 Solution

A bound state has $E<0$ and $\kappa =\sqrt{-2mE}/\hslash >0$, since outside the delta potentials a square-integrable solution must decay exponentially. The wavefunction is continuous at each delta: a jump would produce a delta derivative in ${\psi }^{\prime \prime }$ with no matching term in the equation. Integrating the Time-independent Schrodinger equation across either $x_0=\pm a$ givesThe symmetry of the symmetric double-delta potential permits even and odd solutions. Up to a nonzero normalization factor, the even candidate is $\cosh (\kappa x)$ for $|x|\le a$, and $\cosh (\kappa a)e^{-\kappa (|x|-a)}$ outside. Its derivative jump at $a$ givesThe odd candidate is $\sinh (\kappa x)$ inside and $\mathrm{sgn}(x)\sinh (\kappa a)e^{-\kappa (|x|-a)}$ outside. Its jump givesParity ensures the jump at $-a$ as well. Put $t=\kappa a$ and $L=2m\lambda a/{\hslash }^2$. The continuous strictly increasing function $t(1+\tanh t)$ ranges from $0$ to infinity, so there is exactly one even bound state for every $L>0$. The continuous strictly increasing function $t(1+\coth t)$ ranges from $1$ to infinity, so the odd bound state of a symmetric double-delta potential exists if and only if $L>1$. Strict increase follows from the allowed monotonicity facts plus the strictly increasing term $t$. ThereforeAt equality the odd limiting solution has $\kappa =0$ and is not square integrable, so it is not a bound state. No nonnegative energy can supply another full-line square-integrable state, since its exterior solutions are oscillatory or affine.