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The ground state of an even-even nucleus has $J^P=0^{+}$, and the emitted alpha particle also has $0^{+}$. In an alpha decay to daughter spin $J_Y$, addition of angular momentum permits total zero only when the relative orbital angular momentum is $L=J_Y$. Parity conservation gives $+1={\pi }_Y(-1{)}^L$, hence the daughter has natural parity, $0^{+},1^{-},2^{+},3^{-},\dots$. This rule does not itself restrict $J_Y$ to even integers. An even-$J$ sequence such as $0^{+},2^{+},4^{+}$ requires additional nuclear structure, for example a positive-parity ground-state nuclear rotational band.

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.

For a state of integer total angular momentum $J$, natural parity means $\pi =(-1{)}^J$: $0^{+},1^{-},2^{+},3^{-},\dots$. The other sign is called unnatural parity. The terminology is useful when a spin-zero positive-parity parent emits a spin-zero positive-parity particle: angular momentum conservation fixes relative orbital angular momentum $L=J$, and parity conservation then forces natural parity for the daughter.

Awarded jointly to Chen Ning Yang and Tsung-Dao Lee for investigating parity conservation laws, leading to discoveries about elementary particles.^[ref]

The strong-interaction matrix element between two spin-zero pseudoscalar mesons has only $p^{\mu }$ and $k^{\mu }$ available. The product of the two intrinsic parities is positive. An axial current would require a pseudovector constructed from these momenta, but an expression involving the Levi-Civita symbol needs three independent four-vectors and therefore vanishes. This is a consequence of parity conservation in the hadronic matrix element, not of parity conservation in the weak interaction. The vector current can have the two independent structures $p+k$ and $p-k$. Its coefficients are Lorentz scalars; with $p^2=m_K^2$ and $k^2=m_{\pi }^2$ fixed, their only varying invariant is $s=q^2$. HenceThese are the pseudoscalar-to-pseudoscalar form factors. With relativistically normalized states they are dimensionless. The vanishing axial matrix element and this decomposition explain the two equalities separately.

Write $q_1$ and $q_2$ for the outgoing electron and antineutrino momenta. From the Fermi interaction, an invariant scattering amplitude, up to an irrelevant overall sign or phase, isThe CKM matrix element $V_{us}$ multiplies the quark current in the convention of this interaction. Let ${\ell }_{\mu }=\bar{u}(q_1){\gamma }_{\mu }(1-{\gamma }^5)v(q_2)$. For massless leptons, the massless Dirac equation and chirality matrix anticommutation giveIn the second term move $/q_2$ through the chiral projector before applying $/q_{2}v=0$. Since $p+k=2p-q$, this transverse massless leptonic current givesThe disappearance of $f_{-}$ uses the massless approximation; for a massive charged lepton its contraction is proportional to the lepton mass.

Use the fermion spin sum and the supplied gamma matrix trace identities. The symmetric part of the leptonic tensor isThe Levi-Civita symbol term is antisymmetric and drops out when contracted with $p^{\mu }{p}^{\nu }$. ThusThere is no initial-spin average because the kaon is spinless. For the integrated massless leptonic tensor, keep every factor of $2\pi$ explicit and define the unnormalized two-lepton Lorentz-invariant phase spaceThe leptons are massless, so $q_i^0=|{\mathbf{q}}_i|$. Its trace is $g^{\mu \nu }{I}_{\mu \nu }=\pi s$. ThereforeThe three Lorentz-invariant phase-space measures and their momentum delta function contribute $1/[8(2\pi {)}^5]$, in addition to $1/(2m_K)$ in the decay rate. Combining them with the spin sum givesThis massless semileptonic pseudoscalar decay rate uses a two-lepton integral over future-timelike $q$ and the pion integral is restricted to the physically allowed region. The null endpoint follows by continuity. The coefficient has mass dimension $-5$, so the complete expression has mass dimension one, as a decay rate must in natural units.

In the kaon centre-of-momentum frame, put $E=k^0$ and $\kappa =|\mathbf{k}|$. ThenThe dimensionally consistent Källén function isThe pion-only mass term must have fourth power: the second power printed in the PDF is dimensionally inconsistent. This repair also follows directly from squaring $E=(m_K^2+m_{\pi }^2-s)/(2m_K)$. Angular integration and the change of variable givewhere the negative sign reverses the endpoints. Combining this with the bracket $\lambda /4$ yieldsThus $\boxed{B=G_F^2|V_{us}{|}^2/(192{\pi }^3{m}_K^3),a=0,b=(m_K-m_{\pi }{)}^2}$. The lower limit is the minimum invariant mass of two massless leptons; at the upper limit the pion is at rest. The coefficient has mass dimension $-7$, while $ds{\lambda }^{3/2}$ has dimension eight. The Källén function also shows why the differential decay rate vanishes at zero pion momentum.

