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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 $g\ge 0$, 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.

LetUsing $\sin (2ka)=2ka+O(k^3)$ and $\cos (2ka)=1+O(k^2)$ in the phase-shift formula givesBy the definition of the scattering length from a partial-wave S-matrix,Thus $a_s$ diverges exactly whenAt each such value, an S-wave bound state lies at the continuum threshold: the exterior zero-energy radial solution becomes constant, its logarithmic derivative vanishes, and the system has a zero-energy scattering resonance. As the well deepens through the critical value, a new bound state appears and the scattering length passes through infinity and changes sign.

The scattering length is $a_0=-{\lim }_{k\to 0}\tan {\delta }_0/k$. Since $\sin (kR)\sim kR$ and $\cos (kR)\sim 1$, the exact phase shift gives, away from the pole,Thus plot $a_0/R$ against $g$, so that both axes are dimensionless.

For repulsion, $g>0$, the curve rises from zero to the horizontal asymptote $a_0/R=1$, staying between zero and one. For weak attraction, $-1<g<0$, the scattering length is negative and tends to $-\infty$ as $g\to -1^{+}$. For stronger attraction, $g<-1$, it is positive and greater than $R$, tending to $+\infty$ as $g\to -1^{-}$ and to $R$ as $g\to -\infty$. Its only zero is $g=0$.

The pole at $g=-1$ is the zero-energy threshold of an S-wave bound state. To see this directly, use $u\propto \sinh (\kappa r)$ inside and $u\propto e^{-\kappa r}$ outside for energy $-{\hslash }^2{\kappa }^2/(2m)$. The same derivative jump givesThis approaches $-1$ as $\kappa \downarrow 0$ and admits a positive $\kappa$ for $g<-1$. At the threshold the scattering length diverges; this is a zero-energy scattering resonance, not a divergence of the finite-energy cross-section beyond its partial-wave unitarity bound. These are the main features of the scattering length of a delta-shell potential.

For outer radius $b=2a$ and zero-energy well wavenumber ${\kappa }_0=\sqrt{2m{V}_0}/\hslash$, matching the S-wave scattering logarithmic derivative givesIt diverges when ${\kappa }_{0}a=(n+\frac{1}{2})\pi$, where the well has a zero-energy scattering resonance.