C Violation, Chirality vs Helicity, and the W · Elementary Particle Physics

The W was inside Fermi’s constant from the beginning. G_F is a coupling divided by a mass squared, and the mass it hides is the one found forty years later.

🎯 Why this matters

An effective theory’s constants are composites, so measuring one measures a combination that cannot be unpicked from inside. Separating the coupling from the mass needed an experiment at an energy where the propagator stops looking constant.

Three short sections that close out the structure of the charged current. §7.6 shows that C is broken as badly as P — by an argument that has to be done carefully, because the obvious version of it gives the wrong answer. §7.7 separates chirality from helicity once and for all. §7.8 finally names the thing whose absence has been hanging over the whole chapter: the W boson, and with it the reason $G_F$ has dimensions.

§7.6 The argument that nearly goes wrong

Parity violation was established from the presence of $\boldsymbol\sigma\cdot\mathbf{p}$ in the Lagrangian: it is P-odd, so a rate that depends on it cannot be P-symmetric. Try the same reasoning for charge conjugation and you get a confident wrong answer.

C turns particles into antiparticles and does nothing else. It does not touch space, so momenta are unchanged; it does not touch spin. Therefore $\boldsymbol\sigma\cdot\mathbf{p}$ is C-even, and the naive conclusion is that C is conserved.

It is not. The book gives two routes to the correct answer, and they are worth keeping separate because one is a proof and the other is the physics.

The proof, via CPT. CPT is not up for negotiation — §3.4 established it as a theorem, not an assumption. Work out how PT acts:

C is the only column that leaves it alone — and that is exactly what makes C the one that must be violated.

operator↕	$\mathbf{p}$ ↕	$\boldsymbol\sigma$ ↕	$\boldsymbol\sigma\cdot\mathbf{p}$ ↕
P	$-\mathbf{p}$	$+\boldsymbol\sigma$	odd
T	$-\mathbf{p}$	$-\boldsymbol\sigma$	even — both flip
PT	$+\mathbf{p}$	$-\boldsymbol\sigma$	odd
C	$+\mathbf{p}$	$+\boldsymbol\sigma$	even

The Lagrangian contains a term that is PT-odd. If CPT is a symmetry, that term must be CPT-even overall — so C has to supply the compensating sign. C is violated, and the argument never needed a single measurement.

Notice what went wrong with the naive version. It asked how C transforms a quantity. The right question is what C does to a state that actually exists, which is the second route.

The physics. The weak current couples to left-chirality fields, so beta decay emits neutrinos with $h = -1$ . Apply C: you get an antineutrino with $h = -1$ , because C leaves momentum and spin alone. But every antineutrino ever observed has $h = +1$ . The C-image of a real process is a process that does not happen — which is C violation at its most extreme. Not a small asymmetry: the mirror world is empty.

This is why C is violated maximally, and it is the same “maximally” as for P. Both operators map the one existing state onto one of the three that do not:

P and C each take the one state that exists onto one that does not — that is what “maximal violation” means, and why the two violations are equally extreme. Only the diagonal lands on something real. Landau’s observation in 1957 was that CP survives, restoring matter–antimatter symmetry; Christenson, Cronin, Fitch and Turlay broke that too in 1964, which is Chapter 8.

Direct evidence, rather than the CPT inference, needs two processes that are genuine C-conjugates of each other. A Wu-type experiment cannot do it — you would need the beta decay of an antinucleus. Pions can:

\pi^- \to \mu^- \bar\nu_\mu,\; \mu^- \to e^- \nu_\mu \bar\nu_e \qquad\text{versus}\qquad \pi^+ \to \mu^+ \nu_\mu,\; \mu^+ \to e^+ \bar\nu_\mu \nu_e

The two chains are exact charge conjugates. The measured helicities of the electron in one and the positron in the other come out with opposite expectation values, where C-symmetry demands they be equal. That is the direct proof.

💡 What this really says — why “is this quantity even under C?” was the wrong question

The failed argument was not careless — it is the exact reasoning that works for parity, applied one section later. What changed?

