Re: [Wien] Bi almost cubic
Dear Gerhard, Thank for your comments. I just want to comment that I used your cell parameters but I applied the "sgroup" and I got an hexagonal cell; R LATTICE,NONEQUIV.ATOMS: 2 166 R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM -1: X=0. Y=0. Z=0. Bi1NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.0 ATOM -2: X=0.5000 Y=0.5000 Z=0.5000 Bi2NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.0 And I got the DOS with the "gap" After that Bi1 => Bi Bi2 => Bi and I got the small cell; R LATTICE,NONEQUIV.ATOMS: 1 166 R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 11.207870 90.00 90.00120.00 ATOM -1: X=0. Y=0. Z=0. Bi1NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.0 And I got the DOS without a "gap" Dear Peter Blaha, Thank you for your comments, I did the bands with the HCP and I found that For the reduced cell the bands do not overlap in the Ef region (-6eV, +8eV) For the large cell the bands overlap and they try to avoid the crossing near Ef (-1Ev, +1eV), which could lead to the "gap" Saludos Pablo ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Bi almost cubic
Hallo Peter, you are right, the pseudo gaps should vanish after the change of symmetry, I was just wondering why the transition seemed to be not continouus and did not think too much about the k-points (I am on the road and can't use our cluster to check 100^3 or more) Maybe Pablo can test with smaller changes and much more k-points how the transition and the band splittings develop with the parameters. PS.: I just do not remember that I ever was asked to shift the atom positions after x symmetry. For test reasons I tried several times calculations using P1, in particular for hexagonal structures with only few atoms, or because I forgot something when starting with a cif file in P1. PSS: the k-meshers were identical as I used them long time ago, where I found a solution of the optimisation close to z=1/4 after I used "wrong" start structural parameters in a calculation and finished it more or less for fun to see the energy but not DOS, etc.. Ciao Gerhard DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy: "I think the problem, to be quite honest with you, is that you have never actually known what the question is." Dr. Gerhard H. Fecher Institut of Physics Johannes Gutenberg - University 55099 Mainz Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von Peter Blaha [peter.bl...@tuwien.ac.at] Gesendet: Freitag, 8. Juli 2022 13:07 An: wien@zeus.theochem.tuwien.ac.at Betreff: Re: [Wien] Bi almost cubic Plot the bands and you will "understand" the minimum at EF in the 2-atom structure when you consider how the tetrahedron method works (i.e. by connecting bands according to energy (and not character). Due to the finite k-mesh you get a lot of "avoided" crossings and a pseudogap in the DOS. Try a mesh of 100**3 or even 200**3 k-points for the DOS (probably you still get some artefacts at EF, but they should be "smaller". It is a nice example of Fermi surface nesting and a resulting Peierls distortion. PS: I don't think anything has changed in sgroup. It finds the identical SG for both structures, but with c/2. And in addition, it shifts the origin, so that the Bi atom sits at the origin. PPS: The 2 k-meshes should NOT be identical for the 1 and 2 atom cells, but as close as possible be of similar k-point density. Am 08.07.2022 um 12:22 schrieb Fecher, Gerhard: > Hallo Peter, > > b) I did that for all tested structures P1 and the one after sgroup/symmetry > > the Fermi energy is slightly different (0.4092709547 or 0.4104773792) if one > compares the different structures, > and, if one plots the density of states as Pablo did they are also clearly > different > In the original structure with 2 atoms in the cell, one has a very clear > minimum at the Fermi energy, > in the reduced structure with only one atom and c/2, one has a high density > of states at the Fermi energy. > I performed both calculations with 25x25x25 initial k-mesh and otherwise also > identical parameters > > For 1 atom c/2 one has 2 half filled bands crossing the Fermi energy==> > clearly metallic > For 2 atoms c, the bands are nearly filled up to Ef or empty (within 0.01 > electrons) at the used parameters ==> clearly semimetallic > > I do not remember that sgroup asked to change the structure (warning: !!! > Unit cell has been reduced. sgroup found: 166 (R -3 m) ) > in the possibly very old version that I used in the past for my calculations > with x=1/4 > > Ciao > Gerhard > > DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy: > "I think the problem, to be quite honest with you, > is that you have never actually known what the question is." > > > Dr. Gerhard H. Fecher > Institut of Physics > Johannes Gutenberg - University > 55099 Mainz > > Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von Peter > Blaha [peter.bl...