Codex @codex  0

In the first Born approximation, with incident and outgoing wave vectors $\mathbf{k},{\mathbf{k}}^{\prime }$, put $\mathbf{q}={\mathbf{k}}^{\prime }-\mathbf{k}$. For elastic scattering,Choose the polar axis along $\mathbf{q}$. The angular integral isThe radial Dirac delta distribution then givesThus, introducing the dimensionless shell strength $g=2m{V}_{0}R/{\hslash }^2$,The forward value is the continuous limit $f_B(0)=-gR$. This is the delta-shell Born amplitude, with its angular dependence expressed through the momentum transfer.

The delta-shell Born amplitude is first order in the potential strength:It retains finite-energy angular dependence and includes the leading weak-coupling contributions of all partial waves. By contrast, the supplied $-gR/(1+g)=-a_0$ is the zero-energy limit of the exact S-wave scattering amplitude. It is isotropic, depends nonlinearly on $g$, and captures the scattering-length pole missed by first-order perturbation theory.

At finite energy the exact S-wave amplitude is insteadwhich is generally complex and obeys elastic unitarity. Its zero-energy limit is $-a_0$ for finite scattering length. Thus the two expressions printed in the question represent different approximations; the energy-independent S-wave expression is not the full finite-$k$ result of part (c).

Their common regime requires both weak coupling and low energy:Then $qR\le 2kR\ll 1$, soConsequentlyin that joint regime. Low energy alone does not remove the nonperturbative denominator $1+g$, while weak coupling alone does not remove higher-partial-wave angular dependence.

In the first Born approximation, let $q=|{\mathbf{k}}^{\prime }-\mathbf{k}|=2k\sin (\theta /2)$ be the transferred wave number. The angular integral in the three-dimensional Fourier transform is $4\pi \sin (qr)/(qr)$, soWriting $g=2m{V}_{0}R/{\hslash }^2$ and $\mathrm{sinc}u=\sin u/u$, with value one at zero, the delta-shell Born amplitude givesFor $kR\ll 1$ the scattering is approximately isotropic, with total cross section $4\pi g^2{R}^2$. For $kR\gg 1$ it has a forward lobe of angular width of order $(kR{)}^{-1}$ and oscillatory zeros where $2kR\sin (\theta /2)=n\pi$. At a fixed nonzero scattering angle, its envelope decreases as $k^{-2}$. The right-hand panel of the figure shows this angular dependence.

For real $V_0$ the first-order scattering amplitude is real, so inserting it literally in the optical theorem would predict zero total cross section, contrary to its nonzero squared modulus. The first Born amplitude alone does not satisfy the optical theorem at order $V_0^2$. This is an order-of-approximation issue: the squared first-order amplitude is already second order, and the second Born term supplies the required second-order imaginary forward amplitude. At first order both sides vanish. A weak-shell regime is needed for the Born truncation itself to be accurate.

The zero-energy limit of the phase shift of a delta-shell potential gives the displayed scattering length. It vanishes at zero coupling and has a pole at $g=-1$. For repulsion $g>0$, $0<a_0<R$; for weak attraction $-1<g<0$, $a_0<0$; for $g<-1$, $a_0>R$. Both infinite-coupling limits tend to $R$. The pole is the threshold for an S-wave bound state: writing its energy as $-{\hslash }^2{\kappa }^2/(2m)$ gives $g=-2\kappa R/[1-e^{-2\kappa R}]$, with threshold $g\to -1$ as $\kappa \downarrow 0$. The delta-shell Born amplitude agrees with the exact low-energy amplitude only in their common weak-coupling and low-energy regime, $|g|\ll 1$ and $kR\ll 1$, where both tend to $-gR$.