# Mathematical description of spin 0 state

I was going through the book titled Quantum Computing Explained by David McMohan. I am not able to understand how they arrived at the below mentioned mathematical expression.

A total spin-0 state, which consists of a system of two spin-1/2 particles, can be described in terms of the computational basis with the single state,

$$|\psi\rangle=\frac{|0 1\rangle-|10\rangle}{\sqrt{2}}$$

• Use a different book. – Norbert Schuch Feb 11 at 9:01

## 2 Answers

The addition of angular momentum is the key. Every basic course on Quantum mechanics will explain this to some extent. If you have two systems of spin 1/2, they follow the rules of angular momentum addition. One wants to choose a different basis for the space $$\mathbb{C}^2 \otimes \mathbb{C}^2$$, they new basis has basis vectors labeled by total angular momentum (spin) and total $$z$$-projection. The basis transformation is made via Clebsch-Gordan coefficients, checking tables and using the more traditional notation of spin up being 1/2 (instead of 1) and spin down being -1/2 (instead of 0), the state of total projection in $$z$$ equals to $$| 0, 0 \rangle = \frac{1}{\sqrt{2}}\left( \bigg|-\frac{1}{2},+\frac{1}{2}\bigg\rangle - \bigg|+\frac{1}{2},-\frac{1}{2}\bigg\rangle\right)$$

For a two-level state space, with a basis $$\{|0\rangle,|1\rangle\}$$, the infinitessimal rotation generators are $$J_j = \frac{1}{2}\sigma_j$$ where $$\sigma_j$$ for $$j = 1,2,3$$ are the three Pauli matrices. If you want to rotate a state around the $$x$$-axis by the angle $$\theta$$, you can multiply the state by the $$2 \times 2$$ matrix $$e^{-i \theta J_1}$$.

When you tensor together two of these two-level state spaces, you get a four-dimensional state space with a basis $$\{|00\rangle,|10\rangle\,|01\rangle,|11\rangle\}$$, the infinitessimal rotation generators become

$$J_j = \frac{1}{2}\sigma_j\otimes I+\frac{1}{2}I \otimes \sigma_j$$ where $$I$$ is the identity matrix.

You can explicitly check that for the state $$|\psi\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle) = \frac{1}{\sqrt{2}}(|0\rangle \otimes |1\rangle - |1\rangle \otimes |0\rangle)$$ that

$$J_j |\psi\rangle = 0$$ for all $$j = 1,2,3$$. Therefore when you perform a rotation on this particular state $$|\psi\rangle$$, the state does not change. This means $$|\psi\rangle$$ transforms in the trivial representation of the rotation group, which we call "spin $$0$$."