2

As far as I know the single qubit gate

$$ e^{i\beta\sigma_z} = \begin{bmatrix} e^{i\beta} & 0 \\ 0 & e^{-i\beta} \end{bmatrix} = e^{i\beta} \begin{bmatrix} 1 & 0 \\ 0 & e^{-i2\beta} \end{bmatrix} = e^{i\beta} R_Z(-2\beta). $$

However, I have seen the above gate implemented using $U_1(2\beta)$, where $ U_1(\lambda) = \begin{bmatrix} 1 & 0 \\ 0 & e^{i\lambda} \end{bmatrix} $

Is $e^{i\beta} R_Z(-2\beta)$ equivalent to $U_1(2\beta)$?

**Update:**

As Davit explains below, $e^{i\theta/2} R_Z(\theta) = U_1(\theta)$, so with $\frac{\theta}{2}=\beta$ we have $e^{i\beta} R_Z(2\beta) = U_1(2\beta)$. Note the difference: in this case the $Z$-rotation is positive, whereas in my original question it is negative.

From the notation used in this question $R_z(\theta) = \begin{pmatrix} 1 &0 \ 0&e^{i \theta} \end{pmatrix}$. Although I am not sure about the "conventionality"/widespread of this notation, but If we use this definition for $R_z$ instead of the definition used in the textbook (shown in my answer) then $R_z(\theta) = U_1(\theta)$ without any global phase difference. – Davit Khachatryan – 2020-06-11T06:14:10.800

This can help: https://quantumcomputing.stackexchange.com/questions/11888/why-is-a-different-convention-used-for-the-rz-implementation-on-ibm-q

– Martin Vesely – 2020-06-11T09:34:19.327