# Composition of rotations sign

I'm solving exercise 4.15 from Nielsen and Chuang:

Prove that if a rotation through an angle $$\beta_1$$ about the axis $$\hat{n}_1$$ is followed by a rotation through an angle $$\beta_2$$ about an axis $$\hat{n}_2$$, then the overall rotation is through an angle $$\beta_{12}$$ about an axis $$\hat{n}_{12}$$ given by $$c_{12} = c_1c_2-s_1s_2\hat{n}_1\cdot\hat{n}_2$$ $$s_{12} \hat{n}_{12} = s_1c_2\hat{n}_1+c_1s_2\hat{n}_2-s_1s_2\hat{n}_2\times\hat{n}_1,$$ where $$c_i = \cos \left( \beta_i/2 \right), s_i = \sin \left( \beta_i/2 \right), c_{12} = \cos \left( \beta_{12}/2 \right),$$ and $$s_{12}= \sin \left( \beta_{12}/2 \right)$$.

I tried to solve this by writing $$R_{\hat{n}_2}(\beta_2)R_{\hat{n}_1}(\beta_1) = \left(\cos\frac{\beta_2}{2}I - i\sin\frac{\beta_2}{2}(\hat{n}_2\cdot\vec{\sigma})\right)\left(\cos\frac{\beta_1}{2}I-i\sin\frac{\beta_1}{2}(\hat{n}_1\cdot\vec{\sigma})\right)$$. Using $$(\vec{a}\cdot\vec{\sigma})(\vec{b}\cdot\vec{\sigma}) = (\vec{a}\cdot\vec{b})I + i(\vec{a}\times\vec{b})\cdot\vec{\sigma}$$ the composite rotation becomes: $$R_{\hat{n}_2}(\beta_2)R_{\hat{n}_1}(\beta_1) = (c_2c_1 - s_2s_1(\hat{n}_1\cdot\hat{n}_2))I -i(c_2s_1\hat{n}_1 + s_2c_1\hat{n}_2 + s_2s_1\hat{n}_2\times\hat{n}_1)\cdot\vec{\sigma}$$

However, the sign on the cross product term is flipped from what it should be. Where did I go wrong?

• This might be helpful. The original problem has a sign error. "In Equations (4.20) and (4.22) the minus sign on the right-hand side should be a plus." Aug 5 at 10:56
• Thanks, that solves my issue. Aug 5 at 12:15