Energies of bound state for delta function potential

[ErwinZumer]

Hi
Let's consider a potential of the form

The Schrödinger equation reads

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):

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?

 

[PeroK]

Hi
Let's consider a potential of the form
View attachment 332220
The Schrödinger 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

 

[hilbert2]

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.