Double Delta Function Potential

In summary, this individual solved the homework problem by finding the allowed energies for the potential using the transcendental equation.
  • #1
G01
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Homework Statement


How many stationary states exist for this potential? What are the allowed energies if the strength of the well, [tex]\alpha= \hbar^2/ma[/tex] and [tex]\hbar^2/4ma[/tex] where a= the position of the well(one at a, one at -a)

Homework Equations


[tex] V(x) = -\alpha(\delta(x+a) +\delta(x-a))[/tex]

[tex]E_{one delta well} = -m\alpha^2/2\hbar^2[/tex]

The Attempt at a Solution



OK, i may need to do more work for this but I want to see if I may not need to. First, the potential should have two bound states, one for each well, since one delta well can have one bound state.

Next, this is my question can I use the formula derived for the energy of the bound state of one delta well above and just assume the energies are the same for each state? Or do I have to go and solve for a new energy formula from the shrodinger EQ. If this is the case, please tell me and I'll work on it and put up my work for that. Thanks for your insight.
 
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  • #2
You have to start from scratch.
Find the WF in three regions and apply the BC.
You now can have an oscillating WF between the deltas.
 
  • #3
Okay here's what I got so far

Homework Statement


Consider the Double Delta Function Potential:

[tex]V(x) = -\alpha(\delta(x+a) + \delta(x-a))[/tex]

Homework Equations


Time Independent SEQ:
[tex] \frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} +V(X)\psi = E\psi[/tex]

The Attempt at a Solution



Well, with the given potential the TISEQ becomes:
[tex] \frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} +-\alpha(\delta(x+a) + \delta(x-a))\psi = E\psi[/tex]

Next I considered the following three sections: x<-a, -a<x<a, x>a

In x>a:

[tex]k=\frac{\sqrt{-2mE}}{\hbar}[/tex]

[tex]\psi(x) = Ae^{-kx} + Be^{kx}[/tex] but this has to be normalizable so:

[tex]\psi(x) = Ae^{-kx}[/tex]

In -a<x<a:

[tex]\psi(X)= Ce^{-kx} +De^{kx}[/tex]

In x<- a:

[tex]\psi(x) = He^{kx}[/tex] for the same reasons as for x<-a

Okay now I can set each solution equal to each other at a and -a to eliminate the number of constants:

[tex] \psi = Ce^{-kx} + De^{2ka}e^{-kx} x>a[/tex]
[tex]\psi = Ce^{-kx} + De^{kx}-a<x<a[/tex]
[tex] \psi = Ce^{2ka}e^{-kx} +De^{kx}x<-a[/tex]Next I followed to books method to find [tex]\Delta\frac{d\psi}{dx}[/tex] around the "kinks" in the graph to find the Allowed energy level for the well under that kink. Griffiths did this by integrating the SEQ and then finding the change in derivative in an interval around the well and letting the interval go to 0. Doing this I got the following:

[tex]\Delta\frac{d\psi}{dx}_{at a} = -2Dk(e^{ka}+e^{-ka}) = -2m\alpha/\hbar^2 \psi(a) = -2m\alpha/\hbar^2 * (Ce^{-ka} +
De^{ka})[/tex] Now my problem becomes solving for k to find the allowed energy in the well at a. I can't get rid of the constants. Some help please? Thank you in advance.
 
  • #4
Remember that the wavefunctions of an even potential are either fully even or fully odd, that should reduce the number of constants.

Also to get the allowed energies for the potential (not for the allowed energy in one well) you need to solve a transcendental equation (one equation for even solutions and one for odd solutions).
 
  • #5
I'm sorry I've been working on this for a while and I'm really lost. I'm trying to teach this to myself and this is the only section that has screwed me up so far. Could someone please show me where I should go with the even and odd solutions?
 
  • #6
Ok I've worked on it some more and here's where I'm at:

I've tried to consider the even solutions to the SEQ with this potential. they will be of the form:

[tex]\psi(x) = Ae^{-kx} x>a[/tex]
[tex]\psi(x) = Dcos(kx) 0<x<a[/tex]
[tex] \psi(x) = \psi(-x) -a<x<0[/tex]Now from here I know that:

[tex]Ae^{-kx}=Dcos(kx)[/tex]

And then [tex] \Delta(\frac{d}{dx}\psi(x)) = -Ake^{-ka} - Dksin(ka)[/tex] around the kink at a. Now I'm lost as to where I should go from here... Help?
 
Last edited:
  • #7
Dear All

I have seen these postings concerning the double dirac delta function model. The
problem has been analytically solved. The solutions for the eigenenergies are in
terms of the (standard) Lambert W function for the case of equal charges.

This reference shows you how to solve this problem:

T.C. Scott, J.F. Babb, A. Dalgarno and John D. Morgan III, "The Calculation of Exchange Forces: General Results and Specific Models", J. Chem. Phys., 99, pp. 2841-2854, (1993).

For the case of unequal charges, the eigensolutions involved a generalization of the
Lambert W function.

See the wikipedia side for the Lambert W function (and references herein).

best wishes

Tony
 
  • #8
It is also possible to use the symmetry of the well and the parady of the solution to find the last criteria needed to show c = d after finding
ψ=Ce−kx+De2kae−kxx>a

ψ=Ce−kx+Dekx−a<x<a

ψ=Ce2kae−kx+Dekxx<−a


you can then use the property of the delta function that lim[itex]_{b->0}[/itex] of ∫[−ℏ22md2ψdx2+V(X)ψ=Eψ] for bounds over a small variation [-b to b] over the one of the dirac peaks (your choice) to find a quantisation of the energies through k.
Hint: lim[itex]_{b->0}[/itex] [ ∫δ(x)ψ(x)dx] = ψ(0) evaluated from -b to b
 

1. What is a double delta function potential?

A double delta function potential is a type of potential energy function commonly used in quantum mechanics. It consists of two delta functions, which are mathematical functions used to model point particles, placed at a certain distance apart. This potential is often used to study the behavior of particles in a two-dimensional system.

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

Unlike a single delta function potential, which only has one delta function, a double delta function potential has two delta functions. This means that the potential energy is concentrated at two specific points instead of just one. This can have significant effects on the behavior of particles in the system.

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

A double delta function potential has various applications in quantum mechanics, particularly in the study of two-dimensional systems. It is commonly used to model the behavior of particles in a confined space, such as in quantum dots or semiconductor devices. It is also used in the study of quantum tunneling and scattering processes.

4. How is a double delta function potential mathematically represented?

A double delta function potential is represented by the mathematical expression V(x) = αδ(x - a) + βδ(x - b), where α and β are constants representing the strength of the potential at points a and b, respectively. This potential is often used in the Schrödinger equation to solve for the wave function and energy levels of particles in the system.

5. What are the limitations of using a double delta function potential?

One limitation of using a double delta function potential is that it is a highly idealized model and may not accurately represent real-world systems. It also does not take into account the effects of other forces and interactions between particles. Additionally, the potential is infinite at the points where the delta functions are located, which may lead to mathematical difficulties in some calculations.

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