Collisions with Absorption

Elastic Scattering: $φ_{k, ℓ, m} (\vec{r}) \sim \exp (i k z) + f_{k} (θ, φ) \exp (i k r) / r \sim \sum_{ℓ = 0}^{\infty} i^{ℓ} \sqrt{4 π (2 ℓ + 1)} Y_{ℓ}^{0} (θ) \cdot \frac{1}{2 i k r} (\exp (i (k r - ℓ π / 2) \exp (2 i δ_{ℓ}) - \exp (- i (k r - ℓ π / 2))))$ . Where $\exp (- i (k r - ℓ π / 2))$ is the incoming wave and $\exp (i (k r + ℓ π / 2))$ is the outgoing wave.

(Inelastic with absorption) Now we will consider when $δ_{ℓ}$ is complex, hence it has a lossy imaginary part. Let $η_{ℓ} \equiv \exp (2 i δ_{ℓ})$ now $| η_{ℓ} |^{2} \leq 1 = \exp (- 4 ℑ δ_{ℓ})$ .

Now we have,

\begin{aligned} φ_{k, ℓ, m} (\vec{r}) & \sim \exp (i k z) + f_{k} (θ, φ) \exp (i k r) / r \\ \sim \sum_{ℓ = 0}^{\infty} i^{ℓ} \sqrt{4 π (2 ℓ + 1)} Y_{ℓ}^{0} \frac{\sin (k r - ℓ π / 2)}{k r} + \sum_{ℓ = 0}^{\infty} i^{ℓ} \frac{\sqrt{4 π (2 ℓ + 1)}}{2 i k} Y_{ℓ}^{0} (η_{ℓ} - 1) (- i)^{ℓ} \frac{\exp (i k r)}{r} \\ (1) & \sim \sum_{ℓ = 0}^{\infty} i^{ℓ} \sqrt{4 π (2 ℓ + 1)} Y_{ℓ}^{0} \frac{\sin (k r - ℓ π / 2)}{k r} + \sum_{ℓ = 0}^{\infty} \frac{\sqrt{4 π (2 ℓ + 1)}}{2 i k} Y_{ℓ}^{0} (η_{ℓ} - 1) \frac{\exp (i k r)}{r} \end{aligned}

Then,

\begin{matrix} (2) & f_{k} (θ) = \sum_{ℓ = 0}^{\infty} \sqrt{4 π (2 ℓ + 1)} Y_{ℓ}^{0} (θ) \frac{η_{ℓ} - 1}{2 i k} \end{matrix}

Which gives,

\begin{matrix} (3) & σ_{t o t} = \frac{π}{k^{2}} \sum_{ℓ = 0}^{\infty} (2 ℓ + 1) | 1 - η_{ℓ} |^{2} \end{matrix}

For elastic scattering, $| 1 - η_{ℓ} |^{2} = (1 - \exp (2 i δ_{ℓ})) (1 - \exp (- 2 i δ_{ℓ})) = 2 (1 - \cos (2 δ_{ℓ})) = 4 \sin^{2} δ_{ℓ}$ .

This gives us the optical theorem for elastic scattering,

\begin{matrix} (4) & σ_{t o t}^{(s c a t)} = \frac{4 π}{k} ℑ f_{k} (0) . \end{matrix}

If we consider a fully absorbing target, then we get $σ_{t o t}^{(s c a t)} = \frac{π}{k^{2}} \sum_{ℓ = 0}^{\infty} (2 ℓ + 1)$ .

Now we construct the absorbtion cross section: $σ_{a b s} = \frac{Δ P}{| J_{i} |} = \frac{μ Δ P}{ℏ k}$ . Where $Δ P$ is the total probability current that is absorbed: $Δ P = - \int {\vec{J}}_{a b s} \cdot d \vec{S}$ .

