Subtopic: Orbital Angular Momentum & Spherical Harmonics
Question Statement:
The Δ+ baryon with spin
3
2, at rest, decays to a proton and a pion (Δ+ → p + 𝜋0).
The Δ+ has positive intrinsic parity and 𝜋0 has negative intrinsic parity. The orbital
angular momentum of the proton-pion system (in integer) is _____
Official Answer Key:
1 to 1
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.35)**
**1. Particle Physics Symmetries (Angular Momentum & Parity)**:
- Decay: $\Delta^+ \to p + \pi^0$
- **Initial State $\Delta^+$**:
- Spin: $J_i = 3/2$
- Intrinsic parity: $P_i = +1$
- **Final State $(p + \pi^0)$**:
- Proton: spin $s_p = 1/2$, intrinsic parity $P_p = +1$
- Pion: spin $s_\pi = 0$, intrinsic parity $P_\pi = -1$
- Relative orbital angular momentum between $p$ and $\pi^0$: $l$
**2. Conservation Laws**:
- **Step 1: Parity Conservation in Strong Decays**:
$P_f = P_p \cdot P_\pi \cdot (-1)^l = (+1)(-1)(-1)^l = -(-1)^l = (-1)^{l+1}$
$P_i = P_f \implies +1 = (-1)^{l+1} \implies l+1 \text{ is even} \implies l \text{ must be odd} \ (l = 1, 3, \dots)$
- **Step 2: Total Angular Momentum Conservation**:
$\vec{J}_f = \vec{s}_p + \vec{s}_\pi + \vec{l} = \vec{l} + \frac{1}{2}$
$J_f = \left|l - \frac{1}{2}\right|, \dots, l + \frac{1}{2}$
- For $l = 1$: $J_f \in \{1 - 1/2, 1 + 1/2\} = \{1/2, 3/2\}$. This perfectly matches initial spin $J_i = 3/2$.
- For $l = 3$: $J_f \in \{5/2, 7/2\} \ne 3/2$.
- Hence, the relative orbital angular momentum of the decay products is uniquely **$l = 1$** ($p$-wave decay).
**3. Official Verification**:
The verified official IIT answer is **1** (Numerical range: `1 to 1`).
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