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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 $m_j$).

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    $\begingroup$ The tensor product is distributive with respect to the direct sum. $\endgroup$ – Ryan Unger Feb 13 '15 at 13:14
  • $\begingroup$ somewhat related physics.stackexchange.com/q/29443 though not a duplicate $\endgroup$ – Rob Jeffries Feb 13 '15 at 14:07
  • $\begingroup$ These things are always just successive applications of Clebsch-Gordan coefficients, using that a tensor product distributes over direct sums as 0celo7 indicates. $\endgroup$ – ACuriousMind Feb 13 '15 at 17:22
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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.

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  • $\begingroup$ In the case of three 1/2 particles we have $D^{3/2}\oplus D^{1/2}\oplus D^{1/2} $... How can I then use the Clebsch-Gordan coefficients to construct the final wave function ? $\endgroup$ – Luka8281 Jun 20 '17 at 12:11

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