Symmetric double-delta potential
= Symmetric double-delta potential
The symmetric double-delta potential
$$
V(x)=\frac{\hbar^2U_0}{2m}\bigl(\delta(x-a)+\delta(x+a)\bigr)
$$
separates into even and odd <parity> channels. At either interaction, the <wavefunction> is continuous and its derivative has the jump
$$
\psi'(a^+)-\psi'(a^-)=U_0\psi(a).
$$