[QE-users] Re-use of charge density don't give exactely the same result in scf
Dear all, After a first scf calculation and saving charge density and wave functions, I launch the same scf with these wave functions and charge density as an initialization (using startingpot and startingwfc). I go from 27 scf cycles in the first run to only 4 at in the second run. However, I was expecting that after the 2 first scf cycles, the energy difference is smaller than the conv_thr because it has already been converged during the first run. I suppose that this phenomenon is due to the accuracy (number of digits and number of points) used to saved the electronic density and the wfcs. Is it possible to confirm my assumption and to change this accuracy with a magic parameter? Best regards, Antoine Jay ISAE-SUPAERO Toulouse, France ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users
[QE-users] ASR:?==?utf-8?q? "simple" and "crystal" for huge systems
Dear all, I'am calculating the exponential vibrationnal entropic prefactor k for a NEB after having calulate the phonons at saddle and initial atomic positions. k=product_3N-3(omega_init)/product_3N-4(omega_saddle). As I have more than 1000 frequencies, the ASR "simple" is very fast (few seconds) but the ASR "crystal" is very long (several days) and need lot of memory even for a supercalculator without the restart possibility. Even if the final diference in the frequencies is small (<5cm^-1), its becomming huge when I do the product ratio: simple k=10^-5 crystal k=10^-2 Did anyone already face this problem? What is the better option? System is defects in silicon supercell QE 6.3. Regards, Antoine Jay LAAS-CNRS, Toulouse, France. ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users
[QE-users] dynmat not parallelized?
Dear all, I need to compute the ASR on systems with 1000 frequencies. My first try takes 2 days. Does the dynmat programm has a parallel version? Is it relevant to use a parallel version? Thanks, Antoine Jay LAAS-CNRS, Toulouse France ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users
Re: [QE-users] ?==?utf-8?q? Symmetries in the dynamical matrix
Dear Paolo,
Thank you for this answer,
I should conclude that the small inequalities with respect to the inversion of
indices is due convergeance inaccuracy that "emerge naturally" with the
filling by irreductible representations.
However, if the exchange of indices belongs to the second category, that means
that all the components of the dynamical matrix are calculated if I impose
nosym=.TRUE.!
This seems quite illogical to calculate two times the same thing: only the
upper triangular part of the matrix is needed and the lower is symmetric, the
inversion of indices should should belong to the first category...
Do I say something wrong?
Regards,
Antoine Jay
On Thursday, October 18, 2018 10:25 CEST, Paolo Giannozzi
wrote:
Not sure about this specific case, but only some of the symmetries of the
dynamical matrix are enforced in the calculation (in particular: symmetries wrt
operations of the small group of q). Other symmetries are not enforced and
should "emerge naturally" from the calculation. If I remember correctly, the
symmetry of dynamical matrix with respect to exchange of indices belongs to the
second category. Another famous example is the acoustic sum rule. In the
subsequent processing by auxiliary codes (dynmat, q2r, matdyn) that symmetry is
explicitly imposed. Incomplete scf convergence, for instance, may lead to a
small loss of symmetry. On Tue, Oct 16, 2018 at 5:11 PM JAY Antoine
wrote:Dear all,
I'am trying to understand the symmetries of the dynamical matrix given in the
phonon output :
Dynamical Matrix in cartesian axes
q = ( 0.0 0.0 0.0 )
1 1
0.26846543 0. -0.00021832 0. -0.00074984 0.
-0.00021197 0. 0.26867942 0. -0.00021197 0.
-0.00074984 0. -0.00021832 0. 0.26846543 0.
[...]
which has the form:
A B C
D E D
C B F
Here 6 values A,B,C,D,E and F have been calculated whereas the 3 other force
constants have been filled due to the crystal symmetry operations.
But what I do not understand is why is this dynamical matrix not symmetric?
in the subroutine write_dyn_mat, when phi(i,j,na,nb) is written, its definition
is:
1/sqrt(Mna*Mnb)*
d^2E/dxadxb d^2E/dxadyb d^2E/dxadzb
d^2E/dyadxb d^2E/dyadyb d^2E/dyadzb
d^2E/dzadxb d^2E/dzadyb d^2E/dzadzb
Due to the Schwarz theorem in the case where na=nb,
d^2E/dxadya = d^2E/dyadxa, so in addition to the crystal symmetry operations,
the algorithm should fill exactelly D=b with only one calcul.