@tuwien.ac.at] > Gesendet: Freitag, 8. Juli 2022 11:50 > An: wien@zeus.theochem.tuwien.ac.at > Betreff: Re: [Wien] Bi almost cubic > > This is the effect of: > > > a) a complicated "folding" of the BZ together with a shift of origin. > > b) Plot the band structures (use xcrysden in the hexagonal setting to get > "identical k-mehes". > > c) you will see that some bands at some k-points agree, others do not and > they are very complicated to trace by backfolding. > > d) In any case it should become obvious what the DOS at EF looks so > different. It is most likely a k-mesh problem (you should use enormous > unshifted meshes), but still, the Tetrahedron method has no problems for the > DOS at EF in the small one-atom cell since all bands g
Re: [Wien] Bi almost cubic
Plot the bands and you will "understand" the minimum at EF in the 2-atom structure when you consider how the tetrahedron method works (i.e. by connecting bands according to energy (and not character). Due to the finite k-mesh you get a lot of "avoided" crossings and a pseudogap in the DOS. Try a mesh of 100**3 or even 200**3 k-points for the DOS (probably you still get some artefacts at EF, but they should be "smaller". It is a nice example of Fermi surface nesting and a resulting Peierls distortion. PS: I don't think anything has changed in sgroup. It finds the identical SG for both structures, but with c/2. And in addition, it shifts the origin, so that the Bi atom sits at the origin. PPS: The 2 k-meshes should NOT be identical for the 1 and 2 atom cells, but as close as possible be of similar k-point density. Am 08.07.2022 um 12:22 schrieb Fecher, Gerhard: Hallo Peter, b) I did that for all tested structures P1 and the one after sgroup/symmetry the Fermi energy is slightly different (0.4092709547 or 0.4104773792) if one compares the different structures, and, if one plots the density of states as Pablo did they are also clearly different In the original structure with 2 atoms in the cell, one has a very clear minimum at the Fermi energy, in the reduced structure with only one atom and c/2, one has a high density of states at the Fermi energy. I performed both calculations with 25x25x25 initial k-mesh and otherwise also identical parameters For 1 atom c/2 one has 2 half filled bands crossing the Fermi energy==> clearly metallic For 2 atoms c, the bands are nearly filled up to Ef or empty (within 0.01 electrons) at the used parameters ==> clearly semimetallic I do not remember that sgroup asked to change the structure (warning: !!! Unit cell has been reduced. sgroup found: 166 (R -3 m) ) in the possibly very old version that I used in the past for my calculations with x=1/4 Ciao Gerhard DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy: "I think the problem, to be quite honest with you, is that you have never actually known what the question is." Dr. Gerhard H. Fecher Institut of Physics Johannes Gutenberg - University 55099 Mainz Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von Peter Blaha [peter.bl...@tuwien.ac.at] Gesendet: Freitag, 8. Juli 2022 11:50 An: wien@zeus.theochem.tuwien.ac.at Betreff: Re: [Wien] Bi almost cubic This is the effect of: a) a complicated "folding" of the BZ together with a shift of origin. b) Plot the band structures (use xcrysden in the hexagonal setting to get "identical k-mehes". c) you will see that some bands at some k-points agree, others do not and they are very complicated to trace by backfolding. d) In any case it should become obvious what the DOS at EF looks so different. It is most likely a k-mesh problem (you should use enormous unshifted meshes), but still, the Tetrahedron method has no problems for the DOS at EF in the small one-atom cell since all bands go through EF in a straight line and you get a metal. For the doubled