# Invariance Principles and Conservation Laws

### Recall from last lecture:

• The proton is I=1/2, I3=1/2
• The neutron is I=1/2, I3=-1/2
• The deuteron is an isosinglet, I=0
• The pions form an isotriplet, I=1, I3=pion charge.
• Application of isospin symmetry to strong interactions.

## Isospin Symmetry

#### 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)
where
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+).

#### Example: 3H and 3He

At a given CM energy, what is the ratio of cross-sections for the reactions

(a) p d --> 3He π0 (b) p d --> 3H π+

3H and 3He are an isodoublet, I=1/2, I3=+1/2 (-1/2) for 3He (3H). The initial state has |I=1/2, I3=1/2>. The final states are sqrt{2/3}|3/2,1/2> - sqrt{1/3}|1/2,1/2> for (a) and sqrt{1/3}|3/2,1/2> + sqrt{2/3}|1/2,1/2> for (b). Since the reaction can proceed through the I=1/2 channel only, the ratio of cross sections is σ(pd --> 3Heπ0)/σ(pd --> 3+) = 1/2.