Hadrons: quantum numbes and excited states


Chapter 5

Recall from last lecture:

Hadron Quantum Numbers

Hadrons (an assemblage of quarks bound by the strong force) are characterized by their mass and a set of quantum numbers which we divide into those that characterize space-time symmetries, and those that describe internal degrees of freedom. The space-time symmetries are spin, parity, and C-parity, denoted by JP or JPC, as appropriate. The internal degrees of freedom:

For the quarks, JP=1/2+, and for the anti-quarks JP=1/2-. The remaining quantum numbers for the quarks are:
We can equally well define the quantum numbers in terms of some well established hadrons, and then work out the quantum numbers for all the remaining hadrons from the interactions that are measured for them. In practice, this is what is actually done to determine the quantum numbers for a newly discovered hadron.

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.

Isospin of the deuteron

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. The wave function for a non-relativistic two nucleon state (a state of two identical fermions) can be decomposed into a spatial part, a spin part, and an isospin part:

ψ = φ(r) α(S) χ(I)
In a relativistic state, the orbital and spin angular momenta cannot be separated -- one must use the full Dirac equation. Recall that the deuteron has spin (total angular momentum) J=1, and it is known that L=0, therefore S=1. The parity of the spatial part is (-1)L = 1, and the spin 1 state has parity 1 also. The overall wave function must be antisymmetric (identical fermions), therefore the deuteron must be in the I=0 state, called an isosinglet (isospin singlet state).

Application to nucleon-nucleon interactions

Now let's apply isospin to some strong interactions. It turns out that we can break a reaction down into separate isospin components (channels), and we find that, generally speaking, reactions with the same I but different I3 have the same amplitude (matrix element) at a given energy. Consider the following reactions

(1) pp --> dπ+
(2) pn --> dπ0
The reactions differ in that the first has I3 = 1, while the second has I3 = 0. The initial state of the first reaction must be pure I=1, while the second is a 50-50 mixture of I=1 and I=0. The final state in each case is the combination of an I=0 deuteron with an I=1 pion yielding an I=1 state. Therefore, the reaction proceeds through an I=1 channel, and we expect that the cross sections differ because of the mixed initial state of reaction 2:
σ(1)/σ(2) = 1/2
which is observed. (This is not a high precision prediction since electromagnetic effects which break isospin symmetry are ignored. It should be correct at the level of a few percent.)

Pion-nucleon scattering

Next consider the strong scattering of a pion and a nucleon to a pion and a nucleon. The initial and final states are by the rules for combining angular momenta. (The above notation means if I combine states of isospin 1 and 1/2, then I'll get states of total isospin 3/2 or 1/2. The exact amounts of each are determined by the appropriate Clebsch-Gordon coefficients.)

There are 3×2 different scattering processes (listed below). The hypothesis of isospin symmetry leads to the conclusion that these 6 processes can be described using only two amplitudes, one for the I=3/2 channel and one for the I=1/2 channel.

Two of the processes are I3 = ±3/2 and therefore only occur through the I=3/2 channel:

(a) π+p --> π+p , and
(b) π-n --> π-n
For identical energies, these process will have identical cross sections (to the precision allowed by electromagnetic breaking of isospin).

The remaining processes can proceed through both channels:

(c) π-p --> π-p
(d) π-p --> π0n
(e) π+n --> π+n
(f) π+n --> π0p
We use a table of Clebsch-Gordon coefficients to find the relative amounts of I=1/2 and I=3/2 in the initial and final states of these reactions. (This will be done interactively in lecture.)

Recall that the cross section depends on what I called phase space factors and the square of the matrix element:

σ(i-->f) = |Mif|2 × (phase space)
Mif = <ψf | H | ψi>
Our hypothesis is that these reactions are dominated by strong interactions, and that the strong interaction conserves isospin. Therefore, we can separate the total Hamiltonian into parts that only operates between states of equal isospin, H = H1 + H3, where H1 operates between states with I=1/2 and H3 operates between states with I=3/2. When the initial or final state do not have the correct isospin for the Hamiltonian, the result is zero.

Writing the wave functions as separate I=1/2 and I=3/2 parts,

ψi = ψi(1/2) + ψi(3/2), and
ψf = ψf(1/2) + ψf(3/2)
and inserting these pieces into the formula for Mif we find
Mif = <ψf(1/2) + ψf(3/2) | H1 + H3 | ψi(1/2) + ψi(3/2)>
= <ψf(1/2) | H1 | ψi(1/2)> + <ψf(3/2) | H3 | ψi(3/2)>
= k1 M1 + k3 M3
Where k1 and k3 are constants that arise from the Clebsch-Gordon coefficients, and M1 and M3 are matrix elements for the I=1/2 and I=3/2 channels.

Now we can work out the constants, k1 and k2, for reactions (a) through (f). Again, this will be done interactively in class.

Then consider the limiting cases where one or the other amplitude dominates, M1>>M3, or M3>>M1. The results for reactions (a) through (c) are:

M3>>M1 σabc = 9 : 1 : 2
M1>>M3 σabc = 0 : 2 : 1
The ratio of the cross sections for π+p, and π-p are displayed in Fig. 3.8 of Perkins. The ratios of the cross sections vary depending on the CM energy, implying that the relative strengths of the I=1/2 and I=3/2 interactions vary as well. In particular, at the first resonance (the peak at about 1.2GeV), the ratio of 195/(65-15) = 3.9, not so far from the 9/2=4.5 prediction for a dominance of the I=3/2 channel for this particular resonance.

By the way, the resonance is called the Δ(1232) and is assigned the quantum numbers I(JP) = 3/2(3/2+).

Copyright © Robert Harr 2005