§7.4–7.5Charged Pion Decay and the Helicity of Leptons

§7.3 fixed the structure of the charged current by fiat of experiment: V − A, left chirality only. This section spends it. The next two results are both numbers that come out wrong by orders of magnitude if you reason from phase space alone, and both come out right the moment chirality is taken seriously.

Start with the one the book calls an example of a V − A phenomenon. Charged pions decay overwhelmingly to a muon,

and the electron channel is smaller by

That is the wrong way round. The muon mass is 106 MeV against the pion’s 140, so the muon channel is squeezed hard against threshold while the electron channel has almost the whole pion mass to play with. Phase space should favour the electron:

mpi, mmu, me = 139.57039, 105.6583755, 0.51099895   # MeV

def pstar(M, m):                 # two-body CM momentum, Eq. (1.74)
    return (M*M - m*m) / (2*M)

pe, pmu = pstar(mpi, me), pstar(mpi, mmu)
print(f"p*_e  = {pe:.2f} MeV")
print(f"p*_mu = {pmu:.2f} MeV")
print(f"(7.31) p*_e / p*_mu = {pe/pmu:.3f}          book: 2.3")
print()
print("so two-body phase space favours the ELECTRON by a factor 2.3,")
print("and the electron channel is nevertheless 10^4 times rarer.")
print(f"the matrix element must therefore be suppressed by ~10^4 x 2.3 = {1e4*pe/pmu:.1e}")

p*_e  = 69.78 MeV
p*_mu = 29.79 MeV
(7.31) p*_e / p*_mu = 2.342          book: 2.3

so two-body phase space favours the ELECTRON by a factor 2.3,
and the electron channel is nevertheless 10^4 times rarer.
the matrix element must therefore be suppressed by ~10^4 x 2.3 = 2.3e+04

So the suppression is not kinematic. This is helicity suppression helicity suppression the factor m_ℓ² by which a decay to a light lepton is suppressed when angular momentum forces it into the wrong helicity state. It is why π → μν beats π → eν by 10⁴ despite worse phase space. defined in §7.4-7.5 — open in glossary , it lives in the matrix element, and finding it is a piece of detective work that uses only Lorentz invariance.

What the matrix element is allowed to be

The book does not compute the decay. It asks a narrower question — what could the matrix element possibly be built from? — and the answer turns out to be enough.

The ingredients are fixed. For the leptons, any of the five bilinear covariants of §2.9 sandwiched between $\psi_\ell$ and $\psi_{\nu}$. For the hadron, only two things: the pion field $\phi_\pi$, which is a pseudoscalar, and its four-momentum $p^\mu$. Plus a constant, the pion decay constant pion decay constant f_π, the factor absorbing the strong-interaction physics of the pion's internal structure into its leptonic decay rate. Not calculable perturbatively; f_K/f_π ≈ 1.19 comes from the lattice (§6.10). defined in §7.4-7.5 — open in glossary $f_\pi$, which hides all the strong-interaction physics of what the pion is made of.

$\mathcal{M}$ must be a Lorentz scalar overall. Since $\phi_\pi$ is a pseudoscalar, pairing it with a pseudoscalar or axial lepton bilinear gives a scalar, and pairing it with a scalar or vector bilinear gives a pseudoscalar. Both are allowed if parity is not conserved, so all four survive — and the tensor is excluded because there is nothing to contract its second index with.

Now the step that does the work. Take the vector term, $\mathcal{M}_V \propto f_\pi \,\bar\psi_\ell \,p^\mu\gamma_\mu\, \psi_\nu$, and use $p^\mu = p_\ell^\mu + p_\nu^\mu$. Each piece then hits a spinor that obeys the Dirac equation, Eq. (7.35):

so the whole thing collapses to

The amplitude is proportional to the mass of the charged lepton. Repeat with the axial term and (7.39) gives the same factor. The scalar and pseudoscalar terms do not produce it — so if either of those were present, they would dominate the electron channel completely and the ratio would be near 1. The observed $10^{-4}$ is therefore direct evidence that the current is V, or A, or a mixture. That is a remarkable amount of structure extracted without computing a single integral.

The angular-momentum picture

The $m_\ell$ in (7.36) is an algebraic fact, but there is a geometric argument behind it, and it is the one to keep.

