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?

  • $\begingroup$ 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." $\endgroup$
    – narip
    Aug 5 at 10:56
  • $\begingroup$ Thanks, that solves my issue. $\endgroup$
    – Iliad
    Aug 5 at 12:15


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.