For P, $\boldsymbol\sigma\cdot\mathbf{p}$ being odd is enough, because P maps the set of physical states onto itself; it just relabels them, and a rate that depends on a P-odd quantity therefore differs between two states that both exist. You can measure both and compare.

C does not map the physical state space onto itself here. It sends $\nu(h{=}{-}1)$ to $\bar\nu(h{=}{-}1)$ , which is not in the space at all. So asking whether a particular operator is C-even is asking the wrong thing — the violation is not in the sign of a term, it is in the fact that the image of the transformation is outside the theory.

The lesson generalises: checking that an operation preserves a symmetry means checking it maps the state space to itself, not just that some observable keeps its sign. A transformation that produces states your theory does not contain has already broken the symmetry, whatever the algebra says.

§7.7 Chirality is a property; helicity is a measurement

These two words have been doing separate jobs since §2.8, and §7.7 finally states the relationship cleanly. It is worth being precise, because almost every confusion in this chapter comes from conflating them.

Chirality is what the Lagrangian couples to. It is Lorentz-invariant, defined by $\gamma_5$ , and it is a property of a field. Helicity is $\mathbf{s}\cdot\hat{\mathbf{p}}$ — an observable, frame-dependent for a massive particle (overtake it and the momentum reverses while the spin does not), and a property of a state of motion.

They coincide only for massless particles. For everything else, a field of definite chirality is a superposition of both helicities, in the proportion Eq. (2.66) gives:

\htmlClass{t-h}{\langle h \rangle} = -\htmlClass{t-b}{\beta} \quad\text{for a left-chirality fermion}

(2.66)

The bridge between what the Lagrangian couples to and what a detector can measure. Chirality and helicity are different things, and this is exactly how different.

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.

which is the line Fig. 7.7 confirmed with no free parameter.

The consequence §7.7 draws is about antiparticles, and it comes from CPT again. Following Pal’s argument: if $e^- p \to \nu_e n$ exists, CPT guarantees that $\bar\nu_e n \to e^+ p$ exists with the same amplitude, with each particle in the second replaced by the CP conjugate of its partner in the first. Both are described by the same current, $\bar\psi_{e,L}\gamma^\mu\psi_{e,L}\dots$ — so the same field operator that annihilates a left-chirality electron must create a right-chirality positron.

The chirality column is exact and frame-independent. The helicity column is what a detector sees, and it depends on how fast the particle is going.

particle↕	chirality the CC couples to↕	dominant helicity↕	how dominant↕
electron in β decay	left	$h = -1$	⟨h⟩ = −β; at β = 0.5 the wrong component still carries a quarter of the probability
positron (same field, creation part)	right	$h = +1$	the CP mirror of the line above — C flips the charge, P flips the helicity
nucleon, non-relativistic	left	neither	β ≈ 0, so the two helicity components are almost equal. There is no useful sense in which a slow neutron is "left-handed"
antinucleon, non-relativistic	right	neither	same, mirrored: h = +½ very slightly favoured
neutrino (massless in the SM)	left	$h = -1$	exactly — β = 1, so chirality and helicity coincide and there is no small component at all
antineutrino	right	$h = +1$	exactly, for the same reason. This is the row Goldhaber measured

⚠️ “Left-handed” is two different claims, and only one of them is exact

When a textbook says the weak interaction couples to left-handed particles, it means chirality, which is exact. When it says the neutrino is left-handed, it means helicity, which is exact only because the neutrino is (treated as) massless.

For anything with mass the two statements come apart, and the gap is $\beta$ . A slow electron from tritium decay is a left-chirality field with a mean helicity of only −0.2 — it is barely polarized, and calling it “left-handed” without qualification is wrong by a factor of five. The Fig. 7.7 plot is exactly a picture of the two words separating.

The site follows the book: chirality for the Lorentz-invariant property in the Lagrangian, helicity for the measurable $\mathbf{s}\cdot\hat{\mathbf{p}}$ , and never “handedness” on its own.

Note also that Chapter 10 will make the neutrino massive, at which point the bottom two rows of the table stop being exact and the whole distinction starts doing real work.

