= Solution
The <characteristic polynomial> of the given <Jones matrix> is
$$
\det(J-\lambda I)=(a-\lambda)^2-b^2,
$$
so a convenient choice of <eigenvalue> labels is
$$
\boxed{\lambda_+=a+b,\quad \mathbf u_+=\frac1{\sqrt2}\begin{pmatrix}1\\-i\end{pmatrix};\qquad
\lambda_-=a-b,\quad\mathbf u_-=\frac1{\sqrt2}\begin{pmatrix}1\\i\end{pmatrix}.}
$$
Direct multiplication gives $J\mathbf u_\pm=\lambda_\pm\mathbf u_\pm$, and $\mathbf u_+^\dagger\mathbf u_-=0$. They represent opposite <circular polarizations>. Since $a,b$ are real, $J$ is a <Hermitian matrix> and both <eigenvalues> are real, but they need not be positive. A negative <eigenvalue> adds a <wave phase> reversal to that mode. The labels are algebraic: $\lambda_+$ is the larger numerical <eigenvalue> only when $b\geq0$. If $b=0$, the <eigenvalue> is degenerate and every nonzero <Jones vector> is an <eigenvector>; the circular pair remains a valid basis.
Back to article page
