The Double Dirac Delta Function Potential wave functions

1. Aug 19, 2013

Positron137

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

2. Relevant equations
Time-Independent Schrodinger's Equation: Eψ = (-h^2)/2m (∂^2/∂x^2)ψ + V(x)ψ. (for the single-variable function ψ(x))

3. 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!

2. Aug 19, 2013

TSny

You should be able to write out the condition of continuity of the wave function at x = a and also x = -a. You will get a relation between A, C, D, and e2ka and a similar relation between F, C, D, and e2ka.

Since the potential function is symmetric about x = 0, you can assume the wavefunctions are either even functions or odd functions of x. This will allow you to easily relate A to F and C to D for the even and odd cases. If you don't want to assume the wavefunctions are even or odd, then the algebra is a little longer and you will end up showing that the wavefunction is either even or odd.

The derivative of the wavefunction is not continuous at the location of the delta functions. Hopefully, you've covered how to handle this. If not, you can find a discussion here .

3. Aug 19, 2013

Positron137

Ok. Thanks! So I can equate the wave functions for the x < -a to the -a < x < a regions first; then equate the wave functions for the -a < x < a and the x > a regions right? And it's perfectly fine of the factors e^(ka) and e^(-ka) remain there right?

4. Aug 19, 2013

TSny

Yes, that's right.

5. Aug 20, 2013

Positron137

Ok. Thanks! Yah, I was getting a bit confused whether to include the e^(ka) and e^(-ka) terms in the first place, but it makes sense now. :)