Born Approximation

Equation (1266) is not particularly useful, as it stands, because the quantity $f({\bf k},{\bf k}')$ depends on the, as yet, unknown wavefunction $\psi({\bf r})$ [see Eq. (1261)]. Suppose, however, that the scattering is not particularly strong. In this case, it is reasonable to suppose that the total wavefunction, $\psi({\bf r})$, does not differ substantially from the incident wavefunction, $\psi_0({\bf r})$. Thus, we can obtain an expression for $f({\bf k},{\bf k}')$ by making the substitution $\psi({\bf r})\rightarrow\psi_0({\bf r}) = \sqrt{n} \exp( {\rm i} {\bf k}\cdot{\bf r})$ in Eq. (1261). This procedure is called the Born approximation.

The Born approximation yields

$$f({\bf k},{\bf k}') \simeq \frac{m}{2\pi \hbar^2} \int {\rm... ...} ({\bf k}-{\bf k'})\cdot{\bf r}'} V({\bf r'}) d^3{\bf r}'.$$ (1267)

Thus, $f({\bf k},{\bf k}')$ is proportional to the Fourier transform of the scattering potential $V({\bf r})$ with respect to the wavevector ${\bf q} = {\bf k}-{\bf k}'$.

For a spherically symmetric potential,

$$f({\bf k}', {\bf k}) \simeq - \frac{m}{2\pi \hbar^2} \int\... ...a') V(r') r'^{ 2} dr' \sin\theta' d\theta' d\phi',$$ (1268)

giving

$$f({\bf k}', {\bf k}) \simeq - \frac{2 m}{\hbar^2 q} \int_0^\infty r' V(r') \sin(q r') dr'.$$ (1269)

Note that $f({\bf k}', {\bf k})$ is just a function of $q$ for a spherically symmetric potential. It is easily demonstrated that

$$q \equiv \vert{\bf k} - {\bf k}'\vert = 2 k \sin (\theta/2),$$ (1270)

where $\theta$ is the angle subtended between the vectors ${\bf k}$ and ${\bf k}'$. In other words, $\theta$ is the scattering angle. Recall that the vectors ${\bf k}$ and ${\bf k}'$ have the same length, via energy conservation.

Consider scattering by a Yukawa potential

$$V(r) = \frac{V_0 \exp(-\mu r)}{\mu r},$$ (1271)

where $V_0$ is a constant, and $1/\mu$ measures the ``range'' of the potential. It follows from Eq. (1269) that

$$f(\theta) = - \frac{2 m V_0}{\hbar^2 \mu} \frac{1}{q^2 + \mu^2},$$ (1272)

since

$$\int_0^\infty \exp(-\mu r') \sin(q r') dr' = \frac{q}{q^2+ \mu^2}.$$ (1273)

Thus, in the Born approximation, the differential cross-section for scattering by a Yukawa potential is

$$\frac{d\sigma}{d \Omega} \simeq \left(\frac{2 m V_0}{ \hb... ...u}\right)^2 \frac{1}{[2 k^2 (1-\cos\theta) + \mu^2]^{ 2}},$$ (1274)

given that

$$q^2 = 4 k^2 \sin^2(\theta/2) = 2 k^2 (1-\cos\theta).$$ (1275)

The Yukawa potential reduces to the familiar Coulomb potential as $\mu \rightarrow 0$, provided that $V_0/\mu \rightarrow Z Z' e^2 / 4\pi \epsilon_0$. In this limit, the Born differential cross-section becomes

$$\frac{d\sigma}{d\Omega} \simeq \left(\frac{2 m Z Z' e... ..._0 \hbar^2}\right)^2 \frac{1}{ 16 k^4 \sin^4( \theta/2)}.$$ (1276)

Recall that $\hbar k$ is equivalent to $\vert{\bf p}\vert$, so the above equation can be rewritten

$$\frac{d\sigma}{d\Omega} \simeq\left(\frac{Z Z' e^2}{16\pi \epsilon_0 E}\right)^2 \frac{1}{\sin^4(\theta/2)},$$ (1277)

where $E= p^2/2 m$ is the kinetic energy of the incident particles. Of course, Eq. (1277) is the famous Rutherford scattering cross-section formula.

The Born approximation is valid provided that $\psi({\bf r})$ is not too different from $\psi_0({\bf r})$ in the scattering region. It follows, from Eq. (1258), that the condition for $\psi({\bf r}) \simeq \psi_0({\bf r})$ in the vicinity of ${\bf r} = {\bf0}$ is

$$\left\vert \frac{m}{2\pi \hbar^2} \int \frac{ \exp( {\rm i} k r')}{r'} V({\bf r}') d^3{\bf r'} \right\vert \ll 1.$$ (1278)

Consider the special case of the Yukawa potential. At low energies, (i.e., $k\ll \mu$) we can replace $\exp( {\rm i} k r')$ by unity, giving

$$\frac{2 m}{\hbar^2} \frac{\vert V_0\vert}{\mu^2} \ll 1$$ (1279)

as the condition for the validity of the Born approximation. The condition for the Yukawa potential to develop a bound state is

$$\frac{2 m}{\hbar^2} \frac{\vert V_0\vert} {\mu^2} \geq 2.7,$$ (1280)

where $V_0$ is negative. Thus, if the potential is strong enough to form a bound state then the Born approximation is likely to break down. In the high-$k$ limit, Eq. (1278) yields

$$\frac{2 m}{\hbar^2} \frac{\vert V_0\vert}{\mu k} \ll 1.$$ (1281)

This inequality becomes progressively easier to satisfy as $k$ increases, implying that the Born approximation is more accurate at high incident particle energies.