For two isospin j particles, itrl ' vzl in triplet and singlet states respectively are
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particle of spin 3/2, at rest in the laboratory, disintegrates into two particles, one of spin ½ and one of spin 0.
(a) What values are possible for the relative orbital angular momentum of the two particles? Show that there is only one possible value if the parity of the relative orbital state is fixed.
(b) Assume the decaying particle is initially in the eigenstate of Sz with eigenvalue  Is it possible to determine the parity of the final state by measuring the probabilities of finding the spin ½ particle either in the |+> or |-> state?
Solution:
(a) The initial value of j is j=3/2. We need the final value to be j=3/2, since angular momentum of an isolated system is conserved. Let us call the initial particle particle a, the spin ½ paticle particle b, and the spin 0 particle particle c. Then the total spin of the two particles in the final state is S=Sb+Sc=Sb. Therefore the spin quantum number is s=½. The possible values for the orbital angular momentum quantum number are l=1 and l=2. (j=l+s, l-s; j=l+½, l-½, implies l=1 or l=2.) The parity of the orbital state is (-1)l. If the parity is odd, we have l=1, if the parity is even, we have l=2.
(b) Let P(+) be the probabilities of finding the spin ½ particle in the |+>state. Do states with different l give us different P(+)? Assume that the final state has l=1. It may be written as |j1,j2;j,m>=|1,½;3/2,m>. To find P(+) we need to write it as a linear combination of |j1,j2;m1,m2>=|1,½;ml,ms>.
m can take on the values 3/2, ½, -½, -3/2. Using a table of Clebsch-Gordan coefficients we find




Now assume that the final state has l=2. It may be written as |j1,j2;j,m>=|2,½;3/2,m>.
We may write it as a linear combination of |j1,j2;m1,m2>=|2,½;ml,ms>.
m can take on the values 3/2, ½, -½, -3/2. We find




Yes, it is possible to determine the parity of the final state by measuring the probabilities of finding the spin ½ particle either in the |+> or |-> state
(a) What values are possible for the relative orbital angular momentum of the two particles? Show that there is only one possible value if the parity of the relative orbital state is fixed.
(b) Assume the decaying particle is initially in the eigenstate of Sz with eigenvalue  Is it possible to determine the parity of the final state by measuring the probabilities of finding the spin ½ particle either in the |+> or |-> state?
Solution:
(a) The initial value of j is j=3/2. We need the final value to be j=3/2, since angular momentum of an isolated system is conserved. Let us call the initial particle particle a, the spin ½ paticle particle b, and the spin 0 particle particle c. Then the total spin of the two particles in the final state is S=Sb+Sc=Sb. Therefore the spin quantum number is s=½. The possible values for the orbital angular momentum quantum number are l=1 and l=2. (j=l+s, l-s; j=l+½, l-½, implies l=1 or l=2.) The parity of the orbital state is (-1)l. If the parity is odd, we have l=1, if the parity is even, we have l=2.
(b) Let P(+) be the probabilities of finding the spin ½ particle in the |+>state. Do states with different l give us different P(+)? Assume that the final state has l=1. It may be written as |j1,j2;j,m>=|1,½;3/2,m>. To find P(+) we need to write it as a linear combination of |j1,j2;m1,m2>=|1,½;ml,ms>.
m can take on the values 3/2, ½, -½, -3/2. Using a table of Clebsch-Gordan coefficients we find




Now assume that the final state has l=2. It may be written as |j1,j2;j,m>=|2,½;3/2,m>.
We may write it as a linear combination of |j1,j2;m1,m2>=|2,½;ml,ms>.
m can take on the values 3/2, ½, -½, -3/2. We find




Yes, it is possible to determine the parity of the final state by measuring the probabilities of finding the spin ½ particle either in the |+> or |-> state
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