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Symmetric delta potential

  1. Sep 25, 2009 #1
    This problem is a symmetric delta potential problem that I was given a few days ago and I cant seem to get the gist of it.


    Find the spectrum and wave functions of a particle in the potential V(x)=G[d(x-a)+d(x-a)] Calculate the transmission and reflection amplitude. Where G can be positive or negative.


    In essence I believe this is 3 problems. When G is negative I must consider E>0 and E<0 and when G is positive I must consider E>0

    I have solved the problem for negative G and E<0. It wasnt that bad. I set up decaying exponentials outside of -a<x<a and had a symmetric and antisymmetric wave functions designated by cosh(x) and sinh(x) inbetween.

    The problem comes when I attempt the solutions with positive energy. From my undergrad course in QM I remember that there are incident and reflected waves in each region except for one (the outgoing wave).

    1)Does this mean that i can set up sin/cos waves in two of the 3 regions to represent symmetric/antisymmetric solutions?

    2)If this is so then do I need some kind of phase shift in order to meet boundary conditions?

    3)How do I decompose the sin/cos wave that represents the incoming wave to delineate the incoming/reflected wave. It seems to me that the single coefficient in front of the the sin/cos wave would remove the amplitude characteristic you need for reflection and transmission coefficients.

    I have attempted to plug-and-chug my way through with all 5 coefficients in front of the incoming/outgoing waves but this is giving me a headache. If there is anyway to simplify the setup I would really like to know.

    thanks in advance
  2. jcsd
  3. Sep 25, 2009 #2


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    You need to be a bit careful with the symmetry. Assuming that the wave is incident from the left (region I), I would write the wavefunction in that region as the sum of an incident wave with amplitude 1 and a reflected wave with amplitude R (generally complex) and use complex exponentials, namely


    The complex exponentials allow you to sort out quite easily what is moving to the left and what is moving to the right.

    In the in-between region you will have


    and in the last region


    Yes, you have to match boundary conditions and use them to eliminate complex constants A and B. The reflection coefficient will be R*R and the transmission coefficient T*T.
  4. Sep 25, 2009 #3

    I have attempted this method, and it gets messy quick. The main problem I have is attempting to sort through the amplitudes of 5 waves to get to one. I assumed my problem was setting up the wavefunctions and that if I could find a simplier representation then it would only be a page or so of work to deduce the relevant equations.

    I then considered the fact that using 5 different waves does not take advantage of the symmetry of the potential at all. So maybe I should do that.

    Now that I think about it again, could I not set one of the delta functions at x=0. then when I attempt to satisfy boundary conditions I will get exactly half as many.

    thanks again

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  5. Sep 25, 2009 #4


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    Like I said, be careful when you use symmetry. Although the potential is symmetric and therefore you have even and odd solutions, because you only have an outgoing wave in region III, you will have to find the appropriate linear combination of even and odd that will give that. Personally, I prefer the brute force messy algebra than the nagging doubt of whether I oversimplified or not. Good luck.
  6. Sep 25, 2009 #5
    It only took till noon, but i found the problem in my derivation.

    The real problem did not lie in the symmetry of the problem, but of defining psi(-a) and psi(a)

    For an easy review we have the delta function at the origin:

    -There are decaying exponentials each side:

    region 1 is x<0
    region 2 is x>0


    due to the continuity of psi we have A=B

    then when we evaluate the discontinuity of the slope we show


    this is from integrating the shrodinger eqn.

    and psi=Aexp(k|x|)

    so you end up with your psi(0) being easy to evalute.


    Back to the double delta function, our continuity of psi does not simplify or define psi(-a) or psi(a) because on both sides there are incoming and outgoing waves.

    so I'm not sure how to define this psi at all.


    and at least I'm not staring off into space anymore. thats frustrating.
  7. Sep 25, 2009 #6


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    kuruman explained how to do it. But here is a method that simplifies the algebraic manipulations you need to do.

    Symmetry is no help in the scattering problem, because the boundary condition (incoming and outgoing waves on the left, outgoing only on the right) breaks the symmetry.

    I find it is easiest to use the most general wave function in each region, and impose the boundary condition of no outgoing wave on the right only at the very end.

    So, generalizing what kuruman said, I would start with




    Then, matching at x=-a will give you two equations that you can write in matrix form,

    [tex]\left( {\begin{array}{cc}
    a_{11} & a_{12} \\
    a_{21} & a_{22}
    \end{array} } \right)
    \left( {\begin{array}{cc}A \\ B
    \end{array} } \right)
    \left( {\begin{array}{cc}
    b_{11} & b_{12} \\
    b_{21} & b_{22}
    \end{array} } \right)
    \left( {\begin{array}{cc}C \\ D
    \end{array} } \right), [/tex]

    where the matrix elements aij will involve factors like eika.

    Similarly, matching at x=+a will give you two more equations of the form

    [tex]\left( {\begin{array}{cc}
    c_{11} & c_{12} \\
    c_{21} & c_{22}
    \end{array} } \right)
    \left( {\begin{array}{cc}C \\ D
    \end{array} } \right)
    \left( {\begin{array}{cc}
    d_{11} & d_{12} \\
    d_{21} & d_{22}
    \end{array} } \right)
    \left( {\begin{array}{cc}E \\ F
    \end{array} } \right). [/tex]

    What you want to do is find the matrix elements eij, where

    [tex]\left( {\begin{array}{cc}A \\ B
    \end{array} } \right)
    \left( {\begin{array}{cc}
    e_{11} & e_{12} \\
    e_{21} & e_{22}
    \end{array} } \right)
    \left( {\begin{array}{cc}E \\ F
    \end{array} } \right). [/tex]

    Note that the matrix e can be written as e = a-1bc-1d, and that it's easy to invert a 2x2 matrix:

    [tex]\left( {\begin{array}{cc}
    a_{11} & a_{12} \\
    a_{21} & a_{22}
    \end{array} } \right)^{\!\!-1} =
    {1\over a_{11} a_{22}- a_{12} a_{21} }
    \left( {\begin{array}{cc}
    a_{22} & -a_{12} \\
    -a_{21} & a_{11}
    \end{array} } \right)[/tex]

    Now, set F=0. The reflection probability is then R=|B|2/|A|2 and the transmission probability is T=|E|2/|A|2. You can also just set E=1, since E will cancel out in R and T.
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