With $\hslash =1$, angular momentum eigenstates have $j=0,\frac{1}{2},1,\dots$ and $m=-j,-j+1,\dots ,j$. Their $J^2$ and $J_3$ eigenvalues are $j(j+1)$ and $m$. Adding spins $1/2$ and $j$ gives $J=j+1/2$ and, if $j\ge 1/2$, $J=j-1/2$. For the Clebsch-Gordan coefficients to be nonzero one requires $\boxed{M=r+m}$.

For $j=2$, write $|\pm ,m⟩$ for $|1/2,\pm 1/2⟩|2,m⟩$. The highest state and one application of the angular momentum lowering operator giveThe lowering coefficient of the highest $J=5/2$ state is $\sqrt{5}$, while the two product-state coefficients before division are $1$ and $2$. The last line is its normalized orthogonal complement. Its coefficient with maximal $m=2$ is positive, fixing the requested phase convention. All other coefficients for the allowed $M\ge 3/2$ vanish.

For the decay, the final total intrinsic spin is $1/2$. Combining it with integer orbital angular momentum $\ell$ to obtain $J=3/2$ permits only $\ell =1,2$. Parity conservation gives ${\eta }_X={\eta }_A{\eta }_B(-1{)}^{\ell }=(-1{)}^{\ell }$. Thus $\boxed{{\eta }_X=+1:\ell =2;{\eta }_X=-1:\ell =1}$.

For $\ell =2$ and initial $M=3/2$, the last displayed coupled state shows that the spin-up probability, after ignoring the orthogonal orbital states, is $1/5$. For $\ell =1$, $J=3/2$ is the maximal coupled angular momentum and its highest state is $|1/2,1/2⟩|1,1⟩$. Therefore $\boxed{P=1/5for{\eta }_X=+1,P=1for{\eta }_X=-1}$.

In ordinary perturbative Standard Model reactions all three interactions conserve electric charge, four-momentum, total angular momentum, baryon number, and total lepton number. The last two are accidental conservation laws of these perturbative processes, not exact nonperturbative statements: electroweak sphalerons violate $B+L$ while preserving baryon-minus-lepton number. Total color charge is respected throughout.

The strong interaction and electromagnetic interaction also conserve each net quark flavor number, including strangeness, charm and bottom flavour. The weak charged current can change them through the Cabibbo-Kobayashi-Maskawa matrix. Isospin is an approximate strong interaction symmetry in the small light-quark mass difference limit; the different electric charges of $u$ and $d$ break it in the electromagnetic interaction. Approximate flavor symmetry is distinct from exact net-flavour-number conservation by those interactions.

Parity conservation and charge conjugation symmetry hold for the electromagnetic interaction and, assuming a negligible strong CP parameter, the strong interaction. The weak interaction violates both, and also violates CP symmetry in suitable processes. CPT symmetry holds under the usual local Lorentz invariant quantum field theory assumptions. Individual lepton family numbers are conserved in the conventional massless-neutrino approximation but not in neutrino oscillations; an assertion of exact family conservation in all observed reactions would therefore be false.

The radiative $X^0$ decay conserves strangeness and proceeds electromagnetically. A strong $\Lambda \pi$ decay is closed because even the lighter neutral-pion threshold exceeds the supplied $X^0$ mass. The $Y^{-}$ decay has about $65\mathrm{MeV}$ of available energy, but changes strangeness from $-2$ to $-1$, so it requires the weak interaction. Its lifetime is therefore ordinarily much longer than the electromagnetic lifetime of $X^0$; no numerical lifetime follows from these masses alone.

Both particles are baryons with half-integer spin. A radiative transition to spin $1/2$ can couple through photon multipoles $L\ge 1$, giving $J_X=L\pm 1/2$. Electromagnetic parity is conserved, but without the electric/magnetic multipolarity neither $J_X$ nor its parity is fixed. In the weak $Y^{-}$ decay, coupling orbital $L$ to the final $\Lambda$ spin gives $J_Y=L\pm 1/2$; parity conservation cannot be imposed on that weak amplitude. The supplied modes therefore do not determine unique spin-parity assignments.

