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In a book named "INTRODUCTION TO QUANTUM ALGORITHMS VIA LINEAR ALGEBRA", the authors say:

For any complex $N×N$ matrix $U$, we can uniquely write $$U = R + iQ$$ Assume we have $$ U' = R \otimes I + Q \otimes R_x(2\pi) $$

$R_x(2\pi)$ is of the form $$ \begin{bmatrix} 0 & -1 \\\ -1 & 0\end{bmatrix} $$

My question is, why can we simulate any measurements involving $U$ by measurements involving $U'$ instead? Simulate any measurements means that if we measure outcome $|x⟩$ in sate $U|ψ⟩$ , it equivalen when we measure outcome $|x⟩$ in state $U′|ψ⟩|0⟩$

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    $\begingroup$ Can you more clearly explain your question? What do you exactly mean by measurement involving $U$ and also simulating it by measurement involving $U'$? Also, please link relevant text/material/resources if any. $\endgroup$
    – FDGod
    Aug 28, 2023 at 4:02
  • $\begingroup$ @FDGod i'm reading a book called "INTRODUCTION TO QUANTUM ALGORITHMS VIA LINEAR ALGEBRA" - Richard J. Lipton and Kenneth W. Regan. It's a problems without answer in the book (problem number 7.8, page 115). Actally i'm not so sure about what simulate any measurements mean. Maybe any circuits involve $U$, we can somehow use $U'$ instead, and the result after measuring is the same somehow? $\endgroup$
    – Huy By
    Aug 28, 2023 at 4:10
  • $\begingroup$ Sorry, I don't have access to that book. Is $U$ a unitary or a hermitian matrix? A unitary matrix does not correspond to a measurement afaik. Or do you mean first apply $U$ to your state and then perform a $Z$-measurement? even if that's the case, $U'$ seem to have a higher dimension than $U$. I am sorry, your question does not make any sense to me. $\endgroup$
    – FDGod
    Aug 28, 2023 at 4:56
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    $\begingroup$ please edit the question to include any additional contextualising information, such as the book you're taking this from. Also, I'm guessing there should be additional assumptions on $R$ and $Q$ here, such as whether you assume them to be real, or whether you assume them to be Hermitian. Is $U$ Hermitian? More generally, you should clarify what "measurement involving $U$" means here $\endgroup$
    – glS
    Aug 28, 2023 at 16:38
  • $\begingroup$ @glS Not Hermitian, just unitary. $\endgroup$
    – Huy By
    Aug 29, 2023 at 3:03

1 Answer 1

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In my version of the book $\tilde{U}$ is defined as \begin{align} \tilde{U} = R \otimes I + Q \otimes R_x(\pi), \end{align} where \begin{align} R_x(\theta) = \begin{pmatrix} \cos\left(\frac{\theta}{2}\right) & \sin\left(\frac{\theta}{2}\right) \\ -\sin\left(\frac{\theta}{2}\right) & \cos\left(\frac{\theta}{2}\right)\end{pmatrix}, \end{align} i.e., \begin{align} R_x(\pi) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. \end{align} In this case, \begin{align} \tilde{U} = \begin{pmatrix} R & Q \\ -Q & R \end{pmatrix}. \end{align}

Now, we have \begin{align} \tilde{U} |\psi \rangle |\psi \rangle = \begin{pmatrix}(R + Q)|\psi \rangle \\ (R- Q)|\psi \rangle \end{pmatrix}. \end{align}

By measuring this state, we can reconstruct the corresponding measurement result for $(R+iQ)|\psi \rangle$.

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