# Potential well with 2 delta potentials

I have a problem to derive transcendental equation for eigenenergies of bounded states for potential: $$V(x)= \begin{cases} -\lambda\delta(x+a/4)-\lambda\delta(a-a/4), & |x|a/2 \\ \end{cases} \$$ I am familiar with solution for "classic" potential well, where we start with $$\psi_1, \psi_2, \psi_3$$ _iewith $$\psi_i=A_ie^{\kappa x}+B_ie^{-\kappa x}$$. So there ve divide the x axis into 3 areas $$x<-a/2, -a/2a/2$$ and than calculate coefficients. But how to start when we have additional 2 delta functions in our potential? My initial idea was to define 5 areas on x axis, so, that we will have piecewise defined wavefunction? What is the best approach to start that kind of problem?

• So, I found out, that I actually need 3 areas, since it is infinite deep potential whel, so positions outside aren't allowed. But I am wondering if wavefunction is contineuosus in delta potentials?
– Vid
Commented Jan 26, 2021 at 0:09

The infinite potential in the outer regions $$|x|> a/2$$ means your wavefunction should be zero there. That leaves you with the interval $$-a/2 < x < a/2$$, which has two negative deltas at $$x=-a/4$$ and $$x=a/4$$. These deltas divide your interval in three parts. The potential is zero in these three parts, such that you can use the free particle solutions you mentioned, and enforce the usual boundary conditions of continuity of the wavefunction on the critical points $$x=-a/2$$, $$x=-a/4$$, $$x=a/4$$ and $$x=a/2$$ (note that the usual requirement that the derivative of the wavefunction be continuous does not apply here, as we are dealing with infinite potentials, and the derivative need not be continous in points where the potential is infinite). So your boundary conditions should read:

$$\psi_1(-\frac{a}{2}) = 0$$,

$$\psi_1(-\frac{a}{4}) = \psi_2(-\frac{a}{4})$$,

$$\psi_2(\frac{a}{4}) = \psi_3(\frac{a}{4})$$,

$$\psi_3(\frac{a}{2}) = 0$$

That's 4 equations for 4 of the 6 indeterminate coefficients of your solution. An extra one can be obtained by imposing normalization: $$\int_{-a/2}^{a/2} |\psi(x)|^2 dx = 1$$. That leaves your solution in terms of a single undeterminate coefficient.

To get the final solution, including the transcendental equation for $$E$$, it is essential that you actually consider the effects of the deltas (otherwise you'll just end up with a free-particle solution). The way to do so is to integrate the Schrödinger equation in an infinitesimal interval around the deltas. It's a tricky and specific procedure, but if you're solving the double delta potential, I assume you've solved at least some simple delta potential before, and so you've probably seen it before. If not, the wikipedia entry on Delta Potential features a decent explanation. So, by integrating around one of the deltas you get your final coefficient, and, with the final solution in hands, you integrate around the other delta to get the (in this case, transcendental) equation for E.

• Well, we actually only metioned delta potential, this problem is from couple years before. I checked Wiki and I have a general idea... but... I assume, that I have to integrate around booth deltas, not just one?
– Vid
Commented Jan 26, 2021 at 9:04
• Exactly, you must integrate around both deltas; one of them will give you a relation between the coefficients of your solution (and you'll be able to find out the last remaining constant), and the other one will give you an equation for E. Commented Jan 26, 2021 at 19:05