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<p style="margin-top:0;margin-bottom:0">Hi</p>
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<p style="margin-top:0;margin-bottom:0">Thanks for your reply, but Prof. Blaha says that I have to remove the space group and chose an R lattice; something like this</p>
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<div>R LATTICE,NONEQUIV.ATOMS: 3 </div>
<div>MODE OF CALC=RELA unit=bohr </div>
<div> 9.520440 9.520440 26.025308 90.000000 90.000000120.000000 </div>
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<div>ATOM -1: X=0.10534000 Y=0.10534000 Z=0.10534000</div>
<div> MULT= 2 ISPLIT= 4</div>
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but you are saying that I should change to P (not R as Prof Blaha says) and set to 2 or 4 inequivalent atoms. But<span style="font-family: Calibri, Helvetica, sans-serif, EmojiFont, "Apple Color Emoji", "Segoe UI Emoji", NotoColorEmoji, "Segoe UI Symbol", "Android Emoji", EmojiSymbols; font-size: 16px;"> as
I understand I need at least 3 inequivalent atoms (2 Fe atoms and 1 O atom) with MULT=1 for each Fe atom.</span>
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<div><font size="2">Dr. Israel Pérez</font>
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<div id="divRplyFwdMsg" dir="ltr"><font face="Calibri, sans-serif" style="font-size:11pt" color="#000000"><b>From:</b> Wien <wien-bounces@zeus.theochem.tuwien.ac.at> on behalf of delamora <delamora@unam.mx><br>
<b>Sent:</b> Sunday, May 19, 2019 12:35:58 PM<br>
<b>To:</b> wien@zeus.theochem.tuwien.ac.at<br>
<b>Subject:</b> Re: [Wien] Help on Core hole calculations</font>
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<div dir="ltr">Sorry</div>
<div dir="ltr">It should be</div>
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<pre><span></span><span><pre><span>P LATTICE,NONEQUIV.ATOMS: 4 =></span></pre></span><span></span><span></span><span></span><span>P LATTICE,NONEQUIV.ATOMS: 2 <br><br></span></pre>
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<pre class="x_x_x_western" style="font-variant-numeric:normal; font-variant-east-asian:normal"><font color="#000000">Thank you for your valuable feedback, I appreciate it very much. I still do not know how to create the structure file. Please see below.</font>
<font color="#000000">No. You should not run a supercell for a R structure, since this will </font>
<font color="#000000">even with 1x1x1 create a 3 times (conventional hexagonal) cell.</font>
<font color="#000000">Instead,</font>
<font color="#000000">i) remove the spacegroup and change to "R"-lattice.</font><br><br>*****<br>Change<br><br><span>Title <br></span><div>R LATTICE,NONEQUIV.ATOMS: 2 167_R-3c <br></div><div>MODE OF CALC=RELA unit=ang <br></div><div> 9.519310 9.519310 26.005287 90.000000 90.000000120.000000<br></div><div>ATOM -1: X=0.35500000 Y=0.35500000 Z=0.35500000<br></div><span><br>to<br><br><span>Title <br></span><div>P LATTICE,NONEQUIV.ATOMS: 4 <br></div><div>MODE OF CALC=RELA unit=ang <br></div><div> 10.263863 10.263863 10.263863 55.256000 55.256000 55.256000<br></div><div>ATOM -1: X=0.14500000 Y=0.14500000 Z=0.14500000<br></div><span></span><br>where <pre><span>10.263863 and 55.256000<br><br>you can get it from xcrysden<br><br>In this way you have 4 Fe and 6 O<br><br>*****<br></span></pre><br></span><br><font color="#000000">ii) split the 4 Fe positions into 2 and 2 (Fe1 and Fe2). This splitting </font>
<font color="#000000">can be done in different ways (up-up-dn-dn along z; or in other ways. </font>
<font color="#000000">Please check literature which one is correct). After nn (or sgroup) you </font>
