My PhD thesis - Condensed Matter Theory - Imperial College London
My PhD thesis - Condensed Matter Theory - Imperial College London
My PhD thesis - Condensed Matter Theory - Imperial College London
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CHAPTER 5.<br />
THE JELLIUM SLAB<br />
The DMC results clearly lie above those obtained in DFT. More evidence is provided<br />
by Almeida et al. [3], who studied the jellium sphere system, and were able to match<br />
their DFT surface energies with Sottile and Ballone’s finite-system DMC results over<br />
a range of densities (although the precise value r s = 2.07 was not included in their<br />
calculations). With the correction suggested by Pitarke, the calculation of Acioli<br />
and Ceperley is brought closer to the DFT results, but this correction cannot be<br />
applied to the calculations of Li et al.<br />
In addition, both groups (Li et al. and Acioli and Ceperley) applied only the<br />
independent-particle finite-size correction 4 to the slab energies, and not the Coulomb<br />
correction. The Coulomb finite-size correction would increase the slab energies;<br />
because the corresponding bulk energies were fully finite-size corrected, making the<br />
correction would raise the DMC surface energy still further, increasing the difference<br />
between the DMC and DFT results. Almeida believes the combined DFT-RPA<br />
calculations to be currently the most accurate, putting the surface energy between<br />
−550 and −590 erg cm −2 .<br />
At this density, the separate contributions to the surface energy are individually<br />
sizeable, but cancel as a whole; Pitarke and Eguiluz [72] give the following<br />
breakdown:<br />
• σ s = −4643 erg cm −2 (kinetic);<br />
• σ es = 1072 erg cm −2 (electrostatic);<br />
• σ xc = 3007 erg cm −2 (exchange-correlation).<br />
This is the opposite of the ideal situation, and is partly why jellium surface energy<br />
calculations are particularly difficult.<br />
A relative error of 10% in the surface energy σ corresponds to around 50 erg<br />
cm −2 or 0.03 mHa bohr −2 . Using equation (4.28) to calculate σ, and assuming that<br />
there is no error in the bulk energy gives<br />
∆σ = N 2A ∆ɛ slab =<br />
3s ∆ɛ<br />
8πrs<br />
3 slab . (5.4)<br />
4 These finite-size errors are discussed in chapter 4.<br />
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