Question

The \( \Delta^+ \) baryon with spin \( \frac{3}{2} \), at rest, decays to a proton and a pion (\( \Delta^+ \to p + \pi^0 \)). The \( \Delta^+ \) has positive intrinsic parity and \( \pi^0 \) has negative intrinsic parity. The orbital angular momentum of the proton-pion system (in integer) is:

Hint:

The orbital angular momentum in particle decays can be determined using parity conservation and spin conservation. For the decay \( \Delta^+ \to p + \pi^0 \), the orbital angular momentum must be an even integer, and \( L = 1 \) satisfies both parity and spin conservation.

Answer 1

In this decay process, the total angular momentum of the system is conserved. Let's go through the steps to determine the orbital angular momentum of the proton-pion system. Step 1: Parity Conservation.
The total parity of the system must be conserved during the decay. The parity of the initial state (\( \Delta^+ \)) is the product of the intrinsic parity of the \( \Delta^+ \) and the orbital parity of the system. Similarly, the final state (proton-pion system) has a parity equal to the product of the intrinsic parities of the proton and pion and the orbital parity. The intrinsic parity of the \( \Delta^+ \) is positive, and the intrinsic parity of the \( \pi^0 \) is negative. The proton has positive intrinsic parity. Therefore, for the total parity to be conserved, the orbital angular momentum \( L \) must satisfy the following condition: \[ P_{{total}} = P_{\Delta^+} \times P_{{orbital}} = P_p \times P_{\pi^0} \times P_{{orbital}}. \] \[ P_{{total}} = (+1) \times (-1) \times P_{{orbital}} = (-1) \times P_{{orbital}}. \] \[ P_{{total}} = (+1) { (since parity is conserved)}. \] Thus, for parity to be conserved, \( P_{{orbital}} \) must be \( +1 \), which means that the orbital angular momentum \( L \) must be an even integer. Step 2: Spin Conservation.
The \( \Delta^+ \) has spin \( \frac{3}{2} \), and the final state consists of a proton (spin \( \frac{1}{2} \)) and a pion (spin 0). The total spin \( S_{{total}} \) of the final state must be combined with the orbital angular momentum \( L \) to match the initial spin of the \( \Delta^+ \), which is \( \frac{3}{2} \). The total spin \( S_{{total}} \) of the final state can be \( \frac{1}{2} \) (proton spin) + 0 (pion spin) = \( \frac{1}{2} \), so the total angular momentum of the final system must combine the orbital angular momentum \( L \) and spin \( \frac{1}{2} \) in such a way that it matches the initial spin of the \( \Delta^+ \) (which is \( \frac{3}{2} \)). Step 3: Determining the Orbital Angular Momentum \( L \).
For the total spin to be \( \frac{3}{2} \), the orbital angular momentum \( L \) must be 1, as the total angular momentum is given by: \[ J_{{total}} = L + S_{{total}}, \] where \( J_{{total}} \) is the total angular momentum. Since \( L = 1 \) and \( S_{{total}} = \frac{1}{2} \), the total angular momentum \( J_{{total}} \) can be \( \frac{3}{2} \), satisfying the condition for spin conservation. Thus, the orbital angular momentum \( L \) of the proton-pion system is \( 1 \).