From Griffiths, Introduction to Quantum Mechanics, pg. 73:
Evidently, the delta-function well, regardless of its "strength" $\alpha$, has exactly one bound state
$$\psi(x) = \frac{\sqrt{m \alpha}}{\hbar} e^{-m \alpha |x| / \hbar^2} ; \qquad E = – \frac{m \alpha^2}{2 \hbar^2} \, .$$
I don't understand how the author concludes that the delta-function well has one bound state. I understood every part of the derivation that led to the 2 equations above, but I don't see how all of these conclusions point to the delta-function well having one bound state. Also, I don't understand how the delta-function can be a "well" in the first place. It's a spike!
I understand a bound state as being a state that the particle is in where it's energy is less than the potential that bounds it, so because it doesn't have the energy required to surpass the potential, it can never leave the region of potential.
The parameters of the problem:
\begin{align}
\psi(x)=&Be^{kx} \quad\text{for } x \leq 0 \\
\psi(x)=&Be^{-kx} \quad \text{for } x \geq 0 \\
V=&- \alpha\ \delta(x) \\
k=&\frac{\sqrt{-2mE}}{\hbar} \, .
\end{align}
Best Answer
Here is an argument.
On one hand, in 1D the $n$th bound state has $n\!-\!1$ nodes. But a bound state in the delta function well is in a classically forbidden region (and hence exponentially decaying) for $x\in \mathbb{R}\backslash\{0\}$, so there cannot be any nodes. Hence $n\leq 1$.
On the other hand, any attractive potential in 1D has a bound state, so $n\geq 1$.