Parity violation

Parity violation means that a physical law fails to be invariant under the spatial reflection represented by the parity operator. A process and its fully inverted counterpart can then have different probabilities. For beta-minus decay with polarized parent atomic nuclei, a nonzero correlation between nuclear spin and emitted Electron momentum is such a test. Rotational symmetry alone allows this correlation: a pseudoscalar is unchanged by proper rotations but changes sign under spatial reflection.

In 1956, Tsung-Dao Lee and Chen Ning Yang examined the evidence for parity conservation in the weak interaction. Their paper, Question of Parity Conservation in Weak Interactions, distinguished an untested assumption from an established experimental result and proposed comparing Electron emission along opposite directions relative to oriented nuclear spins, explicitly considering cobalt-60. It proposed a test; it did not report the later Wu experiment.^[ref]

The conceptual improvement is the choice of observable. An angle between two polar vectors is preserved by spatial reflection; its distribution can conceal parity violation. An angle between an axial vector and a polar vector changes to its supplement. Thus a spin-resolved directional measurement asks a different question from an unpolarized total decay rate.

The Wu experiment was performed by Chien-Shiung Wu, Ernest Ambler, Raymond Webster Hayward, Dale D. Hoppes, and Ralph Percy Hudson; its 1957 report was Experimental Test of Parity Conservation in Beta Decay. Their experiment supplied the evidence that the Lee-Yang proposal for a parity test had called for.^[ref]

A cobalt-60 surface layer on a cerium magnesium nitrate crystal was cooled by adiabatic demagnetization; a solenoid's magnetic field polarized its atomic nuclei. An internal anthracene counter detected Electrons by particle scintillation; a light pipe carried flashes to a photomultiplier. Polar and equatorial gamma ray counters monitored nuclear polarization and effective temperature; bulk thermometry could miss surface heating.^[ref]

Reversing nuclear polarization reversed the Electron excess; warming erased it with gamma ray anisotropy. Warm rates were field-independent. Cooling-field reversal checked remanence; poorly polarized sources checked magnetic deflection. Electrons favored the direction opposite spin.^[ref]

The gamma ray anisotropy diagnoses the preparation; the Electron asymmetry diagnoses parity violation. These are different jobs. A directionally uneven gamma ray signal alone need not violate parity conservation: a distribution even in the cosine of the angle can distinguish an axis from its perpendicular without distinguishing its two ends. The spin-momentum correlation instead distinguishes those ends.

The inference requires control of changes in detection efficiency, Electron transport and nuclear polarization. A fixed detector under spin reversal avoids physically exchanging detector positions, but still requires checks for magnetic field effects. The result establishes parity violation in this weak interaction process. A precise microscopic coupling requires more information than the existence or sign of a counting asymmetry; the following model makes that distinction explicit.

Consider a parent atomic nucleus at rest. Write its signed nuclear polarization along a fixed axis $\hat{\mathbf{z}}$ as $P=⟨J_z⟩/(I\hslash )$, where $I\hslash$ is its maximal spin component. Let $\theta$ be the angle of the emitted Electron momentum to that axis and let $\beta =v/c$. Use $\mathcal{P}$ for the parity operator, so it is not confused with $P$.

For an axially symmetric leading-dipole model, retain a constant and the term linear in the polarization vector $\mathbf{P}=P\hat{\mathbf{z}}$. Rotational symmetry makes the directional term proportional to $\mathbf{P}\cdot \hat{\mathbf{p}}$. Define its dimensionless coefficient as $A\beta$. At fixed Electron energy the relative intensity and normalized probability density function per solid angle are thereforeThe factor $\beta$ is the conventional separation of the speed dependence; rotational symmetry alone does not determine that dependence or $A$. The coefficient $A$ contains the decay dynamics. This leading model omits higher angular terms and detailed recoil and radiative effects. A partially oriented higher-spin ensemble can also have alignment that is not specified by its vector nuclear polarization alone.

The unit-integral condition follows without any model of a detector:Thus $f$ has unit angular average, while the probability density function in the polar angle alone contains the Jacobian determinant factor:In particular, $f$ itself is neither a normalized probability density function in $\theta$ nor a detector count. Positivity on $-1\le u\le 1$ gives $|a|\le 1$.

