Why is the Beta-Decay of a Neutron asymmetric?

Question

The Wu-experiment, which originally showed the parity violation of Beta-Decay experimentally, is often used to give an intuitive explanation for the asymmetry of the decay:

$${}^{60}_{27}\text{Co} \rightarrow {}^{60}_{28}\text{Ni}^* + e^- + \bar{\nu}_e$$

Cobalt has $I=+5$, and nickel has $I=+4$ . Due to conservation of angular momentum the electron and the antineutrino both have $S=+1/2$, i.e. in the same direction as the cobalt. The requirement for right-handedness of the antineutrino (which we assume as given for now), gives then the asymmetry in the decay.

However, when I look at the decay of the free neutron,

$$\mathrm{n}\rightarrow\mathrm{p}+\mathrm{e}^-+\bar{\nu}_e$$

I find that the spin of the neutron and proton are equal ($=1/2$). The total angular momentum can be conserved by aligning the spins of the leptons opposite to each other or, I guess, by adding angular momentum to the leptons.

My problem is here, that it is not straightforward to see the asymmetry in this case by just looking at the momentum conservation. It seems that the information of the polarization of the neutron is somehow not conveyed to the leptons: The only requirement is that their $S+L=0$, which allows to arrange the spins of the leptons independently of the neutron. One could also take e.g. the decay of

$${}^{8}_{3}\text{Li} \rightarrow {}^{8}_{4}\text{Be}^* + e^- + \bar{\nu}_e.$$

Here Li has $I=2$ and Be $I=2$, which poses the same difficulty for me.

Is it possible to understand the last cases simply based conservation of angular momentum and the forbidden left-handedness of the antineutrino (similar to the case of cobalt-60)?

Edit:

Thank you very much for the extensive comment!

I found the following diagram in a diploma thesis about the neutron beta decay :

(From https://www.physi.uni-heidelberg.de/Publications/dipl_mund.pdf )

Apparently you can classify the decays on whether they decay into a singlet or triplet of the leptons ("Zerfall" means "decay" in German). You can see that there is a channel, which is actually sensitive to the original polarization of the neutron: In the last one, the proton spin flips, but it is always opposite to the neutron, the leptons align with the neutron. The beta-decay of the neutron seems to be a mixture of all three. This would imply that the first channel ("Fermi") is actually symmetric.

I think this is also the reason, why someone talks about the asymmetry being "full" in the Wu experiment, since it fully decays in an asymmetric channel. Cases like the neutron might be actually less asymmetric than Co60.

Answer 1

There are several different ways that a decay can be "asymmetric." This proposal for an experiment currently collecting data on neutron decay asymmetries describes the dependence of the neutron decay rate on the energy and direction of the electron and neutrino as

\begin{align} \frac{d\Gamma}{dE_e\ d\Omega_e\ d\Omega_\nu} \propto{}& p_e\ E_e\ (E_0-E_e)^2 \times{}\\ &\times\Big[1 + a\frac{\vec p_e\cdot\vec p_\nu}{E_e E_\nu} + A \frac{\vec\sigma_n \cdot \vec p_e}{E_e} + B \frac{\vec\sigma_n \cdot \vec p_\nu}{E_\nu} + D \frac{\vec\sigma_n \cdot \left(\vec p_e\times\vec p_\nu\right)}{E_eE_\nu} \Big] \end{align}

In this expression, $d\Gamma$ is the differential decay rate, the differentials $d\Omega_{(e,\nu)}$ are little bits of solid angle, $\vec p_{(e,\nu)}$ are the electron and neutrino momenta, $E_{(e,\nu)}$ are the electron and neutrino energies after the decay, $E_0$ is the total energy in the decay, and $\vec\sigma_n$ is the neutron spin before the decay.

This set of decay correlations and their coefficients was first written down by Jackson and collaborators in the 1950s, and is basically "what are all the things that a decay rate could depend on? let's give them names." There are some other decay correlation coefficients that aren't relevant for your question. For instance, Jackson had a term $C\ {\vec\sigma_n \cdot\vec p_p}/{E_p}$ depending on the momentum of the proton after the decay, but in the neutron's rest frame the proton's momentum obeys $\vec p_p + \vec p_e + \vec p_\nu = 0$ and the decay coefficient $C$ is therefore some linear combination of $A$ and $B$. There's a $b$ which isn't relevant for your question, so I left it out. There are probably some other possible decay parameters that are omitted from the linked proposal, as well.

What you've derived for the Wu experiment is some reasoning about the correlations corresponding to $A$ and $B$: correlation between the direction of the nuclear polarization and the direction of the electron, or between the nuclear polarization and the direction of the neutrino. That simple analysis doesn't work for the neutron case for the reasons that you cite, but more sophisticated analyses are possible. A typical result is that the correlation coefficient $A$ between the neutron polarization direction and the direction that the electron travels is given, using the Standard Model weak interaction, by $$ A = -2\frac{|\lambda|^2 + \mathrm{Re}(\lambda)}{1 + 3|\lambda|^2} $$ where $\lambda = g_A/g_V$ is the ratio of the axial-vector coupling $g_A$ and the vector coupling $g_V$ in the weak interaction. In extensions to the Standard Model where the weak interaction is not purely $V-A$ ("vee minus aye," "vector minus axial-vector") but contains scalar or tensor terms as well, all of these correlation coefficients are predicted to shift as well.

In case you're curious about values, the current Particle Data Group evaluation gives $A = -0.1184(10)$: the electron is about twelve percent more likely to be emitted from the neutron's south pole than from its north pole. The neutrino correlation $B$ has the opposite sign and is much larger, $B = +0.9807(30)$. Apparently the antineutrinos are emitted almost entirely into the same hemisphere as the neutron's north pole! It seems like there ought to be a good hand-waving analysis for that, but I don't know it.