Suppose I have three spin $s$ particles. What are the possible spins of a symmetric combination of these three particles? Will one of the states always have spin $3s$?

Perhaps the above question is too general to answer, so maybe it's easier in small examples. I think the answer for spin $1/2$ is that the resulting particles fit into a spin $3/2$ quadruplet and a spin $1/2$ doublet. But I'm not sure how to derive this result for the spin $1$ or $3/2$ cases. What happens when the spins of the particles are something other than $1/2$, like $1$ or $3/2$?


There will usually be more than just the sum in the symmetric part.

For instance combining two spin-1 systems will produce symmetric states with $L=2$ and $L=0$ but an antisymmetric state with $L=1$. You can verify this symmetry property under permutation from the symmetry of the corresponding Clebsch-Gordon coefficients.

If you combine $4$ particles all of which have $s=1$, then the symmetric part of the coupling will contain $L=4, L=2$ and $L=0$ states.

What is true is that, if you take $n$ copies of the fundamental (or defining) representation of $U(N)$ or $SU(N)$ $(1,0,\ldots,0)$, then the fully symmetric part will carry the irrep $(n,0,\ldots,0)$. By Schur-Weyl duality it will be the only fully symmetric irrep.

The result applies to taking $n$ copies of $s=1/2$ states since these transform by the defining representation of $SU(2)$. But it does not apply to tensors of irreps for which $s\ne 1/2$, as the example above shows.

In addition, by Schur-Weyl duality the final $j$ value, the permutation symmetry and the number of copies of $j$ for the states in each tensor part of the n-fold tensor product of the fundamental representation is completely determined by Schur-Weyl duality.

Since we can't so easily draw Young diagrams I will use partitions to illustrate. Taking $n=4$ copies of $s=1/2$ spin states or $n=4$ copies of the fundamental $(1,0)$ of $su(3)$ gives \begin{align} \begin{array}{cccccc} \hbox{partition}&\hbox{# of copies}&\hbox{$su(2)$ irrep $j$}&\hbox{$su(3)$ irrep $(\lambda,\mu)$}& su(3)\supset so(3)&\\ \{4\} & 1&2& (4,0) & L=4\oplus 2\oplus 0&\hbox{fully symmetric}\\ \{3,1\} & 3&1& (2,1)& L=3\oplus 2\oplus 1\\ \{2,2\} & 2&0& (0,2)& L= 2\oplus 0\\ \{2,1,1\} &3 &\hbox{does not exist}& (1,0)&L=1 \end{array} \end{align}

The table shows the decomposition of $(1/2)^{\otimes 4}$ or $(1,0)^{\otimes 4}$ for the fundamentals of $su(2)$ or $su(3)$. The partitions are associated with a Young diagram of $S_4$ which uniquely labels an irrep of $S_4$ and thus determines the permutation properties of the resulting spin states or $su(3)$ states.

Sorting out into its properly symmetrized parts the $n$-fold tensor product of an irrep that is not the fundamental requires Schur function techniques.

  • $\begingroup$ Thanks for the great answer, but I do not understand your table. What is the table supposed to show? Are you saying, for example, that a totally symmetric combination of 3 spin-1 particles decomposes into spin-3 and spin-1 multiplets? $\endgroup$ – Kristoll Apr 27 '17 at 15:36
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    $\begingroup$ The symmetric part obtained in combining three $j=1$ guys gives $L=3\oplus L=1$. You can infer this because, at the $su(3)$ level, the symmetric part of $(1,0)^{\otimes 3}$ is $(3,0)$, which contains $L=3$ and $L=1$ states only. $\endgroup$ – ZeroTheHero Apr 27 '17 at 15:59
  • $\begingroup$ Sorry if this seems ignorant, but why are you talking about $su(3)$ in the context of spin $1$? Isn't spin $1$ just the $3$ of $su(2)$? $\endgroup$ – Kristoll Apr 27 '17 at 16:15
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    $\begingroup$ Yes but it is also the only $L$ in the $su(3)$ irrep $(1,0)$, and you can use Schur-Weyl duality for $(1,0)$ of $su(3)$, whereas you cannot use it for $L=1$ of $su(2)$. Thus - if you like - I am using Schur-Weyl duality to solve the $su(3)$ problem knowing how I can then relate this back to the $L=1$ problem of $su(2)$ - well, $so(3)$ really. The only remaining piece is to know what is $L$ content of $(\lambda,0)$ of $su(3)$, which is easy: look up the possible $L$'s in the spherical harmonic oscillator with $\lambda$ excitations. (and this is not an ignorant question...) $\endgroup$ – ZeroTheHero Apr 27 '17 at 16:37
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    $\begingroup$ You are confusing irreps of $su(2)$ and irreps of $so(3)$. This is a very very very very subtle point: the algebras are isomorphic but the embeddings are distinct. I am (and you are) referring to an embedding of $so(3)$ that is irreducible in $su(3)$: $so(3)$ is realized as real antisymmetric matrices inside $su(3)$ whereas $su(2)$ is realized as complex matrices. The best way to think about the difference is to ask: what are the possible values of angular momentum $L$ in the $n=1$ states of a 3d harmonic oscillator? The answer is $L=1$ (only $Y_{1m}$ appears), not $j=1/2$ and $j=0$. $\endgroup$ – ZeroTheHero Apr 27 '17 at 17:24

Yes, the maximum possible spin always adds. Physically, this is because you can just have the particles' spins all pointing in the same direction, and angular momentum adds.

More mathematically, suppose we have two particles with spin $s_1$ and $s_2$, with angular momentum operators $\mathbf{L}_1$ and $\mathbf{L}_2$. Then the definition of the total angular momentum operator is $$\mathbf{L} = \mathbf{L}_1 \otimes I_2 + I_1 \otimes \mathbf{L}_2.$$ Now consider the states of maximum angular momentum along the $z$ axis, satisfying $$L_1^z |m^\text{max}_1 \rangle = (\hbar s_1) |m^\text{max}_1 \rangle, \quad L_2^z |m^\text{max}_2 \rangle = (\hbar s_2) |m^\text{max}_2 \rangle.$$ Then the state $|m_1^\text{max} \rangle \otimes |m_2^\text{max} \rangle$ is an eigenvector of $L^z$ with eigenvalue $\hbar (s_1 + s_2)$. That tells us that there's a set of spin $s_1 + s_2$ states, as desired. (Of course, everything above is literally just a longwinded way of writing 'angular momentum adds'.)


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