Energies of bound state for delta function potential

  • #1
ErwinZumer
1
1
Hi
Let's consider a potential of the form
1695146975780.png

The Schrodinger equation reads
1695147036236.png

as shown in the book 'Introduction to Quantum mechanis' by D.J. Griffiths, Chaper 2 Section 5, the solution of the equation yields (only for bound state, which means E<0):
1695147197115.png

My question:
if i have one particle and i apply this potential on it, how does it take one value of energy?
another question is, if i have one paricle with some energy, E, before applying this potential, does the energy of the particle will change?
 
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  • #2
ErwinZumer said:
Hi
Let's consider a potential of the form
View attachment 332220
The Schrodinger equation reads
View attachment 332222
as shown in the book 'Introduction to Quantum mechanis' by D.J. Griffiths, Chaper 2 Section 5, the solution of the equation yields (only for bound state, which means E<0):
View attachment 332223
My question:
if i have one particle and i apply this potential on it, how does it take one value of energy?
another question is, if i have one paricle with some energy, E, before applying this potential, does the energy of the particle will change?
There are parallels, of course with the hydrogen atom, where an electron may settle into a bound state around the nucleus by emitting a photon that carries away the excess energy. Does this happen for the delta function potential? Good question.

The physical significance of the delta potential is probably in the scattering states and the transmission and reflection coefficients.

There's more here to get you started if you want to investigate further:

https://en.wikipedia.org/wiki/Delta_potential
 
  • #3
That delta function potential is like the limit of a finite square well, ##V(x) = -V_0## when ##|x|<L/2## and ##V(x) = 0## when ##|x|\geq L/2##, if you start making it more narrow by decreasing ##L## and increasing ##V_0## at the same time so that the product ##V_0 L## stays at constant value ##\alpha##.

It's probably difficult to explain in a pictorial way why this system has exactly one bound state in the limit where ##L## is very close to ##0## and ##V_0## is a very large number.

A more difficult problem would be to find out how much more quickly you would have to increase well depth ##V_0## compared to the rate of ##L## decrease so that after some minimum value of ##V_0## it would have exactly two bound states. But it's impossible to define a generalized function similar to ##\delta (x)## that would describe the high-##V_0## limit of this situation.
 

1. What is a bound state in a delta function potential?

A bound state in a delta function potential refers to a quantum state in which a particle is confined within a finite region of space due to the presence of a delta function potential. This potential is a localized potential that has an infinite value at a single point and zero value everywhere else.

2. How are the energies of bound states calculated for a delta function potential?

The energies of bound states for a delta function potential can be calculated using the Schrödinger equation and boundary conditions. The solutions to the Schrödinger equation will give the allowed energy levels for the bound states.

3. What is the significance of the delta function potential in quantum mechanics?

The delta function potential is a useful mathematical tool in quantum mechanics as it allows for the modeling of localized interactions between particles. It is also used to study the behavior of particles in confined spaces, such as in quantum wells or quantum dots.

4. Can a particle in a delta function potential have multiple bound states?

Yes, a particle in a delta function potential can have multiple bound states. The number of bound states depends on the strength of the potential and the mass of the particle. A stronger potential or a lighter particle will result in a larger number of bound states.

5. How does the energy of a bound state change with the strength of the delta function potential?

The energy of a bound state increases as the strength of the delta function potential increases. This is because a stronger potential will confine the particle to a smaller region of space, resulting in a higher energy state. Conversely, a weaker potential will allow the particle to spread out over a larger region of space, resulting in a lower energy state.

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