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When a lambda particle decays into proton and a pion, I am told it does not conserve parity. Why?

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So is the question "How do you measure the parity before and after?" or "What is the underling cause of the violation?", and if the latter will you be disappointed by "That's just the way the weak interaction is."? –  dmckee Feb 21 '11 at 4:49
Yes, I will take everyone at their word that the weak interaction does not conserve parity. Perhaps, what I mean is taking a look at the decay, either mathematically or at a glance, how do you know it does not conserve parity ? –  Cogitator Feb 21 '11 at 4:57
you can't tell that at a glance. You have to compute probabilities of left- and right-handed processes and see whether they differ. They wouldn't have to differ (and indeed people used to think that all theories have $P$-symmetry in the past). So, the answer is: just compute the relevant cross-sections. But perhaps there also exists some intermediate answer between full calculation and just stating that weak interaction is responsible. Let's see... –  Marek Feb 21 '11 at 8:08
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1 Answer

First, an assignment of the parities.

The parity of fermions is a bit ambiguous because one may always redefine parity by $$ P \to P (-1)^{2J}, P(-1)^L, P(-1)^{3B}, P(-1)^{3Q} $$ or one may add the product of several factors of this kind because the second factor is a multiplicatively conserved sign. By this definition, one gets another parity that is still conserved (at least in low-energy processes that also conserve $B,L$).

However, there's a convention that assigns a particular parity to fermions. Note that Weinberg has proved that $$ P^2 = (-1)^{2J} $$ so parity behaves much like the rotation by 180 degrees: its square changes the sign of states with an odd number of fermions. In the standard convention, electron, proton, and neutron are set to have $P=+1$: one is allowed to make three choices like that. The parity of the pion is then determined to be negative, $P=-1$, because a deuteron-pion ground state may decay into two neutrons with $L=1$ - the odd orbital momentum changes the sign of the parity, too.

Strong and electromagnetic interactions preserve parity, so one may assign parity to all hadrons, as determined from various strong and electromagnetic interactions. Lambda then turns out to have a positive parity $P=+1$ much like protons and neutrons because it's just another bound state of three quarks and the change of the quarks' identity doesn't change parity.

The neutral $\Lambda^0$ decays to a nucleon and a pion, $p+\pi^-$ or $n+\pi^0$, so a $P=+1$ state decays into one particle (nucleon) with $P=+1$ and one (pion) with $P=-1$. That violates the parity because $(+1)\neq (+1)(-1)$. It's because the decay is due to the weak interactions. Weak interactions don't preserve the parity because even the spectrum doesn't: for example, a parity-transformed partner of a left-handed neutrino - the right-handed neutrino - doesn't even exist. This asymmetry is confirmed by the weak interactions that contain two-component (fundamentally left-right asymmetric) spinors or, in the four-component spinor formalism, combinations $(1\pm \gamma_5)$ - scalars and pseudoscalars - that maximally violate the parity.

Because of these interactions, any parity-violating process that respects all the other laws is allowed, although it may be slow because it must rely on the weak interactions which are weak. The decays of the neutral Lambda baryon - and similar decays of the other Lambda particles - belong among the parity-violating ones.

The violations of parity conservation, discovered half a century ago, was a shock. But once you appreciate the simple fact that fields may be described by a 2-component spinor that prefers one chirality over the other - may describe left-handed particles without the right-handed ones - it's not so shocking. The two-component spinors work because $SL(2,C)=SO(3,1)$, locally. The fundamental representation of $SL(2,C)$ has a spin-1/2, and it also has a small enough number of components that there's no oppositely spinning particle (with the same other charges). The theory is still Lorentz-invariant.

Not only parity is violated in some processes: at very high energies, those processes become so common that it isn't even possible to define parity accurately. After all, the parity transformation of the neutrino states is ill-defined. That's why it didn't hurt that I could have redefined parity by the signs coming from the lepton and baryon numbers (which are ultimately violated at very high energies, perhaps with the exception of $B-L$) at the beginning: none of the parity operators is actually fully well-defined on the spectrum of particle states.

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