This problem deals with the decay of an unstable Particle X to two daughter part

Question

This problem deals with the decay of an unstable Particle X to two daughter particles A and B. In the decay process X - > AB, total angular momentum is conserved. In the X rest frame, the total angular momentum J = Sx is given just by the spin of Particle X. After the decay, the total angular momentum receives three contributions: J = SA + SB + L, Where SA the spin of particle A, SB is the spin of particle B, and L is the relative orbital angular momentum between A and B. Because angular momentum is conserved in this decay, if the initial state is an eigenstate of J^2 and Jx, then the final state is also an eigenstate with the same eigenvalues. (a) Consider the case where X is spin-0 and both A and B are spin-1/2 (i.e.sx = 0 sA = 1/2, sB = 1/2). What values of the orbital angular momentum l are allowed, consistent with angular momentum conservation? (b) Repeat the above problem for sX = 3/2, sA = 1/2, and sB = 1.

Explanation / Answer

as Sx=0

and Sa=0.5 and Sb=0.5

using conservation of angular moemntum:

Sx=Sa+Sb+L

L=-1

b)Sx=Sa+Sb+L

Sx=3/2

Sa=1/2

Sb=1

then L=Sx-Sa-Sb=1.5-0.5-1=0

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