# Adding angular momenta in quantum mechanics [duplicate]

When there are two spin-1/2 particles, the possible states can be grouped into a singlet and a triplet.

When there are three spin-1/2 particles, there are two possible values for the total angular momentum number, j. These two values are j=s=3/2 and j=s=1/2. How do I determine the possible states for this three particle system in the {j,m} basis, where j is the total angular momentum of the system, and m is the total z component of the angular momentum of the system? Can I treat the j=1/2 case as if there is only one spin 1/2 particle, or is it necessary to still consider all three particles for the j=1/2 case? Note that I am not considering orbital angular momentum at all. Also, for the j=3/2 case, how many particles are involved in this situation? I know that m ranges from -j to j in integer steps, so then for the j=3/2 case, the m value takes on values of -3/2, -1/2, 1/2, and 3/2.

Is there--for the j=3/2 situation of the three particle system--a tricky singlet and triplet grouping as in the case of the two spin-1/2 particle case mentioned at the beginning of this question?

In summary, I am confused about how many particles to consider for each value of j, for the system of three spin-1/2 particles, with j taking on two different values.

Thank you!