# Difference between Surface plasmon and Localised surface plasmon

Here is what I understood from the plasmons:

• Bulk plasmon: we need longitudinal electric wave to excite them. Then it is not possible to excite them with natural light.

• Surface plasmon: we can't "directly" excite them with natural light. Indeed the vacuum dispersion relation of light and the dispersion relation of the surface plasmon don't cross. We thus need to increase the parallel of the surface component of the EM wave (to do it we can use evanescent waves for example).

• Localised surface plasmon: It is a surface plasmon on a small particle (spherical nanoparticle). Using Mie theory we can see that natural light can excite them.

What I don't understand is: Why for localized surface plasmon we don't have the problem of the dispersion relation of light must cross the dispersion relation of the surface plasmon? Does the dispersion relation of the surface plasmon change if the particle start to be very small? I don't understand?

• There is no translation symmetry for a nanoparticle.
– user137289
Commented Jan 3, 2018 at 0:02
• @Pieter I don't understand your remark. Can you explain me more what you mean ? Commented Jan 3, 2018 at 0:13
• Without translation symmetry, there is no reciprocal lattice, no $k$-space, no crystal momentum, no dispersion relation $E$ versus $k$. The nanoparticle is like an isolated atom.
– user137289
Commented Jan 3, 2018 at 0:27

In nanoparticles, like atoms, charge oscillations do not propagate in space (i.e. local). This means they have zero group velocity, or no dispersion ($\omega$ is independent of $k$). So instead of having to match two dispersion relations with different group velocities, you just have to match the localized surface plasmon energy with the vacuum energy.
• Thank you. Just a last question : what exactly changes in the dispersion relation ? Because for me : $k_{SP}=n \frac{\omega}{c} \frac{\epsilon_1 \epsilon_2}{\epsilon_1 + \epsilon_2}$ should be still true for a very small nanosphere because it is obtained from wave propagation on the surface (but the size of the sphere don't really matters). The "on the surface" is just imposed by the fact we have evanescent waves for the E field from both size of this surface. Because when we put the drude permittivity for $2$ and vacuum for $1$ here we don't have an horizontal line ? Commented Jan 3, 2018 at 9:59