cell, there are many "pseudo gaps" and the Tetrahedron method will make a completely different interpolation for the bands and if your k-mesh is not VERY dense, give you a semimetal (or "gap"). Otherwise, the results are "identical" as they should, but you have to be careful with the interpretation. When A is changed to 0.25; R LATTICE,NONEQUIV.ATOMS: 1 166_R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM 1: X=0.2500 Y=0.2500 Z=0.2500 ATOM 1:X= 0.7500 Y=0.7500 Z=0.7500 Bi NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 --- sgroup gives this warning; warning: !!! Unit cell has been reduced. sgroup found: 166 (R -3 m) and the cell is reduced to; R LATTICE,NONEQUIV.ATOMS: 1 166 R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 11.207870 90.00 90.00120.00 ATOM 1: X=0. Y=0. Z=0. Bi1NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.0 - which is "semicubic" with an angle; Angle is 87.539 deg and only one Bi atom in the cell, now in the corners And in this reduced Bi structure the "gap" at Ef in DOS disappears. What I see is that with A=0.249 (0.25-0.001) and A=0.2499 (0.25-0.0001) the DOS have a "gap" and they are quite symilar, but with the addition of the 0.0001 (A=0.25) the cel is reduced and the "gap" disappears. I hope that this answers your questions and becomes clear what I am trying to show. Saludos Pablo ___ Wien mailing list Wien@zeus.theoc
Re: [Wien] Bi almost cubic
Hallo Peter, b) I did that for all tested structures P1 and the one after sgroup/symmetry the Fermi energy is slightly different (0.4092709547 or 0.4104773792) if one compares the different structures, and, if one plots the density of states as Pablo did they are also clearly different In the original structure with 2 atoms in the cell, one has a very clear minimum at the Fermi energy, in the reduced structure with only one atom and c/2, one has a high density of states at the Fermi energy. I performed both calculations with 25x25x25 initial k-mesh and otherwise also identical parameters For 1 atom c/2 one has 2 half filled bands crossing the Fermi energy==> clearly metallic For 2 atoms c, the bands are nearly filled up to Ef or empty (within 0.01 electrons) at the used parameters ==> clearly semimetallic I do not remember that sgroup asked to change the structure (warning: !!! Unit cell has been reduced. sgroup found: 166 (R -3 m) ) in the possibly very old version that I used in the past for my calculations with x=1/4 Ciao Gerhard DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy: "I think the problem, to be quite honest with you, is that you have never actually known what the question is." Dr. Gerhard H. Fecher Institut of Physics Johannes Gutenberg - University 55099 Mainz Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von Peter Blaha [peter.bl...@tuwien.ac.at] Gesendet: Freitag, 8. Juli 2022 11:50 An: wien@zeus.theochem.tuwien.ac.at Betreff: Re: [Wien] Bi almost cubic This is the effect of: a) a complicated "folding" of the BZ together with a shift of origin. b) Plot the band structures (use xcrysden in the hexagonal setting to get "identical k-mehes". c) you will see that some bands at some k-points agree, others do not and they are very complicated to trace by backfolding. d) In any case it should become obvious what the DOS at EF looks so different. It is most likely a k-mesh problem (you should use enormous unshifted meshes), but still, the Tetrahedron method has no problems for the DOS at EF in the small one-atom cell since all bands go through EF in a straight line and you get a metal. For the doubled cell, there are many "pseudo gaps" and the Tetrahedron method will make a completely different interpolation for the bands and if your k-mesh is not VERY dense, give you a semimetal (or "gap"). Otherwise, the results are "identical" as they should, but you have to be careful with the interpretation. When A is changed to 0.25; R LATTICE,NONEQUIV.ATOMS: 1 166_R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM 1: X=0.2500 Y=0.2500 Z=0.2500 ATOM 1:X= 0.7500 Y=0.7500 Z=0.7500 Bi NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 --- sgroup gives this warning; warning: !!! Unit cell has been reduced. sgroup found: 166 (R -3 m) and the cell is reduced to; R LATTICE,NONEQUIV.ATOMS: 1 166 R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 11.207870 90.00 