Weak interactions

Question 1

Parity and charge conjugation violations

What’s the proof of the weak interaction violating parity and charge conjugation?

Answer 1

Two experiments: decay of polarised cobalt-60, decay of polarised muons

Parity:

Cobalt decay:

• the left- and right-handed neutrinos don’t appear with the same probability (fundamental process: decay of a neutron)
• the spins of the products of the decay are aligned: the electron shouldn’t have a preferred direction compared to the spin direction
• preferred direction of electrons when emitted: opposite to the nuclear spin —» violation of parity
• explanation: right-handedness (helicity = 1) of the anti-neutrino

Muon decay:

• the direction of the emitted electron depends on the angle of it compared to the muon polarisation
• preferred direction of the electron: opposite to muon spin

Charge conjugation:

• for antinichel and antimuons: positrons would be emitted in the preferred direction of the spin of the nucleus/muon
• reason: helicity = –1 for the electron neutrino
• this is different from the electron —» violation of charge conjugation

Question 2

Neutral kaon oscillations and CP violation

What happens in the decay of kaons if CP is assumed to be a symmetry?

Answer 2

Let us assume that CP is a symmetry of nature.

Decay of kaons:

• the K0 and ¯K0 both decay into two pions, so they can oscillate into each other
• due to the decay, the time evolution leads out of the kaon subspace
• H(eff) (Weisskopf-Wigner approximation): complex eigenvalues (real parts: masses, imaginary parts: decay width) + K(S) and K(L) eigenvectors (not necessarily orthogonal)
• two types of kaons: short and long (reffering to their lifetimes) whose eigenstates don’t coincide with the strong int. eigenstates
• [CP, H(eff)] = 0 for now, so K(S) and K(L) should be identifiable as K1 and K2 CP eigenstates
• K1 = K(S), short component decays first (into two pions), K2 = K(L), long component decays second (into three pions)

Question 3

Neutral kaon oscillations and CP violation

How do kaons still violate CP symmetry?

Answer 3

Violation of CP:

• if CP isn’t a symmetry, the K1 and K2 states mix: K(S) and K(L) will be linearcombinations
• two pions decays of the K(L) state were observed
• indirect CP violation: physical states aren’t CP eigenstates
• direct CP violation: a CP = –1 (K2) eigenstate can decay into two pions as well

The Standard Model requires an at least 3x3 CKM matrix for CP violation and thus 3 families of quarks. There’s a CP-violating phase in the matrix.