§7.2–7.3Parity Violation and the Charged Current Lagrangian

§4.5 left a puzzle on the table. Two particles, the $\theta$ and the $\tau$, agreed on every property anyone could measure — same mass, same lifetime, same charge — and differed only in what they decayed into. The $\theta$ went to two pions, the $\tau$ to three, and the Dalitz analysis of the three-pion state gave $J^P = 0^-$ where the two-pion state has to be $0^+$. Two particles or one? That is the θ–τ puzzle θ–τ puzzle the same particle decaying to 2π (needing J^P = 0⁺) and to 3π (Dalitz analysis giving 0⁻). Resolved by Lee and Yang's proposal that the weak interaction violates parity. defined in §4.3-4.4 — open in glossary , and if the answer is one, then a single particle decays to states of opposite parity and parity is not conserved.

That last sentence was the problem. Parity is not a property of a particle; it is a symmetry of space itself, in the same family as rotations. Giving it up to tidy a branching ratio felt, in the book’s phrase, “almost like blasphemy”. The experimentalists were more relaxed about it than the theorists, which is usually how these things go.

What Lee and Yang noticed in 1956 was subtler and much harder to argue with: not that parity was violated parity violation the weak interaction distinguishes left from right. Not a small effect: it is maximal, in the sense that only one chirality participates at all. defined in §7.2-7.3 — open in glossary , but that nobody had ever tested it in a weak decay. The symmetry was assumed, not measured. And once you ask how you would measure it, the shape of the required experiment falls out immediately.

Lee and Yang’s argument is a type check

Take beta decay of a nucleus at rest, $N \to N' + e^- + \bar\nu_e$. Write down everything you could possibly measure. You have three momenta — $\mathbf{p}_{N'}$, $\mathbf{p}_e$, $\mathbf{p}_\nu$ — and nothing else.

From three vectors you can build scalar products like $\mathbf{p}_{N'} \cdot \mathbf{p}_e$. Those are even under P, so no rate built from them can ever reveal a violation. The one pseudoscalar available is the triple product $\mathbf{p}_{N'} \cdot (\mathbf{p}_e \times \mathbf{p}_\nu)$ — and it is identically zero, because momentum conservation forces the three vectors into a plane.

That is not a small effect or a hard measurement. It is zero by geometry:

import numpy as np
rng = np.random.default_rng(7)

worst = 0.0
for _ in range(20000):
    p_e, p_nu = rng.normal(size=3), rng.normal(size=3)
    p_N = -(p_e + p_nu)              # decay at rest: total momentum is zero
    worst = max(worst, abs(np.dot(p_N, np.cross(p_e, p_nu))))

print(f"20000 random decays, max |p_N . (p_e x p_nu)| = {worst:.1e}")
print("that is machine epsilon, not physics -- the three vectors are coplanar")

20000 random decays, max |p_N . (p_e x p_nu)| = 5.3e-15
that is machine epsilon, not physics -- the three vectors are coplanar

So a beta-decay experiment that measures only momenta cannot test parity, no matter how precise it is. You need a genuine axial vector in the problem, and the only one on offer is the spin of the decaying nucleus. You have to polarize the sample and measure something proportional to $\mathbf{J} \cdot \mathbf{p}_e$.

The Wu experiment

Wu and her collaborators at the National Bureau of Standards had the answer within months, using $^{60}\mathrm{Co}$. The decay chain is chosen so that the polarization survives long enough to be useful:

followed by two gamma decays, $4^+ \to 2^+ \to 0^+$, emitting 1.173 MeV and 1.332 MeV. The daughter keeps the parent’s polarization, which matters for the reason given below: those gammas are the instrument that measures how polarized the sample still is.

Why millikelvin

The polarization is a Boltzmann problem, and it is worth doing because it explains why this experiment had to wait for cryogenics rather than for an idea. A moment $\mu$ in a field $B$ has energy $-\boldsymbol{\mu}\cdot\mathbf{B}$, so the sublevel populations go as $\exp(\mu B / kT)$ and the competition is between $\mu B$ and $kT$:

import numpy as np

k    = 8.617333e-5                  # eV/K
mu_N = 3.15245e-8                   # eV/T, the nuclear magneton
mu   = 3.799 * mu_N                 # Co-60 moment
J    = 5.0
m    = np.arange(-J, J + 1)         # the 11 magnetic sublevels

def polarization(B, T):
    E = -mu * B * m / J             # evenly spaced Zeeman ladder
    w = np.exp(-E / (k * T)); w /= w.sum()
    return (w * m).sum() / J

