<html><head><meta http-equiv="Content-Type" content="text/html; charset=utf-8"></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; line-break: after-white-space;" class="">Hello Daniel,<div class=""><br class=""></div><div class="">Thanks for your answer. I have the value for z=0:</div><div class="">P = 0.0000049 (mod 14.7398000) (e/Omega).bohr</div><div class=""><br class=""></div><div class="">Pascal</div><div class=""><br class=""></div><div class=""><div><blockquote type="cite" class=""><div class="">Le 27 févr. 2018 à 07:09, Daniel Stoeffler <<a href="mailto:daniel.stoeffler@ipcms.unistra.fr" class="">daniel.stoeffler@ipcms.unistra.fr</a>> a écrit :</div><br class="Apple-interchange-newline"><div class="">
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<div class=""><p class="">Hello Pascal,</p><p class="">Don't forget the polarization quantum : the jump at z=0.5 looks like coming from a quantum of polarization and can be removed (I mean by adding 1 to your numbers for z > 0.5). In principle, variation of P should remain small between points on your path.</p><p class="">Why don't you have the value for z=0 ?</p><p class="">Best wishes,</p><p class="">Daniel</p><div class=""> <br class="webkit-block-placeholder"></div>
<div class=""> </div><p class="">Le 2018-02-26 22:13, pboulet a écrit :</p>
<blockquote type="cite" style="padding-left:5px; border-left:#1010ff 2px solid; margin-left:5px; width:100%" class=""><!-- html ignored --><!-- head ignored --><!-- meta ignored -->Dear all,
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<div class="">Does anyone have tried to used Jiang, Levchenko and Rappe’s method to calculate oxidation state of atoms? (Phys. Rev Letters, 108, 2012, 166403.</div>
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<div class="">In brief, the formula is N=V/e \Delta(<span style="text-decoration: underline;" class="">P</span>) x <span style="text-decoration: underline;" class="">R</span> /<span style="text-decoration: underline;" class="">R</span>^2,</div>
<div class="">where N, V, e, \Delta(<span style="text-decoration: underline;" class="">P</span>) and <span style="text-decoration: underline;" class="">R</span> are the oxidation state, cell volume, electron charge, change in polarisation and lattice vector, respectively. \Delta(<span style="text-decoration: underline;" class="">P</span>) x <span style="text-decoration: underline;" class="">R</span> corresponds to the dot product.</div>
<div class="">The idea is just to displace an atom along a path traversing the cell in a way that the atom goes from its location to that in the next cell.</div>
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<div class=""> I have tried with the PbTiO3 example of QE for the calculation of polarization via the Berry phase. If I am right I should obtain +2 as an oxidation state. </div>
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<div class="">I have displaced the Pb atom (crystal position at 0. 0. 0.) along the z axis from 0 to 1. </div>
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<div class="">The enclosed picture depicts N versus z for z=0..1. I used the polarization P given by QE in units of (e/Omega).bohr, since it is similar to the formula above.</div>
<div class="">I have just divided P by z since Delta(P) and R are collinear… and I guess V/e of the formula cancels with e/Omega… I am not sure of these assumptions, however…</div>
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<div class="">First the variation is -2 instead of +2. In addition the evolution is not smooth and the missing points on the curve are calculated to be negative. I have removed them but it is nonsense to me. </div>
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<div class="">Does anyone know what is wrong?</div>
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<div class="">Thank you for your suggestions,</div>
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<div class="">Pascal</div>
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