How to implement this single-qubit unitary?

Question

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.

Answer 1

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 $-iR_{z}(\pi/2)$.

Answer 2

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.