Born Approximation

Equation (938) is not particularly useful, as it stands, because the quantity $f({\bf k}', {\bf k})$ depends on the unknown ket $\vert\psi\rangle$ . Recall that $\psi({\bf x})=\langle {\bf x}\vert\psi\rangle$ is the solution of the integral equation

$$\psi({\bf x}) = \phi({\bf x})-\frac{m}{2\pi\,\hbar^2} \frac{\exp(... ...x'\, \exp(- {\rm i} \,{\bf k}' \cdot {\bf x}')\, V({\bf x}')\, \psi ({\bf x}'),$$ (939)

where $\phi({\bf x})$ is the wavefunction of the incident state. According to the above equation, the total wavefunction is a superposition of the incident wavefunction and lots of spherical waves emitted from the scattering region. The strength of the spherical wave emitted at a given point is proportional to the local value of the scattering potential, $V$ , as well as the local value of the wavefunction, $\psi$ .

Suppose that the scattering is not particularly strong. In this case, it is reasonable to suppose that the total wavefunction, $\psi({\bf x})$ , does not differ substantially from the incident wavefunction, $\phi({\bf x})$ . Thus, we can obtain an expression for $f({\bf k}', {\bf k})$ by making the substitution

$$\psi({\bf x}) \rightarrow \phi({\bf x}) = \frac{\exp(\,{\rm i}\,{\bf k} \cdot {\bf x} ) }{(2\pi)^{3/2}}.$$ (940)

This is called the Born approximation.

The Born approximation yields

$$f({\bf k}', {\bf k}) \simeq - \frac{m}{2\pi\, \hbar^2} \int d^3 x'\,\exp\left[\, {\rm i}\, ({\bf k} - {\bf k}')\cdot {\bf x}'\right] V({\bf x}').$$ (941)

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

For a spherically symmetric potential,

$$f({\bf k}', {\bf k}) \simeq - \frac{m}{2\pi\, \hbar^2} \int_0^\in... ...,d\phi'\,r'^{\,2}\,\sin\theta' \,\exp(\,{\rm i} \, q \,r'\cos\theta') \, V(r'),$$ (942)

giving

$$f({\bf k}', {\bf k}) \simeq - \frac{2\,m}{\hbar^2\,q} \int_0^\infty dr'\,r' \,V(r') \sin(q \,r').$$ (943)

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

$$q \equiv \vert{\bf k} - {\bf k}'\vert = 2\, k \,\sin (\theta/2),$$ (944)

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

Consider scattering by a Yukawa potential

$$V(r) = \frac{V_0\,\exp(-\mu \,r)}{\mu \,r},$$ (945)

where $V_0$ is a constant, and $1/\mu$ measures the ``range'' of the potential. It follows from Equation (943) that

$$f(\theta) = - \frac{2\,m \,V_0}{\hbar^2\,\mu} \frac{1}{q^2 + \mu^2},$$ (946)

because

$$\int_0^\infty dr'\, \exp(-\mu \,r') \,\sin(q\,r') = \frac{q}{q^2+\mu^2}.$$ (947)

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

$$\frac{d\sigma}{d {\mit\Omega}} \simeq \left(\frac{2\,m \,V_0}{ \hbar^2\,\mu}\right)^2 \frac{1}{[4\,k^2\,\sin^2(\theta/2) + \mu^2]^{\,2}}.$$ (948)

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

$$\frac{d\sigma}{d{\mit\Omega}} \simeq \left(\frac{2\,m \,Z\, Z'\, e^2}{4\pi\,\epsilon_0\,\hbar^2}\right)^2 \frac{1}{ 16 \,k^4\, \sin^4( \theta/2)}.$$ (949)

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

$$\frac{d\sigma}{d{\mit\Omega}} \simeq\left(\frac{Z \,Z'\, e^2}{16\pi\,\epsilon_0\,E}\right)^2 \frac{1}{\sin^4(\theta/2)},$$ (950)

where $E= p^2/2\,m$ is the kinetic energy of the incident particles. Equation (950) is identical to the classical Rutherford scattering cross-section formula.

The Born approximation is valid provided that $\psi({\bf x})$ is not too different from $\phi({\bf x})$ in the scattering region. It follows, from Equation (922), that the condition for $\psi({\bf x}) \simeq \phi({\bf x})$ in the vicinity of ${\bf x} = {\bf0}$ is

$$\left\vert \frac{m}{2\pi\, \hbar^2} \int d^3 x'\,\frac{ \exp(\,{\rm i}\, k \,r')}{r'} \,V({\bf x}') \right\vert \ll 1.$$ (951)

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

$$\frac{2\,m}{\hbar^2} \frac{\vert V_0\vert}{\mu^2} \ll 1$$ (952)

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

$$\frac{2\,m}{\hbar^2} \frac{\vert V_0\vert} {\mu^2} \geq 2.7,$$ (953)

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, Equation (951) yields

$$\frac{2\,m}{\hbar^2} \frac{\vert V_0\vert}{\mu \,k} \ll 1.$$ (954)

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