# Non-relativistic decay of the $\Lambda^{0}$

The $\Lambda^{0}$ baryon has two possible decays, the first is $\Lambda^{0} \to \pi^{0} + n^{0}$ and the second $\Lambda^{0} \to \pi^{-} + p^{+}$.

I've been asked to determine the ratio of protons to neutrons observed in the decay.

Properties: $$\begin{array}{cccc} \text{Particle} & \text{Spin} & \text{Charge} & \text{total isospin} \\\hline \Lambda^0 & \frac12 & 0 & 0 \\ \rm n^0 & \frac12 & 0 & \frac12 \\ \rm p^+ & \frac12 & +1 & \frac12 \\ \pi^- & 0 & -1 & 1 \\ \pi^0 & 0 & 0 & 1 \\ \end{array}$$

Charge and angular momentum are conserved in the decay, but the total isospin changes by $\frac{1}{2}$. I'm supposed to ignore all relativistic effects.

How do I even begin with this question? I don't understand how these conservations lead to the number of particles observed in the decay...

• – Qmechanic Nov 16 '16 at 7:36

Isospin gets its name because the isospin of a composite system is related to its components using the same rules as for combining spin angular momenta. The strong force is, to good approximation, blind to the $z$-component of isospin (a.k.a. electric charge); your two final states are the components of a composite state whose isospin is well-defined. (You even say what it is.) What are its Clebsch-Gordan coefficients?
• As I understand it, is this what you are telling me? The $\Lambda^{0}$ starts out with isospin $0$ and decays into a superposition of states whose (z-isospin values must add up to $\frac{1}{2}$). So in terms of isospins: – QuantumEyedea Nov 16 '16 at 4:19
• $\left| 0, 0 \right> \mapsto \sqrt{1/3} \left|1,-1 \right>\otimes \left|1/2,1/2\right> + \sqrt{2/3} \left|1,1\right>\otimes\left|1/2,-1/2 \right>$ – QuantumEyedea Nov 16 '16 at 4:22