Peter,
Thank you very much.
Most likely it is not necessary to accept the changes of the unit cell
vectors suggested by sgroup.
You can just try the combination of nn and symmetry, and when you get the
same number of atoms/multiplicities/symmetry
operations/point symmetries (case.outputs) there is no need to follow the
sgroup suggestion.
For low symmetry, monoclinic or triclinic cells there are many equivalent
possibilities to define a unit cell.
I went back and checked. At least for the primitive cell, when I just do
nn and symmetry, the multiplicities, the number of symmetry operations, and
the point groups of each atom do agree. But the case.struct_st has the
lattice type as "H" even though the space group is monoclinic.
Case.struct_sgroup has "CXZ LATTICE,NONEQUIV.ATOMS: 10 5 C2" The warning
from sgroup was "bravais lattice has changed". Can I avoid accepting the
sgroup struct file in this case?
---------------- from case.struct_st
berlin_3
H 10
RELA
9.558538 9.558538 21.105402 90.000000 90.000000120.000000
---------------- from case.struct_sgroup
berlin_3
CXZ LATTICE,NONEQUIV.ATOMS: 10 5 C2
RELA
16.555873 26.824148 9.558538 90.000000 90.000000128.111975
------------------ case.struct_init is below
Thanks,
David
------------------ case.struct_init (The first Al atom has been labeled Al1.
Before editing Atom -1 had multiplicity 3.)
berlin_3
H 6 152_P3121
RELA
9.558538 9.558538 21.105402 90.000000 90.000000120.000000
ATOM -1: X=0.47199740 Y=0.00000000 Z=0.33333333
MULT= 1 ISPLIT= 8
Al1 NPT= 781 R0=0.00010000 RMT= 1.5900 Z: 13.00000
0.0000000-0.5000000 0.8660254
0.0000000-0.8660254-0.5000000
1.0000000 0.0000000 0.0000000
-5: X=0.00000000 Y=0.47199740 Z=0.66666666
MULT= 1 ISPLIT= 8
Al NPT= 781 R0=0.00010000 RMT= 1.5900 Z: 13.00000
0.0000000-0.5000000 0.8660254
0.0000000-0.8660254-0.5000000
1.0000000 0.0000000 0.0000000
-6: X=0.52800260 Y=0.52800260 Z=1.00000000
MULT= 1 ISPLIT= 8
Al NPT= 781 R0=0.00010000 RMT= 1.5900 Z: 13.00000
0.0000000-0.5000000 0.8660254
0.0000000-0.8660254-0.5000000
1.0000000 0.0000000 0.0000000
ATOM -2: X=0.47324468 Y=0.00000000 Z=0.83333333
MULT= 3 ISPLIT= 8
-2: X=0.00000000 Y=0.47324468 Z=0.16666666
-2: X=0.52675532 Y=0.52675532 Z=0.50000000
P NPT= 781 R0=0.00010000 RMT= 1.3300 Z: 15.00000
0.0000000-0.5000000 0.8660254
0.0000000-0.8660254-0.5000000
1.0000000 0.0000000 0.0000000
ATOM -3: X=0.41577315 Y=0.28157150 Z=0.40057705
MULT= 6 ISPLIT= 8
-3: X=0.71842850 Y=0.13420165 Z=0.73391038
-3: X=0.28157150 Y=0.41577315 Z=0.59942294
-3: X=0.86579835 Y=0.58422685 Z=0.06724372
-3: X=0.58422685 Y=0.86579835 Z=0.93275628
-3: X=0.13420165 Y=0.71842850 Z=0.26608961
O NPT= 781 R0=0.00010000 RMT= 1.4700 Z: 8.00000
1.0000000 0.0000000 0.0000000
0.0000000 1.0000000 0.0000000
0.0000000 0.0000000 1.0000000
ATOM -4: X=0.41765928 Y=0.24912454 Z=0.88745047
MULT= 6 ISPLIT= 8
-4: X=0.75087546 Y=0.16853474 Z=0.22078381
-4: X=0.24912454 Y=0.41765928 Z=0.11254952
-4: X=0.83146526 Y=0.58234072 Z=0.55411714
-4: X=0.58234072 Y=0.83146526 Z=0.44588285
-4: X=0.16853474 Y=0.75087546 Z=0.77921619
O NPT= 781 R0=0.00010000 RMT= 1.4700 Z: 8.00000
1.0000000 0.0000000 0.0000000
0.0000000 1.0000000 0.0000000
0.0000000 0.0000000 1.0000000
6 NUMBER OF SYMMETRY OPERATIONS
0 1 0 0.00000000
1 0 0 0.00000000
0 0-1-0.00000001
1
1 0 0 0.00000000
0 1 0 0.00000000
0 0 1 0.00000000
2
-1 1 0 0.00000000
-1 0 0 0.00000000
0 0 1 0.66666667
3
0-1 0 0.00000000
1-1 0 0.00000000
0 0 1 0.33333333
4
-1 0 0 0.00000000
-1 1 0 0.00000000
0 0-1 0.33333333
5
1-1 0 0.00000000
0-1 0 0.00000000
0 0-1 0.66666666
6
Precise positions
0.471997399095264 0.000000000000000 0.333333330000000
0.000000000000000 0.471997399095264 0.666666663333333
0.528002600904736 0.528002600904736 -0.000000003333333
0.473244676107465 0.000000000000000 0.833333330000000
0.000000000000000 0.473244676107465 0.166666663333333
