Given $\vec{r'}.\vec{\sigma}=\hat{U}(\vec{r}.\vec{\sigma})\hat{U}^{-1}$ where $\vec{r'}=(x',y',z'),\vec{r}= (x,y,z),\vec{\sigma}=(\sigma_x,\sigma_y,\sigma_z)$, ($\sigma_k$ the Pauli matrices) and $\hat{U}=\exp{(i\theta \sigma_z/2)}, \quad \theta$ being a constant. How can I calculate $\vec{r'}$ in terms of $\vec{\sigma}$ and $\vec{r}$? I used anti-commutation relations between the Pauli matrices, but did not get the answer.
[Math] Transformation of Pauli matrices
linear algebra
Best Answer
One way is to use that fact that $\vec r'\cdot\vec\sigma = \hat{U}(\vec{r}\cdot\vec{\sigma})\hat{U}^{-1}$ will be $$\left(\begin{matrix} z' & x'-iy' \\ x'+iy' & -z' \end{matrix}\right)$$ From this it's easy to identify $x,y,z$.
Another, more algebraic way, is to use that $\sigma_i \sigma_j + \sigma_j \sigma_i = 2 \delta_{ij}$: $$(x \sigma_x + y \sigma_y + z \sigma_z) \sigma_x + \sigma_x (x \sigma_x + y \sigma_y + z \sigma_z) \\ = x(\sigma_x \sigma_x + \sigma_x \sigma_x) + y (\sigma_y \sigma_x + \sigma_x \sigma_y) + z (\sigma_z \sigma_x + \sigma_x \sigma_z) \\ = x \cdot 2 + y \cdot 0 + z \cdot 0 = 2x$$ Thus, generally, $$x_i = (\vec r \cdot \vec\sigma) \sigma_i + \sigma_i (\vec r \cdot \vec\sigma)$$