How do you add angular momentum of three or more particles in quantum mechanics?

I'm trying to find some information on how to add the angular momentum of three or more particles, but all the sources I look at deal with only two. In this case I understand that if the angular momentum numbers of the two particles are $j_1$ and $j_2$, then the possible total angular momentum numbers are $J=(j_1+j_2),(j_1+j_2-1)+...+|j_1-j_2|$. However, I don't see how to combine this to three particles.

For example, if I have three protons in a $1d_{5/2}$ nuclear energy level (for example), then the protons all have angular momentum $j=5/2$. However, how do I then find the possible total angular momentum of the state? I appreciate that the particles cannot occupy the same state, and hence must have different $m_j$ values which range from $5/2,3/2...-5/2$, and then the $m_j$ is the sum of these, but then how could this be used to find the possible total angular momentum of the state (not just the total $m_j$).

Answer

For every half-integer $j=n/2,n\in\mathbb{Z}$, there is an irreducible representation of $SU(2)$

$$D^j=\exp(-i\vec\theta\cdot\vec J^{(j)})$$ in which the three generators $J_i^{(j)}$ are $(2j+1)\times(2j+1)$ square hermitian matrices. As you probably know, $D^j$ describes states with angular momentum $-j$ to $j$ in integer steps. Given two particles with angular momenta $j$ and $\ell$, we write the total angular momentum of the system as the tensor product $D^j\otimes D^\ell$. We have the standard result $$D^j\otimes D^\ell=\bigoplus_{k=|j-\ell|}^{j+\ell}D^k$$ where on the right it is (conventionally) understood that we write the rotation matrices in order of decreasing angular momentum. Consider, for instance, a meson. This is a composite particle of two spin-half particles. In its ground state, the meson spin representation is $$D^{1/2}\otimes D^{1/2}=D^1\oplus D^0$$ We can thus predict that there are spin-1 and spin-0 mesons, which has been experimentally verified.

Suppose then we wish to find the angular momentum of a baryon. We need $D^{1/2}\otimes D^{1/2}\otimes D^{1/2}$. Convince yourself of the following: Given three matrices $A,B,C$ we have

$$A\otimes(B\oplus C)=A\otimes B\oplus A\otimes C$$ Using this, we have $$D^{1/2}\otimes D^{1/2}\otimes D^{1/2}=D^{1/2}\otimes(D^1\oplus D^0)=D^{1/2}\otimes D^1\oplus D^{1/2}\otimes D^0=D^{3/2}\oplus D^{1/2}\oplus D^{1/2}$$ Again, we find baryons with identical quark content in both $3/2$ and $1/2$ states. (There is a slight technicality with the second $D^{1/2}$ involving the Pauli principle.)

The obvious generalization of this is

$$D^m\otimes D^j\otimes D^\ell=D^m\otimes\left(\bigoplus_{k=|j-\ell|}^{j+\ell}D^k\right)=\bigoplus_{k=|j-\ell|}^{j+\ell}D^m\otimes D^k=\bigoplus_{k=|j-\ell|}^{j+\ell}\bigoplus_{n=|m-k|}^{m+k}D^n$$ and so on for more particles. From here you can use the Clebsch-Gordan method to construct the actual state vectors for your system.