Consider the double Dirac delta function V(x) = -α(δ(x+a) + δ(x-a)). Using this potential, find the (normalized) wave functions, sketch them, and determine the # of bound states.
Time-Independent Schrodinger's Equation: Eψ = (-h^2)/2m (∂^2/∂x^2)ψ + V(x)ψ. (for the single-variable function ψ(x))
The Attempt at a Solution
First, I substituted the relevant potential: so the TISE (Time-Independent Schrodinger Equation) became: Eψ = (-h^2)/2m (∂^2/∂x^2)ψ - α(δ(x+a) + δ(x-a))ψ.
Next, I considered the points x = -a and x = a, since those are the points where V(x) contains an infinite discontinuity.
Then, I divided this problem into 3 sections: x < -a, -a < x< -a, and x > a. Unfortunately, I wasn't able to finish the solution for the bound states, but nevertheless I started with bound-state solutions (E < 0).
I solved the general solution for x < -a, which was ψ(x) = Ae^(kx) + Be^(-kx). Since this blows up when x -> -∞, the physical solution would be ψ(x) = Ae^(kx).
For the section -a < x < a, I got as a general sol'n, ψ(x) = Ce^(kx) + De^(-kx). However, this time, ψ(x) does not blow up since x is constrained between -a and a.
For the section x > a, after I got the general solution and modified it (to take care of the infinite part), I got ψ(x) = Fe^(-kx).
Then I applied the continuity conditions. The first one states that ψ(x) must be continuous at the discontinuity points. But I wasn't sure how to patch up the discontinuity here. I couldn't just simply match coefficients, because the points of discontinuity weren't at x = 0, but at x = ±a. From this point, I knew that if I could satisfy the first condition, then by integrating the TISE, I could satisfy the second boundary condition (which was that ψ'(x) had to be continuous). From there, I could normalize ψ(x) (and obtain the wave function(s) for bound states), and determine the allowed energies. But I got stuck at the point where I had to patch up the discontinuities by satisfying the 2 boundary conditions. If anyone could help from this point onwards, I would really appreciate it. Thanks!