Double delta-function potential

mattmatt321

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.

vela

[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.