1
$\begingroup$

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

$$U_{\frac{7\pi}{16}}=\cos\left(\frac{7\pi}{8}\right)\sigma_{z}+\sin\left(\frac{7\pi}{8}\right)\sigma_{x}$$

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.

$\endgroup$
7
  • $\begingroup$ It may be helpful if you link the paper (and mention the section where that transformation is introduced) you're talking about $\endgroup$ – epelaaez Jun 22 at 1:33
  • 1
    $\begingroup$ @epelaaez Edited the question with the link $\endgroup$ – Jasmine Jun 22 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 Jun 22 at 1:44
  • $\begingroup$ @narip But then there would be a factor of i, I cant find that in the paper $\endgroup$ – Jasmine Jun 22 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 Jun 22 at 2:05
2
$\begingroup$

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.

$\endgroup$
0

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.