Parity violation

Hold any law of physics up to a mirror. Newton's laws survive the reflection unchanged; so do Maxwell's equations and the whole of the strong force. For three decades everyone assumed the same of the weak interaction — that nature cannot tell its own reflection from the original, that parity is conserved. In 1956 Lee and Yang noticed that this had never actually been tested for weak decays, and within months Chien-Shiung Wu proved it false: cobalt-60 nuclei, spun up and cooled to near absolute zero, emit their beta electrons preferentially against their spin, a pattern the mirror image reverses. The weak force is left-handed, and its mirror image is a different physics. The reflection asymmetry is written into the structure of the weak vertex. Where QED and QCD couple through a pure vector current $\bar u\gamma^\mu u$ — whose scalar products stay invariant under spatial inversion — the weak charged current couples through $\bar u\,\gamma^\mu(1-\gamma^5)\,u$ , a fixed one-to-one mixture of a vector ( $\gamma^\mu$ ) and an axial-vector ( $\gamma^\mu\gamma^5$ ). Those two pieces transform oppositely under parity, so their sum has no definite reflection behaviour, and the cross term in any rate is a genuine pseudoscalar that flips sign in the mirror. This is the "V−A" law, and the coefficient is not some small tunable $\epsilon$ — experiment fixes it to exactly $-1$ , so the violation is maximal: the weak force couples to left-chiral fermions and ignores right-chiral ones completely. This article builds parity from the Dirac algebra, classifies the bilinears by how they reflect, and lands on the experiments — Wu's cobalt and Goldhaber's neutrino — that made a mirror-asymmetric universe a fact.

Key takeaways

Parity $\hat P$ is spatial inversion $\mathbf{x}\to-\mathbf{x}$ ; on Dirac spinors it acts through $\gamma^0$ , is its own inverse ( $\hat P^2=1$ ), and has eigenvalues $P=\pm1$ .
The Dirac bilinears split by reflection: the vector $\bar\psi\gamma^\mu\psi$ transforms as a polar vector, the axial $\bar\psi\gamma^\mu\gamma^5\psi$ as an axial one — opposite signs on their space parts (and scalar $P=+$ , pseudoscalar $P=-$ ).
QED and QCD conserve parity because their pure vector currents give scalar products that survive inversion; the weak vertex $\tfrac{-ig_w}{2\sqrt2}\gamma^\mu(1-\gamma^5)$ mixes vector with axial-vector and cannot.
Because $1-\gamma^5=2P_L$ , the weak current is purely left-chiral, $\bar u_L\gamma^\mu u_L$ : right-handed fermions are untouched, so the violation is maximal (coefficient $-1$ , not a small $\epsilon$ ).
Wu's 1957 cobalt-60 experiment saw beta electrons emitted preferentially opposite the nuclear spin — a nonzero pseudoscalar $\langle\mathbf{S}\cdot\mathbf{p}\rangle$ , hemisphere asymmetry $\beta/2\approx0.30$ at $\beta=0.6$ .
V−A forces $\nu$ left-handed ( $h=-1$ ) and $\bar\nu$ right-handed (Goldhaber 1958); a right-handed neutrino has never been seen.
Key idea: Mirror symmetry is broken by the weak force because its vertex is a fixed mixture of a vector and an axial-vector — a combination whose scalar products do not all stay invariant under spatial inversion, unlike the pure vector currents of QED and QCD.

Where this lives today

Parity violation graduated from a shocking discovery to a precision tool. Because a longitudinally polarized electron beam is, at collider energies, essentially a beam of definite chirality, the left–right asymmetry $A_{LR}$ became one of the sharpest probes of electroweak theory: the SLD experiment at SLAC collided polarized electrons on positrons at the $Z$ pole and read off $\sin^2\theta_W^{\text{eff}}=0.23098\pm0.00026$ from the difference in $Z$ production between left- and right-handed beams. The same physics runs through low-energy precision experiments: parity-violating electron scattering (Qweak at Jefferson Lab, and the ongoing MOLLER and P2 programs) measures the parts-per-billion parity-odd asymmetry in electron–proton scattering to test the running of $\sin^2\theta_W$ far below the $Z$ mass. It even reaches into atomic physics: the weak neutral current mixes a small opposite-parity admixture into atomic states, producing the atomic parity-violation signal in cesium that constrains the nuclear weak charge $Q_W$ . And every neutrino experiment rests on the fact Wu uncovered — that the weak interaction handles left- and right-handed particles differently — down to the definition of a "sterile" neutrino that feels no weak force at all.

