# Why can we simulate any measurement involving $U=R+iQ$ using $U'=R\otimes I+Q\otimes R_x(2\pi)$ instead?

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 $$U′|ψ⟩|0⟩$$

• 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. Aug 28, 2023 at 4:02
• @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? Aug 28, 2023 at 4:10
• 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. Aug 28, 2023 at 4:56
• 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
– glS
Aug 28, 2023 at 16:38
• @glS Not Hermitian, just unitary. Aug 29, 2023 at 3:03

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 $$(R+iQ)|\psi \rangle$$.