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

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

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?

• $\mathbb{l}$ is not a 3 dimensional vector above. Unless you intended to put some commas? Jun 2 at 8:09
• Very sorry about it. I've added the commas. Jun 2 at 8:34
• – glS
Jun 2 at 10:55

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.

• It's persuasive to do the things in SU(2) first and then change the eigenvector of SU(2) element into one 3D vector. And the 3D vector should be the eigenvector of the corresponding SO(3) element. Unluckily, the last belief is not such right which will add a lot of computation therefore, while still simpler than SO(3) way. For example, $\left(\begin{array}{cc} 2 & 3 \\ 4 & 6 \end{array}\right)$ has eigenvector $(3,-2)^T$ while $(3,-2e^{i\phi})^T$ is not an eigenvector of $\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{i\phi} \end{array}\right)\left(\begin{array}{cc} 2 & 3 \\ 4 & 6 \end{array}\right)$ Jun 2 at 10:28
• While I freely admit to not having had time to do the calculating fully yet (and so I could easily be wrong), your counter-example is not valid because the matrix you're using does not have the structure (based on parameter $\beta$) that we're relying on. Jun 2 at 11:06