Re-mount the pion decay with that vocabulary and the §7.4 argument reads differently: the antineutrino’s helicity is exactly $+1$ because it is massless, and the charged lepton’s forced $+1$ is a helicity that its left-chirality field can only supply through a component of size $m_\ell/(E+p)$ .

Angular momentum in K⁻ → μ⁻ ν̄

parentlepton

The μ⁻ is forced into h = +1, and for a left-chirality field that is the small component, amplitude m/(E + p). So M ∝ m and the rate carries m².

CM momentum p*

235.5 MeV

lepton β

0.9124

wrong-helicity amplitude

0.2140

its square

4.58e-2

Γ(K⁻ → e ν̄) / Γ(K⁻ → μ ν̄), the two factors separated

helicity m_e²/m_μ² = 2.34e-5 × phase space (p*_e/p*_μ)² = 1.098 = 2.57e-5

For the kaon the phase-space factor is nearly 1 — the muon is no longer near threshold — so the ratio is almost pure helicity suppression.

The same widget as §7.4, opened on the kaon. Chirality is what the vertex demands; helicity is what angular momentum demands; the mismatch between them is the mass.

§7.8 The boson that was there all along

Fermi’s contact vertex was always a fiction, and §7.1 said why: a coupling with dimensions is a theory announcing that something has been integrated out. Here it is.

Fig. 7.8 — neutron beta decay, before and after

Click a vertex or an internal line.

Bettini Fig. 7.8(b). Click the W line: everything that separates this from Fermi's point is in that propagator.

The matrix element is now the product of two couplings and a propagator:

\mathcal{M} \propto \frac{\htmlClass{t-g}{g\,g}}{\htmlClass{t-P}{M_W^2 - t}} \qquad\xrightarrow{\;\;|t|\,\ll\, M_W^2\;\;}\qquad \mathcal{M} \propto \frac{\htmlClass{t-g}{g^2}}{\htmlClass{t-M}{M_W^2}}

(7.53)

Bettini p. 292, Eqs. (7.53) and (7.54). The arrow is where Fermi's theory lives.

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.

⚙️ Engineer’s bridge — Fermi’s theory is a lumped-element model

There is a precise engineering statement of what is going on, and it is not an analogy — it is the same argument.

A momentum transfer $q$ probes with a reduced wavelength $\hbar c/q$ . The W’s range is $\hbar/M_W c = 0.0025$ fm. In a beta decay $q \sim 1$ MeV, so the probe’s wavelength is about 200 fm — eighty thousand times longer than the structure it is looking at. Nothing about the two-vertex structure is resolvable, so it is not merely convenient to draw it as one point: at that resolution it is one point.

This is exactly why you model a transmission line as a lumped capacitor when the signal wavelength is much longer than the line, or a distributed RC network as a single pole. The lumped model is not an approximation you tolerate — it is correct, until the frequency rises far enough to resolve the structure, at which point it fails abruptly and completely.

And you can say where. The contact approximation errs by 1 % at a momentum transfer of 8 GeV, by 5 % at 18 GeV, and by 50 % at $M_W$ itself. Every measurement made before the 1970s sat many orders of magnitude below the first of those, which is why a theory that was structurally wrong survived fifty years without a single discrepancy.

The dimensional argument of §7.1 is the same statement read backwards. $G_F$ has units of GeV⁻² because it is $g^2/M_W^2$ with the mass scale hidden inside it — a lumped parameter always hides the scale it lumped. Reading the scale back out of the units is how you find out what your model threw away.

Where it breaks: a lumped circuit model degrades gracefully and tells you it is degrading — the phase error creeps up, the fit worsens, you add an element. Fermi’s theory does not. It stays exact to within your error bars for fifty years and then violates unitarity outright: the predicted probability exceeds 1, which is not an inaccuracy but a contradiction.

