# Why does parity violation in weak decay imply decay asymmetry?

I googled the sentence in the title of this question and found the famous experiment by Wu et al demonstrating that electrons in weak decay are emitted in the direction of motion of a left-handed screw rotating with the nuclear spin''.

This show that: asymmetry in decay -> parity violation.

How to prove the opposite? (parity violation -> asymmetry ?)

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## 1 Answer

Dear Pie86, the emission of particles in a weak decay is a complicated reaction, and Gell-Mann's totalitarian principle applies: every process or effect that is not prohibited by a symmetry will occur at a nonzero probability. The asymmetry or the spin-momentum correlation for the electrons is such an effect.

In this case, it is infinitely unlikely that the asymmetry will be exactly zero unless the symmetry is implied by a valid symmetry. Because parity is not a valid symmetry, there's no reason for the asymmetry - the correlation between the spin and direction of the electron, among other similar correlation coefficients - to be exactly zero, so it will probably not be exactly zero.

For the particular case of the beta decay, one may calculate the corresponding probability amplitude and the asymmetry - or correlation coefficients - out of it. If the Lagrangian has the $(a+b\gamma_5)$ matrix to guarantee an asymmetry, the asymmetry of the particular decay - or the spin-momentum correlation coefficient - will be proportional to something like $ab$ or $(a^2+b^2)$ times something. More generally, it will be a function of $a,b$ that is manifestly nonzero if both $a,b$ are nonzero. And in weak interactions, they are nonzero; in fact, weak interactions are maximally parity violating so $a=\pm b$.

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