However, it seems that two calculs have been performed in this case giving
slighly different values...
Another example is given here
1 2
-0.06563923 0. -0.04490934 0. -0.04404086 0.
-0.04493368 0. -0.06590828 0. -0.04493368 0.
-0.04404086 0. -0.04490934 0. -0.06563923 0.
[...]
2 1
-0.06528947 0. -0.04495765 0. -0.04411243 0.
-0.04490286 0. -0.06585596 0. -0.04490286 0.
-0.04411243 0. -0.04495765 0. -0.06528947 0.
I was expecting d^2E/dx1dx2 = d^2E/dx1dx2 and so on...
but two calculs have been done and -0.06563923 .neq. -0.06528947!
What I am missing there?
System is a simple vacancy in a supercell of 215 silicon atoms (ibrav=1), QE
V6.0, input given bellow
Regards,
Antoine Jay
LAAS-CNRS, Toulouse France
scf.in:
calculation = 'scf'
prefix = 'Si_1V_D2d',
tstress = .true.
tprnfor = .true.
pseudo_dir = '/pseudo/',
outdir = 'tmp_Si_1V_D2d/',
etot_conv_thr=1.0d-6
forc_conv_thr=1.0d-4
nstep=1400
/
ibrav = 1,
celldm(1) = 30.613571,
celldm(2) = 1 ,
celldm(3) = 1 ,
nat = 215,
ntyp = 1,
ecutwfc = 20.0,
/
diagonalization='david'
mixing_beta = 0.7,
conv_thr = 1.0d-9,
electron_maxstep=800,
/
ATOMIC_SPECIES
Si 28.086 Si.pz-vbc.UPF
ATOMIC_POSITIONS {crystal}
Si 0.000162336 0.00014327 0.000162336
Si 0.08330307 0.084189225 0.08330307
Si 0.166992459 0.16800145 0.00058752
Si 0.250912336 0.250721016 0.084924085
Si 0.166156275 -0.000106871 0.166156275
Si 0.249895433 0.082528066 0.249895433
Si 0.00058752 0.16800145 0.166992459
Si 0.084924085 0.250721016 0.250912336
Si 0.333438217 0.001346258 -0.00050172
Si 0.417071186 0.084081282 0.083872754
Si 0.499961127 0.168420495 0.001124771
Si 0.582907731 0.251646989 0.085458254
Si 0.500336958 -0.000934751 0.16550848
Si 0.584924657 0.08068021 0.248592644
Si 0.333277507 0.16668747 0.168141141
Si 0.417190875 0.24867044 0.25351339
Si 0.001507258 0.333858164 0.001507258
Si 0.086913471 0.416162509 0.086913471
Si 0.1673
[QE-users] Symmetries in the dynamical matrix
Dear all,
I'am trying to understand the symmetries of the dynamical matrix given in the
phonon output :
Dynamical Matrix in cartesian axes
q = ( 0.0 0.0 0.0 )
1 1
0.26846543 0. -0.00021832 0. -0.00074984 0.
-0.00021197 0. 0.26867942 0. -0.00021197 0.
-0.00074984 0. -0.00021832 0. 0.26846543 0.
[...]
which has the form:
A B C
D E D
C B F
Here 6 values A,B,C,D,E and F have been calculated whereas the 3 other force
constants have been filled due to the crystal symmetry operations.
But what I do not understand is why is this dynamical matrix not symmetric?
in the subroutine write_dyn_mat, when phi(i,j,na,nb) is written, its definition
is:
1/sqrt(Mna*Mnb)*
d^2E/dxadxb d^2E/dxadyb d^2E/dxadzb
d^2E/dyadxb d^2E/dyadyb d^2E/dyadzb
d^2E/dzadxb d^2E/dzadyb d^2E/dzadzb
Due to the Schwarz theorem in the case where na=nb,
d^2E/dxadya = d^2E/dyadxa, so in addition to the crystal symmetry operations,
the algorithm should fill exactelly D=b with only one calcul.
However, it seems that two calculs have been performed in this case giving
slighly different values...
Another example is given here
1 2
-0.06563923 0. -0.04490934 0. -0.04404086 0.
-0.04493368 0. -0.06590828 0. -0.04493368 0.
-0.04404086 0. -0.04490934 0. -0.06563923 0.
[...]
2 1
-0.06528947 0. -0.04495765 0. -0.04411243 0.
