1
$\begingroup$

I' m trying to figure out how to decompose unitary operator U using only Rx and Ry rotations. I understand that I must use result of exercise 4.8 ($ U = e^{i\alpha}R_n(\theta)$). But I don't understand how, my thoughts is use Bloch vector coordinates and place it instead n-vector coordinate and try to solve using this, but I think it is not correct way. Can anyone help me or give some insights how to solve this exercise?

$\endgroup$
1
  • $\begingroup$ Please add the text of the exercise in full. $\endgroup$ Commented Apr 11, 2023 at 18:04

1 Answer 1

2
$\begingroup$

Problem 4.10 of Nielsen and Chuang's text asks the reader to produce an $XY$-decomposition of a single-qubit unitary $U$, in parallel with the $ZY$-decomposition given in Theorem 4.1 of the text. That is, the problem asks to show that any single-qubit unitary $U$ can be written as $$ U = e^{i \alpha} R_x(\beta) R_y(\gamma) R_x(\delta) $$ for appropriate choices of $\alpha, \beta, \gamma, \delta$.

Certainly, we could proceed along similar lines as the text does for the $ZY$-decomposition of Theorem 4.1, producing the result essentially from scratch. Alternatively, we could simply use Theorem 4.1 and do a "change of reference", to interchange the $x$ and $z$ axes. Let's proceed the latter route and see how this works.

We know that the Hadamard $H$ (which we quickly note is unitary and self-inverse) implements the appropriate swap between $x$ and $z$, while sending $y$ to $-y$. So instead of working with $U$ directly, consider $$ U' = H U H^{-1} = H U H $$ which is the frame-rotated $U$. Using Theorem 4.1 of the text, we can produce a $ZY$-decomposition of $U'$. $$ U' = e^{i\alpha} R_z(\beta) R_y(\gamma) R_z(\delta) $$ What does this imply for $U$? We will see that the above implies an $XY$-decomposition for $U$. \begin{align} U = HU'H &= e^{i\alpha} (H R_z(\beta) H)(H R_y(\gamma)H)(HR_z(\delta) H) \\ &= e^{i\alpha} R_x(\beta) R_y(-\gamma) R_x(\delta). \end{align} Notice the change in sign for $\gamma$ that occurs since $H Y H = -Y$. Regardless, we have $U$ written in the required form.

As a final elaboration, this argument could be generalized. We could choose any two orthogonal axes $\hat{n}, \hat{m}$ for rotations in our decomposition, using the fact that we can always frame-rotate to the $z$ and $x$ axes.

$\endgroup$
2
  • $\begingroup$ Thanks, understood! $\endgroup$ Commented Apr 12, 2023 at 7:09
  • $\begingroup$ I tried to change $R_z$ rotation with $R_x$ rotations but I always got $HR_x(\beta)R_y(-\gamma)R_x(\delta)H$ And stopped at this moment without knowing how transform this in desired view.)) I didn't figure it out before this trick) $\endgroup$ Commented Apr 12, 2023 at 7:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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