Recently I have read a paper in which they have used a unitary transformation as follows:


Here $ \sigma_{x} $ and $\sigma_{z}$ are the Pauli operators. I didn't understand where this came from? Also are any other combinations of sigma operators with any angles is a Unitary rotation? Do anyone know of any general formula for Unitary rotation? Any references would be great. Please see Eq. (3) in the paper: Experimental test of local observer-independence.

Why I am concerned about the above unitary operator is: d Please see the below definition for the rotation operators. Here there is an imaginary $i$ coming which is not in the paper I have mentioned.

  • $\begingroup$ It may be helpful if you link the paper (and mention the section where that transformation is introduced) you're talking about $\endgroup$
    – epelaez
    Commented Jun 22, 2021 at 1:33
  • 1
    $\begingroup$ @epelaaez Edited the question with the link $\endgroup$
    – Jasmine
    Commented Jun 22, 2021 at 1:36
  • $\begingroup$ $e^{-i\frac{\theta}{2}n\cdot\sigma}$ is rotation around $\vec{n}$ about $\theta$ angle. You can find this in Nielsen's book, chapter 4. $\endgroup$
    – narip
    Commented Jun 22, 2021 at 1:44
  • $\begingroup$ @narip But then there would be a factor of i, I cant find that in the paper $\endgroup$
    – Jasmine
    Commented Jun 22, 2021 at 1:45
  • $\begingroup$ You only need to see that $UU^\dagger=I$, hence it's unitary. As for the reason why there is no imaginary part because it's a specific angle(e.g., when $\theta$ in your rotation is 2$\pi$, there will also miss the imaginary part). $\endgroup$
    – narip
    Commented Jun 22, 2021 at 2:05

1 Answer 1


Setting $\theta=\pi/2$ for a general rotation, you get $$ \mathrm{e}^{-\mathrm{i}\pi/2 A} = \mathrm{i} A\,, $$ which is unitary and up to a global phase $\mathrm{i}$ the operator $A$ itself. Now define a new operator $U=-\mathrm{i}A$ and it follows that $U$ is unitary $$ U^\dagger U = \mathrm{i}(-\mathrm{i}) A^\dagger A = I\,. $$

You can decompose $A$ into $$ A = n_x \sigma_x + n_y \sigma_y + n_z \sigma_z $$ with $$ n_x^2 + n_y^2 + n_z^2 = 1 $$ and $n_i \in[-1,1]$, $i=x,y,z$. In your example $$ n_x = \sin\left(\frac{7\pi}{8}\right) \qquad n_y = 0 \qquad n_z = \cos\left(\frac{7\pi}{8}\right)\,. $$ For more details see Nielsen and Chuang, Quantum Computation and Quantum Information.


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