A clean way to demonstrate how a Feynman diagram becomes equations is to use the tree-level scattering of two distinguishable charged fermions through exchange of a virtual photon:

[ e^-(p_1)+\mu^-(p_2)\longrightarrow e^-(p_3)+\mu^-(p_4). ]

The diagram is a shorthand for a perturbative contribution to the quantum amplitude—not literally a picture of particles’ paths. Each vertex and internal line maps to a factor prescribed by QED. arxiv

1. Start from QED

Use natural units, (\hbar=c=1), and metric convention (g^{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)).

The QED Lagrangian for electrons and muons coupled to electromagnetism is

[ \mathcal L = \bar\psi_e(i\gamma^\mu\partial_\mu-m_e)\psi_e + \bar\psi_\mu(i\gamma^\mu\partial_\mu-m_\mu)\psi_\mu -\frac14 F_{\mu\nu}F^{\mu\nu} -\underbrace{e\,\bar\psi_e\gamma^\mu A_\mu\psi_e -e\,\bar\psi_\mu\gamma^\mu A_\mu\psi_\mu}{\mathcal L{\mathrm{int}}}, ]

where

[ F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu. ]

The interaction term,

[ \mathcal L_{\mathrm{int}}=-e\bar\psi\gamma^\mu A_\mu\psi, ]

says that a charged fermion can emit or absorb a photon. It produces the QED vertex factor

[ \boxed{-ie\gamma^\mu}. ]

The coupling satisfies

[ \alpha \equiv \frac{e^2}{4\pi}\approx \frac{1}{137}, ]

where (\alpha) is the fine-structure constant.

2. The diagram

With momenta labelled as above, the exchange channel is

electron:  p1  ───►●────────►  p3
                   │
                   │ q = p1 − p3
                   │
muon:      p2  ───►●────────►  p4
  • Straight directed external lines: incoming/outgoing electron or muon states.
  • Wavy internal line: a virtual photon.
  • Each dot: an electromagnetic interaction vertex.
  • Momentum conservation at the vertices implies

[ p_1+p_2=p_3+p_4, ]

and the momentum through the virtual photon can be chosen as

[ q=p_1-p_3=p_4-p_2. ]

Since the photon is internal, it need not obey the real-photon relation (q^2=0). Its off-shell four-momentum appears in the propagator denominator.

3. Translate diagram to factors

For a basic QED calculation in Feynman gauge, use:

Diagram element Mathematical factor
Incoming fermion with momentum (p), spin (s) (u(p,s))
Outgoing fermion (\bar u(p,s))
Electron/photon or muon/photon vertex (-ie\gamma^\mu)
Internal photon carrying momentum (q) (\displaystyle \frac{-ig_{\mu\nu}}{q^2+i\epsilon})
Four-momentum conservation (\displaystyle (2\pi)^4\delta^{(4)}!\left(\sum p_{\rm in}-\sum p_{\rm out}\right))

The (i\epsilon) prescription means (q^2\to q^2+i\epsilon); it specifies how the propagator’s pole is handled and encodes causal time ordering. The scalar propagator has the analogous form (i/(p^2-m^2+i\epsilon)). southampton.ac

4. Derive the amplitude

Apply the factors in the order of the fermion lines. For the electron line,

[ \bar u_e(p_3)\,(-ie\gamma^\mu)\,u_e(p_1). ]

For the muon line,

[ \bar u_\mu(p_4)\,(-ie\gamma^\nu)\,u_\mu(p_2). ]

For the exchanged photon,

[ \frac{-ig_{\mu\nu}}{q^2+i\epsilon}. ]

Multiplying them gives the (S)-matrix contribution:

[ i\mathcal M = \left[\bar u_e(p_3)(-ie\gamma^\mu)u_e(p_1)\right] \left[\frac{-ig_{\mu\nu}}{q^2+i\epsilon}\right] \left[\bar u_\mu(p_4)(-ie\gamma^\nu)u_\mu(p_2)\right]. ]

