Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?)

share|cite|improve this question
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 $a=\pm b$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.