Quantum Theory of Scattering: Partial Waves, Born Approximation & Green's Functions

1. Asymptotic Wavefunction and the Scattering Amplitude

Consider a beam of monoenergetic particles of mass $m$ and incident wavevector $\vec{k} = k\hat{z}$ ($E = \frac{\hbar^2 k^2}{2m}$) incident upon a localized scattering potential $V(\vec{r})$ of finite range $a$. In the asymptotic region far from the scattering center ($r \gg a$), the stationary scattering wavefunction $\psi(\vec{r})$ is an exact superposition of the incident plane wave and an outgoing spherical wave:

$$\psi(\vec{r}) \xrightarrow{r \to \infty} e^{ikz} + f(\theta, \phi)\frac{e^{ikr}}{r}$$

where $f(\theta, \phi)$ is the scattering amplitude having dimensions of length.

2. Differential and Total Scattering Cross-Sections

The incident probability current density is:

$$\vec{J}_{\text{inc}} = \frac{\hbar}{2mi}\left(\psi_{\text{inc}}^* \nabla \psi_{\text{inc}} - \psi_{\text{inc}} \nabla \psi_{\text{inc}}^*\right) = \frac{\hbar k}{m}\hat{z} = v\hat{z}$$

The outgoing scattered radial current density into solid angle $d\Omega = \sin\theta\,d\theta\,d\phi$ through surface element $dA = r^2 d\Omega$ is:

$$J_{\text{scatt}, r} = \frac{\hbar}{2mi}\left(\psi_{\text{scatt}}^* \frac{\partial}{\partial r}\psi_{\text{scatt}} - \psi_{\text{scatt}} \frac{\partial}{\partial r}\psi_{\text{scatt}}^*\right) = \frac{\hbar k}{m}\frac{|f(\theta, \phi)|^2}{r^2} = v \frac{|f(\theta, \phi)|^2}{r^2}$$

The differential cross-section $\frac{d\sigma}{d\Omega}$ is defined as the number of particles scattered into solid angle $d\Omega$ per unit time, divided by the incident flux $J_{\text{inc}}$:

$$\frac{d\sigma}{d\Omega} \equiv \frac{J_{\text{scatt}, r} r^2 d\Omega}{J_{\text{inc}} d\Omega} = |f(\theta, \phi)|^2$$

The total scattering cross-section $\sigma_{\text{tot}}$ is the integral over all solid angles:

$$\sigma_{\text{tot}} = \int_{4\pi} \frac{d\sigma}{d\Omega}\,d\Omega = \int_0^{2\pi} d\phi \int_0^\pi |f(\theta, \phi)|^2 \sin\theta\,d\theta$$

3. Laboratory vs Center-of-Mass (CM) Frame Kinematics

For a projectile of mass $m_1$ striking a target particle of mass $m_2$ initially at rest in the Laboratory frame:

$$\tan\theta_{\text{lab}} = \frac{\sin\theta_{\text{cm}}}{\cos\theta_{\text{cm}} + \gamma_m}, \quad \gamma_m \equiv \frac{m_1}{m_2}$$

The differential cross-sections relate by:

$$\left(\frac{d\sigma}{d\Omega}\right)_{\text{lab}} = \left(\frac{d\sigma}{d\Omega}\right)_{\text{cm}} \frac{(1 + 2\gamma_m\cos\theta_{\text{cm}} + \gamma_m^2)^{3/2}}{|1 + \gamma_m\cos\theta_{\text{cm}}|}$$

1. Partial Wave Expansion

For a spherically symmetric potential $V(r) = V(|\vec{r}|)$, angular momentum $\vec{L}$ is conserved, and the scattering amplitude exhibits azimuthal symmetry: $f(\theta, \phi) = f(\theta)$. We can expand both the incident plane wave and the scattering wavefunction in terms of Legendre polynomials $P_l(\cos\theta)$:

$$e^{ikz} = e^{ikr\cos\theta} = \sum_{l=0}^\infty i^l (2l+1) j_l(kr) P_l(\cos\theta)$$

where $j_l(kr)$ are the spherical Bessel functions. Using the asymptotic identity $j_l(\rho) \xrightarrow{\rho\to\infty} \frac{\sin(\rho - l\pi/2)}{\rho}$:

$$e^{ikz} \xrightarrow{r\to\infty} \sum_{l=0}^\infty i^l (2l+1) \frac{\sin(kr - l\pi/2)}{kr} P_l(\cos\theta) = \sum_{l=0}^\infty \frac{2l+1}{2ik r}\left[ e^{ikr} - (-1)^l e^{-ikr} \right] P_l(\cos\theta)$$

