Symmetric double-delta potential

The symmetric double-delta potentialseparates into even and odd parity channels. At either interaction, the wavefunction is continuous and its derivative has the jump

For $V(x)=g[\delta (x-a)+\delta (x+a)]$ with $a>0$ and $\nu =-mg/{\hslash }^2>0$, an even bound state obeys $\kappa =\nu (1+e^{-2a\kappa })$, and an odd bound state obeys $\kappa =\nu (1-e^{-2a\kappa })$, with energy $E=-{\hslash }^2{\kappa }^2/(2m)$ and $\kappa >0$. The even right-hand side is strictly decreasing from $2\nu$, so it meets $\kappa$ once. For the odd equation, $h(\kappa )=\nu (1-e^{-2a\kappa })-\kappa$ is strictly concave, starts at zero with slope $2a\nu -1$, and tends to negative infinity. It has one positive equation root precisely when $2a\nu >1$. Thus there is one bound state for $-{\hslash }^2/(2ma)\le g<0$, and two for $g<-{\hslash }^2/(2ma)$. For $g\ge 0$ positivity of the Hamiltonian operator excludes negative energy, and square integrability excludes nonnegative-energy exterior solutions. At $g=g_c$ the odd zero-energy solution is bounded but not square integrable: it is a zero-energy scattering resonance, not a second bound state.

For positive wavenumber $k$, the odd channel of the symmetric double-delta potential hasFor real $k$ its numerator and denominator are complex conjugates, making the channel unitary. Its poles are the zeros of $k{e}^{-ika}+U_0\sin (ka)$.

For $U_0<0$, an odd bound state has $k=i\kappa$ with $\kappa >0$ and obeysA nonzero positive solution exists exactly when $-U_{0}a>1$; equality is the zero-energy threshold.

For $U_{0}a\gg 1$, the lowest odd scattering resonance pole lies near the first odd mode trapped between the barriers:in its real part and to leading nonzero order in its imaginary part. Writing $E={\hslash }^2{k}^2/(2m)=E_R-i\Gamma /2$, the survival probability decays as $e^{-\Gamma t/\hslash }$.