Nor could it be repaired the way a lumped model is, by adding the next term — no resummation of a contact interaction produces a propagator, because the missing object is a new field rather than a correction to an existing one. And the ordinary engineering move of measuring the response near the cut-off to identify what you lumped was unavailable: reaching 8 GeV of momentum transfer took four decades of accelerator construction. The analogy is right about the structure of the approximation and wrong about how you find out you have exceeded it.

what 'point-like' means, quantitatively

import numpy as np
GF, MW, mmu, mp = 1.1663788e-5, 80.377, 0.1056583755, 0.93827209
hbarc = 0.1973269804                      # GeV fm

g2 = 8*MW**2*GF/np.sqrt(2)                # invert Eq. (7.55b)
print("the couplings, from G_F = (sqrt2/8) g^2/M_W^2   -- Eq. (7.55b)")
print(f"  g^2 = {g2:.4f}   ->  alpha_W = g^2/4pi = {g2/(4*np.pi):.4f} = 1/{4*np.pi/g2:.1f}")

print("\nresolution: a momentum transfer q probes a wavelength hbar c / q")
for q, what in [(0.001, 'beta decay, ~1 MeV '), (0.1, 'muon decay, ~100 MeV'), (10, 'DIS, Sec. 6.2      ')]:
    print(f"  q = {q:6.3f} GeV  ({what})  ->  {hbarc/q:8.3f} fm")
print(f"  the W's own range, hbar / M_W c        =    {hbarc/MW:.5f} fm")
print(f"  a beta decay probes with a wavelength {MW/0.001:.0f}x longer than that.")

print("\nwhere the contact approximation actually fails:")
for dev in (0.01, 0.05, 0.50):
    t = dev/(1-dev) * MW**2
    print(f"  {dev*100:2.0f}% error at |t| = {t:5.0f} GeV^2,  sqrt|t| = {np.sqrt(t):5.1f} GeV")

Eth = ((mmu + MW + mp)**2 - mp**2) / (2*mp)
print("\nQuestion 7.2 -- threshold for nu_mu p -> mu- W+ p")
print(f"  E_nu = [(m_mu + M_W + m_p)^2 - m_p^2] / 2 m_p = {Eth:.0f} GeV = {Eth/1e3:.1f} TeV")
print( "  1960s-70s neutrino beams reached a few tens of GeV.")
print(f"  Short by a factor of ~{Eth/50:.0f} -- which is why the W was never going")
print( "  to turn up in a neutrino beam, and why 1983 needed a collider.")

prints

the couplings, from G_F = (sqrt2/8) g^2/M_W^2   -- Eq. (7.55b)
g^2 = 0.4263   ->  alpha_W = g^2/4pi = 0.0339 = 1/29.5

resolution: a momentum transfer q probes a wavelength hbar c / q
q =  0.001 GeV  (beta decay, ~1 MeV )  ->   197.327 fm
q =  0.100 GeV  (muon decay, ~100 MeV)  ->     1.973 fm
q = 10.000 GeV  (DIS, Sec. 6.2      )  ->     0.020 fm
the W's own range, hbar / M_W c        =    0.00246 fm
a beta decay probes with a wavelength 80377x longer than that.

where the contact approximation actually fails:
 1% error at |t| =    65 GeV^2,  sqrt|t| =   8.1 GeV
 5% error at |t| =   340 GeV^2,  sqrt|t| =  18.4 GeV
50% error at |t| =  6460 GeV^2,  sqrt|t| =  80.4 GeV

Question 7.2 -- threshold for nu_mu p -> mu- W+ p
E_nu = [(m_mu + M_W + m_p)^2 - m_p^2] / 2 m_p = 3532 GeV = 3.5 TeV
1960s-70s neutrino beams reached a few tens of GeV.
Short by a factor of ~71 -- which is why the W was never going
to turn up in a neutrino beam, and why 1983 needed a collider.

M_W²/(M_W² + |t|) — the true propagator
Fermi: a constant

Fermi's theory is the flat line. It is not approximately right at low momentum transfer — on this scale it is indistinguishable, and stays so for five decades of energy. The whole of experimental physics up to the 1970s lived to the left of the '1% error' mark.