2. Radial Phase Shifts $\delta_l(k)$

In the presence of the potential $V(r)$, the asymptotic radial wavefunction undergoes a phase shift $\delta_l(k)$ relative to the free spherical Bessel function:

$$R_l(r) \xrightarrow{r\to\infty} A_l \frac{\sin(kr - l\pi/2 + \delta_l)}{kr}$$

The total asymptotic wavefunction becomes:

$$\psi(\vec{r}) \xrightarrow{r\to\infty} \sum_{l=0}^\infty i^l (2l+1) e^{i\delta_l} \frac{\sin(kr - l\pi/2 + \delta_l)}{kr} P_l(\cos\theta)$$

Subtracting the incident plane wave $e^{ikz}$ from $\psi(\vec{r})$ and equating the coefficient of $\frac{e^{ikr}}{r}$ yields the exact partial wave expansion of the scattering amplitude:

$$f(\theta) = \sum_{l=0}^\infty (2l+1) f_l(k) P_l(\cos\theta)$$

where the $l$-th partial wave amplitude $f_l(k)$ is:

$$f_l(k) = \frac{e^{2i\delta_l} - 1}{2ik} = \frac{e^{i\delta_l}\sin\delta_l}{k} = \frac{1}{k(\cot\delta_l - i)}$$

1. The Optical Theorem

Integrating $|f(\theta)|^2$ over the sphere and using the orthogonality of Legendre polynomials $\int_{-1}^1 P_l(u) P_{l'}(u)\,du = \frac{2}{2l+1}\delta_{ll'}$:

$$\sigma_{\text{tot}} = 2\pi \int_0^\pi |f(\theta)|^2 \sin\theta\,d\theta = \frac{4\pi}{k^2}\sum_{l=0}^\infty (2l+1)\sin^2\delta_l$$

Evaluating the forward scattering amplitude at $\theta = 0$ (where $P_l(1) = 1$):

$$f(0) = \frac{1}{k}\sum_{l=0}^\infty (2l+1) e^{i\delta_l}\sin\delta_l = \frac{1}{k}\sum_{l=0}^\infty (2l+1)(\cos\delta_l\sin\delta_l + i\sin^2\delta_l)$$

Taking the imaginary part:

$$\text{Im}[f(0)] = \frac{1}{k}\sum_{l=0}^\infty (2l+1)\sin^2\delta_l = \frac{k}{4\pi}\sigma_{\text{tot}}$$

This is the celebrated Optical Theorem:

$$\sigma_{\text{tot}} = \frac{4\pi}{k}\text{Im}[f(0)]$$

It expresses conservation of probability: the total scattered flux equals the quantum interference depletion of the forward beam (the "shadow" cast by the scatterer).

2. Low-Energy Scattering Length & Effective Range

As the incident energy approaches zero ($k \to 0$), centrifugal barrier terms $\frac{\hbar^2 l(l+1)}{2mr^2}$ suppress all higher partial waves ($l \ge 1$): $\delta_l \propto k^{2l+1}$. Only the $s$-wave ($l = 0$) contributes:

$$\lim_{k\to 0} k\cot\delta_0 = -\frac{1}{a_0} + \frac{1}{2}r_0 k^2 + \mathcal{O}(k^4)$$

where $a_0$ is the scattering length and $r_0$ is the effective range.

In the extreme low-energy limit ($k \to 0$):

$$f_0 \to \frac{1}{-1/a_0 - ik} \to -a_0 \implies \sigma_{\text{tot}} = 4\pi a_0^2$$

The scattering is completely isotropic ($d\sigma/d\Omega = a_0^2$).

1. Hard Sphere Scattering

Consider a hard sphere of radius $a$: $V(r) = \infty$ for $r \le a$, and $V(r) = 0$ for $r > a$. The radial wavefunction must vanish at $r = a$:

$$R_l(a) = 0 \implies j_l(ka)\cos\delta_l - n_l(ka)\sin\delta_l = 0 \implies \tan\delta_l = \frac{j_l(ka)}{n_l(ka)}$$

For low energy ($ka \ll 1$), considering $l = 0$:

$$\tan\delta_0 = \frac{\sin(ka)}{-\cos(ka)} = -\tan(ka) \implies \delta_0 = -ka$$

The scattering length is $a_0 = -\lim_{k\to 0}\frac{\tan\delta_0}{k} = a$. The low-energy cross-section is:

$$\sigma_{\text{tot}} = 4\pi a^2$$

Key Physical Insight: The quantum total cross section $\sigma = 4\pi a^2$ is exactly four times the classical geometric cross-section $\sigma_{\text{class}} = \pi a^2$! The factor of 4 arises from wave diffraction around the shadow edge (the optical theorem).

