- The proton is I=1/2, I
_{3}=1/2 - The neutron is I=1/2, I
_{3}=-1/2 - The deuteron is an isosinglet, I=0
- The pions form an isotriplet, I=1, I
_{3}=pion charge. - Application of isospin symmetry to strong interactions.

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 I_{3} = ±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).
(b) π

The remaining processes can proceed through both channels:

(c) π^{-}p --> π^{-}p

(d) π^{-}p --> π^{0}n

(e) π^{+}n --> π^{+}n

(f) π^{+}n --> π^{0}p

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.)
(d) π

(e) π

(f) π

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

σ(i-->f) = |M_{if}|^{2} × (phase space)

where
M_{if} = <ψ_{f} | H | ψ_{i}>

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

ψ_{i} = ψ_{i}(1/2) + ψ_{i}(3/2), and

ψ_{f} = ψ_{f}(1/2) + ψ_{f}(3/2)

and inserting these pieces into the formula for Mψ

M_{if} = <ψ_{f}(1/2) + ψ_{f}(3/2) | H_{1} + H_{3} | ψ_{i}(1/2) + ψ_{i}(3/2)>

= <ψ_{f}(1/2) | H_{1} | ψ_{i}(1/2)> + <ψ_{f}(3/2) | H_{3} | ψ_{i}(3/2)>

= k_{1} M_{1} + k_{3} M_{3}

Where k= <ψ

= k

Now we can work out the constants, k_{1} and k_{2}, 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, M_{1}>>M_{3}, or M_{3}>>M_{1}.
The results for reactions (a) through (c) are:

M_{3}>>M_{1} σ_{a}:σ_{b}:σ_{c} = 9 : 1 : 2

M_{1}>>M_{3} σ_{a}:σ_{b}:σ_{c} = 0 : 2 : 1

The ratio of the cross sections for πM

By the way, the resonance is called the Δ(1232) and is assigned the quantum numbers I(J^{P}) = 3/2(3/2^{+}).

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

(a) p d --> ^{3}He π^{0}
(b) p d --> ^{3}H π^{+}

^{3}H and ^{3}He are an isodoublet, I=1/2, I_{3}=+1/2 (-1/2) for ^{3}He (^{3}H).
The initial state has |I=1/2, I_{3}=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 --> ^{3}Heπ^{0})/σ(pd --> ^{3}Hπ^{+}) = 1/2.

Copyright © Robert Harr 2003