How to implement this single-qubit unitary? I was reading this paper on qubit state preperation, and encountered an interesting type of single-qubit gate:
\begin{align}
 U_\theta = \left(\begin{matrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{matrix}\right) = \cos\theta\, \sigma_z + \sin\theta\, \sigma_x
\end{align}
and more generally,
\begin{align}
 U = \frac{1}{\sqrt{|a|^2+|b|^2}}\left(\begin{matrix} a & b \\ b^* & -a^*\end{matrix}\right) 
\end{align}
I would like to try and decompose these gates as compositions of standard rotations, i.e. 
\begin{align}
 R_x(\theta) = \exp(-i\theta \sigma_x/2)\ \ ,\ \ \sigma_x = \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right) \\
 R_y(\theta) = \exp(-i\theta \sigma_y/2)\ \ ,\ \ \sigma_y = \left(\begin{matrix} 0 & -i \\ i & 0 \end{matrix}\right) \\
 R_z(\theta) = \exp(-i\theta \sigma_z/2)\ \ ,\ \ \sigma_z = \left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right)
\end{align}
However, I'm not really sure how to go about it, mostly due to the minus sign in the (2,2) matrix element. I've tried solving the simultaneous equations e.g. $R_X(\theta)R_Z(\phi) = U$ but I end up getting no solutions. 
 A: The most general form of unitary matrix which can be decomposed into standard rotation is $$U=exp(-i\vec{\sigma}.\hat{n}\frac{\phi}{2})$$
Where $\hat{n}$ is the unit vector corresponding to axis of rotation and $\phi$ is angle of rotation.
Determinant of $U$ is 1. But the given matrix has determinant -1. So, I think, it can not be decomposed into rotation.
A: Consider the most general rotation as 
$$
R_{\mathbf{n}}(\theta)=e^{i \theta (\mathbf{n}.\mathbf{\sigma})}
 = \begin{pmatrix}
\cos \theta + i n_{z} \sin \theta & i n_{x} \sin \theta + n_{y} \sin \theta \\
in_{x} \sin \theta -n_{y} \sin \theta & \cos \theta - i n_{z} \sin \theta
\end{pmatrix},
$$
which we have obtained using $exp(i \theta (\mathbf{n}.\mathbf{\sigma}) )= \cos \theta \mathbb{1} + i \mathbf{n}.\mathbf{\sigma} \sin \theta$. To keep this rotation unitary, we should also set the norm of $\mathbf{n}$ to one which is equivalent to devide the matrix by its determinant. 
Putting everything together, we would end up with a general form of
$$
R_{\mathbf{n}}(\theta) =
\frac{1}{ \sqrt{|a|^{2} + |b|^{2}}}
\begin{pmatrix}
a & b \\
-b^{*} &  a^{*}
\end{pmatrix}.
$$ 
As you may have noticed, the minus sign is not on the right component which implies that we should act by one more $\sigma_{z}$ on this rotation. Finally,
$$
\sigma_{z} R_{\mathbf{n}}(\theta) =
\frac{1}{ \sqrt{|a|^{2} + |b|^{2}}}
\begin{pmatrix}
a & b \\
b^{*} &  -a^{*}
\end{pmatrix}.
$$
You can also view $\sigma_{z}$ as $-iR_{z}(\pi/2)$.