2. The Ramsauer-Townsend Effect

For scattering by an attractive square well of depth $V_0$, as the energy varies, the phase shift $\delta_0$ may pass through integer multiples of $\pi$ ($\delta_0 = n\pi$). When this occurs, $\sin\delta_0 = 0 \implies \sigma_0 \approx 0$. At specific low electron energies ($\approx 0.7\text{ eV}$ in noble gases like Argon and Krypton), the scattering cross-section vanishes, making the gas effectively transparent to electrons—the Ramsauer-Townsend effect.

3. Breit-Wigner Resonance Scattering

When the incident particle energy $E$ matches a quasi-bound state energy $E_R$ inside a potential well surrounded by a barrier, the phase shift $\delta_l$ increases rapidly by $\pi$, passing through $\pi/2$:

$$\delta_l(E) = \delta_{\text{bg}} + \arctan\left(\frac{\Gamma/2}{E_R - E}\right)$$

where $\Gamma$ is the resonance width. The partial wave cross-section displays a Lorentzian peak:

$$\sigma_l = \frac{4\pi}{k^2}(2l+1)\sin^2\delta_l = \frac{4\pi}{k^2}(2l+1)\frac{\Gamma^2/4}{(E - E_R)^2 + \Gamma^2/4}$$

The resonance width $\Gamma$ is directly related to the lifetime $\tau$ of the decaying quasi-bound state by the time-energy uncertainty relation: $\tau = \hbar / \Gamma$.

1. The Scattering Matrix (S-Matrix)

The asymptotic radial partial wave function can be decomposed into incoming and outgoing spherical waves:

$$R_l(r) \propto \frac{1}{2ikr}\left[ S_l(k) e^{i(kr - l\pi/2)} - e^{-i(kr - l\pi/2)} \right]$$

where the $S$-matrix element is defined by:

$$S_l(k) \equiv e^{2i\delta_l(k)}$$

Conservation of probability (elastic scattering) requires that the modulus of the outgoing flux equals the incoming flux:

$$|S_l(k)| = |e^{2i\delta_l}| = 1 \iff S_l^\dagger S_l = \hat{I}$$

This proves that the $S$-matrix is strictly unitary.

2. Analytic Continuation and Bound State Poles

Continuing the wavevector $k$ into the complex $k$-plane ($k \to \kappa = i\kappa_B$, with $\kappa_B > 0$):

$$e^{ikr} = e^{i(i\kappa_B)r} = e^{-\kappa_B r}$$

For a true bound state, the wavefunction must decay exponentially as $r \to \infty$ without any incoming flux. In the complex $k$-plane, bound states of the Hamiltonian correspond precisely to simple poles of the $S$-matrix on the positive imaginary axis ($k = i\kappa_B$), with binding energy:

$$E_{\text{bound}} = \frac{\hbar^2 k^2}{2m} = -\frac{\hbar^2 \kappa_B^2}{2m} < 0$$

Resonances correspond to poles in the lower half of the complex $k$-plane at $k = k_R - i\frac{\gamma}{2}$.

1. The Inhomogeneous Helmholtz Equation

The stationary Schrödinger equation can be rewritten as an inhomogeneous Helmholtz differential equation:

$$\left(\nabla^2 + k^2\right)\psi(\vec{r}) = U(\vec{r})\psi(\vec{r}), \quad U(\vec{r}) \equiv \frac{2m}{\hbar^2}V(\vec{r})$$

The free-particle Green's function $G_0(\vec{r}, \vec{r}')$ satisfies:

$$\left(\nabla^2 + k^2\right)G_0(\vec{r}, \vec{r}') = \delta^3(\vec{r} - \vec{r}')$$

Imposing the boundary condition of outgoing spherical waves at infinity selects the causal Green's function $G_0^+(\vec{r}, \vec{r}')$:

$$G_0^+(\vec{r}, \vec{r}') = -\frac{1}{4\pi}\frac{e^{ik|\vec{r} - \vec{r}'|}}{|\vec{r} - \vec{r}'|}$$

