# How does Equation (9) follows from these definitions?

How does Equation (9) follows from these definitions?

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

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

• 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? Sep 26 '21 at 18:10
• @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. Sep 26 '21 at 18:22

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.

• 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. Sep 26 '21 at 18:38
• @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. Sep 26 '21 at 18:43
• You may write to the authors to verify with them, if wrong they may add an errata. Sep 26 '21 at 18:46

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.

• Very nice stuff, thanks! Sep 28 '21 at 16:05