How to compute the eigenvector of this complex matrix in Grover's algorithm?

Question

We know that $\mathsf{SO}(3)$ matrix stands for the proper rotation in 3D space. But when I read this paper, there is a $\mathsf{SO}(3)$ matrix stands for the general query matrix of Grover's algorithm in $\mathsf{SO}(3)$ form: $$ \left(\begin{array}{ccc} R_{11} & R_{12} & R_{13} \\ R_{21} & R_{22} & R_{23} \\ R_{31} & R_{32} & R_{33} \end{array}\right), $$ where \begin{align*} R_{11}&=\cos \phi\left(\cos ^{2} 2 \beta \cos \theta+\sin ^{2} 2 \beta\right)+\cos 2 \beta \sin \theta \sin \phi\\ R_{12}&=\cos 2 \beta \cos \phi \sin \theta- \cos \theta \sin \phi\\R_{13}&=-\cos \phi \sin 4 \beta \sin ^{2} \frac{\theta}{2}+\sin 2 \beta \sin \theta \sin \phi\\R_{21}&=-\cos (2 \beta) \cos \phi \sin \theta+ \left(\cos ^{2} \frac{\theta}{2}-\cos 4 \beta \sin ^{2} \frac{\theta}{2}\right) \sin \phi\\R_{22}&=\cos \theta \cos \phi+\cos 2 \beta \sin \theta \sin \phi\\ R_{23}&=-\cos \phi \sin 2 \beta \sin \theta- \sin 4 \beta \sin ^{2} \frac{\theta}{2} \sin \phi\\R_{31}&=-\sin 4 \beta \sin ^{2} \frac{\theta}{2}\\R_{32}&=\sin 2 \beta \sin \theta\\R_{33}&=\cos ^{2} 2 \beta+\cos \theta \sin ^{2} 2 \beta\,. \end{align*} The paper says that the eigenvector of this matrix is $(\cot \frac{\phi}{2},1,-\cot 2 \beta \cot \frac{\phi}{2}+\cot \frac{\theta}{2} \csc 2 \beta)^{T}$.

I know this question is very basic and I've tried to use Matlab to calculate it. But I just can't figure out how can the author got the eigenvector of such a simple form? Can it be manually calculated? Is there a better way to compute the eigenvector of this kind of parameterized matrix?

Answer 1

If I were you, I'd ignore the matrix $R$ and instead work with the matrix $Q$. They give you a conversion between vectors in the two different representations.

First, I'm going to simplify things a bit by working with $$ \tilde Q=\left(\begin{array}{cc} e^{i\phi/2} & 0 \\ 0 & e^{-i\phi/2} \end{array}\right)Q. $$ You'll have to compensate for this in the final analysis. Now, if I want to find the eigenvectors of $\tilde Q$, note that I can remove any amount of the identity and the eigenvectors don't change. So, remove $-\cos\frac{\theta}{2}I$. You're left with $$ -\sin\frac{\theta}{2}\left(\begin{array}{cc} \cos2\beta & \sin2\beta \\ \sin2\beta & -\cos2\beta \end{array}\right) $$ Again, for the sake of the eigenvector, you can ignore the overall multiplicative factor ($-\sin\frac{\theta}{2}$). Your eigenvector will be for the form $$ \left(\begin{array}{c} \cos\beta \\ \sin\beta \end{array}\right) $$

I believe that, ultimately, when you incorporate the adjustment between $\tilde Q$ and $Q$, you'll find the eigenvector is $$ \left(\begin{array}{c} \sqrt{1-\sin^2\frac{\phi}{2}\cos^22\beta}+\cos\frac{\phi}{2}\cos2\beta \\ \sin2\beta e^{-i\phi/2} \end{array}\right). $$

However, if you want to analyse the $R$ matrix directly, there must be an equivalent to each of my steps.