90.00120.00 ATOM 1: X=0. Y=0. Z=0. Bi1NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.0 - which is "semicubic" with an angle; Angle is 87.539 deg and only one Bi atom in the cell, now in the corners And in this reduced Bi structure the "gap" at Ef in DOS disappears. What I see is that with A=0.249 (0.25-0.001) and A=0.2499 (0.25-0.0001) the DOS have a "gap" and they are quite symilar, but with the addition of the 0.0001 (A=0.25) the cel is reduced and the "gap" disappears. I hope that this answers your questions and becomes clear what I am trying to show. Saludos Pablo ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at<mailto:Wien@zeus.theochem.tuwien.ac.at> http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html -- --- Peter Blaha, Inst. f. Materials Chemistry, TU Vienna, A-1060 Vienna Phone: +43-158801165300 Email: peter.bl...@tuwien.ac.at<mailto:peter.bl...@tuwien.ac.at> WWW: http://www.imc.tuwien.ac.at WIEN2k: http://www.wien2k.at - ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Bi almost cubic
This is the effect of: a) a complicated "folding" of the BZ together with a shift of origin. b) Plot the band structures (use xcrysden in the hexagonal setting to get "identical k-mehes". c) you will see that some bands at some k-points agree, others do not and they are very complicated to trace by backfolding. d) In any case it should become obvious what the DOS at EF looks so different. It is most likely a k-mesh problem (you should use enormous unshifted meshes), but still, the Tetrahedron method has no problems for the DOS at EF in the small one-atom cell since all bands go through EF in a straight line and you get a metal. For the doubled cell, there are many "pseudo gaps" and the Tetrahedron method will make a completely different interpolation for the bands and if your k-mesh is not VERY dense, give you a semimetal (or "gap"). Otherwise, the results are "identical" as they should, but you have to be careful with the interpretation. When A is changed to 0.25; R LATTICE,NONEQUIV.ATOMS: 1 166_R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM 1: X=0.2500 Y=0.2500 Z=0.2500 ATOM 1:X= 0.7500 Y=0.7500 Z=0.7500 Bi NPT= 781 R0=0.0500 RMT= 2.5000 Z: 83.000 --- sgroup gives this warning; warning: !!! Unit cell has been reduced. sgroup found: 166 (R -3 m) and the cell is reduced to; R LATTICE,NONEQUIV.ATOMS: 1 166 R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 11.207870 90.00 90.00120.00 ATOM 1: X=0. Y=0. Z=0. Bi1 NPT= 781 R0=0.0500 RMT= 2.5000 Z: 83.0 - which is "semicubic" with an angle; Angle is 87.539 deg and only one Bi atom in the cell, now in the corners And in this reduced Bi structure the "gap" at Ef in DOS disappears. What I see is that with A=0.249 (0.25-0.001) and A=0.2499 (0.25-0.0001) the DOS have a "gap" and they are quite symilar, but with the addition of the 0.0001 (A=0.25) the cel is reduced and the "gap" disappears. I hope that this answers your questions and becomes clear what I am trying to show. Saludos Pablo ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at:http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html -- --- Peter Blaha, Inst. f. Materials Chemistry, TU Vienna, A-1060 Vienna Phone: +43-158801165300 Email:peter.bl...@tuwien.ac.at WWW:http://www.imc.tuwien.ac.at WIEN2k:http://www.wien2k.at - ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Bi almost cubic
Hallo Pablo, you are right, something strange happens if one accepts the suggestion of sgroup and/or symmetry. Unfortunately I can not clearly find out anymore which version of Wien2k I used some years ago (only hint, the .struct files did not contain the part "Precise positions"). Peter (or someone else who knows the subroutines well) should check what is wrong with the new structure or symmetry operations. When accepting the suggested changes the result is a metal as you observed. I checked by using the structure in P1 (struct is below), in that case the semimetallic character is kept if I strictly use P1 (note in hexagonal setup one has 6 atoms in the cell, if interested I can send it in P1, too). (Note one has to ignore some warnings during initialisation. Even if a calculation in P1 may be in general a bad idea with respect to calculation time, I do not see why I "MUST" shift the atomic positions while using P1 ) I also played another -- possibly nasty -- trick, I accepted everything of sgroup and symmetry (resulting in .struct with only 1 atom and half c), then I inserted the symmetry operations into the initial structure file (with 2 atoms and original c) and continued with the initialisation with this .struct file, in that case, the semimetallicity of Bi was kept. PS.: the program findsym (see https://stokes.byu.edu/iso/findsym.php) reduces the structure with z=1/4 in the same way as sgroup (just with the atom at 3b and not 3a, what should principally not matter), however, it uses consistently hexagonal set up and not a mixture of hexagonal lattice parameters with rhombohedral atomic positions, which is indeed an unlucky choice. PSS.: The symmetry checker of Eneavor (a commercial crystallographic program) did not reduce to a one atom structure, for unknown reason. Note the following structure has experimenta a and c from the Pearson database. blebleble P LATTICE,NONEQUIV.ATOMS: 2 1_P1 MODE OF CALC=RELA unit=bohr 8.966184 8.966184 8.966184 57.227000 57.227000 57.227000 ATOM -1: X=0.2500 Y=0.2500 Z=0.2500 MULT= 1 ISPLIT= 8 Bi1NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -2: X=0.7500 Y=0.7500 Z=0.7500 MULT= 1 ISPLIT= 8 Bi2NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 2 NUMBER OF SYMMETRY OPERATIONS 1 0 0 0. 0 1 0-0. 0 0 1-0. 1 -1 0 0 0.5000 0-1 0 0.5000 0 0-1 0.5000 2 Precise positions 0.250 0.250 0.250 0.750 0.750 0.750 Ciao Gerhard DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy: "I think the problem, to be quite honest with you, is that you have never actually known what the question is." Dr. Gerhard H. Fecher Institut of Physics Johannes Gutenberg - University 55099 Mainz Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora [delam...@unam.mx] Gesendet: Freitag, 8. Juli 2022 01:15 An: A Mailing list for WIEN2k users Betreff: [Wien] Bi almost cubic Dear Gerhard, The optimized cell is; --- R LATTICE,NONEQUIV.ATOMS: 1 166_R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM 1: X=0.23592070 Y=0.23592070 Z=0.23592070 ATOM 1:X= 0.76407930 Y=0.76407930 Z=0.76407930 Bi NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 with two atoms in the cell XYZ = A and 1-A A=0.23592070 and 1-A=0.76407930 When A is changed to 0.25; R LATTICE,NONEQUIV.ATOMS: 1 166_R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM 1: X=0.2500 Y=0.2500 Z=0.2500 ATOM 1:X= 0.7500 Y=0.7500 Z=0.7500 Bi NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 --- sgroup gives this warning; warning: !!! Unit cell has been reduced. sgroup found: 166 (R -3 m) and the cell is reduced to; R LATTICE,NONEQUIV.ATOMS: 1 166 R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 11.207870 90.00 90.00120.00 ATOM 1: X=0. Y=0. Z=0. Bi1NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.0 - which is "semicubic" with an
[Wien] Bi almost cubic
Dear Gerhard, The optimized cell is; --- R LATTICE,NONEQUIV.ATOMS: 1 166_R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM 1: X=0.23592070 Y=0.23592070 Z=0.23592070 ATOM 1:X= 0.76407930 Y=0.76407930 Z=0.76407930 Bi NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 with two atoms in the cell XYZ = A and 1-A A=0.23592070 and 1-A=0.76407930 When A is changed to 0.25; R LATTICE,NONEQUIV.ATOMS: 1 166_R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM 1: X=0.2500 Y=0.2500 Z=0.2500 ATOM 1:X= 0.7500 Y=0.7500 Z=0.7500 Bi NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 --- sgroup gives this warning; warning: !!! Unit cell has been reduced. sgroup found: 166 (R -3 m) and the cell is reduced to; R LATTICE,NONEQUIV.ATOMS: 1 166 R-3m MODE OF CALC=RELA unit=ang 8.591340 8.591340 11.207870 90.00 90.00120.00 ATOM 1: X=0. Y=0. Z=0. Bi1NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.0 - which is "semicubic" with an angle; Angle is 87.539 deg and only one Bi atom in the cell, now in the corners And in this reduced Bi structure the "gap" at Ef in DOS disappears. What I see is that with A=0.249 (0.25-0.001) and A=0.2499 (0.25-0.0001) the DOS have a "gap" and they are quite symilar, but with the addition of the 0.0001 (A=0.25) the cel is reduced and the "gap" disappears. I hope that this answers your questions and becomes clear what I am trying to show. Saludos Pablo ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Bi almost cubic
