 Solution

For the symmetric double-delta potential, take an odd state at $E=-{\hslash }^2{\lambda }^2/(2M)$, $\lambda >0$. On $0<x<a$ odd parity selects $B\sinh (\lambda x)$. For $x>a$, decay and continuity give $B\sinh (\lambda a)e^{-\lambda (x-a)}$; extend to negative $x$ by oddness. At $x=a$ the jump condition isFor a nonzero state this means $\lambda \Delta [1+\coth (\lambda a)]=1$, and thereforeOddness makes the jump at $-a$ follow automatically: both its derivative jump and its required wavefunction value have the corresponding opposite signs.

There is a positive solution precisely when $\Delta <a$. To see this, rewrite the equation asThe right side decreases strictly from $a$ to zero as $\lambda$ increases from zero to infinity: its derivative has the sign of $(1+2a\lambda )e^{-2a\lambda }-1<0$. Hence for $0<\Delta <a$ there is exactly one positive solution, while $\Delta =a$ is a nonnormalizable zero-energy threshold and $\Delta >a$ gives no odd bound state. This verifies the stated sufficiently-small-$\Delta$ condition and identifies its exact boundary, consistent with the odd bound state of a symmetric double-delta potential.