[Wien] Which fermi energy for XPS?
David Olmsted
olmsted at berkeley.edu
Mon Apr 13 16:48:53 CEST 2015
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-bounces at zeus.theochem.tuwien.ac.at
[mailto:wien-bounces at 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-bounces at zeus.theochem.tuwien.ac.at
> [mailto:wien-bounces at 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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>> l
>>
>
> --
> -----------------------------------------
> Peter Blaha
> Inst. Materials Chemistry, TU Vienna
> Getreidemarkt 9, A-1060 Vienna, Austria
> Tel: +43-1-5880115671
> Fax: +43-1-5880115698
> email: pblaha at 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: pblaha at theochem.tuwien.ac.at
-----------------------------------------
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