B = 50.0                            # tesla, generated inside the crystal
print(f"mu B / J = {mu*B/J:.2e} eV  <-- the energy that has to beat kT")
for T in (300, 1.0, 0.010, 0.003):
    print(f"  T = {T*1e3:8.1f} mK   kT = {k*T:.2e} eV   P = {polarization(B, T):.4f}")

mu B / J = 1.20e-06 eV  <-- the energy that has to beat kT
T = 300000.0 mK   kT = 2.59e-02 eV   P = 0.0001
T =   1000.0 mK   kT = 8.62e-05 eV   P = 0.0278
T =     10.0 mK   kT = 8.62e-07 eV   P = 0.9336
T =      3.0 mK   kT = 2.59e-07 eV   P = 0.9980

At room temperature the polarization is $10^{-4}$ — the asymmetry would be buried four orders of magnitude below the statistical noise of any 1957 counting experiment. At 10 mK it is 0.93. The entire difficulty of the Wu experiment is in that table.

Reading the asymmetry

If parity is violated, the electron angular distribution is not symmetric under $\theta \leftrightarrow \pi - \theta$, and the counting rate acquires a term linear in $\cos\theta$:

Fig. 7.3 is the whole result, and it is really two measurements of the same decay curve. Panel (a) is the gamma anisotropy — the thermometer. Panel (b) is the beta asymmetry — the signal. Both relax with the same time constant because both are proportional to $P(t)$:

Now extract $\alpha$ from those intercepts. This is where the $\beta_e$ in Eq. (7.21) stops being a footnote:

import numpy as np

up, dn = 1.22, 0.76                       # the two beta intercepts
A = (up - dn) / (up + dn)                 # the measured asymmetry
P = 0.6                                   # Wu's nuclear polarization
print(f"asymmetry A = {A:.3f}   ->   |alpha| * beta_e = A/P = {A/P:.3f}")

m_e = 0.511                               # MeV
print("\nthe Co-60 beta spectrum runs to an endpoint of 318 keV:")
for K in (0.05, 0.10, 0.15, 0.20):        # electron kinetic energy, MeV
    E = m_e + K
    beta = np.sqrt(E*E - m_e*m_e) / E
    print(f"  K = {K*1e3:5.0f} keV   beta = {beta:.3f}   |alpha| = {A/P/beta:.2f}")

asymmetry A = 0.232   ->   |alpha| * beta_e = A/P = 0.387

the Co-60 beta spectrum runs to an endpoint of 318 keV:
K =    50 keV   beta = 0.413   |alpha| = 0.94
K =   100 keV   beta = 0.548   |alpha| = 0.71
K =   150 keV   beta = 0.634   |alpha| = 0.61
K =   200 keV   beta = 0.695   |alpha| = 0.56

The extracted $|\alpha|$ ranges from 0.94 down to 0.56 depending on which electron energy you use, and that is not sloppiness — the detector accepts a whole spectrum, and $\beta_e$ varies across it by a factor of nearly two. Compare what the book quotes two paragraphs later:

The lower bound of 0.7 is sitting right in the middle of the range above. The book presents (7.23) as “taking the uncertainties of the measurement into account” without saying which uncertainty dominates; the arithmetic says it is the electron energy — you cannot pin $\alpha$ tighter than you can pin $\langle\beta_e\rangle$, and the correction from the diluted polarization (the aside above) pushes in the same direction, towards $|\alpha| = 1$.

Fig. 7.4 is the picture that makes the result feel inevitable. The cobalt spin is $5^+$ and the nickel is $4^+$, so the electron and the antineutrino must carry away one unit of angular momentum between them, with both spins along $\mathbf{J}$:

Read the last step again, because it is the one that matters for the rest of the chapter. The experiment measured an electron asymmetry, but what it determined was the helicity of the antineutrino — a particle it never detected. Momentum conservation did the work.

