Definition of 'good quantum number' after parity violation

[metroplex021]

Hi folks. I was always under the impression that the 'good quantum numbers' that we use to classify a particle species were always the eigenvalues of operators that commute with Hamiltonian governing that species. But it just struck me that weakly interacting particles have definite parity and yet the parity operator of course does not commute with the weak Hamiltonian! Are there some considerations I'm missing which means that this isn't a counterexample to the idea that the GQNs of particles are always the eigenvalues of those operators that commute with those particles' Hamiltonians? Thanks!

[Bill_K]

How good is "good"? Just "pretty good", or "really really good"? Parity is conserved by the strong interaction, and therefore for practical purposes it is a good quantum number for nuclei. However nucleons do feel the effect of the weak interaction, and it is possible to detect parity violations in nuclei if you try hard enough. In other words, the true eigenstate of a nucleus may be an admixture of parity states, |Ψ> = |Ψeven> + ε |Ψodd> where ε is quite small. Typically ε ~ 10-6.

A more interesting example is the neutral K meson. They occur in collisions produced by the strong interaction Hamiltonian, such as π- p → K0 Λ and π+ p → K0 K+ Λ (sorry, I'm forced to denote an antiparticle here by an underscore!) The good quantum numbers for K0 and K0 are the good quantum numbers for the strong interactions, namely isospin and hypercharge, I3 and Y. But when K0's decay, the strong interaction does not come into play, and the effective Hamiltonian is the weak Hamiltonian, which does not conserve I3 and Y. It also does not conserve parity P. These are no longer good quantum numbers.

Originally it was thought that the weak Hamiltonian at least conserved the combined operation CP, where C is charge conjugation, CP|K0> = | K0>. And to support this notion, K0's appear to be a mixture of two different eigenstates, KS0 and KL0 with different masses and different lifetimes:

KS0 = (|K0> + |K0>)/√2
KL0 = (|K0> - |K0>)/√2

KS0 has CP = +1, while KL0 has CP = -1. And so CP appears to be a "good" quantum number.

However it was later found that the weak interactions even violate CP conservation, so CP is not a good quantum number after all. The true eigenstates in the decay are really KS0 with a small admixture (about 10-3) of KL0, and vice versa.

[metroplex021]

Thank you for that really full answer - it's very interesting & helpful. I guess that fundamentally what I'm confused about is this. If you have a charged hadron - a K+ say - then it participates in all of the strong, weak and electromagnetic interactions. Now, if you look up a table of the known particles, it will tell you that the K+ has a definite isospin (Iz=1/2). But since the full Hamiltonian of the K+ consists of the sum of the strong and electroweak Hamiltonians, and since the former commutes with strong isospin but the latter does not, the complete Hamiltonian of the K+ presumably does not commute with strong isospin either. Hence the isospin of the kaon is not well-defined along with its mass, assuming that we have

(Hs + Hew) |K+> = m |K+>

in the K+'s rest frame, where m is the mass of the kaon. So why do we use Iz=1/2 as one of the K+'s defining properties, given that it isn't well defined even with its own rest energy? It seems very odd to me now that we routinely classify particles using quantum numbers that aren't even well-defined along with their total governing Hamiltonians. (That damned quantum mechanics!!)