The Interaction as an Exchange of Quanta; The Feynman Diagrams

This is where a drawing turns into an instruction: read off the vertices, read off the internal lines, multiply. Whatever the picture suggested about what happened stops mattering here.

🎯 Why this matters

That split is what lets an entire interaction be summarised by one constant. Comparing two forces means comparing couplings, and their ranges follow from masses the coupling never mentions.

§5.3 drew diagrams and asked you to read them as pictures. This section makes them arithmetic. Two questions get answered: what is an internal line worth, and what is a vertex worth — after which a diagram is a formula.

5.4 A force is an exchange, and the mediator’s mass is its range

Particle a moving alone continually emits and reabsorbs mediators V. Bring particle b close and sometimes a V emitted by a is absorbed by b instead. That is what “they interact” means here.

Emission costs energy the particle has not got — at least $\Delta E = m$ , the mediator’s mass — so it can only last as long as the uncertainty principle permits it to go undetected:

\htmlClass{t-R}{R} = c\,\Delta t \simeq \frac{\hbar}{\htmlClass{t-m}{m}c}

(5.27)

Bettini p. 195. The range of a force is the reciprocal of its mediator's mass.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

Every mediator in this book, and the range it implies

hbarc = 197.3269804     # MeV fm

for name, m in [("photon", 0.0), ("pion (Yukawa 1935)", 139.57), ("rho", 775.3),
                ("W", 80377.0), ("Z", 91187.6)]:
    R = "infinite" if m == 0 else f"{hbarc/m:.4g} fm"
    print(f"  {name:20s} m = {m:9.1f} MeV -> R = {R}")

print("\n  Yukawa ran this backwards in 1935: nuclei are about 1 fm across,")
print("  so he predicted a mediator of about 200 MeV. The pion is 140.")

prints

  photon               m =       0.0 MeV -> R = infinite
pion (Yukawa 1935)   m =     139.6 MeV -> R = 1.414 fm
rho                  m =     775.3 MeV -> R = 0.2545 fm
W                    m =   80377.0 MeV -> R = 0.002455 fm
Z                    m =   91187.6 MeV -> R = 0.002164 fm

Yukawa ran this backwards in 1935: nuclei are about 1 fm across,
so he predicted a mediator of about 200 MeV. The pion is 140.

Where the propagator comes from

The book’s derivation is one of the most useful pages in the chapter, because it gets a relativistic object out of a first-year quantum-mechanics integral.

🪜 From the Yukawa potential to the propagator

Step 1 of 6 — the Born matrix element(5.29)

\langle f|g\varphi(\mathbf{r})|i\rangle = g\int e^{-i\mathbf{p}_2\cdot\mathbf{r}}\,\varphi(\mathbf{r})\,e^{i\mathbf{p}_1\cdot\mathbf{r}}\,\mathrm{d}V = g\int e^{i\mathbf{q}\cdot\mathbf{r}}\varphi(\mathbf{r})\,\mathrm{d}V

Why you may do this: Initial and final states are plane waves, because the particle is free before and after. Their product collapses to a single exponential in the momentum transfer q = p₂ − p₁ — so the amplitude depends on q and nothing else. This is Eq. (1.84) of §1.8, reused verbatim.

Bettini pp. 195–197. Non-relativistic scattering off a fixed centre, Fourier transformed — and the answer turns out to be the thing that sits on every internal line in every diagram in this book.

🔢 Worked example — do the transform numerically, and watch the massless case fail

The step above is the only real calculation in the section, so it is worth checking rather than believing. And the check has a sting in it.

Reproduce it

import numpy as np

# f(q) = int exp(i q.r) phi(r) dV with the Yukawa phi = exp(-m r)/(4 pi r).
# The angular integral leaves  (1/q) * int_0^inf sin(qr) exp(-mr) dr.
def transform(q, m, R=200.0, N=400000):
    r = np.linspace(1e-9, R, N)
    return np.trapezoid(np.sin(q*r) * np.exp(-m*r), r) / q

print("  m (1/fm)  |q| (1/fm)      numeric   1/(q^2+m^2)     ratio")
for m in (2.0, 1.0, 0.5):
    for q in (0.3, 1.0, 3.0):
        print(f"  {m:8.1f} {q:11.1f} {transform(q,m):12.6f} {1/(q*q+m*m):13.6f} {transform(q,m)/(1/(q*q+m*m)):9.6f}")

print("\n  and the massless case, which is the interesting one:")
for q in (0.3, 1.0, 3.0):
    print(f"  {0.0:8.1f} {q:11.1f} {transform(q,0.0):12.6f} {1/(q*q):13.6f}"
          f" {transform(q,0.0)/(1/(q*q)):9.6f}")
print("  the integral does not converge at m = 0 -- sin(qr) never dies away.")
print("  You keep m > 0, transform, and take m -> 0 at the end. Yukawa's screened")
print("  potential is the right starting point even for electromagnetism.")

