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How does Equation (9) follows from these definitions?

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This is my working out for the first identity. It appears that the identity is not true.

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For the second identity, I used sympy to check. The identity also appears to not hold. This is the final simplification.

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  • $\begingroup$ Have you tried to prove it by yourself? Is there anything to suspect that the identity does not hold? Do you block in any particular step? $\endgroup$
    – Mauricio
    Sep 26 at 18:10
  • $\begingroup$ @Mauricio I tried to work through the steps by hand, and I even recheck it on Sympy. However, both attempts appears to show that the identity does not hold. Below I will show my working out for the top identity. $\endgroup$
    – Minh Pham
    Sep 26 at 18:22
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I think $R_y(\theta)$is possibly wrongly defined. If we define it as usual, we write

$$R_y(\theta)=\exp\left(-i\frac\theta2 Y\right)=\begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2) \\ \sin(\theta/2) & \cos(\theta/2) \end{pmatrix}\neq \begin{pmatrix} \cos(\theta/2) & \sin(\theta/2) \\ -\sin(\theta/2) & \cos(\theta/2) \end{pmatrix}$$

The matrix that you show has a different sign, that's why the identities do not follow. Check for example Quantum Inspire $R_y$ gate.

You can easily verify the first identity, with this definition of $R_y$, by swapping the minus signs in your calculation above. I leave to you the second identity.

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  • $\begingroup$ I agree with you. I also think that the Ry gate is a little weird. This actually came from at the end of page 5 of this paper arxiv.org/pdf/quant-ph/0406176.pdf. $\endgroup$
    – Minh Pham
    Sep 26 at 18:38
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    $\begingroup$ @MinhPham interestingly the same definitions are still there in the final version . Eventually, the signs in the rotation matrices are a matter of convention, but the identities (8) and (9) do not follow from those definitions. $\endgroup$
    – Mauricio
    Sep 26 at 18:43
  • $\begingroup$ You may write to the authors to verify with them, if wrong they may add an errata. $\endgroup$
    – Mauricio
    Sep 26 at 18:46
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Probably the main problem is the bad definition of $Ry$ gate in your excercise. However, I will try to give you another way how to solve the problem without many matrix calculations.

Lets assume that we have a qubit in state $|0\rangle$. If we apply $Ry(\theta)$ gate we get the qubit in state $$ \cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle. $$

This is the first column of $Ry$ matrix, or in other words result of $Ry$ acting on basis state $|0\rangle$.

If we now apply $Rz(\varphi)$ gate we have $$\mathrm{e}^{-i\varphi/2}\cos(\theta/2)|0\rangle + \mathrm{e}^{i\varphi/2}\sin(\theta/2)|1\rangle, $$

which is your state $|\psi\rangle$. It is enough to add diagonal elements (phases) to a proper basis state - $\mathrm{e}^{-i\varphi/2}$ to $|0\rangle$ and $\mathrm{e}^{-i\varphi/2}$ to $|1\rangle$.

Gates $Rz(-\varphi)$ and $Ry(-\theta)$ are inverse operations to $Rz(\varphi)$ and $Ry(\theta)$, respectivelly. Applying them on state $|\psi\rangle$ naturally lead to the initial state $|0\rangle$. This proves the identity.

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    $\begingroup$ Very nice stuff, thanks! $\endgroup$
    – Minh Pham
    Sep 28 at 16:05

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