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Double delta-function potential

  1. Mar 1, 2011 #1
    Hello, recently I had a problem in QM involving a double delta-function potential. A bunch of qualitative questions were asked, some of which were obvious to me immediately, some of which I made an educated guess on, and others I totally guessed. I'm following Griffiths' textbook in studying QM, and luckily I found another problem analogous to the one I had before. Any guidance/explanation would be appreciated.

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

    The potential is V(x) = -α[[tex]\delta[/tex](x + a) + [tex]\delta[/tex](x - a)].

    The problem specifies that we are only conserved about bound states, E < 0.

    2. Relevant equations

    In the end, the even wave function solution yields (ħk/mα) = e-2ka + 1, and the odd wave function solution yields (ħk/mα) = 1 - e-2ka.

    3. The attempt at a solution

    Based on these solutions, I'm asked a few questions:

    1) What is the approximate energies of both the even and odd bound states in the limit 2maα/ħ2 >> 1?
    2) Show that as 2maα/ħ2 --> [tex]\infty[/tex], both energies converge to the energy of the bound state of the single delta-function potential.
    3) Show that there is only one bound state in the limit 2maα/ħ2 << 1.
    4) Show that the wave function behaves like the bound-state wave function of a single delta-function potential for |x| >> a.

    Any guidance regarding these questions is appreciated. I might be able to provide more information if anything is unclear.
  2. jcsd
  3. Mar 2, 2011 #2


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    [itex]\hbar[/itex] should be squared in both equations.
    You can write the relations above as

    [tex]\frac{2ka}{2ma\alpha/\hbar^2} = 1\pm e^{-2ka}[/tex]

    Let [itex]c=2ma\alpha/\hbar^2[/itex] and [itex]x=2ka[/itex]. Then you get

    [tex]\frac{x}{c} = 1 \pm e^{-x}[/tex]

    You might find it illuminating to plot both sides of the equation to see where the solutions are and what the effect of varying the parameter c is.
    Last edited: Mar 2, 2011
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