Surface plasmon polariton

Remarks

Check: each half has $\kappa_j^2 = \beta^2 - \varepsilon_j k_0^2$, and continuity of $H_y$ and of $E_x$ gives $\kappa_d / \varepsilon_d + \kappa_m / \varepsilon_m = 0$. With $\kappa_d, \kappa_m > 0$ this forces $\varepsilon_m < 0$, and squaring and solving gives the formula for $\beta$, which is real and larger than $\sqrt{\varepsilon_d}\, k_0$ exactly when $\varepsilon_m + \varepsilon_d < 0$. So on each side the wave is an evanescent field. Example: a metal in the Drude model with $\varepsilon_m = 1 - \omega_p^2/\omega^2$ against vacuum ($\varepsilon_d = 1$) supports the wave for $\omega < \omega_p / \sqrt{2}$. Non-example: a transverse electric field ($\mathbf{E}$ along $\hat{\mathbf{y}}$) cannot be bound to such an interface, because continuity of $E_y$ and of its normal derivative would need $\kappa_d + \kappa_m = 0$. For real metals $\varepsilon_m$ has an imaginary part, $\beta$ becomes complex, and the wave decays along the surface too.