Scattering Theory

For $s$-wave scattering ($l = 0$) at low energy ($ka \ll 1$), only the $l = 0$ phase shift contributes:

$\sigma \approx \frac{4\pi}{k^2}\sin^2\delta_0 \approx 4\pi a_s^2$

Where the scattering length $a_s$ is defined by $k\cot\delta_0 \to -1/a_s$ as $k \to 0$.

For a hard sphere of radius $a$: $\delta_0 = -ka$ (exact), giving $a_s = a$ and $\sigma = 4\pi a^2$ (four times the geometric cross section $\pi a^2$ --- a purely quantum result).

For the $^3$He—$^4$He system: $a_s \approx 1.4$ nm (positive, indicating a repulsive effective potential). For the neutron—proton system (triplet): $a_s \approx 5.4$ fm (positive, with a bound state --- the deuteron). For singlet: $a_s \approx -23.7$ fm (negative, indicating a virtual state).

Consider $V(r) = V_0\,e^{-r^2/(2a^2)}$.

$f(\theta) = -\frac{m}{2\pi\hbar^2}\int e^{-i\mathbf{q}\cdot\mathbf{r}}V_0 e^{-r^2/(2a^2)}\,d^3r$

$= -\frac{m V_0}{2\pi\hbar^2}(2\pi a^2)^{3/2}e^{-q^2 a^2/2} = -\frac{m V_0}{\hbar^2}(2\pi)^{1/2}a^3\,e^{-2k^2a^2\sin^2(\theta/2)}$

The total cross section:

$\sigma = \int |f|^2\,d\Omega = 2\pi\int_0^\pi |f|^2\sin\theta\,d\theta$

At low energy ($ka \ll 1$): $f \approx -\frac{mV_0}{\hbar^2}(2\pi)^{1/2}a^3$ (independent of $\theta$), giving:

$\sigma \approx 4\pi\left(\frac{mV_0}{\hbar^2}\right)^2 2\pi\,a^6 = \frac{8\pi^2 m^2 V_0^2 a^6}{\hbar^4}$

The Born approximation is valid when $|V_0| \ll \hbar^2/(ma^2)$I.e., the potential is weak compared to the kinetic energy associated with the length scale $a$.

Lithium ($Z = 3$) has the electron configuration $1s^2 2s^1$. Using the variational method with $Z_{\text{eff}}$ for the $1s$ electrons:

(a) Calculate $Z_{\text{eff}}$ for the $1s$ electrons, treating the $2s$ electron as a perturbation.

(b) Calculate the ionisation energy (removing the $2s$ electron) and compare with the experimental value of 5.39 eV.

(c) Explain why the $2s$ electron is effectively screened by $Z_{\text{eff} \approx 1.26}$.

Solution:

(a) For the $1s$ electrons, the effective charge is reduced from $Z = 3$ by screening from the other $1s$ electron and partial penetration of the $2s$ electron. The $1s$ electrons screen each other partially: using the helium result, $Z_{\text{eff}(1s) \approx Z - 5/16 = 3 - 0.3125 = 2.69}$.

(b) The $2s$ electron sees an effective nuclear charge of $Z_{\text{eff}(2s) \approx 3 - 2 \times 0.85 = 1.3}$ (Slater’s rules). The energy:

E_{2s} = -\frac{Z_{\text{eff}^2}{n^2}\times 13.6\ \text{eV} = -\frac{1.3^2}{4}\times 13.6 = -\frac{1.69}{4}\times 13.6 = -5.75\ \text{eV}}

The ionisation energy is $|E_{2s}| = 5.75$ eV, close to the experimental 5.39 eV. The discrepancy reflects the crudeness of the Slater screening constants.

(c) The $2s$ electron has significant radial extent beyond the $1s$ core, so it sees a nearly bare nuclear charge at small $r$ but is screened by both $1s$ electrons at large $r$. The effective charge $Z_{\text{eff} \approx 1.26}$ (using Hartree—Fock) represents this average screening.

Consider scattering from the attractive square well $V(r) = -V_0$ for $r < a$ and $V(r) = 0$ for $r > a$.

(a) Show that the $s$-wave phase shift satisfies:

$\delta_0 = -ka + \arctan\!\left(\frac{k}{\kappa}\tan(\kappa a)\right)$

Where $\kappa = \sqrt{2m(V_0 + E)}/\hbar$ and $k = \sqrt{2mE}/\hbar$.

(b) Show that a bound state exists at energy $E = -|E|$ when $\kappa_0 a = \pi/2$ where $\kappa_0 = \sqrt{2m(V_0 - |E|)}/\hbar$.

(c) Show that the scattering length diverges as a new bound state appears.

Solution:

(a) Inside the well ($r < a$), the radial wavefunction for $l = 0$ is $u(r) = A\sin(\kappa r)$. Outside ($r > a$), $u(r) = B\sin(kr + \delta_0)$.

Matching $u$ and $u'$ at $r = a$:

$\kappa\cos(\kappa a) = k\cos(ka + \delta_0)/\sin(ka + \delta_0)\cdot k$

Wait: $\kappa\cot(\kappa a) = k\cot(ka + \delta_0)$.

$\cot(ka + \delta_0) = \frac{\kappa}{k}\cot(\kappa a)$

$ka + \delta_0 = \arccot\!\left(\frac{\kappa}{k}\cot(\kappa a)\right) = \arctan\!\left(\frac{k}{\kappa}\tan(\kappa a)\right)$

$\delta_0 = -ka + \arctan\!\left(\frac{k}{\kappa}\tan(\kappa a)\right)$

(b) A bound state has $E < 0$So $k = i\kappa'$ where $\kappa' = \sqrt{2m|E|}/\hbar$. The bound state condition is that the exterior solution decays exponentially: $u(r) = Be^{-\kappa' r}$. Matching:

$\kappa\cot(\kappa a) = -\kappa'$

As $|E| \to 0$: $\kappa' \to 0$So $\kappa\cot(\kappa a) \to 0$Giving $\kappa a = \pi/2$ (the threshold for the first bound state).

(c) The scattering length $a_s = -\lim_{k \to 0}(\delta_0/k)$. As $\kappa a \to \pi/2$, $\tan(\kappa a) \to \infty$So:

$\delta_0 \approx -ka + \arctan(\infty) = -ka + \pi/2$

$a_s = -\frac{1}{k}\left(-ka + \frac{\pi}{2}\right) = a - \frac{\pi}{2k}$

As $k \to 0$: $a_s \to \pm\infty$ (diverges), changing sign as the bound state appears.