Invariance Principles and Conservation Laws

Recall from last lecture:

CP violation and T violation

While the combination CPT is thought to be absolutely invariant, no other combination seems to have that property. C and P are violated in the weak interaction.

In the weak interaction, the W± and Z0 bosons couple to left-handed neutrinos and right-handed anti-neutrinos only. For massless particles, and neutrinos are for this discussion sufficiently close to be called massless, handedness is conserved. (For a massless particle, like the photon, if the spin points along the direction of motion in one reference frame, it points along the direction of motion in all reference frames.) The charge conjugate of a left-handed neutrino is a left-handed anti-neutrino (charge conjugation doesn't affect the handedness). But the left-handed anti-neutrino doesn't couple to W's or Z's, implying that the weak interaction is not symmetric under charge conjugation.

Under a parity transformation, a left-handed neutrino transforms into a right-handed neutrino. Again, the weak force doesn't couple to right-handed neutrinos implying that it is not symmetric under parity transformation. The weak interaction is observed to violate both C and P.

The combination of CP transforms a left-handed neutrino into a right-handed anti-neutrino which does couple to W's and Z's. The combination CP was believed to be conserved by the weak interaction, until:

CP violation has now been observed in decays of B-mesons as well. We believe the CP violation in kaons and B-mesons arises from the properties of the CKM matrix. Experimenters are now testing this hypothesis. It is predicted that this CP violation is insufficient to account for the matter--anti-matter asymmetry in the universe. The additional source of CP violation, if it exists, is unknown.

Baryon and Lepton Conservation

To the limit that present experiments are able to test, baryon number and lepton number appear to be absolutely conserved quantities. Yet again, the matter--anti-matter asymmetry in the universe screams for a process that violates one or both of these. Several so-called grand unified theories (GUT's) predict that the proton will decay with a lifetime in the ballpark of 1035years! Large Cerenkov water experiments were built to try to observe proton decay. No signal has appeared, and the experiments have placed a lower limit on the proton lifetime of 1033 years (how many m3 of water is needed to have 1033 protons?).

It is interesting to note that while these experiments have failed to observe proton decay, they have seen evidence for neutrino oscillations, garnering a Nobel prize and creating a whole new subfield of particle physics in the process!

Isospin Symmetry

Heisenberg observed that the proton and neutron had such similar mass, and similar properties as regards the strong force, that they might be considered as different states of a single particle dubbed the nucleon. A new quantum number was needed to denote the 2 possible states of a nucleon as either a proton or a neutron. In analogy with the two possible z-spin states of a spin ½ particle, this new quantum number was called isospin. For the proton and neutron, I=½ and the z component of isospin called I3 is I3=+½ for the proton and I3=-½ for the neutron. This is not a "spin" in physical space with a corresponding angular momentum. This is a spin in an abstract space with no associated angular momentum.

Isospin symmetry helps to explain the existence of mirror nuclei, nuclei that differ in the exchange of a neutron and proton. For instance, I will shortly do an example involving tritons (3H) and helium-3 (3He) nuclei.

In the context of the quark model, isospin symmetry arises from the nearly identical masses of the up and down quarks. A proton becomes a neutron when one of the up quarks is replace by a down quark. Therefore, we assign the up quark I3=+½, and the down quark I3=-½.

We can extend isospin symmetry to other particles, for instance, the pions. Being mesons, pions are a quark and an anti-quark pair constructed using only up and down quarks. By analogy with the combination of two spin-½ particles, we can construct 4 states from the combination of two isospin-½ particles. (Careful, there is a phase convention that enters s.t. the I3=0 states come out with the opposite signs, see Sec. 4.6.)

|½,½> = |I=1, I3=1> : u dbar = π+
(|½,-½>-|-½,½>)/sqrt{2} = |I=1, I3=0> : (u ubar - d dbar)/sqrt{2} = π0
|-½,-½> = |I=1, I3=-1> : d ubar = π-
(|½,-½>+|-½,½>)/sqrt{2} = |I=0, I3=0> : (u ubar + d dbar)/sqrt{2} = η
The mass of π± is 140MeV while the mass of the π0 is 135MeV, different by a few MeV just like the masses of the proton and neutron. The η is mentioned in section 4.6 of Perkins. It is a singlet state in isospin, and transforms into itself upon exchange of u and d quarks. It's mass is about 550MeV, significantly greater than the masses of the pions. There is a phenomenological model called the chiral quark model that attempts to explain the mass of the pion in terms of chiral symmetry breaking. I won't have time to go into that in this course.

Isospin in the Two-Nucleon and the Pion-Nucleon Systems

Now we apply isospin to some situations. The first is the deuteron, a nucleus containing a proton and a neutron.

The deuteron consists of two nucleons, so let's consider the possible isospin combinations of two I=½ nucleons. Again we have four possible states, denoting them as being either p (I3=+½) or n (I3=-½) and labeling the particles as (1) or (2):

χ(1,1) = p(1)p(2)
χ(1,0) = [p(1)n(2) + n(1)p(2)]/sqrt{2}
χ(1,-1) = n(1)n(2)
χ(0,0) = [p(1)n(2) - n(1)p(2)]/sqrt{2}
The first state are members of an I=1 triplet, and the last state is an I=0 singlet. Bound nuclei consisting of two protons or two neutrons don't exist, so it is natural to associate the singlet state with the deuteron.

We can further justify this with an analysis of the symmetry of the wave function.

Copyright © Robert Harr 2003