Plane rotations and complex diagonalisation

The matrix is real, so $R^\dagger=R^{\mathsf T}$, and

$$\begin{aligned} R^{\mathsf T}R=\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}=\begin{pmatrix}\cos^2\theta+\sin^2\theta&0\\0&\sin^2\theta+\cos^2\theta\end{pmatrix}=\mathbf{1}. \end{aligned}$$

The determinant is $\cos^2\theta+\sin^2\theta=1$. Finally, the top-left entry of $R(\theta)R(\varphi)$ is $\cos\theta\cos\varphi-\sin\theta\sin\varphi=\cos(\theta+\varphi)$, while its bottom-left entry is $\sin\theta\cos\varphi+\cos\theta\sin\varphi=\sin(\theta+\varphi)$; the other two entries follow in the same way. Two successive rotations compose to give a rotation whose angle is the sum of the two angles.

The characteristic polynomial is $(\cos\theta-\lambda)^2+\sin^2\theta=\lambda^2-2\lambda\cos\theta+1$. Its reduced discriminant is $\cos^2\theta-1=-\sin^2\theta$, and its roots are

$$\lambda_\pm=\cos\theta\pm i\sin\theta=e^{\pm i\theta}.$$

They have unit modulus, as for every unitary matrix. They are real only if $\sin\theta=0$, that is, for $\theta\equiv0$ (the identity) or $\theta\equiv\pi$ (a half-turn rotation, equal to $-\mathbf{1}$). In every other case, the rotation has no real eigenvector, which is geometrically evident: a rotation through an arbitrary angle leaves no direction in the plane invariant. It is therefore not diagonalisable over $\R$.

Let us calculate the action of $R(\theta)$ on $(1,-i)$:

$$R(\theta)\begin{pmatrix}1\\-i\end{pmatrix}=\begin{pmatrix}\cos\theta+i\sin\theta\\ \sin\theta-i\cos\theta\end{pmatrix}=\begin{pmatrix}e^{i\theta}\\-i(\cos\theta+i\sin\theta)\end{pmatrix}=e^{i\theta}\begin{pmatrix}1\\-i\end{pmatrix}.$$

Similarly, $R(\theta)(1,i)=e^{-i\theta}(1,i)$. Thus $\ket+=\frac{1}{\sqrt2}(1,-i)$ is an eigenvector with eigenvalue $e^{i\theta}$, while $\ket-=\frac{1}{\sqrt2}(1,i)$ has eigenvalue $e^{-i\theta}$. These vectors are normalised and orthogonal with respect to the Hermitian inner product: $\braket+-=\frac12\bigl(1\cdot1+(i)(i)\bigr)=0$, since the bra $\bra+$ has components $\frac{1}{\sqrt2}(1,i)$. Therefore $R(\theta)=e^{i\theta}\ket+\bra++e^{-i\theta}\ket-\bra-$. In optics, if the two components represent the horizontal and vertical polarisations of a light wave, the vectors $\ket\pm$ represent the two circular polarisations, which are changed only by a phase when the apparatus is rotated.

We have $-i\sigma_y=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ and $\sigma_y^2=\mathbf{1}$. Separating the even and odd terms of the exponential series gives $e^{-i\theta\sigma_y}=\cos\theta\,\mathbf{1}-i\sin\theta\,\sigma_y$, which is exactly $R(\theta)$. The matrix $\sigma_y$ is Hermitian, with eigenvalues $\pm1$; one checks that $\sigma_y(1,i)=(1,i)$ and $\sigma_y(1,-i)=-(1,-i)$. The eigenvectors of $R(\theta)=e^{-i\theta\sigma_y}$ are those of $\sigma_y$, with eigenvalues $e^{-i\theta\times(\pm1)}$: we recover $e^{-i\theta}$ for $\ket-=\frac{1}{\sqrt2}(1,i)$ and $e^{i\theta}$ for $\ket+$. The Hermitian matrix $\sigma_y$ is the generator of plane rotations, just as angular momentum is the generator of rotations in quantum mechanics.