I used experimental as well as optimized lattice parameters for fixed z=0.25 and optimized z I don't Know what you call semicubic. The lattice parameters you give do not result in a cunic structure. It seems you changed the lattice parameter and not the z-parameter Ciao Gerhard DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy: "I think the problem, to be quite honest with you, is that you have never actually known what the question is." Dr. Gerhard H. Fecher Institut of Physics Johannes Gutenberg - University 55099 Mainz Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora [delam...@unam.mx] Gesendet: Donnerstag, 7. Juli 2022 19:01 An: A Mailing list for WIEN2k users Betreff: Re: [Wien] Bi almost cubic Comments below Some 5 years ago I was interested in the total energy of Bi and checked what happens with Bi if setting z=1/4 using PBE the semimetallic character (or nearly semimetallic with SO) was conserved. The band structure between z=0.23457217 and 1/4 was not much different Is your total energy changing continously ? Z Ene 0.249; -86326.25852144 0.2499; -86326.25475092 0.25; -43163.12738079*2=-86326.25476158 Was, for z=0.25, the cell reduced to a semi-cubic with only one atom in the cell? Ciao Gerhard Saludos Pablo Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora [delam...@unam.mx] Gesendet: Mittwoch, 6. Juli 2022 23:44 An: A Mailing list for WIEN2k users Betreff: [Wien] RV: Bi almost cubic Dear WIEN2k community, Its me again reporting something that does not seem correct; Bi is almost cubic, next to Po, the only element with a true simple cubic unit cell Bi unit cell parameters; R LATTICE,NONEQUIV.ATOMS: 1 MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM -1: X=0.23592070 Y=0.23592070 Z=0.23592070 ATOM -1:X= 0.76407930 Y=0.76407930 Z=0.76407930 Bi1NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 __ If XYZ => 0.25 the cell reduces to half and is almost a cube with a little distortion in the 111 direction with an angle of 90 => 87.54 What atracted me was that the compound almost has a gap at Ef, with the above structure XYZ=0.2359 but if this atomic position is moved towards XYZ=0.2359 => 0.25 the "gap" diminishes, but it does not disappear even at XYZ=0.2499 but if XYZ=0.25 then, as mentioned above, the cell is reduced and the "gap" disapears As can be seen, with XYZ=0.249 and 0.2499 the "gap" is almost the same, but adding 0.0001 it disappears My question is, why it disappears when the cells with XYZ=0.2499 and 0.25 are almost the same but with XYZ=0.25 the cel reduces, but still it should give the same DOS. Cheers Pablo ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Bi almost cubic
Comments below Some 5 years ago I was interested in the total energy of Bi and checked what happens with Bi if setting z=1/4 using PBE the semimetallic character (or nearly semimetallic with SO) was conserved. The band structure between z=0.23457217 and 1/4 was not much different Is your total energy changing continously ? Z Ene 0.249; -86326.25852144 0.2499; -86326.25475092 0.25; -43163.12738079*2=-86326.25476158 Was, for z=0.25, the cell reduced to a semi-cubic with only one atom in the cell? Ciao Gerhard Saludos Pablo Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora [delam...@unam.mx] Gesendet: Mittwoch, 6. Juli 2022 23:44 An: A Mailing list for WIEN2k users Betreff: [Wien] RV: Bi almost cubic Dear WIEN2k community, Its me again reporting something that does not seem correct; Bi is almost cubic, next to Po, the only element with a true simple cubic unit cell Bi unit cell parameters; R LATTICE,NONEQUIV.ATOMS: 1 MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM -1: X=0.23592070 Y=0.23592070 Z=0.23592070 ATOM -1:X= 0.76407930 Y=0.76407930 Z=0.76407930 Bi1NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 __ If XYZ => 0.25 the cell reduces to half and is almost a cube with a little distortion in the 111 direction with an angle of 90 => 87.54 What atracted me was that the compound almost has a gap at Ef, with the above structure XYZ=0.2359 but if this atomic position is moved towards XYZ=0.2359 => 0.25 the "gap" diminishes, but it does not disappear even at XYZ=0.2499 but if XYZ=0.25 then, as mentioned above, the cell is reduced and the "gap" disapears As can be seen, with XYZ=0.249 and 0.2499 the "gap" is almost the same, but adding 0.0001 it disappears My question is, why it disappears when the cells with XYZ=0.2499 and 0.25 are almost the same but with XYZ=0.25 the cel reduces, but still it should give the same DOS. Cheers Pablo ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Bi almost cubic
