The Born Approximation

The hard sphere was solvable exactly because of its symmetry. Most potentials are not. The Born approximation is the workhorse for the rest: a perturbative formula for the scattering amplitude that is accurate whenever the potential is weak or the energy is high.

The integral equation for scattering

The Schrödinger equation with the outgoing boundary condition is equivalent to the Lippmann–Schwinger integral equation, obtained by inverting the operator $E - H_0$ with the free outgoing Green's function $G_0(\mathbf{r},\mathbf{r}') = -\frac{m}{2\pi\hbar^2}\frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}$ :

\psi(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}} - \frac{m}{2\pi\hbar^2}\int \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}\, V(\mathbf{r}')\,\psi(\mathbf{r}')\,d^3r' .

This is exact but implicit — $\psi$ appears on both sides. Reading off the asymptotic $f$ gives

f(\theta,\phi) = -\frac{m}{2\pi\hbar^2}\int e^{-i\mathbf{k}'\cdot\mathbf{r}'}\,V(\mathbf{r}')\,\psi(\mathbf{r}')\,d^3r',

with $\mathbf{k}'$ the outgoing wavevector ( $|\mathbf{k}'| = k$ ).

The first Born approximation

If $V$ is weak, the wave inside the potential is barely disturbed: replace $\psi(\mathbf{r}')$ by the unperturbed plane wave $e^{i\mathbf{k}\cdot\mathbf{r}'}$ . The amplitude becomes a single Fourier transform of the potential:

\boxed{\;f^{(1)}(\theta,\phi) = -\frac{m}{2\pi\hbar^2}\int e^{i\mathbf{q}\cdot\mathbf{r}'}\,V(\mathbf{r}')\,d^3r' \;}\qquad \mathbf{q} = \mathbf{k} - \mathbf{k}'.

Here $\mathbf{q}$ is the momentum transfer, with magnitude $q = 2k\sin(\theta/2)$ for elastic scattering. The whole angular dependence enters through $q$ . Iterating (putting $f^{(1)}$ back into the integral) generates the Born series $f = f^{(1)} + f^{(2)} + \cdots$ , a power series in the potential strength.

Central potentials: a one-dimensional integral

For $V = V(r)$ the angular integrals collapse and only a radial integral remains:

f^{(1)}(\theta) = -\frac{2m}{\hbar^2 q}\int_0^\infty r\,V(r)\,\sin(qr)\,dr .

Worked example: the Yukawa potential

The screened-Coulomb (Yukawa) potential $V(r) = \beta\,\dfrac{e^{-\mu r}}{r}$ has the radial integral $\int_0^\infty e^{-\mu r}\sin(qr)\,dr = q/(\mu^2 + q^2)$ , giving the clean result

f^{(1)}(\theta) = -\frac{2m\beta}{\hbar^2}\,\frac{1}{\mu^2 + q^2}, \qquad q = 2k\sin(\theta/2).

The differential cross-section is $\big(\tfrac{2m\beta}{\hbar^2}\big)^2 (\mu^2 + q^2)^{-2}$ — the celebrated form behind both meson exchange and (via $\mu\to 0$ ) Rutherford scattering: letting the screening length $1/\mu \to \infty$ and $\beta = Z_1 Z_2 e^2/4\pi\varepsilon_0$ recovers the Coulomb result $d\sigma/d\Omega \propto 1/\sin^4(\theta/2)$ exactly.

When is it valid?

The replacement $\psi \to e^{i\mathbf{k}\cdot\mathbf{r}}$ is good when the potential barely perturbs the wave. This holds for weak potentials at any energy and, more usefully, for strong potentials at high energy (the fast particle spends little time in the interaction region). It fails badly near resonances or bound-state thresholds, where multiple scattering — the higher Born terms — dominates.

Try it

For the Yukawa potential with $2m/\hbar^2 = 1$ , $\beta = 1$ , $\mu = 1$ , $k = 1$ , evaluate the first Born amplitude at backscattering $\theta = \pi$ . Use $q = 2k\sin(\theta/2)$ and $f^{(1)} = -(2m\beta/\hbar^2)/(\mu^2 + q^2)$ .

Run your code to see the quantum state.