<font color="#000000">should have only 2 Fe positions.</font>
2 Fe positions for each Fe1 and Fe2 or one for each Fe atom?
The steps are not clear for me. The initial structure has 10 atoms per unit cell (SG=167 file attached), then you say I should select R and then click on “split”. This creates a new inequivalent Fe atom and leaves the previous with 3 positions. So for the second atom I created a new position and removed one position from the first atom. This leaves 2 Fe atoms, Fe1 and Fe2, with two positions each and of course MULT=2 for each. However, when I run nn this suggests a new inequivalent oxygen atom. If accept this and run sgroup, sgroup suggests a new space group R32 (155) and I end up with two inequivalent Fe atoms with MULT=2 for each Fe atom and two inequivalent oxygen atoms. Could you please elaborate in more detail how to create the structure with two inequivalent Fe atoms (Fe1 and Fe2) and MULT=1. How many atoms the final cell should have?</pre>
<pre class="x_x_x_western" style="font-variant-numeric:normal; font-variant-east-asian:normal"><br></pre>
<pre class="x_x_x_western" style="font-variant-numeric:normal; font-variant-east-asian:normal">Much more critical than where to add the extra electron is, that you now <br></pre>
<pre class="x_x_x_western" style="font-variant-numeric:normal; font-variant-east-asian:normal"><font color="#000000">create a big supercell (as big as you can handle), break symmetry and </font></pre>
<pre class="x_x_x_western" style="font-variant-numeric:normal; font-variant-east-asian:normal"><font color="#000000"></font>Cells must always be neutral. Either you have a huge core leakage (NEVER </pre>
<pre class="x_x_x_western" style="font-variant-numeric:normal; font-variant-east-asian:normal"><font color="#000000">change cutt-off to -4Ry, it means huge core leakage), or you have done </font>
<font color="#000000">something wrong when putting the extra electron or the background </font>
<font color="#000000">charge. It could also be that your core-hole atom has MULT=2, then </font>
<font color="#000000">obviously you would have to add 2 electrons - but you should never have </font>
<font color="#000000">core-hole atoms with MULT=2 (see above).</font>
I checked this and I have MULT=2 so I have to make sure that I have MULT=1 for, say, Fe1 and this depends on the structure file, that I do not know how to make (see above). Still I do not understand what is called background charge. </pre>
<pre class="x_x_x_western" style="font-variant-numeric:normal; font-variant-east-asian:normal">I’ll appreciate your comments</pre>
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Best regards</p>
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<div id="x_x_x_divRplyFwdMsg" dir="ltr"><font color="#000000" face="Calibri, sans-serif" style="font-size:11pt"><b>From:</b> Israel Omar Perez Lopez<br>
<b>Sent:</b> Thursday, May 16, 2019 7:51:23 PM<br>
<b>To:</b> wien@zeus.theochem.tuwien.ac.at<br>
<b>Subject:</b> Help on Core hole calculations</font>
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<pre style="white-space:pre-wrap">Dear all</pre>
<pre style="white-space:pre-wrap">I few days ago I sent this email but I noticed that it appears in the mailing list as SING IN ALERT and since I have not received any reply I thought that perhaps it was considered as a malicious mail. So I decided to send it again through this email account. I apologise if this causes inconveniences but I really need some feedback. <br></pre>
<pre style="white-space:pre-wrap">I have read some of the discussions in the mailing list (<a href="https://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/msg11037.html" id="LPlnk952916" class="x_x_x_x_OWAAutoLink">https://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/msg11037.html</a>) and somewhere else but I couldn’t find some useful information to solve my doubts; I actually got more confused. I am doing core hole calculations for alpha-Fe2O3 (hematite) and I want to compute just the DOS for this system for holes from Fe 1s, 2p, and 3s (and possibly 3p). I have some doubts regarding the correct procedure. I would appreciate any feedback on the following points.
1) I am not sure if the crystal structure I made is correct. As a starting point I used the .cif file downloaded from AMCSD web site with space group R-3c (167), then converted it to .struct file with RMT Fe=1.91, RMT O=1.64. This structure is corundum and all Fe atoms are equivalent. However, to reproduce experimental observations, spin-polarized calculations with LDA+U are required. So I need to make Fe atoms nonequivalent and assign the spin orientation along the [0001] direction, with Fe atoms alternating spin in planes {0001}. For this I created a 1X1X1 supercell (just to test) from the original file and later execute x nn, x sgroup and x symmetry. Sgroup suggested a new space group P3c1 (158) with 6 nonquivalent Fe atoms and 3 O equivalent atoms which I accepted and then labelled the Fe atoms as Fe1,... Fe6. Then I called instgen and assigned the spin orientation to each Fe and O atom (oxygen is non-magnetic), and run the rest of the initialization programs with 100 k points, ecut= -6 Ry, RKmax=7, and for the functional PBE. Thus, since now I have six nonequivalent Fe atoms, I would have, after the scf cycle, 6 partial DOS (actually 12, 6 up and 6 down), one for each Fe atom. Is this correct?
