HW-7

The 3 H nucleus (a pnn bound state) and the 3 He nucleus (a ppn bound state) are known to form an isospin 1 2 doublet. Use isospin to predict the ratio of the cross sections for p + d →3H + π+ and p + d →3He + π0

Process 1:

$$P+D\to H^3+{\pi }^{+}$$

Process 2:

$$P+D\to H{e}^3+{\pi }_0$$

The magic equation: (there is another)

$$|j=l+\frac{1}{2},m⟩=\sqrt{\frac{l+m+\frac{1}{2}}{2l+1}}|m-\frac{1}{2},\frac{1}{2}⟩+\sqrt{\frac{l-m+\frac{1}{2}}{2l+1}}|m+\frac{1}{2},-\frac{1}{2}⟩$$ $$H^3=PNN=DN=-\frac{1}{2}$$ $$H{e}^3=PPN=PD=\frac{1}{2}$$

Using the above relations:

$$|H^3+{\pi }^{+}⟩\equiv |1,-\frac{1}{2}⟩,|H{e}^3,{\pi }^0⟩\equiv |0,\frac{1}{2}⟩$$

These are the possible states from this reaction

Starting with the result of the interaction.

$$\frac{1}{2}\otimes 1=\frac{3}{2}\oplus \frac{1}{2}$$

The dimensions of this:

$$2\times 3=4+2$$

We get this relation from the starting state with the total projection of the isospin for the H3 or He3 and the total isospin of the pion.

There will be 6 total states, corresponding to the 6 dimensions above.