Photoemission footprints of extrinsic plasmarons

A prediction of how to experimentally distinguish excitations of extrinsic plasmarons from intrinsic plasmarons is presented. In surface systems where excitations of acoustic surface plasmons are possible, it is shown that the photoelectron yield in normal photoemission should decay according to an inverse square-root dependence with respect to the photon energy. A computational analysis of the system $p(2\times 2)$-K/graphite confirms this prediction.

Figure 1

Calculated $p(2\times 2)$-K/graphite band structure [25]. Red and blue color lines indicate the quantum well band and the lowest and the highest branches of the folded ${\pi }^{*}$ and $\pi$ bands, respectively. $\overline{K^{\prime }}$ and $\overline{M^{\prime }}$, represent the $\overline{K}$ and $\overline{M}$ points of the folded band structure due to the ($2\times 2$) overlayer of potassium.

Figure 2

ASP dispersion of the system $p(2\times 2)$-K/graphite. The arrows indicate the directions $\overline{\mathrm{\Gamma }}-\overline{M}$ and $\overline{\mathrm{\Gamma }}-\overline{K}$. The solid lines are extracted from the calculated $\text{Im}\left[g({\mathbf{q}}_{\parallel },\omega )\right]$ in [23]. The dashed lines indicate the strong damping of the ASP due to incoherent excitations of electron-hole pairs.

Figure 3

Photoelectron intensity as a function of $k_{\parallel }$ and $E=-{\epsilon }_b-\epsilon$, where $\epsilon$ is the plasmaron excitation energy. The photon energy is 10 eV.

Figure 4

Photoelectron intensity in the $\overline{\mathrm{\Gamma }}$ point ($k_{\parallel }=0$) as a function of photon energy and $E=-{\epsilon }_b-\epsilon$, where $\epsilon$ is the plasmaron excitation energy.

Figure 5

Solid line: Calculated photoelectron yield versus photon energy at the plasmaron peak for normal emission ($k_{\parallel }=0$). The peak is located 1.29 eV below the Fermi energy. Filled circles: Fitted intensity versus photon energy $h\nu$ with to functional form given in Eq. (6) with $A=1.99$ and $\alpha =0.49$.