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dipstik
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Homework Statement
A pion is a spin-0 particle with a negative unit charge. A neutron is a spin-1/2
particle with no charge. A proton is a spin-1/2 particle with positive unit charge.
One can construct an unstable version of the hydrogen atom where the nucleus
contains both a proton and a neutron, but uses a pion instead of an electron. (No
electrons in this problem!) This is called a "pionic atom".
A.) In 1951, it was observed that the ground state of a spin-1 "pionic atom" could
spontaneously decay into two neutrons (and nothing else). (The spin-1 refers to
the original total spin of the entire atom, including the original nucleus.)
Given that total angular momentum must be conserved during this decay process, list
ALL of the allowable final values for |L,S>, where S is the total spin of the 2
resulting neutrons and L is the orbital angular momentum of the two resulting
neutrons (orbiting around each other). Ignore symmetry constraints.
B) Now use symmetry arguments to determine which of your answers from A) will
actually happen. Hint: the symmetry of two neutrons with an orbital angular
momentum is (-1)^L. Make sure you show your reasoning.
Homework Equations
J=L+S
|s1-s2|<=s1+s2<=s1+s2 with integer steps
Clebsch–Gordan Table for 1/2 x 1/2 (maybe)
fermions only exist in antisymmetric states: (-1)^L=-1 for fermions
The Attempt at a Solution
A) The initial total angular momentum J0=L0+S0, and it is given that the nucleus is spin 1 (S=1) and the pion is spin 0 and L=0 (ground state). This implies that J0=1. Because this is conserved we require that the final total angular momentum is Jf=1. This requires Lf+Sf=1. This is possible by |1,0> or |0,1> final |L,S> states (the |1/2,1/2> is not allowed because the neutrons are 1/2 which gives S=0,1 only)
B) Only the L=1 state obeys the (-1)^L=-1 fermionic condition. This implies that the |0,1> is not allowed.