prints

  m (1/fm)  |q| (1/fm)      numeric   1/(q^2+m^2)     ratio
     2.0         0.3     0.244499      0.244499  1.000000
     2.0         1.0     0.200000      0.200000  1.000000
     2.0         3.0     0.076923      0.076923  1.000000
     1.0         0.3     0.917431      0.917431  1.000000
     1.0         1.0     0.500000      0.500000  1.000000
     1.0         3.0     0.100000      0.100000  1.000000
     0.5         0.3     2.941176      2.941176  1.000000
     0.5         1.0     0.800000      0.800000  1.000000
     0.5         3.0     0.108108      0.108108  1.000000

and the massless case, which is the interesting one:
     0.0         0.3    21.693478     11.111111  1.952413
     0.0         1.0     0.512812      1.000000  0.512812
     0.0         3.0     0.222114      0.111111  1.999023
the integral does not converge at m = 0 -- sin(qr) never dies away.
You keep m > 0, transform, and take m -> 0 at the end. Yukawa's screened
potential is the right starting point even for electromagnetism.

Six decimal places for every massive case, and garbage for the massless one — the numbers there are not converging to anything, they are wherever the truncation happened to stop. The mass is not a complication in this calculation, it is what makes it exist; you keep it, transform, and take m → 0 last.

Erratum — the cross-reference after Eq. (5.34)

The book introduces the relativistic amplitude with “which is very similar to (5.29)”. Equation (5.29) is the Born matrix element $\langle f|g\varphi|i\rangle = g\int e^{i\mathbf{q}\cdot\mathbf{r}}\varphi\,\mathrm{d}V$ — an integral, not a closed form, and it does not resemble $g_0g/(m^2-t)$ at all.

The expression (5.34) is very similar to is (5.32), $f(q) = g_0g/(|\mathbf{q}|^2 + m^2)$ : same numerator, and a denominator that goes over into the other when $|\mathbf{q}|^2 \to -t$ . That is the whole point being made, and it is worth having the right pointer, because the resemblance is the argument. Confirmed on the page render.

m = 0 — the photon
m = 1 fm⁻¹ ≈ 200 MeV
m = 5 fm⁻¹ ≈ 1 GeV

The propagator of Eq. (5.32), for three mediator masses. Two regimes and one crossover: below |q| ≈ m the massive curves are FLAT at 1/m², and above it they merge with the massless 1/q². The mediator mass is invisible to a probe that pushes harder than m — which is the whole content of the next figure, and the reason the weak interaction stops being weak above 80 GeV.

⚙️ Engineer’s bridge — the propagator is a transfer function, and the mass is a pole

$\Pi(t) = 1/(m^2 - t)$ is a first-order response with a pole at $t = m^2$ . You have met that shape.

Low frequency — here, momentum transfer below the mass — the response is flat at $1/m^2$ . The system cannot resolve any structure; it sees a contact interaction with a fixed strength. Fermi’s four-fermion theory of beta decay is exactly this limit, and Chapter 7 shows it is the low-|q| approximation to a W exchange.
High frequency the pole dominates and the response rolls off as $1/|q|^2$ . A probe pushing harder than the mass sees the mediator as effectively massless.
On the pole, $t \to m^2$ , the response diverges — and in the s channel, where the invariant can actually reach $m^2$ , that divergence is a resonance (§5.6). A pole on the real axis is a state that exists.

So the same plot tells you the range of the force, the strength of its low-energy contact approximation, and where its resonances live. Reading a propagator is reading a Bode plot.

Where the analogy breaks: the “frequency” here is a Lorentz invariant, not a lab quantity, so which regime you are in is not a matter of how fast you drive the input — it is set by $t$ , which mixes energy and momentum transfer. And the pole is at $t = m^2 > 0$ , a value the t channel can never reach; only the s channel gets there, which is why resonances appear in annihilation and not in scattering.

5.5 The rules, and what a diagram is worth

\htmlClass{t-z}{z\sqrt{\alpha}}\;\htmlClass{t-A}{A_\mu}\,\htmlClass{t-j}{\bar\psi_f \gamma^\mu \psi_f}

(5.36)

Bettini p. 199. The electromagnetic vertex — and every vertex in Chapter 5 is this one expression.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

The line vocabulary of Fig. 5.14–5.15, and what each element is worth

element	drawn as	contributes
external leg	a line with one free end	Nothing multiplicative — it is a real particle, on the mass shell, whose four-momentum you specified as part of the initial or final state. The legs are the boundary conditions of the problem.
vertex	a point where three lines meet	One factor of √α (times the charge z). Energy, momentum and every conserved quantum number balance exactly here — which is precisely what forces the internal line off the mass shell.
internal line	a line between two vertices	One propagator, 1/(m² − t). The particle on it is virtual: it has no independent existence and is never observed.
fermion arrow	an arrowhead on a solid line	Nothing numerically — it records the direction in which charge and flavour advance in time. An arrow pointing backwards in time is an antiparticle going forwards, which is §5.6.
loop	a closed circuit of internal lines	Two more vertices, hence one more power of α in the amplitude — and an integral over the undetermined momentum circulating in it. Loops are where the infinities of §5.3 live.