The pion has spin 0 and decays at rest, so the two daughters come out back-to-back with their spin projections on the decay axis cancelling. Opposite momenta and opposite spin projections means the two particles have the same helicity — which is the step that trips people, because the spins point the same way in space while the momenta do not.

The antineutrino is (as far as anyone can tell) massless and V − A gives it helicity $+1$, full stop. So the charged lepton is forced to helicity $+1$ too. But the current couples to the left-chirality electron, and by Eq. (2.66) a left-chirality fermion is found with $h = +1$ only in its small component, of amplitude $m_\ell/(E+p)$. The decay has to run through that component or not happen at all:

Press the toggle. In a V + A world the antineutrino would be left-handed, the charged lepton would be forced into its favoured component, and $\pi\to e\nu$ would be the dominant channel. Nothing about the masses or the phase space changes — only the chirality of the current. That is what makes (7.30) a measurement of the Lagrangian rather than of kinematics.

Putting the two factors together

With V − A assumed, the full result — which the book quotes rather than derives, and so does this page — is:

mpi, mK = 139.57039, 493.677                        # MeV
mmu, me = 105.6583755, 0.51099895

def pstar(M, m): return (M*M - m*m) / (2*M)

def ratio(M):
    hel   = (me/mmu)**2
    phase = (pstar(M, me)/pstar(M, mmu))**2
    return hel, phase, hel*phase

h, f, r = ratio(mpi)
print("(7.41) pion")
print(f"  helicity     m_e^2/m_mu^2         = {h:.3e}")
print(f"  kinematics   (p*_e/p*_mu)^2       = {f:.3f}")
print(f"  product                           = {r:.3e}")
print(f"  book: 0.22e-4 x 2.3^2 = 1.2e-4 ;  measured {1.230e-4:.3e}")
print(f"  agreement with the measurement: {abs(r/1.230e-4-1)*100:.1f}%")

h, f, rK = ratio(mK)
print("\nQuestion 7.1 -- the same formula for the kaon")
print(f"  helicity     m_e^2/m_mu^2         = {h:.3e}")
print(f"  kinematics   (p*_e/p*_mu)^2       = {f:.3f}")
print(f"  predicted                         = {rK:.3e}")
print(f"  measured     1.6e-5 / 0.63        = {1.6e-5/0.63:.3e}")
print(f"  agreement: {abs(rK/(1.6e-5/0.63)-1)*100:.1f}%")

print("\nwhy the kaon is the cleaner test:")
print("  the kaon is heavy enough that the muon is nowhere near threshold,")
print(f"  so its kinematic factor is {f:.2f} instead of the pion's {ratio(mpi)[1]:.2f}.")
print("  the K ratio is therefore almost PURE helicity suppression --")
print(f"  m_e^2/m_mu^2 alone already gives {h:.2e}, within {abs(h/(1.6e-5/0.63)-1)*100:.0f}% of the data.")

(7.41) pion
helicity     m_e^2/m_mu^2         = 2.339e-05
kinematics   (p*_e/p*_mu)^2       = 5.487
product                           = 1.283e-04
book: 0.22e-4 x 2.3^2 = 1.2e-4 ;  measured 1.230e-04
agreement with the measurement: 4.3%

Question 7.1 -- the same formula for the kaon
helicity     m_e^2/m_mu^2         = 2.339e-05
kinematics   (p*_e/p*_mu)^2       = 1.098
predicted                         = 2.569e-05
measured     1.6e-5 / 0.63        = 2.540e-05
agreement: 1.2%

why the kaon is the cleaner test:
the kaon is heavy enough that the muon is nowhere near threshold,
so its kinematic factor is 1.10 instead of the pion's 5.49.
the K ratio is therefore almost PURE helicity suppression --
m_e^2/m_mu^2 alone already gives 2.34e-05, within 8% of the data.

That is Question 7.1 answered, and it is a better test than the pion. In the pion, two large effects — a factor 5.5 one way and 43 000 the other — happen to land on the measured value, and a reader could reasonably suspect a fluke. In the kaon the kinematic factor is switched off almost entirely, and the raw mass ratio $m_e^2/m_\mu^2$ alone lands within 8 % of the data.

