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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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up vote 3 down vote accepted

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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