Feynman Diagrams

Compton scattering has two tree-level diagrams in QED:

Diagram (a): The incoming photon is absorbed, then the outgoing photon is emitted ($s$-channel Intermediate electron).

Diagram (b): The outgoing photon is emitted first, then the incoming photon is absorbed ($u$-channel intermediate electron).

The amplitude for diagram (a) is:

$\mathcal{M}_a = (-ie)^2 \bar{u}(p")\gamma^\mu \frac{i(\slashed{p} + \slashed{k} + m)}{(p+k)^2 - m^2 + i\epsilon}\gamma^\nu u(p)\,\epsilon_\mu^*(k')\,\epsilon_\nu(k)$

Where $\slashed{p} \equiv \gamma^\mu p_\mu$ and $\epsilon_\mu$ are photon polarisation vectors.

The amplitude for diagram (b) is:

$\mathcal{M}_b = (-ie)^2 \bar{u}(p')\gamma^\nu \frac{i(\slashed{p} - \slashed{k}' + m)}{(p-k')^2 - m^2 + i\epsilon}\gamma^\mu u(p)\,\epsilon_\mu^*(k')\,\epsilon_\nu(k)$

The total tree-level amplitude is:

$\mathcal{M} = \mathcal{M}_a + \mathcal{M}_b$

Squaring, summing over final-state polarisations and spins, averaging over initial-state Polarisations, and integrating over phase space yields the Klein—Nishina cross section. In the low-energy limit ($\omega \ll m_e$), this reduces to the classical Thomson cross Section:

$\sigma_T = \frac{8\pi}{3}\left(\frac{e^2}{4\pi m_e}\right)^2 = \frac{8\pi}{3}\frac{\alpha^2}{m_e^2} \approx 0.665 \times 10^{-28}\;\mathrm{m}^2$

Møller scattering has two tree-level $t$- and $u$-channel diagrams. Because the electrons Are identical fermions, the total amplitude must be antisymmetric under exchange:

$\mathcal{M} = \mathcal{M}_t - \mathcal{M}_u$

Where:

$\mathcal{M}_t = \frac{e^2}{t}\,\bar{u}(p_3)\gamma^\mu u(p_1)\,\bar{u}(p_4)\gamma_\mu u(p_2)$

$\mathcal{M}_u = \frac{e^2}{u}\,\bar{u}(p_4)\gamma^\mu u(p_1)\,\bar{u}(p_3)\gamma_\mu u(p_2)$

And $t = (p_1 - p_3)^2$, $u = (p_1 - p_4)^2$ are Mandelstam variables. The minus sign Is a consequence of Fermi-Dirac …/4-statistics-and-probability/2_statistics and ensures the Pauli exclusion principle is Satisfied. When the two electrons scatter at $90^\circ$ in the CM frame, $t = u$ and the Two amplitudes cancel, giving $\mathcal{M} = 0$. This is the expected result: identical Fermions cannot be distinguished in the final state at $90^\circ$ scattering.

At tree level, $e^+e^- \to \mu^+\mu^-$ proceeds via a single virtual photon ($s$-channel). The amplitude is:

$\mathcal{M} = \frac{(-ie)^2}{s}\,\bar{v}(p_2)\gamma^\mu u(p_1)\,\bar{u}(p_3)\gamma_\mu v(p_4)$

Where $s = (p_1 + p_2)^2$ is the Mandelstam variable (the centre-of-mass energy squared).

In the centre-of-mass frame with $s = 4E_{\mathrm{cm}^2 \gg m_\mu^2}$The spin-averaged Squared amplitude is:

$\lvert\overline{\mathcal{M}}\rvert^2 = \frac{e^4}{s^2}\cdot 8(s^2 + 2sm_\mu^2 + m_\mu^4) \xrightarrow{s \gg m_\mu^2} \frac{8e^4}{1}$

The differential cross section in the CM frame is:

$\frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4s}(1 + \cos^2\theta)$

Where $\theta$ is the scattering angle. Integrating over solid angle:

$\sigma = \frac{4\pi\alpha^2}{3s}$

This is the leading-order QED result. At LEP energies, electroweak corrections ($Z^0$ exchange and interference) become significant.

Bhabha scattering has two tree-level diagrams:

Diagram (a): $s$-channel annihilation into a virtual photon, producing a new $e^+e^-$ pair (same topology as muon pair production).

Diagram (b): $t$-channel exchange of a virtual photon between the incoming Electron and outgoing positron (Møller-type scattering).

The amplitude for the $s$-channel diagram is:

$\mathcal{M}_s = \frac{(-ie)^2}{s}\,\bar{v}(p_2)\gamma^\mu u(p_1)\,\bar{u}(p_3)\gamma_\mu v(p_4)$

The amplitude for the $t$-channel diagram is:

$\mathcal{M}_t = \frac{(-ie)^2}{t}\,\bar{u}(p_3)\gamma^\mu u(p_1)\,\bar{v}(p_2)\gamma_\mu v(p_4)$

Where $t = (p_1 - p_3)^2$ is the Mandelstam $t$ variable. The total amplitude is:

$\mathcal{M} = \mathcal{M}_s - \mathcal{M}_t$

The minus sign arises from Fermi …/4-statistics-and-probability/2_statistics (exchange of identical fermions in the Final state). When squaring, there is a cross term $\mathcal{M}_s\mathcal{M}_t^*$ That leads to interference between the two diagrams. This interference is destructive At small angles (forward scattering) and constructive at large angles, producing a Characteristic angular distribution that is essential for calibrating detectors at $e^+e^-$ colliders.