What are the Bound States for a Sum of Two Negative Delta-Function Potentials?

In summary, Griffiths Intro to QM 2nd Ed. explains that for bound/scattering states, ##E<0## corresponds to a bound state and ##E>0## corresponds to a scattering state. However, this changes depending on whether the potential is positive or negative. A positive delta-function potential only has scattering states because it cannot trap a particle into a bound state. On the other hand, a negative delta-function potential has one bound state, which must be solved for separately. If the potential is the sum of two negative delta-function potentials, there will be two bound states, but the problem must be solved in sections for each potential centered at ##\pm a##.
  • #1
Logan Rudd
15
0
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?
 
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  • #2
It does change. If you have a positive delta-function potential, there are no bound states.
 
  • #3
Why is that, and why is it not the case for a negative delta-function potential?
 
  • #4
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
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.
 

1. What is the Delta-Function Potential?

The Delta-Function Potential is a mathematical model used in quantum mechanics to describe the interaction between a particle and a potential barrier. It is represented by a Dirac delta function, which is a function that is zero everywhere except at a single point, where it has an infinite value.

2. How is the Delta-Function Potential used in physics?

The Delta-Function Potential is used to model various physical systems, such as electron scattering, nuclear reactions, and quantum tunneling. It is also used to simplify the calculations in quantum mechanics, as it allows for the description of a potential barrier with a single point in space.

3. What are the properties of the Delta-Function Potential?

The Delta-Function Potential has a few important properties, such as being infinite at a single point, zero everywhere else, and having an area of one under its curve. It is also symmetric, meaning that the potential is the same on both sides of the barrier.

4. How does the strength of the Delta-Function Potential affect the behavior of a particle?

The strength of the Delta-Function Potential determines how much the particle is affected by the barrier. A stronger potential will cause the particle to have a higher probability of being reflected off the barrier, while a weaker potential will allow for more transmission through the barrier.

5. Can the Delta-Function Potential be extended to multiple dimensions?

Yes, the Delta-Function Potential can be extended to multiple dimensions by using a multidimensional Dirac delta function. This allows for the modeling of more complex systems, such as potentials with barriers in multiple directions or potentials with varying strengths in different dimensions.

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