# 12.5: The Differential Cross-Section


We now have all the ingredients, the scattered flux and the incident flux, to compute the cross-section:

$\frac{d\sigma}{ d\Omega} \equiv \frac{\text{scattered flux}}{\text{incident flux}} = \frac{mL^3}{\hbar k'} \frac{2 \pi}{\hbar} |V_{{\bf k'k}}|^2 \frac{L^3}{8\pi^3} \frac{mk}{\hbar^2} \nonumber$

Noting that, for elastic scattering, $$k' = k$$, we obtain finally the so-called Born approximation for the differential cross-section:

$\frac{d\sigma}{ d\Omega} = \frac{m^2}{4\pi^2\hbar^4}L^6 \left| \langle {\bf k'} | \hat{V} | {\bf k} \rangle \right|^2 \nonumber$

where the matrix element $$V_{{\bf k'k}} \equiv \langle {\bf k'} | \hat{V} | {\bf k} \rangle$$ is given by

$\langle {\bf k'} |\hat{V} |{\bf k} \rangle = \frac{1}{L^3} \int \int \int V ({\bf r}) \text{ exp } (−i\chi.{\bf r}) d\tau \nonumber$

with $$\chi \equiv {\bf k' − k}$$, the so-called wave-vector transfer. Thus the required matrix element in the Born approximation is just the 3-dimensional Fourier transform of the potential energy function. The total scattering cross section is simply:

$\sigma_T = \int \frac{d\sigma}{d\Omega} d\Omega = \int \int \frac{d\sigma}{d\Omega} \sin \theta d\theta d\phi \nonumber$

Observe that the final result for the differential cross-section is independent of the box size, $$L$$, which we used to normalise the plane-wave states.

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