# Double and single delta-function potential well energy similarity

I am working through Griffiths QM at present, and I came across a question that asks me to find

"the bound state energies in the limiting cases (i) $$a \to 0$$ and (ii) $$a \to \infty$$ (holding $$\alpha$$ fixed)"

for the double delta-function potential $$V(x)= - \alpha(\delta(x+a)+\delta(x-a))$$. After manipulating the time-independent Schrödinger equation for this potential, I have arrived at the transcendental equation

$$\left( 1- \frac{\hbar^2 \kappa}{m \alpha}\right)^2 = e^{-4 \kappa a},$$

where $$\kappa = \sqrt{-2 m E}/\hbar$$. I omit this computation since I think it distracts from my question, but I can include it if needed. Now, the aforesaid limiting cases are obvious, with $$a \to 0$$ first yielding $$E = - \frac{2 m \alpha^2}{\hbar^2}.$$ I am very willing to accept this intuitively since this is nothing but the bound state energy associated with the single delta-function potential $$V(x) = - 2\alpha \delta(x)$$; we are, in a manner of speaking, doubling the "strength" of the well when we push these two equally "strong" wells together.

What I am not willing to accept however, is when I take the limit $$a \to \infty$$, which implies $$E = - \frac{m \alpha^2}{2 \hbar^2},$$ which is exactly the energy associated with the potential $$V(x) = -\alpha \delta(x)$$. How can it be that two identical potential wells that become increasingly more spatially removed from each other have a bound state energy that increasingly more resembles that of a single delta-function well of equivalent "strength" $$\alpha$$? Is this just a coincidence, or can I get anything inutitive out of this, like I can with the $$a \to 0$$ case?

As the two delta peaks are pulled apart, they influence each other less and less. In the limit $$a \to \infty$$, the bound state energies are completely determined by the bound state energies of a single peak. Maybe it might help you to draw the approximate ground state wavefunction as you pull $$a \to \infty$$.
• Oh, I see. In the limit $a \to \infty$, there is a "traveling" wavefunction that much resembles the solution to the TISE with $V(x)=-\alpha \delta(x)$ and $E=-\frac{m \alpha^2}{2 \hbar^2}$. Thanks! I tried to graph this earlier explicitly (desmos.com/calculator/ztmmqopwox) but was met with a graphing error at $a \approx 89$. Commented Aug 15 at 5:43