# Dimension of spin zero Hilbert subspace

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?

• 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? Mar 23, 2022 at 9:26

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.
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$$^\natural$$.)
$$^\natural$$ The multiplicity of total spin s in the composition of n spin j multiplets is always given by a difference, $$$$M\left( s;n;j\right) =c_{0}\left( s,n,j\right) -c_{0}\left( s+1,n,j\right) \ ,$$$$ where $$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$$ is a semi-integer:
$$$$\Large c_{0}\left( s,n,j\right) =\sum_{k=0}^{\left\lfloor \frac{nj+s} {2j+1}\right\rfloor }\left(-1\right) ^{k}\binom{n}{k}\binom{nj+s-\left( 2j+1\right) k+n-1}{nj+s-\left( 2j+1\right) k}\ .$$$$