Was, for z=0.25, the cell reduced to a semi-cubic with only one atom in the cell? Saludos Pablo I cannot reproduce your result Some 5 years ago I was interested in the total energy of Bi and checked what happens with Bi if setting z=1/4 using PBE the semimetallic character (or nearly semimetallic with SO) was conserved. The band structure between z=0.23457217 and 1/4 was not much different Is your total energy changing continously ? Ciao Gerhard DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy: "I think the problem, to be quite honest with you, is that you have never actually known what the question is." Dr. Gerhard H. Fecher Institut of Physics Johannes Gutenberg - University 55099 Mainz Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora [delam...@unam.mx] Gesendet: Mittwoch, 6. Juli 2022 23:44 An: A Mailing list for WIEN2k users Betreff: [Wien] RV: Bi almost cubic Dear WIEN2k community, Its me again reporting something that does not seem correct; Bi is almost cubic, next to Po, the only element with a true simple cubic unit cell Bi unit cell parameters; R LATTICE,NONEQUIV.ATOMS: 1 MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM -1: X=0.23592070 Y=0.23592070 Z=0.23592070 ATOM -1:X= 0.76407930 Y=0.76407930 Z=0.76407930 Bi1NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 __ If XYZ => 0.25 the cell reduces to half and is almost a cube with a little distortion in the 111 direction with an angle of 90 => 87.54 What atracted me was that the compound almost has a gap at Ef, with the above structure XYZ=0.2359 but if this atomic position is moved towards XYZ=0.2359 => 0.25 the "gap" diminishes, but it does not disappear even at XYZ=0.2499 but if XYZ=0.25 then, as mentioned above, the cell is reduced and the "gap" disapears As can be seen, with XYZ=0.249 and 0.2499 the "gap" is almost the same, but adding 0.0001 it disappears My question is, why it disappears when the cells with XYZ=0.2499 and 0.25 are almost the same but with XYZ=0.25 the cel reduces, but still it should give the same DOS. Cheers Pablo ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Bi almost cubic
I cannot reproduce your result Some 5 years ago I was interested in the total energy of Bi and checked what happens with Bi if setting z=1/4 using PBE the semimetallic character (or nearly semimetallic with SO) was conserved. The band structure between z=0.23457217 and 1/4 was not much different Is your total energy changing continously ? Ciao Gerhard DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy: "I think the problem, to be quite honest with you, is that you have never actually known what the question is." Dr. Gerhard H. Fecher Institut of Physics Johannes Gutenberg - University 55099 Mainz Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora [delam...@unam.mx] Gesendet: Mittwoch, 6. Juli 2022 23:44 An: A Mailing list for WIEN2k users Betreff: [Wien] RV: Bi almost cubic Dear WIEN2k community, Its me again reporting something that does not seem correct; Bi is almost cubic, next to Po, the only element with a true simple cubic unit cell Bi unit cell parameters; R LATTICE,NONEQUIV.ATOMS: 1 MODE OF CALC=RELA unit=ang 8.591340 8.591340 22.415740 90.00 90.00120.00 ATOM -1: X=0.23592070 Y=0.23592070 Z=0.23592070 ATOM -1:X= 0.76407930 Y=0.76407930 Z=0.76407930 Bi1NPT= 781 R0=0.0500 RMT=2.5000 Z: 83.000 __ If XYZ => 0.25 the cell reduces to half and is almost a cube with a little distortion in the 111 direction with an angle of 90 => 87.54 What atracted me was that the compound almost has a gap at Ef, with the above structure XYZ=0.2359 but if this atomic position is moved towards XYZ=0.2359 => 0.25 the "gap" diminishes, but it does not disappear even at XYZ=0.2499 but if XYZ=0.25 then, as mentioned above, the cell is reduced and the "gap" disapears As can be seen, with XYZ=0.249 and 0.2499 the "gap" is almost the same, but adding 0.0001 it disappears My question is, why it disappears when the cells with XYZ=0.2499 and 0.25 are almost the same but with XYZ=0.25 the cel reduces, but still it should give the same DOS. Cheers Pablo ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
