Double delta function potential

In summary, you found that only one solution exists for the double delt potential, and you were not able to solve the equation for k using numerical methods.
  • #1
63
0

Homework Statement


Consider the double delts-function potential
[tex]
V(x)=-\alpha[\delta(x+a)+\delta(x-a)]
[/tex]
How many bound states does this possess? Find the allowed energies for
[tex]\alpha=\frac{\hbar^{2}}{ma^{2}}[/tex]and[tex]\alpha=\frac{\hbar^{2}}{4ma^{2}}[/tex]

Homework Equations


The Attempt at a Solution


I divided the region into three parts x<-a(Region 1) ; -a<x<+a(Region 2) ; x>+a(Region 3)
Since we consider bound states, E<0 and solving the SE yields
[tex]Ae^{kx} (x<-a)[/tex]
[tex]Be^{kx}+Ce^{-kx} (-a<x<a)[/tex]
[tex]De^{-kx}(x>a)[/tex]
where [tex]k=\frac{\sqrt{-2mE}}{\hbar}[/tex]
Continuity at x=-a and x=+a respectively give
[tex]A-B=Ce^{2ka}.....(1)[/tex]
[tex]D-C=Be^{2ka}.....(2)[/tex]

For this infinite potentials at points x=-a and x=+a,
[tex]\Delta(\frac{d\psi}{dx})=-\frac{2m\alpha}{\hbar^{2}}\psi(\underline{+}a)[/tex]
So these give two more BC

[tex]A(1-\frac{2m\alpha}{k\hbar^{2}})=B-Ce^{2ka}...(3)[/tex]
[tex]D(1-\frac{2m\alpha}{k\hbar^{2}})=C-De^{2ka}...(4)[/tex]
So I tried to solve these (eqns 1 to 4)and what I got was A=D and B=C
and taking A/B ratios from 1 and 3, i get [tex]ke^{4ka}=\frac{2m\alpha}{\hbar^{2}}[/tex]

And I am not able to solve this equation for k..i'd be grateful for any help :)
 
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  • #2
A breakthrough!
I tried graphical solution and its clear that there is only one bound state. But how do I calculate energy corresponding to that state without analytically solving this?
 
  • #3
so you obtained [tex] k [/tex]?

[tex]k=\frac{\sqrt{-2mE}}{\hbar}[/tex] ?
 
  • #4
No that's where I am stuck . How do I solve this equation for k?

[tex]
ke^{4ka}=\frac{2m\alpha}{\hbar^{2}}
[/tex]

By taking exp(4ka) to the other side, we can graphically find that only one solution exists. But how can we calculate that solution without using computers and numerical methods?
 
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  • #5
Raze2dust said:
No that's where I am stuck . How do I solve that equation?

By taking exp(4ka) to the other side, we can graphically find that only one solution exists. But how can we calculate that solution without using computers and numerical methods?

ah just solve it nummerical. Newton-Rhapson method is good.

It is quite often that in physics we encounter equations that are not analytically solvable, so we have to solve them by nummerical methods. Same holds for integrals and differential equations.
 
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  • #6
ok..thanks a lot!

actually i have not encountered ant problem in griffiths book SO FAR which requires numerical solution..so i was thinking if there is any way to solve it analytically. haven't studued Newton-Raphson method yet..I'll use MATLAB may be for now..
 
  • #7
I am not 100% sure that a solution to x*exp(x) = konstant, exists analytically.

Newton Raphson is something you do if you only have a pocket calculator. Matlab is easier if you just study on your own.

Good luck!
 

1. What is a double delta function potential?

A double delta function potential is a mathematical representation of a potential energy function in quantum mechanics. It consists of two delta functions, which are point-like interactions that can model the effects of an external force or potential on a particle.

2. How is a double delta function potential used in physics?

A double delta function potential is commonly used to model the behavior of particles in a system where there are two localized interactions or forces acting on the particles. It is often used in the study of quantum mechanical systems, such as in the field of solid state physics.

3. What are the properties of a double delta function potential?

The properties of a double delta function potential include:

  • It is localized at two points in space, represented by the two delta functions.
  • It is symmetric about both points of interaction, meaning the potential is the same at both points.
  • It is infinite at the points of interaction, representing an infinitely strong force or potential.
  • It is zero everywhere else in space.

4. How is a double delta function potential different from a single delta function potential?

A single delta function potential only has one point of interaction, while a double delta function potential has two. Additionally, the potential energy function for a single delta function potential is asymmetric, while the potential energy function for a double delta function potential is symmetric.

5. What are some applications of a double delta function potential?

A double delta function potential has various applications in physics, including:

  • Modeling the behavior of particles in a crystal lattice in solid state physics.
  • Studying the scattering of particles off of a potential barrier or well.
  • Exploring the effects of localized interactions on the behavior of quantum systems.
  • Investigating the behavior of particles in a two-dimensional system.

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