The physical picture

Parity means inverting every spatial coordinate through the origin — looking at the world in a mirror. Reflect a scattering experiment and ask whether the reflected film shows an allowed process. For gravity, electromagnetism, and the strong force the answer is always yes; the weak force is the sole exception, and the cleanest place to see it is a spinning nucleus. Spin is angular momentum, an axial vector: reflect it and it does not reverse the way an ordinary arrow does. Momentum is a polar vector: reflect it and it flips. So the correlation between a nucleus's spin and its beta electron's direction — the quantity $\mathbf{S}\cdot\mathbf{p}$ — changes sign in the mirror. If nature has a preference, the mirror world has the opposite one, and only one of the two is real.

Figure

Parity reflects the Wu experiment through a mirror. The nuclear spin S is an axial vector and survives the reflection, but the electron momentum p is a polar vector and reverses, so the correlation S·p flips sign. Nature emits electrons preferentially opposite to S so that ⟨S·p⟩ is less than zero; the mirror world would emit them along S, and that world is never observed

Wu's data turned that cartoon into a number. With the cobalt spins aligned and the electron speed near $\beta=0.6$ , the emission rate follows $W(\theta)=1-\beta\cos\theta$ from the spin axis: fewer electrons come out along the spin, more against it. Warm the sample so the nuclei randomize and the asymmetry melts away. Integrating over the two hemispheres leaves a forward–backward asymmetry of $\beta/2=0.30$ , which would be identically zero if — and only if — parity were conserved. The second figure is that distribution.

Figure

The beta-decay angular distribution W(θ) = 1 − β cosθ for polarized cobalt-60 with electron speed β = 0.60, plotted against the angle θ from the nuclear spin. More electrons emerge against the spin near θ = 180° than along it near θ = 0°; integrating over each hemisphere gives a forward-backward asymmetry of β/2 = 0.30. A parity-conserving world would give the flat dashed line W = 1

The core equations

Parity is a discrete symmetry represented by an operator whose square returns the identity. Acting on a Dirac spinor it reflects the spatial argument and reshuffles the components through $\gamma^0$ .

\hat P\,\psi(t,\mathbf{x})\,\hat P^{-1}=\gamma^0\,\psi(t,-\mathbf{x}),\qquad \hat P^2=\mathbb{1},\qquad P=\pm1 \qquad (1)

where: $\hat P$ — the unitary parity operator; $\gamma^0$ — the Dirac matrix supplying the spinor's intrinsic-parity flip (why a fermion and its antiparticle carry opposite intrinsic parity); $P=\pm1$ — the eigenvalues forced by $\hat P^2=1$ . Read it as: reflecting a Dirac particle inverts its position and acts with $\gamma^0$ on its spin. Feeding this into the fermion bilinears sorts them by reflection behaviour. The vector and axial-vector differ by one factor of $\gamma^5$ , which makes their space parts transform with opposite signs.

\bar\psi\psi\xrightarrow{\hat P}+\bar\psi\psi,\quad \bar\psi\gamma^5\psi\xrightarrow{\hat P}-\bar\psi\gamma^5\psi,\quad \bar\psi\gamma^\mu\psi\xrightarrow{\hat P}\bar\psi\gamma_\mu\psi,\quad \bar\psi\gamma^\mu\gamma^5\psi\xrightarrow{\hat P}-\bar\psi\gamma_\mu\gamma^5\psi \qquad (2)

where: lowering the index ( $\gamma^\mu\to\gamma_\mu$ ) is shorthand for "time component kept, space components flipped." The vector $V^\mu=\bar\psi\gamma^\mu\psi$ is a polar vector (space part flips, like $\mathbf{x}$ ); the axial $A^\mu=\bar\psi\gamma^\mu\gamma^5\psi$ carries the extra sign and is an axial vector (space part preserved, like spin). The reading: a pure vector current gives a parity-even Lagrangian, a $V$ – $A$ mixture does not. The weak charged current is exactly that forbidden mixture — its vertex welds the vector to the axial-vector with equal weight and a relative minus sign, the "V minus A" structure. Since $1-\gamma^5=2P_L$ , the current is built from left-handed fields alone.

