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For $g\ge 0$, the Hamiltonian operator is nonnegative: its quadratic form is ${\hslash }^2\int |{\psi }^{\prime }{|}^{2}dx/(2m)+g(|\psi (a){|}^2+|\psi (-a){|}^2)\ge 0$. Thus it has no negative-energy bound states. A nonnegative-energy exterior wavefunction cannot be square integrable unless it vanishes identically, so there are no other bound states either.

For $g<0$, set $\nu =-mg/{\hslash }^2>0$. The even equation is the zero of $f(\kappa )=\nu (1+e^{-2a\kappa })-\kappa$. Since $f(0)=2\nu >0$, $f(\kappa )\to -\infty$ and $f^{\prime }(\kappa )=-2a\nu e^{-2a\kappa }-1<0$, the intermediate value theorem and strict monotonicity give exactly one positive equation root.

The odd equation is the zero of $h(\kappa )=\nu (1-e^{-2a\kappa })-\kappa$. Here $h(0)=0$, $h^{\prime }(0)=2a\nu -1$, $h^{\prime \prime }(\kappa )=-4a^2\nu e^{-2a\kappa }<0$, and $h(\kappa )\to -\infty$. If $2a\nu \le 1$, strict concavity makes $h(\kappa )<0$ for every $\kappa >0$. If $2a\nu >1$, $h$ initially increases, has exactly one maximum, and then decreases through zero exactly once. This proves the bound-state count of an attractive symmetric double-delta potential:At the threshold $g=-{\hslash }^2/(2ma)$, the odd zero-energy solution is proportional to $x$ inside $(-a,a)$ and to $a\mathrm{sgn}x$ outside. It is bounded but not a square-integrable function, so it is a zero-energy scattering resonance rather than a second bound state. Each negative-energy bound state is either even or odd by nondegeneracy of one-dimensional bound states; the two branches above exhaust the count.