# Generator of 3D rotations in $\mathbb{C}^2 \otimes \mathbb{C}^2$

Let us consider a system of two spinors. The 3D rotation operator around the $$\vec{n}$$ axis in $$\mathbb{C}^2$$ is clearly $$R(\theta) = \exp(i \frac{\theta}{2}\vec{n}\cdot\vec{\sigma})$$.

If I wish to rotate my system of two spinors around the $$z$$ axis for instance, which one is the generator of rotations, $$S^1_z \otimes S^2_z$$ or $$S^1_z \otimes \mathbb{I}+ \mathbb{I} \otimes S^2_z$$?

• The second suggestion. – Qmechanic Feb 11 at 16:19
• Then can you rotate them through different axes? For example if we took $S^1_{x_1} \otimes \mathbb{I}+ \mathbb{I} \otimes S^2_{x_2}$ for axes $x_1$ and $x_2$? Would it make any sense? It's not really a possible rotation for the system, is it? – Salvador Villarreal Feb 11 at 16:53
• Related : Total spin of two spin- 1/2 particles 1. T H I R D__ A N S W E R equation (62) 2. FOURTH__A N S W E R equations (68), (69), (70). – Frobenius Feb 11 at 17:27

It's the second one. To derive this, consider a finite rotation $$|\psi_i \rangle \to |\psi' \rangle = U_{ij}(\theta) |\psi_j\rangle$$. Acting on a tensor product, we have $$|\psi_i \rangle |\phi_k \rangle \to U_{ij}(\theta) U_{kl}(\theta) |\psi_j\rangle |\phi_l\rangle.$$ If $$\theta$$ is an infinitesimal angle, we can write $$U_{ij}(\theta) = \delta_{ij} + \theta S_{ij} + O(\theta^2)$$ where $$S$$ is a Lie algebra generator. Hence $$|\psi_i \rangle |\phi_k \rangle \to |\psi_i \rangle |\phi_k \rangle + \theta \left[ (S_{ij} |\psi_j \rangle) \otimes |\phi_k \rangle + |\psi_i \rangle \otimes (S_{kl}|\phi_l \rangle) \right] + O(\theta^2).$$ The term inside brackets is precisely of the form $$S^1 \otimes \mathbb{1} + \mathbb{1} \otimes S^2$$.