0.526755323892535 0.526755323892535 0.499999996666667
0.415773150900341 0.281571502463394 0.400577050835227
0.718428497536606 0.134201648436947 0.733910384168560
0.281571502463394 0.415773150900341 0.599422942498107
0.865798351563052 0.584226849099659 0.067243717501893
0.584226849099659 0.865798351563052 0.932756275831440
0.134201648436947 0.718428497536606 0.266089609164773
0.417659284247994 0.249124540269199 0.887450471713884
0.750875459730801 0.168534743978795 0.220783805047217
0.249124540269199 0.417659284247994 0.112549521619450
0.831465256021205 0.582340715752006 0.554117138380550
0.582340715752006 0.831465256021205 0.445882854952783
0.168534743978795 0.750875459730801 0.779216188286116
-----Original Message-----
From: wien-boun...@zeus.theochem.tuwien.ac.at
[mailto:wien-boun...@zeus.theochem.tuwien.ac.at] On Behalf Of Peter Blaha
Sent: Friday, April 10, 2015 10:34 PM
To: A Mailing list for WIEN2k users
Subject: Re: [Wien] Which fermi energy for XPS?
Most likely it is not necessary to accept the changes of the unit cell
vectors suggested by sgroup.
You can just try the combination of nn and symmetry, and when you get the
same number of atoms/multiplicities/symmetry operations/point symmetries
(case.outputs) there is no need to follow the sgroup suggestion.
For low symmetry, monoclinic or triclinic cells there are many equivalent
possibilities to define a unit cell.
PS: We don't count "electrons", we count atoms/unit cell. 18 x 4 atoms
should be a reasonable number. Your k-mesh sounds ok.
PPS: without core hole: the small and large cell should give identical
results when the k-mesh is equivalent (and all other computational
parameters, in particular RMT values, too).
Am 11.04.2015 um 00:45 schrieb David Olmsted:
Ouch! That is too bad. Thank you for letting me know. Am I right in
thinking that in a computational size cell, the missing 1/2 electron
will lower the Fermi energy from what it "should be" for a macroscopic
cell?
That might mean I have to do very large supercells, or some kind of
finite-size scaling. (For these two structures, there are 720
electrons for the supercell, and 180 for the primitive cell.)
The k-meshes I have used are not exactly compatible because the
monoclinic angle is different in the two structures, so the lengths of
the reciprocal lattice vectors are not in simple ratios. After making
one Al atom unique, both structures have space group 5 (C2) but the
smaller cell has a monoclinic angle of 128.1 degrees, and the
supercell has a monoclinic angle of 147.5 degrees. The kpoints meshes
are 8,4,8 with 18 atoms for the primitive cell, giving atoms*kpoints
of 4,608, and 4,2,4 with 72 atoms for the supercell, for kpoints*atoms
of only 2,304. The only test of the kmesh I have made so far is one of
5,2,5 instead of 8,4,8 for the primitive cell.
The Fermi energy is 0.0939 Ry for 5,2,5 compared with 0.0944 for 8,4,8.
This difference is small compared to the difference between these and
the supercell where the Fermi energy is .055 Ry. 5,2,5 and 18 atoms
give kpoints*atoms of 900, so "coarser" than the kmesh for the
supercell, so I think the difference is not simply a matter of
kpoints. (These are for the configurations with the half core-hole.)
Thanks,
David
-----Original Message-----
From: wien-boun...@zeus.theochem.tuwien.ac.at
[mailto:wien-boun...@zeus.theochem.tuwien.ac.at] On Behalf Of Peter
Blaha
Sent: Friday, April 10, 2015 1:24 PM
To: A Mailing list for WIEN2k users
Subject: Re: [Wien] Which fermi energy for XPS?
No, I don't think so.
Every calculation uses its own Energy-zero (the average of the
Coulomb-potential in the interstitial region is set to zero), so
clearly one must use EF and E-2p from the same (half-core hole)
calculation.
Eventually, you can check the k-mesh, as with a small k-mesh, EF could
vary a bit.