2) For core hole calculations there exist at least two options to send the core electron. After removing a core hole (from a core level: 1s, 2s, 2p, or 3s) in the file case.inc we can send this electron to:
a) the valence electrons in case.in2c; or,
b) to the background charge in case.inm
I’m not sure where to add the electron, what physical criterion should I apply? In any case, I decided to try both cases. For a), I removed the electron from the file case.inc for the levels 1s, 2s and 3s (of course, in three different computing sessions). For b), I removed one electron from the same three levels in case.inc. The core electron (for each calculation) was removed from only one Fe atom, say Fe1 (the remaining atoms kept the same configuration), as follows:
Core hole from Fe1 1s
1,-1,1 ( N,KAPPA,OCCUP) 1s electron removed
2,-1,2 ( N,KAPPA,OCCUP)
2, 1,2 ( N,KAPPA,OCCUP)
2,-2,4 ( N,KAPPA,OCCUP)
3,-1,2 ( N,KAPPA,OCCUP)
Core hoe from Fe1 2p
1,-1,1 ( N,KAPPA,OCCUP)
2,-1,2 ( N,KAPPA,OCCUP)
2, 1,1 ( N,KAPPA,OCCUP) 2p electron removed
2,-2,4 ( N,KAPPA,OCCUP)
3,-1,2 ( N,KAPPA,OCCUP)
Core hole from Fe1 3s
1,-1,1 ( N,KAPPA,OCCUP)
2,-1,2 ( N,KAPPA,OCCUP)
2, 1,1 ( N,KAPPA,OCCUP)
2,-2,4 ( N,KAPPA,OCCUP)
3,-1,1 ( N,KAPPA,OCCUP) 3s electron removed
Electron added to the valence electrons
The electron was added to the valence electrons in case.in2. Before we had
-12.0 276.0 0.50 0.05 1 EMIN, NE, ESEPERMIN, ESEPER0, iqtlsave
And after adding the electron the file look like
-12.0 277.0 0.50 0.05 1 EMIN, NE, ESEPERMIN, ESEPER0, iqtlsav
3) Here I have a question, where exactly is the electron added? The 3d band, the 4s band? In the user’s guide, it reads that the -12 stands for the lower energy cut-off for defining the range of occupied states and NE is the number of electrons (per unit cell) in that energy range. Does it mean that those 277 electrons are in states with energies higher than -12 Ry. In other words, is it possible to decide where exactly in the valence to put the excited electron?
Electron added to the background charge
The electron was added to the background charge in case.inm. Before we had
MSR1 0.0 YES (BROYD/PRATT, BG charge (-1 for core hole), norm)
According to Prof. Blaha for core hole we need to type -1, so the file after the change looks like this
MSR1 -1 YES (BROYD/PRATT, BG charge (-1 for core hole), norm)
4) Here I have another question. The user’s guide reads that the second number in the line, bgch, means “background charge to apply to the cell”. Normally, this is set to 0.0, meaning that the cell is not charged??? I do not have clear what is called background charge. Could you please explain what is the nature of this background charge. As I understand, the cell/material is electrically neutral, and we only have charges due to nuclei and electrons, where does this background charge come from? This question is connected with the following.
Since I want LDA+U, I also included the files case.orb and case.indmc. Then I executed:
runsp_lapw -orb -ec 0.001
After the scf cycle, Prof. Blaha suggested to check charge neutrality. I did in all cases using grep :NEC01 in case.scf and obtained:
NEC01: NUCLEAR AND ELECTRONIC CHARGE 456.00000 454.90177
so, I have no charge neutrality. I added the electron to either the valence or background before running the scf cycle, as suggested everywhere, why there is no charge neutrality?
5) After the scf-cycle, for the case in which the electron was added to the valence electrons, I removed the electron from the file case.in2c and computed the DOS. But, what should I do for the case in which the electron was added to the background? Should I reset to 0.0 in case.inm before calculating the DOS?
6) My final question. When we run lstart we define the cut-off energy, say – 6 Ry, that separates core from valence states. In my case, with this energy, I have, in case.outputst, as core states 1s, 2s, 2p, and 3s; and as valence states, 3p, 3d, 4s. However, in the file case.inst core states are considered those of Ar and 3d and 4s as valence states. So, during the computation of the scf, how is the 3p level taken, as a core or a valence state? If I changed cut-off energy=-4 Ry to consider the 3p as valence state (and so do core hole calculations for this level) then I get a charge leakage warning! I am confused about this apparent contradiction in these two files.
I would be happy if you could help me.
Best Regards
Israel Perez</pre>
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<font size="2">Insurgentes Sur No. 1582, </font></p>
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<font size="2">Col. Crédito Constructor, C.P. 03940</font></p>
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<font size="2">Del. Benito Juárez, México D. F.</font></p>
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