I know that an $R_z (\theta)$ gate is equivalent to the unitary transformation $e^{-iZ * \theta/2}$ but I'm not sure how we get there.

I know that for every Hermitian matrix there is a corresponding Unitary matrix as $ U = e^{iH} $ where the eigenvalues are exponentiated and the eigenstates remain the same. But I don't see how in this case, it leads to the matrix $$\left(\begin{array}{cc}e^{-i \frac{\lambda}{2}} & 0 \\ 0 & e^{i \frac{\lambda}{2}}\end{array}\right)$$

and not: $$\left(\begin{array}{cc}e^{i} & 0 \\ 0 & e^{-i}\end{array}\right)$$

I feel like I'm missing something really obvious


1 Answer 1


$$-iZ\frac{\theta}{2} = \begin{bmatrix}-i\frac{\theta}{2} &0\\0&i\frac{\theta}{2} \end{bmatrix}$$

This is alread diagonal, so now you just take the exponenital operation on each of the eigenvalues.

To get the matrix you thought you would get, you would need $$iZ=\begin{bmatrix}i&0\\0&-i\end{bmatrix}$$ which when exponeniated, would give you $$\begin{bmatrix}e^{i}&0\\0&e^{-i}\end{bmatrix}$$

I am assuming you meant to write $\theta$ and not $\lambda$ in your first matrix.


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