$\vec{J} = ℜ {φ_{k}^{*} (\vec{r}) \frac{ℏ}{i μ} \nabla φ_{k} (\vec{r})}$ In the asymtotic limit, $J_{r} = ℜ {φ_{k}^{*} \frac{ℏ}{i μ} \frac{\partial}{\partial r} φ_{k}}$ . Where $φ_{k} \sim \sum_{ℓ = 0}^{\infty} i^{ℓ} \sqrt{4 π (2 ℓ + 1)} Y_{ℓ}^{0} \frac{\exp (i (k r - ℓ π / 2)) η_{ℓ} - \exp (- i (k r - ℓ π / 2))}{2 i k r}$ . Then, $(J_{i n}^{ℓ})_{r} = \frac{ℏ}{μ k} \frac{π (2 ℓ + 1)}{r^{2}} | Y_{ℓ}^{0} |^{2}$ and $(J_{o u t}^{ℓ})_{r} = (J_{i n}^{ℓ})_{r} | η_{ℓ} |^{2}$ . Then, $(J_{a b s}^{ℓ})_{r} = (J_{i n}^{ℓ})_{r} (1 - | η_{ℓ} |^{2}) = \frac{ℏ}{μ k} \frac{π (2 ℓ + 1)}{r^{2}} | Y_{ℓ}^{0} |^{2} (1 - | η_{ℓ} |^{2})$ .

Thus,

\begin{aligned} Δ P & = - \sum_{ℓ} \int \frac{ℏ}{μ k} \frac{π (2 ℓ + 1)}{r^{2}} | Y_{ℓ}^{0} |^{2} (1 - | η_{ℓ} |^{2}) r^{2} d Ω \\ = - \frac{ℏ}{μ k} π \sum_{ℓ} (2 ℓ + 1) (1 - | η_{ℓ} |^{2}) \int | Y_{ℓ}^{0} |^{2} d Ω \\ = - \frac{ℏ}{μ k} π \sum_{ℓ} (2 ℓ + 1) (1 - | η_{ℓ} |^{2}) . \end{aligned}

Therefore,

\begin{aligned} σ_{a b s} & = \frac{Δ P}{| J_{i} |} \\ = \frac{μ}{ℏ k} \frac{ℏ}{μ k} π \sum_{ℓ} (2 ℓ + 1) (1 - | η_{ℓ} |^{2}) & = \frac{π}{k^{2}} \sum_{ℓ} (2 ℓ + 1) (1 - | η_{ℓ} |^{2}) . \end{aligned}

Our total cross section is then, $σ_{t o t} = σ_{t o t}^{(s c a t)} + σ_{a b s} = \frac{4 π}{k} | ℑ f_{k} (0) |$ .

The Attractive Potential

Slow Particle

\begin{align*} u_{k,0}(r) &= \begin{cases}A\sin(kr+\delta_0) & r>r_0 \ B\sin(Kr) & r Imposing continuity,

{(\frac{B}{A})}^{2} = \frac{k^{2}}{k^{2} + k_{0}^{2} \cos^{2} (K r_{0})}, k_{0} = \sqrt{\frac{2 m V_{0}}{ℏ^{2}}}

. If

K r_{0} = \frac{π}{2} + π n

then

{(\frac{B}{A})}^{2} \to 1

and

δ_{0} = \frac{π}{2} - k r_{0} \to \frac{π}{2}

. If

K r_{0} = π + π n

then

{(\frac{B}{A})}^{2} \to \frac{k^{2}}{k^{2} + k_{0}^{2}}

and

δ_{0} \to - k r_{0}

. Further, this goes to zero for slow particles. This is called Ramsauer-Townsend effect - it is as if the particle does not see the target.

The maximal cross section for a particular $ℓ$ is expected to be $\frac{4 π}{k^{2}} (2 ℓ + 1)$ . In general, $\frac{4 π}{k^{2}} \frac{Γ^{2} (E_{R}) / 4}{(E - E_{R})^{2} + Γ^{2} (E_{R}) / 4}$ .

Now if we analyze higher angular momentums, $V_{e f f} = V + \frac{ℏ^{2} ℓ (ℓ + 1)}{2 μ r^{2}}$ . Then we get metastable bound states in our attractive potential with lifetimes $τ = \frac{ℏ}{Γ}$ .