# 12.6: Further Simplification to 1D for Conservative, Central Potential


Consider a central potential $$V ({\bf r}) = V (|{\bf r}|)$$ where energy is conserved $$|{\bf k'}|^{\bf 2} = |{\bf k}|^{\bf 2}$$. Here $$\chi$$ is a vector of length $$2k \sin \frac{\theta}{2}$$ where $$\theta$$ is the scattering angle.

We can make some progress with the matrix element integral if we choose a polar coordinate system with $$\chi$$ along the $$z$$-axis, so that $$\chi.{\bf r} = \chi r \cos \theta$$. Since we are trying to integrate over all space this change does not affect the limits of the integral.

$V_{{\bf k'k}} = \int^{2\pi}_0 d\phi \int^{+1}_{−1} \int^{\infty}_0 V (r)e^{−i\chi r \cos \theta} r^2 drd(\cos \theta ) \\ = 2\pi \int^{\infty}_0 \frac{e^{−i\chi r} − e^{i\chi r}}{−i\chi r} V (r)r^2 dr = \frac{4\pi}{\chi} \int^{\infty}_0 rV (r) \sin(\chi r)dr \nonumber$

But since $$|{\bf k}| = |{\bf k}'|$$, $$|\chi | = 2k \sin \frac{\theta}{2}$$, Whence we obtain the most useful form of the Born approximation:

$\frac{d\sigma}{d\Omega} = \frac{m^2}{(k \sin \frac{\theta}{2} )^2 \hbar^4} \left| \int^{\infty}_0 rV (r) \sin(2kr \sin \frac{\theta}{2} )dr \right|^2 \nonumber$

Thus the scattering cross-section is independent of $$\phi$$ (due to cylindrical symmetry of the problem). Note that this shows that the differential cross section does not depend on scattering angle and beam energy independently, but on a single parameter $$\chi$$. By using a range of energies for the incoming particles, $$k$$, this dependence can be used to test whether experimental data can be well described by the Born Approximation.

The most common use of the Born approximation is, of course, in reverse. Having found $$\frac{d\sigma}{d\Omega}$$ experimentally, a reverse Fourier transform can be used to obtain the form of the potential.

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