-0.04490286 0. -0.06585596 0. -0.04490286 0.
-0.04411243 0. -0.04495765 0. -0.06528947 0.
I was expecting d^2E/dx1dx2 = d^2E/dx1dx2 and so on...
but two calculs have been done and -0.06563923 .neq. -0.06528947!
What I am missing there?
System is a simple vacancy in a supercell of 215 silicon atoms (ibrav=1), QE
V6.0, input given bellow
Regards,
Antoine Jay
LAAS-CNRS, Toulouse France
scf.in:
calculation = 'scf'
prefix = 'Si_1V_D2d',
tstress = .true.
tprnfor = .true.
pseudo_dir = '/pseudo/',
outdir = 'tmp_Si_1V_D2d/',
etot_conv_thr=1.0d-6
forc_conv_thr=1.0d-4
nstep=1400
/
ibrav = 1,
celldm(1) = 30.613571,
celldm(2) = 1 ,
celldm(3) = 1 ,
nat = 215,
ntyp = 1,
ecutwfc = 20.0,
/
diagonalization='david'
mixing_beta = 0.7,
conv_thr = 1.0d-9,
electron_maxstep=800,
/
ATOMIC_SPECIES
Si 28.086 Si.pz-vbc.UPF
ATOMIC_POSITIONS {crystal}
Si 0.000162336 0.00014327 0.000162336
Si 0.08330307 0.084189225 0.08330307
Si 0.166992459 0.16800145 0.00058752
Si 0.250912336 0.250721016 0.084924085
Si 0.166156275 -0.000106871 0.166156275
Si 0.249895433 0.082528066 0.249895433
Si 0.00058752 0.16800145 0.166992459
Si 0.084924085 0.250721016 0.250912336
Si 0.333438217 0.001346258 -0.00050172
Si 0.417071186 0.084081282 0.083872754
Si 0.499961127 0.168420495 0.001124771
Si 0.582907731 0.251646989 0.085458254
Si 0.500336958 -0.000934751 0.16550848
Si 0.584924657 0.08068021 0.248592644
Si 0.333277507 0.16668747 0.168141141
Si 0.417190875 0.24867044 0.25351339
Si 0.001507258 0.333858164 0.001507258
Si 0.086913471 0.416162509 0.086913471
Si 0.167357425 0.499603001 0.000978135
Si 0.250615274 0.581869305 0.085464282
Si 0.173265996 0.334321177 0.173265996
Si 0.262140675 0.414034508 0.262140675
Si 0.000978135 0.499603001 0.167357425
Si 0.085464282 0.581869305 0.250615274
Si -0.00050172 0.001346258 0.333438217
Si 0.083872754 0.084081282 0.417071186
Si 0.168141141 0.16668747 0.333277507
Si 0.25351339 0.24867044 0.417190875
Si 0.16550848 -0.000934751 0.500336958
Si 0.248592644 0.08068021 0.584924657
Si 0.001124771 0.168420495 0.499961127
Si 0.085458254 0.251646989 0.582907731
Si 0.333739196 0.333959699 0.000861911
Si 0.41676146 0.41675827 0.085292632
Si 0.499782156 0.49955788 0.000861393
Si 0.582608779 0.582796164 0.08492405
Si 0.498891265 0.334010796 0.1706163
Si 0.571385857 0.419481169 0.262134574
Si 0.334633728 0.499504498 0.170620451
Si 0.41675 0.584843238 0.253519495
Si 0.000861911 0.333959699 0.333739196
Si 0.085292632 0.41675827 0.41676146
Si 0.170620451 0.499504498 0.334633728
Si 0.253519495 0.584843238 0.41675
Si 0.1706163 0.334010796 0.498891265
Si 0.262134574 0.419481169 0.571385857
Si 0.000861393 0.49955788 0.499782156
Si 0.08492405 0.582796164 0.582608779
Si 0.333252803 -0.000634212 0.333252803
Si 0.416757556
[QE-users] Symmetries in the dynamical matrix
Dear all,
I'am trying to understand the symmetries of the dynamical matrix given in the
phonon output :
Dynamical Matrix in cartesian axes
q = ( 0.0 0.0 0.0 )
1 1
0.26846543 0. -0.00021832 0. -0.00074984 0.
-0.00021197 0. 0.26867942 0. -0.00021197 0.
