Problem 2.27.b.Consider the double delta-function potential $V(x) = -\alpha[\delta(x + a) + \delta(x - a)]$, where $\alpha$ and $a$ are positive constants. How many bound states does it possess? Find the allowed energies, for $\alpha = \hbar^2/ma$ and for $\alpha = \hbar^2/4ma$, and sketch the wave functions. Solution. There are two states that satisfy the time-independent Schroedinger equation for the given potential: $\psi_1 = A(\psi(x + a) + \psi(x - a))$ and $\psi_2 = A(\psi(x + a) - \psi(x - a))$, where $\psi$ is the state corresponding to the Dirac delta distribution potential $\psi = B e^{-\kappa|x|}$.
For $x < -a$ the time-independent Schroedinger equation is $-\frac{\hbar^2}{2m} \psi_{xx} = E\psi$. Inserting $\psi_1$ or $\psi_2$ yields $\kappa^2= -\frac{mE}{2\hbar^2}$, a relation between the unknowns $\kappa$ and $E$. Same holds for $x \in (-a, a)$ and $x > a$.
At $x = -a$, the time-independent Schroedinger equation becomes $-\frac{\hbar^2}{2m} \psi_{xx} - \alpha \delta(0) \psi = E\psi$. Inserting $\psi_1$ and integrating from $-\epsilon$ to $\epsilon > 0$ yields $\psi_x (\epsilon) - \psi_x(-\epsilon) + 2m\alpha/\hbar^2 \psi(0) = 0$. In the limit, $\psi_x (\epsilon) \rightarrow -B\kappa$ and $\psi_x (-\epsilon) \rightarrow B\kappa$ such that $2B \kappa = 2m\alpha/\hbar^2 (\psi(0) + \psi(-2a))$ which reduces to $\kappa = m\alpha/\hbar^2(1 + e^{-2a\kappa})$.
According to the Lambert function wikipedia article, it was reported in a 1993 article "that the Lambert $W$ function provides an exact solution to the quantum-mechanical double-well Dirac delta function model for equal charges". For $q \equiv m\alpha/\hbar^2$ and $R \equiv 2a$, the analytical solution takes the form $\kappa = q + W(qRe^{-qR})/R$.
Choosing $\alpha = \hbar^2/ma$, the equation takes the special form $\kappa = (1 + W(2e^{-2})/2)/a$, and the associated energies are $E_1 = -2\hbar^2/(a^2m)(1 + W(2e^{-2})/2)^2$, and $E_2 = -2\hbar^2/(a^2m)(1 - W(2e^{-2})/2)^2$.
The following (normalized) plots show the single-well and double-well bound states for different $\alpha$.
Energies of bound states for $\alpha=\hbar^2/ma$.
state
energy
$\psi_1$
$-2\hbar^2/(a^2m)(1 + W(2e^{-2})/2)^2$
$\psi_2$
$-2\hbar^2/(a^2m)(1 - W(2e^{-2})/2)^2$
Energies of bound states for $\alpha=\hbar^2/4ma$.
