# Dimension of singlet subspace of $2N$ identical spin-1/2 particles

Consider a quantum system of $$2N$$ identical spin 1/2 particles, each having a 2D Hilbert space $$V$$. The total Hilbert space is the tensor product $$V^{\otimes 2N} \equiv V \otimes \cdots \otimes V$$. Define the total spin operator as

$$\mathbf{S} = \sum_{i=1}^{2N} \mathbf{S}_i$$

Let $$V_0$$ be the subspace of $$V^{\otimes 2N}$$ which has zero total spin:

$$V_0 = \{|\psi\rangle \in V^{\otimes 2N} | \ \mathbf{S}^2 |\psi\rangle = 0\}$$

What is the dimension of $$V_0$$? In this article, it is stated that (below Eq. 23)

$$\dim V_0 = \frac{1}{N+1} \binom{2N}{N} = \frac{(2N)!}{N!(N+1)!}$$

The combination number $$\binom{2N}{N}$$ can be naively understood as requiring the total $$S^z = 0$$ (number of different ways to choose $$N$$ spins with spin-up; the other $$N$$ spins have spin down). But I am having trouble in understanding the additional $$1/(N+1)$$ factor. Can someone please explain it? In addition, from the point of view of reducing the tensor product of $$N$$ copies of the fundamental (spin-1/2) representation of SU(2), what are the basis vectors in $$V^{\otimes 2N}$$ that furnish these $$(\dim V_0)$$ copies of the trivial (spin-0) representation?

The Catalan coefficient for the correct multiplicity, in agreement with your reference, is $$C(2N,N)= \binom{2N}{N}-\binom{2N}{N-1},$$ as explained; or you might just evaluate the easy integral (7), a direct approach.
• Thanks for the references. Wikipedia says $C(x,y)$ is the number of strings consisting of $x$ X's and $y$ Y's such that no initial segment of the string has more Y's than X's. But how is the case $x = 2N$ and $y = N$ related to the singlet state? Oct 24, 2023 at 7:48