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The hard-sphere limit of a repulsive spherical potential makes penetration into a sphere of radius $a$ vanish. Its low-energy S-wave scattering length tends to $a$, and its quantum total cross-section tends to $4\pi a^2$.

For elastic scattering by a central potential,At low momentum, the S-wave normally supplies the leading term.

For an S-wave, put $u(r)=r\psi (r)$. A bound state has $E=-\epsilon <0$, and the radial Schrödinger equation giveswhereContinuity of $u,u^{\prime }$ at $a$ givesThe first bound state appears at threshold $\kappa \downarrow 0$, where $ka=\pi /2$. The zero-energy threshold itself is not normalizable, so a deuteron bound state exists precisely whenThe wavefunction is $\psi =u/r$, with normalization fixing the remaining constant.

The asymptotic scattering wavefunction isIn each partial wave, the Partial-wave S-matrix is the ratio of its outgoing coefficient to its incoming coefficient. With the conventions in the question,For elastic scattering by a real central potential, unitarity gives $|S_l|=1$, so $S_l=e^{2i{\delta }_l}$ for a real scattering phase shift ${\delta }_l$. HenceUsing the Orthogonality of Legendre polynomials in ${\sigma }_T=\int |f(\theta ){|}^{2}d\Omega$ yields the partial-wave total scattering cross-section

For the S-wave, write the reduced radial wavefunction aswhere regularity at $r=0$ selected the hyperbolic sine. Continuity of $u$ and $u^{\prime }$ at $r=a$, equivalently logarithmic-derivative matching, givesorFor $ka\ll 1$, let the scattering length be $a_s=-{\lim }_{k\to 0}\tan {\delta }_0/k$. Expanding the matching relation givesOnly the S wave contributes at leading order, soIn the hard-sphere limit $\gamma a\to \infty$, penetration is suppressed, $a_s\to a$, andThis is four times the geometric area $\pi a^2$, a wave effect associated with diffraction from an impenetrable sphere.