\frac{-ig_w}{2\sqrt2}\,\gamma^\mu\!\left(1-\gamma^5\right),\qquad j^\mu_{\text{weak}}=\bar u\,\gamma^\mu\tfrac12\!\left(1-\gamma^5\right)u=\bar u_L\,\gamma^\mu u_L \qquad (3)

where: $g_w$ — the weak coupling (not small: see Example 2); $\tfrac12(1-\gamma^5)=P_L$ — the left-chiral projector; $\bar u_L\gamma^\mu u_L$ — a current of purely left-handed fermions. The $V$ / $A$ coefficient is fixed to $-1$ by experiment: maximal parity violation. The reading: the $W$ couples to left-chiral fermions and ignores right-chiral ones. What a detector registers is a rate — a modulus-squared containing an interference between the vector and axial pieces. That interference is a pseudoscalar and appears as a spin–momentum correlation in the angular distribution.

|\mathcal M|^2\sim g_V^2\,(V\!\cdot\!V)+g_A^2\,(A\!\cdot\!A)-2g_Vg_A\,(V\!\cdot\!A),\qquad W(\theta)=1+a\,\beta\cos\theta \qquad (4)

where: $V\!\cdot\!V$ and $A\!\cdot\!A$ are parity-even; the cross term $V\!\cdot\!A$ is the parity-odd pseudoscalar; $a$ — the asymmetry parameter ( $a=-1$ for Gamow–Teller cobalt-60), $\beta=v/c$ , $\theta$ from the nuclear spin. A nonzero $a$ means a nonzero $\langle\mathbf{S}\cdot\mathbf{p}\rangle$ . The reading: parity violation lives entirely in the $V$ – $A$ interference.

Derivation AHow the Dirac bilinears transform under a mirror

The goal is to prove equation (2) from the single rule that parity acts on a Dirac field through $\gamma^0$ , and thereby show why the vector and axial-vector currents reflect with opposite signs on their space components — which makes a parity-even Lagrangian impossible from a $V$ – $A$ mixture, since the two halves cannot be reflection eigenstates with the same eigenvalue. We need only the rules $\psi\to\gamma^0\psi$ and $\bar\psi\to\bar\psi\gamma^0$ and the anticommutation facts $\{\gamma^0,\gamma^i\}=0$ and $\{\gamma^0,\gamma^5\}=0$ .

Step 1. Fix the field and its adjoint. Parity sends $\psi(t,\mathbf{x})\to\gamma^0\psi(t,-\mathbf{x})$ , so the adjoint transforms as $\bar\psi=\psi^\dagger\gamma^0\to\psi^\dagger\gamma^0\gamma^0=\bar\psi\gamma^0$ . Why: every bilinear $\bar\psi\,\Gamma\,\psi$ reflects to $\bar\psi\gamma^0\,\Gamma\,\gamma^0\psi$ — the whole problem reduces to pushing $\gamma^0$ through the matrix $\Gamma$ .

Step 2. Do the scalar and pseudoscalar. For $\Gamma=1$ : $\gamma^0\gamma^0=1$ , so $\bar\psi\psi$ is invariant, $P=+1$ . For $\Gamma=\gamma^5$ : since $\gamma^0\gamma^5=-\gamma^5\gamma^0$ , we get $\gamma^0\gamma^5\gamma^0=-\gamma^5$ , so $\bar\psi\gamma^5\psi\to-\bar\psi\gamma^5\psi$ , $P=-1$ . Why: the pseudoscalar carries the hidden sign of $\gamma^5$ — the first place the "extra $\gamma^5$ " flips a sign, and it will do so again for the axial vector.

