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If I have a 2-qubits circuit with a Ry rotation gate acting on each one :

enter image description here

My unitary transformation performed on the 2-qubits state is written as :

$$e^{-i\theta_{1} \sigma_{y}} \otimes e^{-i\theta_{2} \sigma_{y}}$$

I was wondering if I could simplify this product in order to have only one exponential(thus making the tensor product disappear) and what would the result be ?

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  • $\begingroup$ Since the transformation is unitary, this is possible- any unitary can be written as exp(iH) with H an Hermitian operator. $\endgroup$
    – Yaron Jarach
    Commented Apr 19, 2023 at 19:50
  • $\begingroup$ Can you detail how you would do it ? I am still confused about those kind of calculations $\endgroup$
    – Duen
    Commented Apr 19, 2023 at 20:51

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The two gates act on separate qubits, so their generators commute and you can use the identity $$ e^A \otimes e^B = e^{A \otimes \mathbb{I} + \mathbb{I} \otimes B} $$ to get $$ e^{-i\theta_1 \sigma_y} \otimes e^{-i\theta_2 \sigma_y} = e^{-i(\theta_1 \sigma_y \otimes \mathbb{I} + \mathbb{I} \otimes \theta_2 \sigma_y)} $$ if that helps simplify things.

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  • $\begingroup$ Thank you for the great answer ! Beginner question : what are the generators of my gates here ? $\endgroup$
    – Duen
    Commented Apr 20, 2023 at 7:14
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    $\begingroup$ The generator is the operator in the exponent; in your case, the generator of $R_y(\theta)$ is $-\theta \sigma_y$. $\endgroup$
    – Cody Wang
    Commented Apr 20, 2023 at 9:18
  • $\begingroup$ Do I have to also take the "i" or is it useless ? $\endgroup$
    – Duen
    Commented Apr 20, 2023 at 9:20
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    $\begingroup$ Oops, nice catch; I've added the $i$ to the answer! The $i$ is not considered part of the generator though. $\endgroup$
    – Cody Wang
    Commented Apr 20, 2023 at 9:22

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