2. The Lippmann-Schwinger Equation

Integrating using Green's identity converts the differential Schrödinger equation into the exact Lippmann-Schwinger integral equation:

$$\psi(\vec{r}) = \phi_{\text{inc}}(\vec{r}) + \int G_0^+(\vec{r}, \vec{r}') U(\vec{r}') \psi(\vec{r}')\,d^3r'$$ $$\psi(\vec{r}) = e^{i\vec{k}\cdot\vec{r}} - \frac{m}{2\pi\hbar^2}\int \frac{e^{ik|\vec{r} - \vec{r}'|}}{|\vec{r} - \vec{r}'|} V(\vec{r}') \psi(\vec{r}')\,d^3r'$$

In the asymptotic limit $r \gg r'$:

$$|\vec{r} - \vec{r}'| \approx r - \hat{r}\cdot\vec{r}', \quad \frac{e^{ik|\vec{r} - \vec{r}'|}}{|\vec{r} - \vec{r}'|} \approx \frac{e^{ikr}}{r} e^{-i\vec{k}'\cdot\vec{r}'}$$

where $\vec{k}' \equiv k\hat{r}$ is the scattered wavevector. Comparing with $\psi \to e^{i\vec{k}\cdot\vec{r}} + f(\theta, \phi)\frac{e^{ikr}}{r}$ yields the exact expression for the scattering amplitude:

$$f(\theta, \phi) = -\frac{m}{2\pi\hbar^2}\int e^{-i\vec{k}'\cdot\vec{r}'} V(\vec{r}') \psi(\vec{r}')\,d^3r'$$

1. The Born Series Expansion

Iteratively substituting the Lippmann-Schwinger equation into itself generates the Neumann-Born series:

$$\psi = \phi + \hat{G}_0 \hat{V}\phi + \hat{G}_0 \hat{V}\hat{G}_0 \hat{V}\phi + \dots$$

The First Born Approximation replaces the unknown internal wavefunction $\psi(\vec{r}')$ inside the integral by the incident unperturbed plane wave $\phi(\vec{r}') = e^{i\vec{k}\cdot\vec{r}'}$:

$$f^{(1)}(\theta, \phi) = -\frac{m}{2\pi\hbar^2}\int e^{-i\vec{k}'\cdot\vec{r}'} V(\vec{r}') e^{i\vec{k}\cdot\vec{r}'}\,d^3r' = -\frac{m}{2\pi\hbar^2}\int V(\vec{r}') e^{i\vec{q}\cdot\vec{r}'}\,d^3r'$$

where $\vec{q} \equiv \vec{k} - \vec{k}'$ is the momentum transfer wavevector, with magnitude:

$$q = |\vec{k} - \vec{k}'| = 2k\sin\left(\frac{\theta}{2}\right)$$

Thus, the First Born scattering amplitude is proportional to the 3D spatial Fourier transform of the scattering potential evaluated at momentum transfer $\vec{q}$.

For a spherically symmetric potential $V(r)$:

$$f^{(1)}(\theta) = -\frac{2m}{\hbar^2 q}\int_0^\infty r V(r)\sin(qr)\,dr$$

2. Application to the Screened Coulomb (Yukawa) Potential

The Yukawa potential models the screened Coulomb or nuclear force:

$$V(r) = V_0 \frac{e^{-\mu r}}{r}$$

Evaluating the Fourier integral:

$$f^{(1)}(\theta) = -\frac{2m V_0}{\hbar^2 q}\int_0^\infty e^{-\mu r}\sin(qr)\,dr = -\frac{2m V_0}{\hbar^2 q}\left(\frac{q}{\mu^2 + q^2}\right) = -\frac{2m V_0}{\hbar^2(\mu^2 + q^2)}$$

Substituting $q^2 = 4k^2\sin^2(\theta/2)$:

$$\frac{d\sigma}{d\Omega} = |f^{(1)}(\theta)|^2 = \frac{4m^2 V_0^2}{\hbar^4\left[\mu^2 + 4k^2\sin^2(\theta/2)\right]^2}$$

Taking the unscreened Coulomb limit ($\mu \to 0$, $V_0 = \frac{z Z e^2}{4\pi\varepsilon_0}$):

$$\frac{d\sigma}{d\Omega} = \frac{4m^2}{\hbar^4}\left(\frac{z Z e^2}{4\pi\varepsilon_0}\right)^2 \frac{1}{16k^4\sin^4(\theta/2)} = \left(\frac{z Z e^2}{16\pi\varepsilon_0 E}\right)^2 \frac{1}{\sin^4(\theta/2)}$$

This recovers the classic Rutherford Scattering Formula identically!