Consider a system of 4 different spin-5/2 particles, each transforming in the dimension-6 irreducible representation of $SU(2)$. The dimension of the total Hilbert space is $6^4$. Could anyone show how to compute the dimension of a subspace of total spin zero?

I would like to calculate the number of ways 4 particles combine to spin 0. Is there a general procedure?

  • 1
    $\begingroup$ Hint: Try first with 2 different spin-5/2 particles. Dimension of total Hilbert space is then $6^2=36$. What do you then get? $\endgroup$
    – Qmechanic
    Mar 23, 2022 at 9:26

1 Answer 1


As suggested by @Qmechanic, your problem is addressable by inspection, $$ 5/2 \otimes 5/2= 5\oplus 4\oplus 3\oplus 2\oplus 1\oplus 0 , ~~~\leadsto \\ ( 5\oplus 4\oplus 3\oplus 2\oplus 1\oplus 0)\otimes (5\oplus 4\oplus 3\oplus 2\oplus 1\oplus 0). $$ But you can only get singlets out of the diagonal terms in the distribution of this, 5⊗5, 4⊗4, ...,0⊗0, so, then, six of them. That is, of your 1296 states, 6 are spinless.

Is there a general procedure?

Yes, eqn (6) & (11) of Curtright, Van Kortryk, and Zachos 2016, with explicit evaluations which have been around forever...(However, admittedly, your eyes will pop out at the $M(0,4;5/2)=146-140$. cf footnote$^\natural$.)

Can you now supplant your 4 with n tending to infinity, should your spirit move you?

$^\natural$ The multiplicity of total spin s in the composition of n spin j multiplets is always given by a difference, \begin{equation} M\left( s;n;j\right) =c_{0}\left( s,n,j\right) -c_{0}\left( s+1,n,j\right) \ , \end{equation} where $2s$ is any integer such that $0\leq2s\leq2nj$, and where $s=0$ is always allowed when $j$ is an integer but is only allowed for even $n$ when $j$ is a semi-integer:
\begin{equation} \Large c_{0}\left( s,n,j\right) =\sum_{k=0}^{\left\lfloor \frac{nj+s} {2j+1}\right\rfloor }\left(-1\right) ^{k}\binom{n}{k}\binom{nj+s-\left( 2j+1\right) k+n-1}{nj+s-\left( 2j+1\right) k}\ . \end{equation}


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