Step 3. Do the vector. Time component $\Gamma=\gamma^0$ : $\gamma^0\gamma^0\gamma^0=\gamma^0$ , unchanged. Space component $\Gamma=\gamma^i$ : $\{\gamma^0,\gamma^i\}=0$ gives $\gamma^0\gamma^i\gamma^0=-\gamma^i$ , flipped. Together $\bar\psi\gamma^\mu\psi\to\bar\psi\gamma_\mu\psi$ — a polar vector, transforming like $\mathbf{x}$ . Why: this is why QED and QCD conserve parity — the scalar product of two polar vectors is reflection-invariant.

Step 4. Do the axial vector by adding one $\gamma^5$ . The time component $\Gamma=\gamma^0\gamma^5$ gives $\gamma^5\gamma^0=-\gamma^0\gamma^5$ , which flips; the space component $\Gamma=\gamma^i\gamma^5$ commutes with $\gamma^0$ (two anticommutations cancel), giving $\gamma^i\gamma^5$ , unchanged. So $\bar\psi\gamma^\mu\gamma^5\psi\to-\bar\psi\gamma_\mu\gamma^5\psi$ — an axial vector, the opposite sign pattern to the vector. Hence in $V^\mu-A^\mu$ the two pieces reflect oppositely, no common eigenvalue exists, and a $V$ – $A$ interaction cannot conserve parity. Why: this single sign mismatch between $V$ and $A$ is the whole of parity violation — everything measurable follows from it. *In summary: $\gamma^0\gamma^5\gamma^0=-\gamma^5$ and $\gamma^0\gamma^i\gamma^0=-\gamma^i$ ; the vector is polar (time $+$ , space $-$ ), the axial is axial (time $-$ , space $+$ ) — opposite space parts.*

Go deeper · Derivation B · 1 minV minus A means the weak force is purely left-chiral

Derivation BV minus A means the weak force is purely left-chiral

The aim is to show that the vertex factor $\gamma^\mu(1-\gamma^5)$ is precisely twice the left-chiral projection of the current, so the weak interaction acts on left-handed fermions alone and right-handed ones are entirely blind to it — which is what elevates the violation from "present" to "maximal" and why the mirror image of a weak process does not exist in the Standard Model. We use $1-\gamma^5=2P_L$ , the projector identity $P_R\gamma^\mu=\gamma^\mu P_L$ , and the bar rule $\overline{P_L\psi}=\bar\psi P_R$ .

Step 1. Rewrite the vertex as a projector. Since $P_L=\tfrac12(1-\gamma^5)$ , the current factor is $\gamma^\mu\tfrac12(1-\gamma^5)=\gamma^\mu P_L$ , so $j^\mu=\bar u\,\gamma^\mu P_L\,u$ . Why: recasting $1-\gamma^5$ as a projector shows the interaction only ever sees the left-chiral part of the field.

Step 2. Move the projector onto the barred spinor. Using $\gamma^\mu P_L=P_R\gamma^\mu$ and $\bar u P_R=\overline{P_L u}=\bar u_L$ , we get $j^\mu=\bar u P_R\gamma^\mu P_L u=\bar u_L\gamma^\mu u_L$ — both spinors left-chiral. Why: the current is built from $u_L$ and $\bar u_L$ only — the right-chiral field $u_R$ never appears, so it does not couple to the $W$ .

Step 3. Read off the maximal violation. Parity turns a left-handed particle into a right-handed one, so the parity image of the vertex would require a right-chiral coupling $\bar u_R\gamma^\mu u_R$ — which is exactly zero. There is no reflected process at all. Why: a partial coupling of both chiralities would give partial violation; the complete absence of the right-chiral coupling is what "maximal" means.

Step 4. Contrast with the general parametrization. A trial current $\bar u\gamma^\mu(g_V-g_A\gamma^5)u$ conserves parity only in the pure limits $g_A=0$ (vector, QED) or $g_V=0$ (axial); any other ratio violates it, with magnitude $2g_Vg_A/(g_V^2+g_A^2)$ maximized at $g_V=g_A$ . Experiment gives $g_A/g_V=1$ — the peak. Why: this puts the weak interaction at the exact top of the scale, and explains why history briefly entertained a small $\epsilon$ before the data forced $\epsilon=-1$ . *In summary: $\gamma^\mu(1-\gamma^5)=2\gamma^\mu P_L$ ; $\bar u\gamma^\mu P_L u=\bar u_L\gamma^\mu u_L$ ; the right-chiral current $\bar u_R\gamma^\mu u_R$ is absent; and the violation $\propto 2g_Vg_A/(g_V^2+g_A^2)$ peaks at $g_V=g_A$ .*