Collecting constants and contracting the Lorentz indices:

[ i\mathcal M = i\,\frac{e^2}{q^2+i\epsilon} \left[\bar u_e(p_3)\gamma^\mu u_e(p_1)\right] \left[\bar u_\mu(p_4)\gamma_\mu u_\mu(p_2)\right]. ]

Therefore, with the common convention in which the (S)-matrix contains (i\mathcal M),

[ \boxed{ \mathcal M = \frac{e^2}{q^2+i\epsilon} \left[\bar u_e(p_3)\gamma^\mu u_e(p_1)\right] \left[\bar u_\mu(p_4)\gamma_\mu u_\mu(p_2)\right] } ]

together with the overall conservation factor

[ (2\pi)^4\delta^{(4)}(p_1+p_2-p_3-p_4). ]

This has a useful interpretation:

[ J_e^\mu=\bar u_e(p_3)\gamma^\mu u_e(p_1), \qquad J_{\mu}^\nu=\bar u_\mu(p_4)\gamma^\nu u_\mu(p_2), ]

so that

[ \mathcal M = \frac{e^2}{q^2+i\epsilon}\,J_e^\mu J_{\mu,\mu}. ]

In words: the electron current produces a virtual photon, the photon propagates, and the muon current absorbs it.

5. From amplitude to an observable

A single diagram yields an amplitude, not directly a probability. For an unpolarized scattering experiment, one averages over initial spins and sums over final spins:

[ \overline{|\mathcal M|^2} = \frac14\sum_{\text{spins}}|\mathcal M|^2. ]

The spin sums are reduced using

[ \sum_s u(p,s)\bar u(p,s)=\slashed p+m, \qquad \slashed p\equiv \gamma^\mu p_\mu. ]

Thus,

[ \overline{|\mathcal M|^2} = \frac{e^4}{4(q^2)^2} \operatorname{Tr} \left[ (\slashed p_3+m_e)\gamma^\mu (\slashed p_1+m_e)\gamma^\nu \right] \operatorname{Tr} \left[ (\slashed p_4+m_\mu)\gamma_\mu (\slashed p_2+m_\mu)\gamma_\nu \right]. ]

The differential cross section for a generic (2\to2) process is then

[ d\sigma = \frac{1}{4\sqrt{(p_1\cdot p_2)^2-m_e^2m_\mu^2}} \, \overline{|\mathcal M|^2} \, d\Phi_2, ]

where the two-body Lorentz-invariant phase-space element is

[ d\Phi_2 = (2\pi)^4\delta^{(4)}(p_1+p_2-p_3-p_4) \prod_{f=3,4} \frac{d^3\mathbf p_f}{(2\pi)^3\,2E_f}. ]

This workflow—draw all permitted diagrams at a chosen perturbative order, write (\mathcal M), sum amplitudes, square the total, then integrate phase space—is the standard diagram-to-prediction pipeline. arxiv

Why this is a good demonstration

Electron–muon scattering is simpler than electron–electron scattering because the final particles are distinguishable. For (e^-e^-\to e^-e^-), there is an additional exchange diagram, and the total amplitude becomes

[ \mathcal M_{\text{total}} = \mathcal M_t-\mathcal M_u, ]

with the relative minus sign coming from exchanging identical fermions. That interference is physically important, but it obscures the core “vertex × propagator × vertex” idea for a first example.

At low momentum transfer, the photon factor (1/q^2) becomes the momentum-space origin of the familiar long-range Coulomb interaction. In the nonrelativistic static limit, (q^0\simeq0), so (q^2\simeq-\mathbf q^2), and Fourier transforming a (1/\mathbf q^2) dependence yields a (1/r) potential. Thus the diagrammatic QED calculation connects directly back to

[ V(r)=\frac{e^2}{4\pi r} = \frac{\alpha}{r}, ]

for like charges, with the appropriate sign determined by the charges involved.