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GATE 2023 PH – Question 58

Nuclear and Particle Physics · conservation laws, isospin symmetry, charge conjugation, parity and time-reversal invariance. · 2 marks · Multiple select

A slow moving π− particle is captured by a deuteron (d) and this reaction produces two neutrons (n) in the final state, i.e., π−+ d→n+ n. Neutron and deuteron have even intrinsic parities, whereas π− has odd intrinsic parity. L and S are the orbital and spin angular momenta, respectively of the system of two neutrons. Which of the following statements regarding the final two-neutron state is(are) CORRECT?

  1. It has odd parity
  2. L+ S is odd
  3. L= 1, S= 1
  4. L= 2, S= 0

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Show answer and explanation

Correct answer: (A) It has odd parity; (C) L= 1, S= 1

Explanation

**Conservation laws in $\pi^-+d\to n+n$** for a slow pion captured from an atomic $s$-orbit (orbital angular momentum 0).

**Total angular momentum.** The pion has spin 0 and the deuteron has spin 1, so $J=1$ initially. The final two-neutron state must have $J=1$.

**Parity.** The intrinsic parities are: pion $-$, deuteron $+$ (as the problem states for the neutron and deuteron). With zero orbital angular momentum in the initial state, the initial parity is $-1$. So the final nn state has **odd parity** (A is true): $(-1)^L=-1$, so $L$ is odd.

**Pauli principle.** The two neutrons are identical fermions with isospin $T=1$ (symmetric), so their space-spin wave function must be antisymmetric: $(-1)^{L+S+1}=-1$, so $L+S$ must be **even**.

With $L$ odd, $S$ must be odd too, so $S=1$. For $J=1$ with $S=1$ and odd $L$: $L=1$.

Answer **A and C**.