Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.10.

The first Born approximation for the scattering amplitude comes from the integral form of the Schrödinger equation:

This equation is valid for all , even for positions near to the origin where the scattering potential could be significantly different from zero. In a scattering problem, the detector is usually situated far from the scattering region, so for all of interest, and we’re well outside the region where . In such cases, we can approximate (see Griffiths, section 11.4.2 for details) the integral equation by

where is a vector pointing from the origin to the detector (that is, parallel to ). The first term on the RHS represents the incoming plane wave, as usual.

Since the scattering amplitude is the coefficient of inside the square brackets, we have

This formula still doesn’t help us much, since we still need to know the wave function inside the scattering region where . The Born approximation assumes that the potential is weak, so that the incoming plane wave doesn’t change much after it scatters. That is, we assume that, for all points where the integrand is non-zero:

The incident plane wave has a wave vector of magnitude that is parallel to , which we can write as

For some position with component :

so the assumption above amounts to saying that

This assumption gives us an approximation for :

It’s important to remember that, from the point of view of the integral, and are constant, with the direction of being how the polar angles and are specified. The vector is always the same as it specifies the direction of the incident particle.

For a spherically symmetric potential, the integral can be simplified a bit by defining the vector

The vector is the base of an isosceles triangle with sides and , so since the angle between and is ( is the direction to the detector and is the direction of the incident particle, so is the scattering angle), we can divide the isosceles triangle into two symmetric right angled triangles by drawing a line from the origin to the the midpoint of . The length of the base is which is also , so

Letting the polar axis in the integral 9 lie along we get [where is the polar angle of integration, *not* !] and

ExampleSoft-sphere scattering. Asoft sphereis defined by the potentialwhere is a constant. [The hard sphere takes .] From 13, we can get the Born approximation for the scattering amplitude:

with the dependence given by the definition of in 11.

For low energy scattering and we can expand the sin and cos.

To this order, the scattering amplitude is

which agrees with equation 11.82 in Griffiths.

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