I am trying to understand the following statement. "Suppose a single qubit has a state represented by the Bloch vector $\vec{\lambda}$. Then the effect of the rotation $R_{\hat{n}}(\theta)$ on the state is to rotate it by an angle $\theta$ about the $\hat{n}$ axis of the Bloch sphere. This fact explains the rather mysterious looking factor of two in the definition of the rotation matrices." I could work out that the rotation operators $R_x(\theta)$, $R_y(\theta)$ and $R_z(\theta)$ are infact rotations about the $X,Y$ and $Z$ axis. I know the rotation matrices in terms of the Pauli matrices, i.e $R_x(\theta) = e^{-i\sigma_x /2}$ and the rotation matrices for $\sigma_y$ and $\sigma_z$ follows in the same manner. I could also prove that $R_{\hat{n}}(\theta) = cos(\frac{\theta}{2})I - i sin(\frac{\theta}{2})(n_x\sigma_x+n_y\sigma_y+n_z\sigma_z)$ using the Taylor expansion. But the difficulties for me start from here. How do I show that $R_\hat{n}(\theta)$ is infact a rotation about $\hat{n}$ axis by $\theta$. How can I construct a concrete proof?

  • $\begingroup$ It would be useful if you could write down explicitly the state as well as the rotation operator, so we can use the same notation. That would also help us understand what factor of two you are talking about (it is right now rather mysterious indeed...). $\endgroup$
    – Adam
    Commented Mar 22, 2016 at 8:28
  • $\begingroup$ @Adam Thanks, I have made the required changes. However I am still not sure if it is clear enough. $\endgroup$ Commented Mar 22, 2016 at 9:22
  • $\begingroup$ It is indeed a better (and self-contained) question now. How do you know that $R_x$, etc. are rotation matrices ? Then maybe that the same arguments can help you show that $R_n$ is also a rotation. $\endgroup$
    – Adam
    Commented Mar 22, 2016 at 14:08
  • $\begingroup$ @Adam For $R_x(\theta)$ etc, it was easy to visualize the result, hence I just rotated the $\vec{\lambda}$ by $\theta$, and then applied the operator $R_x(\theta)$etc, and thereby determined that they are infact rotation matrices. However I cant seem to be able to do that same for $R_{\hat{n}}(\theta)$, because its a bit difficult. There surely has to be a better way. $\endgroup$ Commented Mar 22, 2016 at 14:17
  • $\begingroup$ What exactly are you after? The generic R you wrote down is the exponential of SU(2) ~ SO(3) Lie algebra generators, and as such the doublet representation of the rotation group, whose generic composition law is provided in that very WP article. You know it leaves eigenvectors of $\hat{n}\cdot\vec{\sigma}$ invariant, and by isomorphic equivalence redefinitions of the Pauli matrices you may trivially probe its action in the plane perpendicular to that of $\hat{n}$. Are you familiar with the rotation group and Pauli-matrix algebra? $\endgroup$ Commented Mar 28, 2016 at 15:10

1 Answer 1


You can in fact construct a concrete proof by direct computation:

  1. Take a mixed quantum state represented by a density operator, $\rho = \frac{1}{2}(I + \vec{r} \cdot \vec{\sigma})$, where $I$ is the identity operator on a Hilbert space of two dimensions (representing your quantum state). $\vec{r}$ is the Bloch vector representing the mixed state.
  2. Conjugate this quantum state by your unitary operation $U = R_{\hat{n}}(\theta)$ to compute $\rho' = U \rho U^{\dagger}$.
  3. You will find that $\rho' = \frac{1}{2}(I + \vec{r'} \cdot \vec{\sigma})$ for some $\vec{r'}$, and you will note that this $\vec{r'}$ is obtained by rotating $\vec{r}$ around $\hat{n}$ by $\theta$ .

This link should help: http://www.vcpc.univie.ac.at/~ian/hotlist/qc/talks/bloch-sphere-rotations.pdf


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