 Solution

The characteristic polynomial of the given Jones matrix isso a convenient choice of eigenvalue labels isDirect multiplication gives $J{\mathbf{u}}_{\pm }={\lambda }_{\pm }{\mathbf{u}}_{\pm }$, and ${\mathbf{u}}_{+}^{†}{\mathbf{u}}_{-}=0$. 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 $b\ge 0$. If $b=0$, the eigenvalue is degenerate and every nonzero Jones vector is an eigenvector; the circular pair remains a valid basis.