# First Born approximation: soft-sphere scattering

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:

$\displaystyle \psi\left(\mathbf{r}\right)=\psi_{0}\left(\mathbf{r}\right)-\frac{m}{2\pi\hbar^{2}}\int\frac{e^{ik\left|\mathbf{r}-\mathbf{r}_{0}\right|}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (1)$

This equation is valid for all ${\mathbf{r}}$, even for positions near to the origin where the scattering potential ${V}$ could be significantly different from zero. In a scattering problem, the detector is usually situated far from the scattering region, so for all ${\mathbf{r}}$ of interest, ${r\gg r_{0}}$ and we’re well outside the region where ${V\ne0}$. In such cases, we can approximate (see Griffiths, section 11.4.2 for details) the integral equation by

 $\displaystyle \psi\left(\mathbf{r}\right)$ $\displaystyle \cong$ $\displaystyle Ae^{ikz}-\frac{m}{2\pi\hbar^{2}}\frac{e^{ikr}}{r}\int e^{-i\mathbf{k}\cdot\mathbf{r}_{0}}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\left[e^{ikz}-\frac{m}{2\pi\hbar^{2}A}\frac{e^{ikr}}{r}\int e^{-i\mathbf{k}\cdot\mathbf{r}_{0}}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}\right] \ \ \ \ \ (3)$

where ${\mathbf{k}\equiv k\hat{\mathbf{r}}}$ is a vector pointing from the origin to the detector (that is, parallel to ${\mathbf{r}}$). The first term on the RHS represents the incoming plane wave, as usual.

Since the scattering amplitude ${f}$ is the coefficient of ${e^{ikr}/r}$ inside the square brackets, we have

$\displaystyle f\left(\theta,\phi\right)=-\frac{m}{2\pi\hbar^{2}A}\int e^{-i\mathbf{k}\cdot\mathbf{r}_{0}}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (4)$

This formula still doesn’t help us much, since we still need to know the wave function ${\psi}$ inside the scattering region where ${V\ne0}$. The Born approximation assumes that the potential is weak, so that the incoming plane wave ${Ae^{ikz}}$ doesn’t change much after it scatters. That is, we assume that, for all points where the integrand is non-zero:

$\displaystyle \psi\left(\mathbf{r}_{0}\right)\approx\psi_{0}\left(\mathbf{r}_{0}\right) \ \ \ \ \ (5)$

The incident plane wave has a wave vector of magnitude ${k}$ that is parallel to ${\hat{\mathbf{z}}}$, which we can write as

$\displaystyle \mathbf{k}'\equiv k\hat{\mathbf{z}} \ \ \ \ \ (6)$

For some position ${\mathbf{r}_{0}}$ with ${z}$ component ${z_{0}}$:

$\displaystyle kz_{0}=\mathbf{k}'\cdot\mathbf{r}_{0} \ \ \ \ \ (7)$

so the assumption above amounts to saying that

$\displaystyle \psi\left(\mathbf{r}_{0}\right)\approx Ae^{i\mathbf{k}'\cdot\mathbf{r}_{0}} \ \ \ \ \ (8)$

This assumption gives us an approximation for ${f}$:

$\displaystyle f\left(\theta,\phi\right)\approx-\frac{m}{2\pi\hbar^{2}}\int e^{i\left(\mathbf{k}'-\mathbf{k}\right)\cdot\mathbf{r}_{0}}V\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (9)$

It’s important to remember that, from the point of view of the integral, ${\mathbf{k}}$ and ${\mathbf{k}'}$ are constant, with the direction of ${\mathbf{k}=k\hat{\mathbf{r}}}$ being how the polar angles ${\theta}$ and ${\phi}$ are specified. The vector ${\mathbf{k}'=k\hat{\mathbf{z}}}$ 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

$\displaystyle \boldsymbol{\kappa}\equiv\mathbf{k}'-\mathbf{k} \ \ \ \ \ (10)$

The vector is the base of an isosceles triangle with sides ${\mathbf{k}'}$ and ${\mathbf{k}}$, so since the angle between ${\mathbf{k}'}$ and ${\mathbf{k}}$ is ${\theta}$ (${\mathbf{k}'}$ is the direction to the detector and ${\mathbf{k}}$ is the direction of the incident particle, so ${\theta}$ 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 ${\boldsymbol{\kappa}}$. The length of the base is ${\kappa/2}$ which is also ${k\sin\frac{\theta}{2}}$, so

$\displaystyle \kappa=2k\sin\frac{\theta}{2} \ \ \ \ \ (11)$

Letting the polar axis in the integral 9 lie along ${\boldsymbol{\kappa}}$ we get ${\left(\mathbf{k}'-\mathbf{k}\right)\cdot\mathbf{r}_{0}=\kappa r_{0}\cos\theta_{0}}$ [where ${\theta_{0}}$ is the polar angle of integration, not ${\theta}$!] and

 $\displaystyle f\left(\theta\right)$ $\displaystyle \approx$ $\displaystyle -\frac{m}{2\pi\hbar^{2}}\int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}e^{i\kappa r_{0}\cos\theta_{0}}V\left(r_{0}\right)r_{0}^{2}\sin\theta_{0}d\phi_{0}d\theta_{0}dr_{0}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2m}{\hbar^{2}\kappa}\int_{0}^{\infty}V\left(r_{0}\right)r_{0}\sin\left(\kappa r_{0}\right)dr_{0} \ \ \ \ \ (13)$

Example Soft-sphere scattering. A soft sphere is defined by the potential

$\displaystyle V\left(\mathbf{r}\right)=\begin{cases} V_{0} & r\le a\\ 0 & r>a \end{cases} \ \ \ \ \ (14)$

where ${V_{0}>0}$ is a constant. [The hard sphere takes ${V_{0}=\infty}$.] From 13, we can get the Born approximation for the scattering amplitude:

 $\displaystyle f\left(\theta\right)$ $\displaystyle \approx$ $\displaystyle -\frac{2mV_{0}}{\hbar^{2}\kappa}\int_{0}^{a}r\sin\left(\kappa r\right)dr\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2mV_{0}}{\hbar^{2}\kappa^{3}}\left[\sin\left(\kappa a\right)-a\kappa\cos\left(\kappa a\right)\right] \ \ \ \ \ (16)$

with the ${\theta}$ dependence given by the definition of ${\kappa}$ in 11.

For low energy scattering ${\kappa a\ll1}$ and we can expand the sin and cos.

 $\displaystyle \sin\left(\kappa a\right)-a\kappa\cos\left(\kappa a\right)$ $\displaystyle =$ $\displaystyle \kappa a-\frac{\left(\kappa a\right)^{3}}{3!}+\ldots-\kappa a\left(1-\frac{\left(\kappa a\right)^{2}}{2!}+\ldots\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(\kappa a\right)^{3}}{3}+\ldots \ \ \ \ \ (18)$

To this order, the scattering amplitude is

$\displaystyle f\left(\theta\right)\approx-\frac{2mV_{0}a^{3}}{3\hbar^{3}} \ \ \ \ \ (19)$

which agrees with equation 11.82 in Griffiths.