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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.

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  • $\begingroup$ 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. $\endgroup$ – Davit Khachatryan Jun 11 at 6:14
  • $\begingroup$ This can help: quantumcomputing.stackexchange.com/questions/11888/… $\endgroup$ – Martin Vesely Jun 11 at 9:34
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$R_z$ gate from M. Nielsen and I. Chuang textbook (page 174):

$$R_z(\theta) = e^{-i\theta Z/2} =\begin{pmatrix} e^{-i \theta /2} &0 \\ 0&e^{i \theta /2} \end{pmatrix}$$

If we use this definition for $R_z(\theta)$ then:

$$R_z(\theta) = e^{-i \theta/2} \begin{pmatrix} 1 &0 \\ 0&e^{i \theta} \end{pmatrix} = e^{-i \theta/2} U_1(\theta)$$

Therefore, $R_z(\theta)$ and $U_1(\theta)$ can be regarded as equivalent gates, because the global phase $e^{-i \theta/2}$ can be neglected. Note that this comparison is not true for their controlled versions: $cR_z(\theta)$ and $cU_1(\theta)$ can't be regarded as equivalent gates.

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  • $\begingroup$ Thanks @DavitKhachatryan, but the angle seems to have the wrong sign. I have update the OP. $\endgroup$ – John Jun 10 at 23:12
  • $\begingroup$ @John, I have looked in the textbook and I think that there is no sign inconsistency with the textbook's definition for $R_z$ gate. Also, I don't see a sign problem in the $R_z(\theta) = e^{-i \theta/2}U_1(\theta)$ (note, that I have added some details in the answer). $\endgroup$ – Davit Khachatryan Jun 11 at 6:00
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    $\begingroup$ This can interesting for you: quantumcomputing.stackexchange.com/questions/11888/… $\endgroup$ – Martin Vesely Jun 11 at 9:34
  • $\begingroup$ The link provided by MartinVesely and the answer from @DavitKhachatryan resolved this for me $\endgroup$ – John Jun 12 at 11:21

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