⚠️ Three symbols on this page collide with earlier ones

$g$ — the fifth meaning. After the metric tensor $g_{\mu\nu}$ (§5.1), the gyromagnetic ratio (§5.9a), the strong coupling $g_s$ (§6.3), and the gluon distribution $g(x)$ (§6.2), $g$ here is the weak charge. It is dimensionless, and $g^2 = 0.426$ , so $\alpha_W = g^2/4\pi = 1/29.5$ .

The book estimates $\mathcal{M} \sim 10^{-6}$ GeV⁻² by supposing ” $g^2$ of the same order of magnitude as the fine-structure constant”. That is explicitly a hypothetical, but it is worth knowing it understates: the real $g^2$ is 58 times $\alpha$ , so $g^2/M_W^2 = 6.6\times10^{-5}$ GeV⁻², which is $5.7\,G_F$ . The conclusion — feeble because $M_W$ is large, not because $g$ is small — is the same one §7.1 reached from the other direction.

$t$ is the Mandelstam momentum transfer of §1.6, not time. It is negative in scattering, so $M_W^2 - t$ is always larger than $M_W^2$ .

$\mathcal{M}$ is the matrix element; $M_W$ is a mass. The book uses both in Eq. (7.53).

Universality, and how you test it

Lepton universality is the claim that $g$ is one number — the same at every vertex, for every fermion. That is a strong statement and it is testable, because the only thing that should distinguish two leptonic decays is the phase space.

Figs. 7.10 and 7.11 — the τ's two leptonic decays, with the couplings named separately

Click a vertex or an internal line.

Bettini Figs. 7.10 and 7.11 combined. Two decays of the same parent, differing in one leg.

Both widths go as $g_\tau^2 g_\ell^2 m_\tau^5 / M_W^4$ , so everything except $g_\ell^2$ and the phase space cancels from the ratio. The same trick run between the muon’s and the tau’s beta decays gives $g_\tau/g_\mu$ , this time needing two lifetimes and two masses because the parents differ.

the two universality tests, reproduced from the measurements

import numpy as np
mtau, mmu, me = 1.77686, 0.1056583755, 0.51099895e-3
tau_mu, tau_tau, BRe = 2.1969811e-6, 290.3e-15, 0.1782      # s, s, --

def f(x):                       # muon-decay phase-space function
    return 1 - 8*x + 8*x**3 - x**4 - 12*x*x*np.log(x)

ps = f((mmu/mtau)**2) / f((me/mtau)**2)
br = 17.36/17.84
print("e-mu universality, from the tau's two leptonic decays (7.59)-(7.60)")
print(f"  BR(tau->mu) / BR(tau->e) = 17.36/17.84 = {br:.5f}       book: 0.974")
print(f"  phase-space ratio f(x_mu)/f(x_e)       = {ps:.5f}")
print(f"  -> (g_mu/g_e)^2 = {br:.5f}/{ps:.5f}      = {br/ps:.5f}")
print(f"     g_mu/g_e = {np.sqrt(br/ps):.5f}                                   book: 1.001 +- 0.002")

meas = tau_tau/(tau_mu*BRe)
pred = (mmu/mtau)**5
print("\nmu-tau universality, from the two beta decays (7.61)-(7.64)")
print(f"  measured  Gamma(mu->e)/Gamma(tau->e) = tau_tau/(tau_mu BR) = {meas:.4e}")
print(f"  predicted at g_tau = g_mu:  (m_mu/m_tau)^5              = {pred:.4e}")
print(f"  -> g_tau/g_mu = {np.sqrt(pred/meas):.5f}                                 book: 1.001 +- 0.003")
print("\nboth couplings equal to one part in a thousand, over three families")
print(f"spanning a factor {mtau/me:.0f} in mass.  The W does not care what it couples to.")

prints

e-mu universality, from the tau's two leptonic decays (7.59)-(7.60)
BR(tau->mu) / BR(tau->e) = 17.36/17.84 = 0.97309       book: 0.974
phase-space ratio f(x_mu)/f(x_e)       = 0.97256
-> (g_mu/g_e)^2 = 0.97309/0.97256      = 1.00055
   g_mu/g_e = 1.00027                                   book: 1.001 +- 0.002

mu-tau universality, from the two beta decays (7.61)-(7.64)
measured  Gamma(mu->e)/Gamma(tau->e) = tau_tau/(tau_mu BR) = 7.4150e-07
predicted at g_tau = g_mu:  (m_mu/m_tau)^5              = 7.4345e-07
-> g_tau/g_mu = 1.00131                                 book: 1.001 +- 0.003

both couplings equal to one part in a thousand, over three families
spanning a factor 3477 in mass.  The W does not care what it couples to.