The masses and flavours identify the familiar candidates as ${\Sigma }^0$ and ${\Xi }^{-}$. Their established ground-state assignments are both $1/2^{+}$, consistent with the lowest-orbital quark model, but this is additional empirical/model information rather than a consequence of the two decay channels alone.^[ref]

Alpha lifetimes and competing beta decays. The Geiger-Nuttall law follows approximately from quantum tunnelling through the daughter’s Coulomb barrier. Increasing $Q_{\alpha }$ shortens the barrier and greatly increases the decay rate; increasing the daughter charge makes the barrier harder to penetrate. A useful local parametrization isUsing the supplied nuclear masses, no electron-mass subtraction is needed for alpha decay. The calculated Q values areCalibrate the two constants to the two supplied even-even thorium half-lives. Their daughter charges are both 88, soThis gives the order-of-magnitude estimatesHere ${\lambda }_{\alpha }=\ln 2/T_{\alpha }$. The known beta-minus decay rates are approximately $3.8\times 10^{-9}{\mathrm{s}}^{-1}$ and $3.2\times 10^{-5}{\mathrm{s}}^{-1}$, respectively. Thus the reference alpha branching fractions are only about $1.6\times 10^{-8}$ and $1.4\times 10^{-8}$. Beta decay overwhelmingly removes the parents before alpha emission, explaining why the alpha branches are not seen in the stated chain.

These estimates are local extrapolations, not precision predictions. In particular, actinium is odd-odd, so extrapolating an even-even Geiger-Nuttall law neglects nuclear-structure and angular-momentum hindrance. Such effects can further suppress its alpha branch.

Allowed daughter spin and parity. The even-even parent in its ground state and the alpha particle both have $J^P=0^{+}$. The total final angular momentum must be zero, so coupling daughter spin $J_Y$ to relative orbital momentum $L$ requires $J_Y=L$. Parity conservation gives $+1={\pi }_Y(+1)(-1{)}^L$. The alpha-decay angular-momentum and parity selection rule is thereforeThese are the natural parity states. Restricting the list to even spin would be too strong; even-spin rotational bands are a further structural specialization.

Ground-state line and recoil. The nuclear alpha-decay recoil correction givesFor the thorium ground-state decay, this predicts $T_{\alpha }\simeq 4.0004\mathrm{MeV}$, with daughter recoil about $70.2\mathrm{keV}$. This is close to the highest measured line, $4.0112\mathrm{MeV}$: the difference is $10.8\mathrm{keV}$, or $0.27\%$. Conversely that line implies $Q_{\alpha }\simeq 4.0816\mathrm{MeV}$. The quoted nuclear masses have only five decimal places in atomic mass units; rounding three such entries can shift their mass difference by up to about $14\mathrm{keV}$. Thus the requested approximate agreement is supported, but the rounded table does not reproduce the line energy exactly. Dropping daughter recoil would give an error around $70\mathrm{keV}$ instead.

Excited levels. Relative to the highest line, the two alpha-energy deficits are $62.7$ and $201.2\mathrm{keV}$. Multiplying by the common recoil factor $1+m_{\alpha }/M_d\simeq 232/228$ givesA deformed even-even nucleus has a ground-state nuclear rotational band with $E_J\simeq J(J+1)/(2\mathcal{J})$ and $J^P=0^{+},2^{+},4^{+},\dots$. Its $4^{+}$ to $2^{+}$ excitation ratio is $20/6=10/3=3.333$, with the observed ratio about $3.7\%$ below this ideal value. The likely assignments are therefore $2^{+}$ and $4^{+}$ rotational excitations, with modest departures from a rigid constant-inertia rotor. The ratio supports this interpretation; it is not a unique spin measurement by itself.

Sargent comparison. For beta-minus decay using nuclear masses, subtract the emitted electron rest mass:The simple $\lambda \propto Q^5$ version of Sargent's rule would predictThe measured rate ratio is the inverse half-life ratio,The naive fifth-power estimate is about $2.4\times 10^4$ times larger. It correctly predicts the much faster actinium decay but fails quantitatively. The small radium endpoint is far below the electron rest mass, violating the negligible-mass approximation behind the fifth power; Coulomb distortion of the emitted electron, different nuclear matrix elements and different daughter excitations also matter. The rounded mass difference makes the very small radium $Q$ especially uncertain. The appropriate comparison of individual branches uses the actual endpoint-dependent statistical factor and comparative half-life, not a universal fifth power of the ground-state release.

