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?
