I have done intense research, but could not find a satisfying answer. Therefore, pls don't type shallow explanations, such as "due to edge effects etc". I will appreciate if you put some reference as well. Thanks!
It is already in the name. "localized surface plasmons" are not propagating. They are in a simplified view nothing more than an oscillating stationary dipole, a "nano-antenna". There is not dispersion relation, ie. k vs E. Hence, there is nothing to match to.
I hope that helps.
I will expand a bit on the answer from Michael Rüsing which has already captured the key principle.
The k-vector distribution of a state is linked to the Fourier transform of its spatial distribution. Therefore a localised state in space has a very broad k-vector distribution. The extreme of this is a delta function in space, which would have a uniform k-distribution, but the principle is the same for other line shapes as well, e.g. Gaussia or Lorentzian. Therefore the localised plasmon has a very broad k-vector distribution and a wide range of incident k-vector can couple to it.
You can think also as having a large radius of curvature in a nanoparticle allows you to excite the plasmon, and no k vector mismatch exists
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