Go deeper · Derivation C · 1 minWhy the neutrino comes out left-handed

Derivation CWhy the neutrino comes out left-handed

The goal is to trace the V−A structure to its sharpest observable consequence: the weak interaction emits neutrinos in a single helicity state, left-handed ( $h=-1$ ), and antineutrinos right-handed, with the "wrong" helicity forbidden for a massless neutrino — so the mere existence of a left-handed-only neutrino is, by itself, proof that mirror symmetry is broken. We use that the weak current projects $\nu$ with $P_L$ , and that for a massless fermion chirality and helicity coincide exactly.

Step 1. Note where the neutrino enters. In the leptonic current $\bar u_\ell\gamma^\mu P_L u_\nu$ , the neutrino spinor always appears as $P_L u_\nu$ . Why: the neutrino is created only through its left-chiral component, so the interaction can never produce a right-chiral neutrino.

Step 2. Use masslessness to convert chirality into helicity. For $m_\nu\to0$ the chiral projector $P_L$ is the negative-helicity projector, so $P_Lu_\nu$ has helicity $h=-1$ exactly. Why: for a massless particle chirality equals the measurable helicity with no admixture, so "left-chiral" becomes the observable "left-handed."

Step 3. Flip everything for the antineutrino. The spinor $v_\nu$ carries the opposite correspondence between $\gamma^5$ and spin, so the same $P_L$ selects the right-handed antineutrino, $h=+1$ . Why: beta decay therefore pairs a left-handed electron-side current with a right-handed antineutrino — exactly what Wu's cobalt emits.

Step 4. Draw the parity conclusion. Parity would map the left-handed neutrino to a right-handed one; since that state does not exist as a weakly-interacting one, the reflected process has no counterpart in nature. The single-handedness of the neutrino is the cleanest single-particle statement of maximal parity violation — what Goldhaber measured in 1958. Why: no averaging over spins is needed; one particle's fixed handedness already breaks the mirror. *In summary: the weak current contains $P_Lu_\nu$ ; for $m_\nu=0$ , $P_L$ selects $h=-1$ (neutrino) and, via the $v$ -spinor flip, $h=+1$ (antineutrino); the right-handed neutrino is absent from the charged current.*

Limiting cases

Case	Result	Meaning
Pure vector coupling ( $g_A=0$ )	parity conserved	QED and QCD; the current is a polar vector, scalar products reflection-invariant
Pure axial coupling ( $g_V=0$ )	parity conserved	axial current alone is also a parity eigenstate — it is the mix that breaks it
Equal mix ( $g_V=g_A$ , V−A)	parity maximally violated	the weak charged current; couples only to left-chiral fermions
Massless fermion ( $m/E\to0$ )	chirality $=$ helicity	the V−A left-chirality becomes a fixed measurable handedness (neutrino $h=-1$ )
Massive fermion at low $\beta$	asymmetry $\propto\beta$	$W(\theta)=1+a\beta\cos\theta$ shrinks as $\beta\to0$ ; a slow electron shows little asymmetry
Spin-averaged / unpolarized	asymmetry averages to zero	the pseudoscalar $\langle\mathbf{S}\cdot\mathbf{p}\rangle$ needs an oriented spin to reveal itself

Common trapThinking the weak force is weak because its coupling is small

It is not. The weak coupling $g_w$ is actually larger than the electromagnetic one: $\alpha_w=g_w^2/4\pi\approx1/29.5$ against $\alpha\approx1/137$ (Example 2). Weak interactions look feeble at everyday energies only because the massive $W$ propagator $1/(q^2-M_W^2c^4)$ is enormous when $q^2\ll M_W^2$ , throttling every low-energy amplitude to $G_F$ . The second half of this trap is confusing an axial-vector with a polar vector: neither one alone violates parity — a pure axial theory conserves it just as a pure vector one does. Parity violation is a statement about the interference of the two, the cross term $V\!\cdot\!A$ , maximal only because the weak force mixes them with equal weight.

