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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?

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    $\begingroup$ There is no translation symmetry for a nanoparticle. $\endgroup$
    – user137289
    Commented Jan 3, 2018 at 0:02
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    $\begingroup$ @Pieter I don't understand your remark. Can you explain me more what you mean ? $\endgroup$
    – StarBucK
    Commented Jan 3, 2018 at 0:13
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    $\begingroup$ 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. $\endgroup$
    – user137289
    Commented Jan 3, 2018 at 0:27

1 Answer 1

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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.

Pictorially, you can imagine the dispersion of a nano particle plasmon as a horizontal line which is guaranteed to intersect the light cone.

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  • $\begingroup$ Thank you a lot for your answer. I suppose that then for nanoparticle there are localized surface plasmons at the surface of the particle (surface charge oscillate in time not in space). But there also should be localized bulk plasmon (oscillation of charge in the volume of the nanoparticle). But i have never read such a term "LOCALIZED" bulk plasmon so I would like to check with you that indeed bulk plasmon still exist for nanoparticles ? (With a différent behavior as it doesnt propagate but just locally oscillate) $\endgroup$
    – StarBucK
    Commented Jan 3, 2018 at 2:14
  • $\begingroup$ I would expect the existence of a bulk plasmon for nanoparticles to highly depend on the size of the particles and carrier density. For small enough particles you can't really distinguish between the bulk and surface anyways. $\endgroup$
    – KF Gauss
    Commented Jan 3, 2018 at 5:26
  • $\begingroup$ 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 ? $\endgroup$
    – StarBucK
    Commented Jan 3, 2018 at 9:59
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    $\begingroup$ The equation you mention is no longer accurate for nanoparticles because the plasmon is not planewave like across the surface, since the surface is small and highly curved. Instead you have to study the exact eigenvalues, and will find only wavelengths that are integer fractions of the particle size etc. $\endgroup$
    – KF Gauss
    Commented Jan 3, 2018 at 14:39
  • $\begingroup$ Thank you a lot for everything. I am a beginner in this field and I found very hard to find the information you gave me. Plus your explanation was very clear. $\endgroup$
    – StarBucK
    Commented Jan 3, 2018 at 14:49

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