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My PhD thesis - Condensed Matter Theory - Imperial College London

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CHAPTER 8. APPLYING THE PLASMON NORMAL MODE THEORY TO<br />

SLAB SYSTEMS<br />

The cut-off wave vector k c is related to the electron density [73]:<br />

k c ∼ 1 √<br />

rs<br />

. (8.86)<br />

In a homogeneous system, plasmons with wave vector greater than k c are no longer<br />

well-defined; at k ∼ k c , the (almost flat) plasmon dispersion curve merges with the<br />

particle-hole continuum [17]. In general, it is not obvious how to apply the cut-off<br />

in an inhomogeneous system. However, in the slab system under consideration here,<br />

the electron density is approximately constant within the slab and zero outside; it is<br />

therefore reasonable to apply the same cut-off as would be used in a homogeneous<br />

system of equivalent density.<br />

Substituting these functions in place of the {f i } in equation (7.81) gives<br />

χ bulk (r) =<br />

e2 ∑(<br />

∫<br />

φ 1k (r) φ 1k (r ′ )¯n(z ′ ) d 3 r ′<br />

ω p ɛ 0<br />

k<br />

V<br />

∫<br />

)<br />

+ φ 2k (r) φ 2k (r ′ )¯n(z ′ ) d 3 r ′<br />

= e2<br />

ω p ɛ 0<br />

∑<br />

k z<br />

4<br />

k 2 zs sin k zz<br />

V<br />

∫ s<br />

while substitution in equation (7.82) gives<br />

k<br />

0<br />

¯n(z ′ ) sin k z z ′ dz ′ Θ(z)Θ(s − z) (8.87)<br />

u bulk (r, r ′ ) =<br />

e2 ∑(<br />

)<br />

φ 1k (r)φ 1k (r ′ ) + φ 2k (r)φ 2k (r ′ )<br />

ω p ɛ 0<br />

= e2<br />

ω p ɛ 0 V<br />

∑<br />

k<br />

4<br />

k 2 cos k ‖ · (r ‖ − r ′ ‖) sin k z z sin k z z ′<br />

× Θ(z)Θ(s − z)Θ(z ′ )Θ(s − z ′ ). (8.88)<br />

The sum is understood to include the term with k ‖ = 0. The Heaviside functions<br />

appear because the bulk plasmons have zero amplitude outside the slab. Note that<br />

only the modes with k ‖ = 0 contribute to the χ-function; all others integrate to<br />

zero. Taking the density to be constant (to be consistent with the derivation of the<br />

plasmon modes), ¯n(z) = n 0 , giving<br />

χ bulk (r) =<br />

e2 ∑ 4n 0<br />

ω p ɛ 0 k 3 k z<br />

zs sin k zz(1 − cos k z s)Θ(z)Θ(s − z). (8.89)<br />

138

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