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

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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 ?

Duen
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  • Since the transformation is unitary, this is possible- any unitary can be written as exp(iH) with H an Hermitian operator. – Yaron Jarach Apr 19 '23 at 19:50
  • Can you detail how you would do it ? I am still confused about those kind of calculations – Duen Apr 19 '23 at 20:51

1 Answers1

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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.

Cody Wang
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