Under full spatial reflection, $\mathbf{p}\mapsto -\mathbf{p}$ but $\mathbf{J}\mapsto \mathbf{J}$. The latter follows already for orbital angular momentum from $\mathbf{r}\times \mathbf{p}\mapsto (-\mathbf{r})\times (-\mathbf{p})=\mathbf{r}\times \mathbf{p}$; spin has the same axial vector transformation. Hence $\mathbf{P}\cdot \hat{\mathbf{p}}$ is a pseudoscalar, and $\cos \theta \mapsto -\cos \theta$. If parity conservation held for this model, equal probabilities for the inverted process would force $a=0$. Conversely, $a=0$ in an unpolarized sample proves nothing about $A$.

Integrating over the two hemispheres gives a useful original check:Choose $A=-0.8$, $P=0.5$, and $\beta =0.75$, so $a=-0.3$. Then the forward and backward hemisphere probabilities are $0.425$ and $0.575$, and their counting asymmetry is $-0.15$. These are chosen teaching values, not measured values from the Wu experiment. At a single energy, the angular shape determines the product $AP\beta$; it cannot separately determine $A$ without a polarization calibration.

Consider a teaching extension with two opposite detectors, labelled $+$ and $-$ by fixed laboratory directions, and two preparation states $s=+1,-1$ with equal and opposite nuclear polarization. This is an original two-detector model; the historical Wu experiment used a fixed internal Electron counter with reversal.^[ref] Let $C_s>0$ describe the exposure in each state and ${\eta }_d>0$ the stable detection efficiency of detector $d$.

For matched opposite acceptances, write their common mean absolute cosine as $k$. The four expected positive counts areThe same-state ratios each retain the unequal detection efficiencies:Dividing these ratios gives the spin-reversal cross-ratio for beta decay:Both the exposure ratio $C_{+}/C_{-}$ and the detection efficiency ratio ${\eta }_{+}/{\eta }_{-}$ cancel. The positive square root is essential: positive expected counts and $|ϵ|<1$ make $(1+ϵ)/(1-ϵ)$ positive.

For the chosen $a=-0.3$, opposite hemispheres give $k=1/2$, hence $ϵ=-0.15$. Choose $C_{+}=10000$, $C_{-}=8000$, ${\eta }_{+}=0.8$, and ${\eta }_{-}=0.5$, with their common acceptance factor absorbed into $C_s$. The four expected counts in the displayed order are $6800$, $5750$, $7360$, and $3400$. ThusThe first raw detector ratio exceeds one despite negative $ϵ$: detector imbalance can overwhelm the physical directional preference. Reversal separates the stable imbalance from the term that changes sign.

A detector collects a range of solid angles, not a single direction. If its nonnegative angular response is $w(\Omega )$, defineIntegrating the polarized beta-decay angular distribution gives a count proportional to $G(1+ak)$. For a uniform forward cone with half-angle ${\theta }_0$, put $c_0=\cos {\theta }_0$. Direct integration givesThe opposite cone has mean cosine $-k$. A hemisphere gives $k=1/2$; a narrow cone tends to $k=1$. Hence a spin-reversal cross-ratio for beta decay measures $ak$, and an acceptance correction is needed to obtain $a$.

With unequal angular responses, let the two laboratory mean cosines be $k_{+}$ and $k_{-}$, including their signs. Then the actual expectation iswhich reduces to the simple square only when $k_{-}=-k_{+}$. The cancellation of stable multiplicative detection efficiencies survives, but their angular shapes still matter. An energy-integrated response likewise averages $A(E)P\beta (E)\cos \theta$ with the accepted spectrum; a mean speed cannot in general be substituted independently of acceptance.

Additive backgrounds, unequal polarization magnitudes or detection efficiencies that change on reversal do not have the assumed factorization. For example, a reversal-dependent efficiency cross-ratio multiplies $R$ by ${\eta }_{+}^{+}{\eta }_{-}^{-}/({\eta }_{-}^{+}{\eta }_{+}^{-})$. Background subtraction also creates statistical complications: positive expected signal counts are assumed in this derivation, whereas finite samples with Poisson distributions can contain zero counts. Taking square roots of negative background-subtracted counts is not a valid estimator. These limits explain why reversal and independent preparation checks belong together.