The dominant decay modes of the K mesons are the two-pion channels K^0 rightarro

Question

The dominant decay modes of the K mesons are the two-pion channels K^0 rightarrow pi+ pi-, K^0 rightarrow pi^0 pi^0, K^plusminus rightarrow pi^plusminus rightarrow pi^plusminus pi^0 Using the fact that prions are spineless and are therefore bosons, show that the final two-pion state resulting from the decay of a kaon must have either isospin zero or two. Since udo carries I = 1 while su carries I = 1/2 the strangeness-changing non-leptonic Hamiltonian H_w G/squareroot suud involves a linear combination of I = 1/2 and I = 3/2 components, leading to K rightarrow 2 pi decay amplitudes a_1 and a_3. Show that the decay amplitude can be written as A (K^+ rightarrow pi+ pi0) = 3/2 squareroot 2 a_3 A (K^0 rightarrow pi+ pi-) = -a_1 + 1/2 a3 A (K^0 rightarrow pi^0 pi^0 = a_1 + a_3 Compare this parametrization with the experimental lifetimes for the following channels tau K^0 pi+ pi- tau K^0 rightarrow pi^0 pi^0

Explanation / Answer

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 = +
(|½,-½>-|-½,½>)/sqrt{2} = |I=1, I3=0> : (u ubar - d dbar)/sqrt{2} = 0
|-½,-½> = |I=1, I3=-1> : d ubar = -
(|½,-½>+|-½,½>)/sqrt{2} = |I=0, I3=0> : (u ubar + d dbar)/sqrt{2} =

The mass of ± is 140MeV while the mass of the 0 is 135MeV, different by a few MeV just like the masses of the proton and neutron. The 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.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.