§7.5 Measuring a neutrino’s helicity

V − A predicts the antineutrino is purely right-handed. Wu’s experiment inferred that from momentum conservation, without ever detecting one. Goldhaber, Grodzins and Sunyar measured it directly, in 1958 at Brookhaven, in an experiment the book fairly calls a matter of luck — the number of conditions that had to hold simultaneously is absurd.

The strategy is indirect in a way worth stating up front: you cannot measure a neutrino’s polarization, so transfer it to a photon and measure the photon’s. Everything else is the machinery for making that transfer clean.

Three conditions, all of which had to be lucky

The experiment is built on resonant absorption: a nucleus emits a gamma and an identical nucleus reabsorbs it. That process is much more delicate than it sounds, because the emitting nucleus recoils, so the photon leaves with less than the transition energy — and the absorber needs more than the transition energy, since it recoils too. The photon is short by twice the recoil energy:

import numpy as np

M_Sm = 1.52e11        # eV, the book's mass for Sm-152 (152 u x ~1 GeV/u)
E_gam = 963e3         # eV, the Sm* -> Sm transition: the PHOTON's energy
E_Eu  = 911e3         # eV, released in the Eu K-capture: sets the RECOIL
kT    = 26e-3         # eV, room temperature
hbar  = 6.582119569e-16   # eV s

EK = E_gam**2 / (2*M_Sm)
deficit = 2*EK
width_nat = hbar/10e-15
width_dop = 2*np.sqrt(2*np.log(2)*kT/M_Sm) * E_gam

print("condition 1 -- Sm* AT REST cannot drive the resonance")
print(f"  recoil energy E_K,Sm = E_gamma^2 / 2 M_Sm = {EK:.2f} eV      book: 3 eV")
print(f"  photon is short of resonance by 2 E_K     = {deficit:.2f} eV      book: 6 eV")
print(f"  natural width from the quoted 10 fs lifetime = {width_nat*1e3:.1f} meV  book says 'about 20 meV'")
print(f"  thermal Doppler width at 300 K               = {width_dop:.2f} eV   book: 1 eV")
print(f"  -> the deficit is {deficit/width_dop:.1f} resonance widths.  No resonance.   book: 'six times'")

EKs  = E_Eu**2 / (2*M_Sm)
beta = np.sqrt(2*EKs/M_Sm)
boost = beta * E_gam
print("\ncondition 2 -- Sm* IN FLIGHT can, but only just")
print( "  Eu K-capture releases 911 keV, so the Sm* recoils with")
print(f"  E_K,Sm* = {EKs:.2f} eV                                        book: 2.7 eV")
print(f"  beta_Sm* = {beta:.2e}                                      book: 5.8e-6")
print(f"  forward Doppler BOOST = beta x E_gamma = {boost:.2f} eV")
print(f"  needed to close the deficit                = {deficit:.2f} eV")
print(f"  -> short by {deficit-boost:.1f} eV, i.e. a third of a width.  It works.")

flight = beta * 2.998e8 * 10e-15
print("\ncondition 3 -- the Sm* must not scatter before it decays")
print(f"  Sm* lifetime 10 fs; at beta = {beta:.1e} it travels {flight*1e9:.3f} nm")
print( "  -- less than a tenth of an atomic spacing, so it decays")
print( "  essentially where it was made, still moving forward.")

condition 1 -- Sm* AT REST cannot drive the resonance
recoil energy E_K,Sm = E_gamma^2 / 2 M_Sm = 3.05 eV      book: 3 eV
photon is short of resonance by 2 E_K     = 6.10 eV      book: 6 eV
natural width from the quoted 10 fs lifetime = 65.8 meV  book says 'about 20 meV'
thermal Doppler width at 300 K               = 0.94 eV   book: 1 eV
-> the deficit is 6.5 resonance widths.  No resonance.   book: 'six times'

condition 2 -- Sm* IN FLIGHT can, but only just
Eu K-capture releases 911 keV, so the Sm* recoils with
E_K,Sm* = 2.73 eV                                        book: 2.7 eV
beta_Sm* = 5.99e-06                                      book: 5.8e-6
forward Doppler BOOST = beta x E_gamma = 5.77 eV
needed to close the deficit                = 6.10 eV
-> short by 0.3 eV, i.e. a third of a width.  It works.

condition 3 -- the Sm* must not scatter before it decays
Sm* lifetime 10 fs; at beta = 6.0e-06 it travels 0.018 nm
-- less than a tenth of an atomic spacing, so it decays
essentially where it was made, still moving forward.