Beta selection rules and the four observed branches. The radium parent has $0^{+}$, so the ground-state $0^{+}\to 3^{+}$ transition is a unique forbidden beta decay of second order: it preserves parity and requires three units of angular momentum. For the increasing daughter excitation energies, the endpoints decrease. The assignments areFor $0^{+}\to 1^{+}$, zero lepton orbital angular momentum and lepton-pair spin one give an allowed beta decay of Gamow-Teller transition type. The $0^{+}\to 1^{-}$ branches need odd lepton orbital angular momentum, with lowest value one, and are forbidden beta decays of first order; they are not unique because they do not have the maximal transfer $\Delta J=2$ at that order.

The absence of a listed ground-state branch is consistent with its stronger forbiddenness. The two allowed branches carry $70\%$ in total, and the nearly equal-endpoint pair favours the allowed branch by four to one. These trends support the classification. However, first-forbidden branches still carry $30\%$, and the lower-endpoint first-forbidden branch is stronger than the higher-endpoint one. Thus selection-rule labels alone do not explain all intensities: nuclear overlaps and the endpoint-dependent phase space are indispensable. “Forbidden” means suppressed at leading order, not impossible.

In the Wu experiment, Chien-Shiung Wu and collaborators studied beta-minus decay of polarized cobalt-60,A thin cobalt-60 layer on a cerium magnesium nitrate crystal was cooled by adiabatic demagnetization to a few millikelvin. A polarizing magnetic field produced nuclear polarization. An anthracene scintillation counter and photomultiplier detected the emitted electrons; angular gamma ray observations from the daughter de-excitation monitored whether the nuclei remained polarized.^[ref]

The electrons were emitted preferentially opposite the parent nuclear spin. Reversing the nuclear polarization reversed the asymmetry, while warming the sample removed both polarization and the effect. Those controls distinguished a spin-correlated asymmetry from a fixed detector bias.

The relevant angular correlation is a polarized beta-decay angular distribution of the formUnder parity, ${\mathbf{p}}_e\to -{\mathbf{p}}_e$ but $\mathbf{J}\to \mathbf{J}$, so the spin-momentum correlation changes sign. Parity conservation would require equal rates for these two directions relative to the same spin. The observed nonzero coefficient therefore demonstrated parity violation in beta decay. An unpolarized source would average this correlation to zero and could not provide the same test.

Assume unpolarized beams, negligible electron and muon masses, a single scattering s-channel spin-one mediator with equal left and right vector couplings, and leading-order QED. Z boson exchange and radiative corrections are omitted. The angular factors are the Wigner D-matrix elements $d_{m{m}^{\prime }}^1({\theta }^{\star })$, with $m,m^{\prime }=\pm 1$. Up to common phases and normalization they are $(1+\cos {\theta }^{\star })$ for equal pair-spin projections and $(1-\cos {\theta }^{\star })$ for opposite projections. The four orthogonal helicity channels add incoherently, givingWith $s$ the squared center-of-mass energy, the normalized differential scattering cross-sections areThe second expression includes the solid angle Jacobian: a distribution per angle is different from one per $\cos {\theta }^{\star }$. Integration gives $\sigma =4\pi {\alpha }^2/(3s)$.

Parity reverses all momenta but leaves the spin axial vector unchanged, so it interchanges $L\leftrightarrow R$. The $LR\to LR$ and $RL\to RL$ amplitudes have equal magnitudes, as do $LR\to RL$ and $RL\to LR$. Thus every configuration has the same rate as its parity transform, proving parity conservation. The summed distribution is also forward-backward symmetric. Strictly, parity reverses both the beam and outgoing momenta, leaving their relative angle unchanged; forward-backward reflection by itself is not the definition of parity.

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.

A meson with zero spin angular momentum and negative parity. The pion and kaon are familiar examples. Between two pseudoscalar mesons, parity conservation in the strong interaction forces the axial current matrix element to vanish, while the vector current has pseudoscalar-to-pseudoscalar form factors. The weak decay as a whole need not conserve parity.

The weak interaction changes particle species through the charged current and also acts through the neutral current. Familiar examples include beta-minus decay and neutrino scattering. It is short-range and chiral: its W boson and Z boson couplings distinguish chirality. Above the electroweak scale it belongs to the unbroken $SU(2{)}_L\times U(1{)}_Y$ electroweak interaction. At much lower energies, W boson exchange is approximated by the Fermi interaction.

A particularly direct window into the weak interaction is the Wu experiment: a sample with oriented nuclear spins emits electrons preferentially toward one end of its spin axis. The comparison tests parity conservation, because spin is an axial vector and momentum is a polar vector. The polarized beta-decay angular distribution translates that geometric distinction into a measurable counting asymmetry.

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.