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Double Delta Function Potential

  1. Feb 6, 2007 #1


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    1. The problem statement, all variables and given/known data
    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)

    2. Relevant equations
    [tex] V(x) = -\alpha(\delta(x+a) +\delta(x-a))[/tex]

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

    3. 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.
  2. jcsd
  3. Feb 7, 2007 #2

    Meir Achuz

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    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.
  4. Feb 7, 2007 #3


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    Okay here's what I got so far

    1. The problem statement, all variables and given/known data
    Consider the Double Delta Function Potential:

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

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

    3. 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]\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.
  5. Feb 8, 2007 #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).
  6. Feb 11, 2007 #5


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    I'm sorry ive 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?
  7. Feb 11, 2007 #6


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    Ok I've worked on it some more and heres 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:


    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: Feb 11, 2007
  8. Jan 18, 2009 #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

  9. Dec 8, 2011 #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



    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
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