if the potential I am working with is the sum of two negative delta potentials then would there be two bound states? I'm trying to work it out in a similar fashion as the text works it out for a a single negative delta potential centered at 0 but since both of mine are centered about ##\pm a## how do I break it up into sections. I tried analyzing from ##x<-a##, ##-a<x<0## to get ##\psi(x) = Be^{-\kappa x}## and for ##x>a##, ##0<x<a## to get ##\psi(x) = Be^{-\kappa x}## but it seems like there is more to it than that.
     
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