The Delta-Function Potential

1. Oct 12, 2014

Logan Rudd

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. Oct 12, 2014

Matterwave

It does change. If you have a positive delta-function potential, there are no bound states.

3. Oct 12, 2014

Logan Rudd

Why is that, and why is it not the case for a negative delta-function potential?

4. Oct 12, 2014

Matterwave

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.

5. Oct 12, 2014

Logan Rudd

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.