Two independent tests, two different observables, and the same answer to three decimal places.

test↕	from↕	needs↕	result↕
$g_\mu/g_e$	the τ's two leptonic decays	one branching-ratio ratio, plus a calculable phase-space factor	1.001 ± 0.002
$g_\tau/g_\mu$	the μ and τ beta decays	two lifetimes, two masses, one branching ratio	1.001 ± 0.003

Both to 0.2 %, across masses spanning a factor of 3500. Whatever the W couples to, it is not flavour.

That is worth pausing on, because it is the only place in this chapter where the weak interaction is simple. It violates P maximally, violates C maximally, distinguishes left from right absolutely — and then treats an electron, a muon and a tau as completely interchangeable. The complexity is all in the Lorentz structure; the flavour structure is trivial.

For leptons.

🔢 Worked example — g_μ/g_e from two branching ratios

The τ’s two leptonic decays share everything but one vertex, so their ratio is a direct measurement of $g_\mu/g_e$ — provided you keep the phase space.

\frac{\mathrm{BR}(\tau\to\mu\nu\nu)}{\mathrm{BR}(\tau\to e\nu\nu)} = \frac{17.36\,\%}{17.84\,\%} = 0.9731

The phase-space factor is $f(x_\mu)/f(x_e) = 0.97256$ , with $f(x) = 1-8x+8x^3-x^4-12x^2\ln x$ and $x = (m_\ell/m_\tau)^2$ . Dividing it out,

\left(\frac{g_\mu}{g_e}\right)^{\!2} = \frac{0.9731}{0.97256} = 1.00055, \qquad \frac{g_\mu}{g_e} = \mathbf{1.0003}

against the book’s quoted $1.001\pm0.002$ . Do not drop the phase-space factor: it is a 2.7 % effect and the answer it is being compared against has a 0.2 % error bar, so omitting it would shift the result by more than ten standard deviations and make universality look broken.

🔑 If you remember only three things

A contact interaction is a propagator you cannot resolve. At low momentum transfer the exchange looks like a constant, and a constant hides whatever structure produced it.
Reasoning does not transfer freely between symmetries. The argument that settles parity gives the wrong answer for C, because the operator is acting on different things.
Universality is a claim about vertices, not about particles. Testing it means comparing two decays of one particle that differ in exactly one vertex.

Where this goes next

§7.9 asks the same question of quarks and gets a different answer. The coupling that serves $u \to d$ is measurably weaker than the one serving $\mu \to \nu_\mu$ — the hint already dropped in §7.1, where the neutron’s effective coupling came out smaller than the muon’s. Cabibbo’s repair is to keep universality and give up the idea that the quark entering the current is a quark of definite mass, and that single move generates quark mixing, predicts charm via GIM (§7.10), and hands Chapter 8 the one complex phase that is the Standard Model’s only source of CP violation.

Chapter 9 then supplies what this section quietly assumed: why there is a W at all, why it has the mass it has, and where Eq. (7.55)‘s $\sqrt2/8$ comes from.

✅ Check yourself — C violation, chirality, and the W

0/6 answered · 0 correct

1.σ·p is even under C, since C touches neither momentum nor spin. Why does that not prove C is conserved?
2.The book also gives a CPT argument for C violation. How does it run?
3.A tritium beta-decay electron has β ≈ 0.2. Is it left-handed?
4.Why was Fermi's point-like theory not merely a good approximation but effectively exact for everything measured before the 1970s?
5.Question 7.2: what neutrino beam energy would be needed to make a real W off a proton target, and what does the answer explain?
6.Lepton universality is tested to 0.2 %. What makes the τ's two leptonic decays such a clean test?