Worked examples

1 · The cobalt-60 asymmetry, from pseudoscalar to number (fully worked)

Polarized ⁶⁰Co undergoes the Gamow–Teller decay ⁶⁰Co → ⁶⁰Ni + e⁻ + ν̄ₑ with asymmetry parameter $a=-1$ . The beta electrons have $\beta=v/c=0.60$ . Find the emission rate along and against the spin, and the forward–backward hemisphere asymmetry. Show that the same quantity vanishes if parity is conserved.

Step 1. Write the distribution. With $a=-1$ , equation (4) gives $W(\theta)=1-0.60\cos\theta$ , $\theta$ from the nuclear spin. Along the spin ( $\theta=0$ ): $W=\mathbf{0.40}$ . Against it ( $\theta=180^\circ$ ): $W=\mathbf{1.60}$ — four times as many electrons against the spin as along it, a blatantly mirror-asymmetric pattern.

Step 2. Integrate over each hemisphere. With $x=\cos\theta$ and measure $2\pi\,dx$ , the along-spin hemisphere gives $N_\uparrow\propto\int_0^1(1-\beta x)\,dx=1-\tfrac{\beta}{2}$ , and the backward one $N_\downarrow\propto\int_{-1}^0(1-\beta x)\,dx=1+\tfrac{\beta}{2}$ . The linear $\cos\theta$ term is the pseudoscalar that survives asymmetrically.

Step 3. Form the asymmetry. $A=\dfrac{N_\downarrow-N_\uparrow}{N_\downarrow+N_\uparrow}=\dfrac{\beta}{2}=\mathbf{0.30}$ . So $30\%$ more electrons emerge against the spin than along it. The asymmetry $A=\beta/2=0.30$ is a direct measure of parity violation.

Step 4. Check the parity-conserving control. Parity conservation would forbid the $\cos\theta$ term ( $a=0$ ), leaving $W=1$ flat, so $N_\uparrow=N_\downarrow$ and $A=0$ . A slow electron ( $\beta\to0$ ) also gives $A\to0$ — not because parity is restored but because $\langle\mathbf{S}\cdot\mathbf{p}\rangle$ needs relativistic electrons to show. Wu saw the asymmetry appear as the sample polarized and vanish as it warmed, exactly as $A\propto\beta\,P$ demands. A nonzero $A$ is impossible in a parity-conserving world; measuring $A=0.30$ overthrew parity.

2 · The weak force is strongly coupled but short-ranged (fully worked)

Show that the weak coupling is larger than the electromagnetic one by computing $\alpha_w$ from the Fermi constant $G_F=1.166\times10^{-5}\ \mathrm{GeV^{-2}}$ and the $W$ mass $M_W=80.4\ \mathrm{GeV}$ , then explain why beta decay is nonetheless feeble.

Step 1. Relate $G_F$ to $g_w$ . Matching the $W$ -exchange amplitude to the low-energy four-fermion contact interaction gives $\dfrac{G_F}{\sqrt2}=\dfrac{g_w^2}{8M_W^2}$ , so $g_w^2=4\sqrt2\,G_F M_W^2$ . The $W$ mass in the denominator is the whole reason the low-energy interaction is small.

Step 2. Put in the numbers. $g_w^2=4\sqrt2\times(1.166\times10^{-5})\times(80.4)^2=\mathbf{0.426}$ , so $\alpha_w=\dfrac{0.426}{4\pi}=\mathbf{1/29.5}$ . Compared with $\alpha=1/137$ , the weak coupling is about $4.6$ times stronger.

Step 3. Locate the "weakness" in the propagator. A dimensionless measure of weak strength at energy $E$ is $G_F E^2$ : at a beta-decay scale $E\sim1\ \mathrm{MeV}$ , $G_F E^2=1.17\times10^{-11}$ , whereas $\alpha\approx7.3\times10^{-3}$ . The weak process is suppressed by roughly $10^{-9}$ — a factor $(E/M_W)^2$ from the heavy propagator, not from a small vertex. Weak means short-ranged, not weakly coupled: at $q^2\sim M_W^2$ the force is electroweak-strength.