<strong>The direction an internal line is drawn is meaningless.</strong> Fig. 5.17(a) and (b) are the same diagram — whether the photon is emitted by the electron and absorbed by the muon or the reverse is not a question with an answer, because the amplitude sums over all of it. Only the topology matters.

Fig. 5.17 — electron–muon scattering at tree level

Click a vertex or an internal line.

Two vertices, one internal line — the simplest scattering diagram there is. Press the netlist button: amplitude ∝ √α × √α × propagator, so amplitude ∝ α ≈ 1/137 and cross-section ∝ α² ≈ 5 × 10⁻⁵.

💡 What this really says — why truncating an infinite sum is legitimate

The amplitude for any process is a sum over infinitely many diagrams. That should be alarming, and the reason it is not is one number.

Each additional vertex costs a factor $\sqrt\alpha$ in the amplitude, and diagrams come with vertices in pairs, so each successive order costs $\alpha \approx 1/137$ . The tree diagram above carries $\alpha$ ; the next order carries $\alpha^2$ , a hundred and thirty-seven times smaller; the one after that another factor of 137.

Two consequences, and both matter later.

Where α is small the series converges fast, which is why QED can be computed to ten decimal places and why the electron magnetic moment of §5.9 is the most precisely predicted number in physics. It also means “the” diagram for a process is a useful fiction that is right to about 1 %.

Where the coupling is not small the whole scheme collapses. Chapter 6’s α_s is of order 1 at low energy, so no truncation converges — which is why the hadronic contribution to the muon anomaly needs lattice QCD instead of diagrams, and why the strong interaction was hard for thirty years after QED was finished.

Fig. 5.18 — one of the next-order diagrams

Click a vertex or an internal line.

Four vertices and two internal lines. The netlist view makes the counting explicit — this contributes at order α² to the amplitude, so about 1 % of the tree diagram, and its interference with the tree term is what a per-cent-level measurement is actually sensitive to.

📏 A diagram is not a picture of what happened

It is tempting, and wrong, to read Fig. 5.17 as an event: the electron went along, emitted a photon there, the muon absorbed it here. Three reasons that reading fails.

The photon’s direction is undetermined. The external legs are fixed by the initial and final states, but the internal four-momentum is not — the amplitude integrates over all of it. Drawing the line at some particular angle is a convention, and Fig. 5.17(a) and (b) are the same diagram.

Every diagram happens. The amplitude is a sum over topologies, added as complex numbers before squaring, so asking which one occurred is like asking which slit the electron went through.

Internal lines are not observable. A virtual particle is off the mass shell; no detector can register it, and §5.6 will show that even its time ordering is frame-dependent.

What a Feynman diagram is, precisely, is a mnemonic for a term in a series — a notation so good that it is easy to mistake for a photograph.

🔑 If you remember only three things

The vertex is a pure number and the propagator carries the dimensions. One is the same wherever it appears; the other is where a particular force’s physics actually lives.
The infinite sum is safe because every extra vertex costs a factor of the coupling. Truncation is not an approximation anyone chose — it is what a small number does to a series.
The massless case is not a limit of the massive one. The transform that gives a finite range fails outright at zero mass, so Coulomb has to be reached by another route.

Where this goes next

§5.6 takes the internal line seriously: it can run backwards in time, and that is what forces antiparticles to exist.
§5.7 computes a real cross-section with these rules — e⁺e⁻ → μ⁺μ⁻, where the 1/s energy dependence and the (1 + cos²θ) shape both come out of the counting above.
§5.8 adds the loops: once you sum the vacuum-polarization diagrams, α itself starts depending on |q|.
Chapter 7 is what happens to Eq. (5.35) when the mediator is heavy — the flat low-|q| regime of the plot above is Fermi’s theory, and the crossover is at 80 GeV.

✅ Check yourself — propagators, vertices and what a diagram is worth

0/5 answered · 0 correct

1.The numerical check of the Yukawa transform reproduces $1/(|q|^2+m^2)$ to six decimals for every massive case and fails completely at $m = 0$ . Why?
2.Look at the propagator plot. What happens below $|q| \approx m$ , and what is that regime called elsewhere in the book?
3.Why is truncating an infinite sum of diagrams after the first term legitimate in QED?
4.Fig. 5.17(a) and (b) differ in whether the photon is drawn going up or down. What is the difference between them?
5.What does the vector current $\bar\psi_f\gamma^\mu\psi_f$ do at a vertex, and why does it matter beyond QED?