In the classic Wu experiment (https://doi.org/10.1103/PhysRev.105.1413) parity violation was discovered in the weak interaction through the asymmetry in the distribution of electrons in the beta decay of Co-60.

In the experiment the Co-60 nuclei are cooled and polarized in a magnetic field. Since the magnetic field and magnetic moment are axial vectors they do not change sign with a parity transformation. Therefore if the electrons are not emitted isotropically then this indicates that the beta decay does not conserve parity.

In the beta decay of Co-60, two gamma rays are also emitted which, when the nuclei are polarized, are also emitted non-isotropically. The paper writes "The observed gamma-ray anisotropy was used as a measure of polarization, and, effectively, temperature.". It also says that this "has been known for some time".

My question is this: Why does the observed anisotropy of electrons indicate that parity is not conserved in the weak interaction while the observed anisotropy of gamma rays DOES NOT indicated parity violation in the electromagnetic interaction?


1 Answer 1


Here's a level scheme for the cobalt-60 decay, from the evaluated nuclear structure data file (ENSDF), with the 99.9% pathway highlighted:

cobalt/nickel level scheme

In the beta decay, the spin-parity of the nucleus goes from $5^+\to 4^+$. In the limit where the parity violation is small, parity near-conservation requires the final state to have orbital angular momentum $L=\text{even}$, with $L=0$ strongly preferred. The missing unit of angular momentum must be carried away by the electron and neutrino spins. This is known as a Gamow-Teller transition.

However, in the lepton sector the parity violation is not small. Parity violation is maximal for the antineutrino, whose "north pole" always points in the same direction as its momentum. The electron's "south pole" always points forward in the ultrarelativistic limit, but the electron's polarization is reduced when the electron is at lower energy. To carry away the missing unit of spin, the electron and neutrino "north poles" must both point parallel to the cobalt "north pole." In the cobalt decay path to the nickel excited state, only about $300\,\mathrm{keV}\approx \frac{m_e c^2}{2}$ is available to be shared between the electron and neutrino, so the electron polarization is strong but not overwhelmingly strong. But if you pretend that both the electron and the neutrino are completely polarized, then to carry away the missing unit of angular momentum, their momenta are determined by their spins, with the neutrino traveling along the cobalt's "north axis" and the electron along its "south axis."

Your question is about the electromagnetic transitions in the nickel. Those transitions, with spin-parity $4^+\to 2^+$, are "electric quadrupole (E2)" transitions, where the photon carries away two units of angular momentum but the total parity is unchanged. In that case, the orbital wavefunction is the $L=2$ wavefunction, and the photon angular distribution will go like

$$ Y^\ell_m(\theta,\phi) = Y^2_2(\theta,\phi) = \sqrt{\frac{\text{constant}}{2\pi}} \cdot e^{2i\phi} \sin^2\theta $$

This distribution is strongly peaked in the direction of the cobalt's "equator" at $\theta=0$. That's not a parity-violating distribution, because the "equator" points in the same direction under a parity transformation, while electron preference for the "south pole" would be reversed under a parity transformation.


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