Physics of the portrait

Parity violation is a rare case where a single, decisive experiment settled a question the whole field had assumed answered — and the story of how its author was passed over for the prize is nearly as famous as the physics.

Origins · 1 minChien-Shiung Wu and how this was discovered

Chien-Shiung Wu

1912–1997 · the experiment that overturned parity

Chien-Shiung Wu took her PhD at Berkeley under Ernest Lawrence and Emilio Segrè and became the foremost experimental beta-decay physicist of the twentieth century — colleagues at Columbia knew her as the "First Lady of Physics." In 1956 Tsung-Dao Lee and Chen-Ning Yang, wrestling with the "τ–θ puzzle" — two particles identical in every measured property yet decaying to final states of opposite parity — realized that no experiment had ever tested whether the weak interaction respected parity. Wu understood at once that the beta decay of oriented nuclei was the cleanest test, and abandoned a planned trip to launch it immediately. Working with the National Bureau of Standards, she polarized cobalt-60 nuclei cooled to about $0.003\ \mathrm{K}$ and measured whether the electrons preferred the nuclear spin direction. By late 1956 the answer was unmistakable: the electrons streamed out preferentially opposite to the spin, an asymmetry that reversed with the field and melted away as the sample warmed. Parity was not merely broken but maximally broken. The result, announced in January 1957 and confirmed within days by Garwin, Lederman, and Weinrich in muon decay, was one of the fastest and most complete overturnings of a physical law on record. Lee and Yang received the 1957 Nobel Prize; Wu, whose experiment made their proposal real, was not included — an omission widely regarded as one of the committee's clearest injustices.

History

1927

— Wigner formalises parity as a conserved quantum number in atomic and nuclear physics

1956

— Lee and Yang show parity has never been tested in weak interactions, sharpening the τ–θ puzzle

1957

— Wu's cobalt-60 experiment observes maximal parity violation in beta decay

1957

— Garwin, Lederman and Weinrich confirm it in the π→μ→e chain; Lee and Yang receive the Nobel Prize

1958

— Goldhaber, Grodzins and Sunyar measure the neutrino helicity to be −1, fixing the current as V−A

Going further

Texts. Thomson, Modern Particle Physics, §11.1–§11.4 (parity, V−A) — the primary source this article grounds · Griffiths, Introduction to Elementary Particles, 2nd ed., §9.1–§9.3 (weak vertex factor, muon and neutron decay) · Halzen & Martin, Quarks and Leptons, §12.1–§12.4 · Peskin & Schroeder, An Introduction to QFT, §3.6 and §20.1–§20.2 (discrete symmetries, chiral gauge structure) · Schwartz, QFT and the Standard Model, Ch. 11 and Ch. 29. The key relations used throughout this article are: $\gamma^0\gamma^5\gamma^0=-\gamma^5$ and $\gamma^0\gamma^i\gamma^0=-\gamma^i$ (the polar/axial split); $\gamma^\mu(1-\gamma^5)=2\gamma^\mu P_L$ and $\bar u\gamma^\mu P_L u=\bar u_L\gamma^\mu u_L$ ; the cobalt asymmetry $A=\beta/2=0.30$ at $\beta=0.60$ ; $\alpha_w=g_w^2/4\pi=1/29.5$ from $g_w^2=4\sqrt2\,G_FM_W^2$ , versus $\alpha=1/137$ ; and the neutrino helicity $h=-1$ from $P_Lu_\nu$ in the massless limit.

Prerequisites. The Dirac Equation · Dirac Adjoint and Bilinear Covariants · Weyl Spinors and Chirality · Symmetries in Quantum Mechanics · Intrinsic Parity of Fermions. Leads to. The V−A Interaction · The Wu Experiment · Chiral Structure of the Weak Force · Helicity in Pion Decay · The Weak Charged Current.

Ask the tutor: Ask me to reflect any weak process through a mirror and name the pseudoscalar that gives it away, or to work the cobalt-60 asymmetry for a beam speed of your choosing.

What exactly is parity, and how is it different from a mirror reflection?