(I hope you have used "comparable k-meshes". This means the mesh for
the
2x2x1 supercell should be by by a factor of two smaller in x,y than
for the primitive cell
(eg. 2x2x2 vs 4x4x2)
Am 10.04.2015 um 19:33 schrieb David Olmsted:
I am modeling XPS binding energy using a half core-hole, offset by
background charge. As I understand the method that has been
explained here recently, one computes the binding energy as the
energy of the state from case.scfc minus the Fermi energy from ':FER' in
case.scf.
Should the Fermi energy be for the configuration with the half
core-hole, or a configuration without the core-hole? As explained
below, from my results it looks as if it should be the same
configuration,
but without the core hole.
Some details:
Version 14.2
I am computing the differences in the XPS binding energy for Al-2p
for cyrstals in the Al-P-O-H system to see how the binding energy
changes between hydrated and non-hydrated configurations. This is
for comparison with experimental results. (The actual material is
amorphous, but I am hoping the effects of on the spectra will be at
least qualitatively
similar.)
The simplest structure is AlPO4, berlinite. I have run two
configurations, the primitive cell with 18 atoms, including 3 Al
atoms, and a 2x2x1 supercell. In each case I have made one Al
unique, then added one-half core-hole in case.inc and offset it with
-0.5
background charge in case.inm.
For simplicity I will show the results just for the triplet state.
Lines are from case.scf and case.scfc.
-------- 2x2x1 supercell, no core-hole
:LABEL4: using the command: run_lapw -ec 0.00001 -p <skip>
:FER : F E R M I - ENERGY(TETRAH.M.)= 0.0547409802
:NEC01: NUCLEAR AND ELECTRONIC CHARGE 720.00000 720.00112
:NEC02: NUCLEAR AND ELECTRONIC CHARGE 720.00000 720.00000
:NEC03: NUCLEAR AND ELECTRONIC CHARGE 720.00000 720.00000
-------- primitive cell, no core-hole
:LABEL4: using the command: run_lapw -ec 0.00001 -p -NI <skip>
:FER : F E R M I - ENERGY(TETRAH.M.)= 0.0564539224
:NEC01: NUCLEAR AND ELECTRONIC CHARGE 180.00000 180.00073
:NEC02: NUCLEAR AND ELECTRONIC CHARGE 180.00000 180.00000
:NEC03: NUCLEAR AND ELECTRONIC CHARGE 180.00000 180.00000
-------- 2x2x1 supercell, half core-hole
:LABEL4: using the command: run_lapw -ec 0.00001 -p <skip>
:WARN : CHARGED CELL with -0.500
:FER : F E R M I - ENERGY(TETRAH.M.)= 0.0609755546
:NEC01: NUCLEAR AND ELECTRONIC CHARGE 719.50000 719.50115
:NEC02: NUCLEAR AND ELECTRONIC CHARGE 719.50000 719.50000
:NEC03: NUCLEAR AND ELECTRONIC CHARGE 719.50000 719.50000
<case.scfc>
:2P 001: 2P -5.274530454 Ry
------- primitive cell, half core-hole
:LABEL4: using the command: run_lapw -ec 0.00001 -p -NI
:WARN : CHARGED CELL with -0.500
:FER : F E R M I - ENERGY(TETRAH.M.)= 0.0944258517
:NEC01: NUCLEAR AND ELECTRONIC CHARGE 179.50000 179.50067
:NEC02: NUCLEAR AND ELECTRONIC CHARGE 179.50000 179.50000
:NEC03: NUCLEAR AND ELECTRONIC CHARGE 179.50000 179.50000
<case.scfc>
:2P 001: 2P -5.268297265 Ry
--------------
The energy of the state differs by 6 mRy (85 meV) between the
supercell and the primitive cell, making me hopeful that the
supercell is reasonably converged as to size. The Fermi energy,
though differs by 40 mRy (540 meV), so probably the supercell is not
converged with respect to size for the Fermi energy. In the limit of
a large supercell, it would seem that the Fermi energy should
converge to the Fermi energy for the configuration without the core
hole. So it seems to me that I should use the Fermi energy from the
configuration without the core-hole and compute the binding energy as
-5.2745 - 0.0547 =
-5.329 Ry. Is this correct?
Thanks,
David
David Olmsted
Assistant Research Engineer
Materials Science and Engineering
210 Hearst Memorial Mining Building
University of California
Berkeley, CA 94720-1760
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--
-----------------------------------------
Peter Blaha
Inst. Materials Chemistry, TU Vienna
Getreidemarkt 9, A-1060 Vienna, Austria
Tel: +43-1-5880115671
Fax: +43-1-5880115698
email: pbl...@theochem.tuwien.ac.at
-----------------------------------------
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--
-----------------------------------------
Peter Blaha
Inst. Materials Chemistry, TU Vienna
Getreidemarkt 9, A-1060 Vienna, Austria
Tel: +43-1-5880115671
Fax: +43-1-5880115698
email: pbl...@theochem.tuwien.ac.at
-----------------------------------------
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