Notice which width matters. The natural width of the samarium level is tens of meV, and it is irrelevant: the thermal Doppler broadening at room temperature is about 1 eV, fifteen to fifty times larger, and that is the width the 6 eV deficit has to be compared against. (The book quotes the natural width as “about 20 meV” while also quoting a 10 fs lifetime, which gives $\hbar/\tau = 66$ meV. Since both are swamped, nothing downstream depends on which is right.)

Transferring the helicity

The physics content of the experiment is a chain of angular-momentum bookkeeping in which the neutrino’s helicity is copied onto a photon. It needs a nuclide that undergoes K-capture K-capture the absorption of an atomic S-wave electron by its own nucleus. Goldhaber's experiment needed one, because it produces a two-body final state whose neutrino direction is fixed by the recoil. defined in §7.4-7.5 — open in glossary — absorption of an atomic S-wave electron by its own nucleus — from spin 0 to spin 1, and whose daughter’s gamma happens to sit at the right energy. $^{152}\mathrm{Eu} \to {}^{152}\mathrm{Sm}$ does all of it, in the K-capture of Eq. (7.45):

Take the neutrino direction as $z$; the Sm* recoils along $-z$. The captured electron is in an S wave, so it brings no orbital angular momentum, and the initial state has $J_z = s_z(e)$ alone. That single constraint fixes everything:

The deletion is the elegant part. A massless spin-1 particle has only two polarization states, $\pm 1$ along its momentum — the longitudinal state does not exist — so the two rows with $s_z(\gamma) = 0$ are not physical. Of the two rows left, both have $h_\gamma = h_\nu$.

So: measure the circular polarization of the forward gamma, and you have measured the helicity of a neutrino you never touched. The two are locked together by angular momentum, and the experiment does not need to know which of the two rows any particular event took.

The prediction with nothing left to adjust

§7.5 closes by returning to Eq. (2.66): a left-chirality fermion has expectation value of helicity $\langle h\rangle = -\beta$. Not approximately, and with no free parameter — the entire content of V − A for a beta-decay electron is that one line.

Measuring it is awkward, because longitudinal polarization is hard to analyse directly. The trick is to convert it: bend the electrons through a quarter circle in a magnetic field, which rotates the momentum by 90° while leaving the spin alone, turning longitudinal polarization into transverse. Transverse polarization is measurable, via the spin dependence of scattering off a thin high-$Z$ foil.

Three nuclides were chosen to span the velocity range — tritium slow, cobalt intermediate, phosphorus fast:

A prediction curve with no free parameter, and data sitting on it across the whole range, is a rarer thing than it looks. Most agreement between theory and experiment in this book involves at least one number that was measured elsewhere and inserted — $f_\pi$, $\Lambda_{QCD}$, a PDF. Here there is nothing to insert. $\langle h\rangle = -\beta$ follows from the Dirac equation and the assumption that the current is left-chirality, and the plot either agrees or it does not.

Where this goes next

The helicity structure is now established from three directions: the electron asymmetry in a polarized nucleus (§7.2), the branching ratio of a decay that angular momentum nearly forbids (§7.4), and the neutrino’s own polarization (§7.5). §7.6 shows that C is violated too, and in the same maximal way — with the subtlety that $\boldsymbol\sigma\cdot\mathbf{p}$ is even under C, so the naive argument gives the wrong answer and you have to go through CPT to get it right. §7.7 then separates chirality left-chirality projection ψ_L = ½(1 − γ⁵)ψ. The CC weak interaction couples to nothing else, which is why it violates P and C maximally and CP only slightly. defined in §7.2-7.3 — open in glossary from helicity properly, which is what makes all of the above consistent.

After that the chapter changes subject: §7.8 onwards is about universality — whether the same coupling really does serve every fermion — and the discovery that it does for leptons and does not, at first sight, for quarks.