Parity $\hat P$ inverts all three spatial coordinates through the origin, $\mathbf{x}\to-\mathbf{x}$ . A mirror is a reflection in a single plane; the two differ by a $180^\circ$ rotation, so testing invariance under a mirror is equivalent to testing full parity — which is why the "hold it up to a mirror" picture is exact. Under $\hat P$ , polar vectors like momentum reverse, while axial vectors like spin do not. The quantity that decides parity violation is any pseudoscalar — the dot product of a polar with an axial vector, such as $\mathbf{S}\cdot\mathbf{p}$ — because it changes sign under inversion. A nonzero average value of a pseudoscalar is the unambiguous signature that parity is broken.

Why is a mixture of vector and axial-vector the thing that violates parity, when each alone does not?

Because each pure current is a parity eigenstate. A pure vector interaction $\bar\psi\gamma^\mu\psi$ contracts two polar vectors into a reflection-invariant scalar, so parity holds — this is QED. A pure axial interaction contracts two axial vectors, also reflection-invariant, so a pure-axial force would conserve parity too. The trouble arrives only when you add them: the rate $|V-A|^2=V\!\cdot\!V+A\!\cdot\!A-2V\!\cdot\!A$ contains the cross term $V\!\cdot\!A$ , a polar-times-axial contraction that is a pseudoscalar and flips sign under parity. That interference is parity violation, largest when $V$ and $A$ enter with equal weight — the V−A structure of the weak force.

What does "maximal" parity violation mean — could it have been partial?

Yes, and for a few anxious months in 1957 people wondered if it was. The general weak current $\bar\psi\gamma^\mu(g_V-g_A\gamma^5)\psi$ violates parity by an amount $2g_Vg_A/(g_V^2+g_A^2)$ — zero for a pure vector or pure axial coupling, and maximal (equal to $1$ ) when $g_V=g_A$ . Nature chose the maximum: the leptonic current has $g_A/g_V=1$ exactly, equivalent to the coefficient in $1-\gamma^5$ being precisely $-1$ rather than some small $\epsilon$ . So the $W$ couples to left-chiral fermions with full strength and to right-chiral ones not at all. Had the ratio been $1.1$ , right-handed fermions would feel a small weak coupling. They don't — the violation is complete.

If the weak coupling is actually larger than the electromagnetic one, why is beta decay so slow?

Because "weak" refers to the range and low-energy strength of the force, not the coupling at the vertex. The vertex carries $g_w$ with $\alpha_w=g_w^2/4\pi\approx1/29.5$ , comfortably bigger than $\alpha=1/137$ . What suppresses low-energy weak processes is the mass of the $W$ : its propagator $1/(q^2-M_W^2c^4)$ is enormous at nuclear energies $q^2\ll M_W^2$ , so the amplitude collapses to the tiny Fermi constant $G_F\sim g_w^2/M_W^2$ . The weak-to-electromagnetic ratio at a beta-decay scale is about $10^{-9}$ , almost all from $(E/M_W)^2$ . Push the energy up to the $W$ and $Z$ masses and the "weak" force is as strong as electromagnetism — the point of calling them a single electroweak interaction.

How does the single handedness of the neutrino prove parity is violated?

Parity maps a left-handed particle to a right-handed one, because it reverses momentum while leaving spin alone. If the weak interaction produced neutrinos of both helicities equally, its mirror image would look the same. But Goldhaber's 1958 measurement found neutrinos always left-handed ( $h=-1$ ) and antineutrinos always right-handed. The mirror image of a left-handed neutrino is a right-handed one, and that state does not participate in the weak interaction — in the Standard Model it does not exist as a weakly-coupled state at all. So the reflected process has no counterpart in nature: a single particle's fixed handedness is enough to break the symmetry.

Is parity violated by the other forces, or only by the weak interaction?

Only by the weak interaction. Electromagnetism and the strong force conserve parity exactly, because their currents are pure vectors — the photon and gluon couple through $\bar\psi\gamma^\mu\psi$ with no axial admixture — and gravity conserves it too. This is why parity remains an excellent symmetry for atoms and nuclei: the parity-violating weak contribution is buried under the parity-conserving electromagnetic one and shows up only in exquisitely sensitive experiments or in processes that are purely weak, like beta decay and neutrino interactions. Parity is not a broken symmetry of physics in general; it is a symmetry that exactly one of the four forces declines to respect.