If the potential is weak, the wave inside it is nearly the incident plane wave, and the exact formula for $f$ can be evaluated with $\psi$ replaced by $e^{i\mathbf{k}\cdot\mathbf{r}}$. That is the Born approximation, first order in the potential: the amplitude is the Fourier transform of the potential at the momentum transfer $\hbar\mathbf{q}$, where $\mathbf{q}=\mathbf{k}'-\mathbf{k}$ and $q=2k\sin(\theta/2)$. It turns scattering into Fourier analysis. A potential of range $a$ scatters strongly only for $q\lesssim1/a$, so fast particles are thrown mostly forward, and the width of the forward peak reads off the size of the target. For the screened potential $\beta e^{-\mu r}/r$ the amplitude is $-(2m\beta/\hbar^{2})/(q^{2}+\mu^{2})$. The approximation holds when the potential barely changes the wave inside it: $m|V_{0}|a^{2}/\hbar^{2}\ll1$, or $m|V_{0}|a^{2}/\hbar^{2}\ll ka$ for fast particles.

A pane of slightly uneven glass blurs a distant light, and the blur pattern is a Fourier picture of the unevenness. A weak potential scatters a wave the same way, the amplitude in each direction measuring one Fourier component of the potential.

The Born cross section of the screened potential $\beta e^{-\mu r}/r$ against the scattering angle, relative to its forward value, for $k=\mu/2$ and $k=2\mu$. The faster particles are scattered into a narrow forward peak; the slower ones spread over every angle.

Replacing $\psi$ by $e^{i\mathbf{k}\cdot\mathbf{r}}$ inside the potential, with $\mathbf{q}=\mathbf{k}'-\mathbf{k}$:

• For a central potential $f_{B}=-\frac{2m}{\hbar^{2}q}\int_{0}^{\infty}r\,V(r)\sin(qr)\,dr$, a function of alone.

Put the plane wave into the exact formula: the amplitude becomes the Fourier transform of $V$ at $\mathbf{q}$. For a central potential the angular part of the integral gives $\sin(qr)/qr$, leaving one radial integral, which for the screened potential is elementary. Elastic scattering fixes $q$ from $$ and .

The exact formula with $\psi\approx e^{i\mathbf{k}\cdot\mathbf{r}}$.

Two vectors of length $k$ at the angle $\theta$.

Take $\mathbf{q}$ as the polar axis and integrate over directions.

The imaginary part of $\int_{0}^{\infty}e^{(iq-\mu)r}dr=\frac{1}{\mu-iq}$.

Collect the factors of $\int r^{2}dr\,\frac{\beta e^{-\mu r}}{r}\,\frac{4\pi\sin qr}{qr}$.

In elastic scattering through $\theta$, the wave vector changes by $\mathbf{q}=\mathbf{k}'-\mathbf{k}$, of length $2k\sin(\theta/2)$.

In the Born approximation the amplitude is the Fourier transform of the potential at the momentum transfer, times $-m/2\pi\hbar^{2}$.

The Born approximation keeps the potential to first order: the incident wave stands in for the true wave inside it.

A potential of range $a$ has Fourier components only up to $q$ of about $1/a$, so faster particles scatter into narrower forward cones.

Great work! You now know how to:

• derive the Born amplitude from the integral equation
• compute it for the screened potential $\beta e^{-\mu r}/r$
• judge when the approximation holds