§7.3 The charged current Lagrangian

Within the same year, Sudarshan and Marshak — and independently Feynman and Gell-Mann — pinned down what structure the current must have. Fermi had written a vector current, $\bar\psi\gamma^\mu\psi$, by analogy with electromagnetism. Wu’s result says it must be a specific mixture of vector and axial — the structure called V − A V−A the structure γ^μ(1 − γ⁵) of the charged weak current: an equal mixture of vector and axial-vector covariants, which is the same as saying only left-chirality fields participate. defined in §7.2-7.3 — open in glossary :

The reason V − A is special, rather than one mixture among many, is a single line of gamma-matrix algebra. Because $\gamma_5$ anticommutes with every $\gamma_\mu$,

and a $(1-\gamma_5)$ has appeared on both sides of the $\gamma_\mu$. Writing the left projections of the initial and final fields as in (7.26), each current in (7.24) collapses to

Every one of those steps is checkable with four 4×4 matrices, so check them:

import numpy as np

I2, Z2 = np.eye(2), np.zeros((2, 2))
sx = np.array([[0, 1], [1, 0]], complex)
sy = np.array([[0, -1j], [1j, 0]])
sz = np.array([[1, 0], [0, -1]], complex)
blk = lambda a, b, c, d: np.block([[a, b], [c, d]])

g0 = blk(I2, Z2, Z2, -I2)
g  = [g0] + [blk(Z2, s, -s, Z2) for s in (sx, sy, sz)]
g5 = 1j * g[0] @ g[1] @ g[2] @ g[3]
I4 = np.eye(4)

print("gamma5^2 = 1                :", np.allclose(g5 @ g5, I4))
print("{gamma_mu, gamma5} = 0      :", all(np.allclose(g[m] @ g5 + g5 @ g[m], 0) for m in range(4)))
print("Eq. (7.25), all four mu     :",
      all(np.allclose(g[m] @ (I4 - g5), 0.5 * (I4 + g5) @ g[m] @ (I4 - g5)) for m in range(4)))

P_L, P_R = (I4 - g5) / 2, (I4 + g5) / 2
print("P_L is a projector          :", np.allclose(P_L @ P_L, P_L))
print("P_L P_R = 0, P_L + P_R = 1  :", np.allclose(P_L @ P_R, 0) and np.allclose(P_L + P_R, I4))
print("(1 - gamma5) = 2 P_L        :", np.allclose(I4 - g5, 2 * P_L))

rng, ok = np.random.default_rng(3), True
for _ in range(200):
    pf = rng.normal(size=4) + 1j * rng.normal(size=4)
    pi = rng.normal(size=4) + 1j * rng.normal(size=4)
    bf = pf.conj() @ g0
    for m in range(4):
        ok &= np.allclose(bf @ g[m] @ (I4 - g5) @ pi, 2 * (bf @ P_R) @ g[m] @ (P_L @ pi))
print("Eq. (7.27), 200 random pairs:", ok)

gamma5^2 = 1                : True
{gamma_mu, gamma5} = 0      : True
Eq. (7.25), all four mu     : True
P_L is a projector          : True
P_L P_R = 0, P_L + P_R = 1  : True
(1 - gamma5) = 2 P_L        : True
Eq. (7.27), 200 random pairs: True

Eq. (7.27) is the punchline of the whole section. In that form the weak charged current looks exactly like the electromagnetic one — a fermion field, a $\gamma_\mu$, another fermion field — with one difference: the fields are the left-chirality projections 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 and nothing else.

That single asymmetry is the source of everything the chapter has left to explain. P is violated because P exchanges left and right chirality, and only one of them is in the Lagrangian. C is violated for the same reason, and maximally (§7.7). The $\pi \to e\nu$ suppression of $10^{-4}$ (§7.4) is angular momentum colliding with that restriction. And CP is nearly conserved, because CP maps a left-handed particle to a right-handed antiparticle — which is in the theory — leaving room for only the small violation Chapter 8 is about.

Where this goes next

The V − A structure is now fixed by experiment, and the next four sections do nothing but spend it. §7.4 works out charged pion decay and finds a factor $10^{-4}$ that phase space says should have been a factor 2.3 the other way. §7.5 and §7.6 turn the argument around and measure a neutrino’s helicity directly. §7.7 shows that C is broken in the same maximal way, and that the combination CP nearly survives.

One backward link worth making now: §5.9b’s muon $g-2$ experiment depends on parity violation twice — the pions deliver 100 % polarized muons because of it, and the decay electrons are emitted asymmetrically because of it, which is the only reason the wiggle plot wiggles. That page promised this one. This is the promise.