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?


1 Answer 1


You can already see you miscounted for N = 1, just two particles. There is only one singlet in their tensor product, not two! You need the denominator of 2 to cut down on the superfluous/fake multiplicity you inferred. For N =2, 4 particles, you need to divide by 3 to get the answer, one singlet as a composition of two intermediate singlets, and one singlet as a composition of two intermediate triplets.

Your reference's well-known formula is just equation (12) in this reference, for s =0 and n = 2N, following from (7), or, easier, the Catalan triangle structure in the bottom formula of this section in Wikipedia, an ancient result: Zachos, 1992.

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.

  • $\begingroup$ 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? $\endgroup$ Oct 24, 2023 at 7:48
  • $\begingroup$ Think of the random walk utilized... $\endgroup$ Oct 24, 2023 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.