Hadrons: quantum numbes and excited states


Chapter 5

Recall from last lecture:

The sigma (Σ) baryons

Baryons with nonzero strangeness are called hyperons. The Σ baryons are produced in strong interactions, such as

K- p --> π- Σ+.
The charged Σ's decay weakly (with a lifetime of order 10-10s), while the neutral Σ0 decays electromagnetically (with a lifetime of order 10-20s). From the reaction above, we see that the Σ+ must have baryon number B=1, and strangeness S=-1. Charm, beauty, and truth quantum numbers are zero.

We can apply conservation of isospin to the production reaction, but not to the weak or electromagnetic decays. The K-=subar and therefore has I,I3=1/2, -1/2. The proton is I,I3=1/2, 1/2 while the π- is I,I3=1, -1. The production reaction has

| K-; 1/2, -1/2 > | p; 1/2, 1/2 > --> | π-; 1, -1 > | Σ+; I, 1 >.
where the I3 of the Σ+ is clearly 1. The value of I must be greater than or equal to 1. Excluding the possibility that the Σ's are exotic, the only allowed value for I consistent with the above requirement is 1. The fact that two additional states (and only two) with the quantum numbers consistent with the Σ+ supports this conclusion.

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+).

Exotic Hadrons

We can assign all (or almost all) of the known hadrons to one of the two classes of mesons and baryons. That is, the quantum numbers derived for the known hadrons are consistent with quantum numbers possible for mesons or baryons. But not all possible combinations of quantum numbers can be assigned to mesons and baryons. Particles with such quantum numbers are termed exotics, and have been searched for since the inception of the quark model. Recent claims for evidence of several exotic states has sparked renewed interest in this area.

An example of a set of exotic quantum numbers is:

Q=1, B=1, S=1, C=~B=T=0
This state is exotic because all baryons have zero or negative stangeness -- anti-baryons can have positive strangeness. Baryons consist of 3 quarks. The presence of strangeness implies that the baryon contains one or more strange quarks, but strange quarks carry strangeness -1; strange anti-quarks carry strangeness +1. The above quantum numbers are conjectured to belong to an exotic hadron called the Θ+ with quark content uudd-sbar. This state looks like the union of a neutron and K+ or proton and anti-K0.

Copyright © Robert Harr 2005