-0.00074984 0. -0.00021832 0. 0.26846543 0.
[...]
which has the form:
A B C
D E D
C B F
Here 6 values A,B,C,D,E and F have been calculated whereas the 3 other force
constants have been filled due to the crystal symmetry operations.
But what I do not understand is why is this dynamical matrix not symmetric?
in the subroutine write_dyn_mat, when phi(i,j,na,nb) is written, its definition
is:
1/sqrt(Mna*Mnb)*
d^2E/dxadxb d^2E/dxadyb d^2E/dxadzb
d^2E/dyadxb d^2E/dyadyb d^2E/dyadzb
d^2E/dzadxb d^2E/dzadyb d^2E/dzadzb
Due to the Schwarz theorem in the case where na=nb,
d^2E/dxadya = d^2E/dyadxa, so in addition to the crystal symmetry operations,
the algorithm should fill exactelly D=b with only one calcul.
However, it seems that two calculs have been performed in this case giving
slighly different values...
Another example is given here
1 2
-0.06563923 0. -0.04490934 0. -0.04404086 0.
-0.04493368 0. -0.06590828 0. -0.04493368 0.
-0.04404086 0. -0.04490934 0. -0.06563923 0.
[...]
2 1
-0.06528947 0. -0.04495765 0. -0.04411243 0.
-0.04490286 0. -0.06585596 0. -0.04490286 0.
-0.04411243 0. -0.04495765 0. -0.06528947 0.
I was expecting d^2E/dx1dx2 = d^2E/dx1dx2 and so on...
but two calculs have been done and -0.06563923 .neq. -0.06528947!
What I am missing there?
System is a simple vacancy in a supercell of 215 silicon atoms (ibrav=1), QE
V6.0, input given bellow
Regards,
Antoine Jay
LAAS-CNRS, Toulouse France
scf.in:
calculation = 'scf'
prefix = 'Si_1V_D2d',
tstress = .true.
tprnfor = .true.
pseudo_dir = '/pseudo/',
outdir = 'tmp_Si_1V_D2d/',
etot_conv_thr=1.0d-6
forc_conv_thr=1.0d-4
nstep=1400
/
ibrav = 1,
celldm(1) = 30.613571,
celldm(2) = 1 ,
celldm(3) = 1 ,
nat = 215,
ntyp = 1,
ecutwfc = 20.0,
/
diagonalization='david'
mixing_beta = 0.7,
conv_thr = 1.0d-9,
electron_maxstep=800,
/
ATOMIC_SPECIES
Si 28.086 Si.pz-vbc.UPF
ATOMIC_POSITIONS {crystal}
Si 0.000162336 0.00014327 0.000162336
Si 0.08330307 0.084189225 0.08330307
Si 0.166992459 0.16800145 0.00058752
Si 0.250912336 0.250721016 0.084924085
Si 0.166156275 -0.000106871 0.166156275
Si 0.249895433 0.082528066 0.249895433
Si 0.00058752 0.16800145 0.166992459
Si 0.084924085 0.250721016 0.250912336
Si 0.333438217 0.001346258 -0.00050172
Si 0.417071186 0.084081282 0.083872754
Si 0.499961127 0.168420495 0.001124771
Si 0.582907731 0.251646989 0.085458254
Si 0.500336958 -0.000934751 0.16550848
Si 0.584924657 0.08068021 0.248592644
Si 0.333277507 0.16668747 0.168141141
Si 0.417190875 0.24867044 0.25351339
Si 0.001507258 0.333858164 0.001507258
Si 0.086913471 0.416162509 0.086913471
Si 0.167357425 0.499603001 0.000978135
Si 0.250615274 0.581869305 0.085464282
Si 0.173265996 0.334321177 0.173265996
Si 0.262140675 0.414034508 0.262140675
Si 0.000978135 0.499603001 0.167357425
Si 0.085464282 0.581869305 0.250615274
Si -0.00050172 0.001346258 0.333438217
Si 0.083872754 0.084081282 0.417071186
Si 0.168141141 0.16668747 0.333277507
Si 0.25351339 0.24867044 0.417190875
Si 0.16550848 -0.000934751 0.500336958
Si 0.248592644 0.08068021 0.584924657
Si 0.001124771 0.168420495 0.499961127
Si 0.085458254 0.251646989 0.582907731
Si 0.333739196 0.333959699 0.000861911
Si 0.41676146 0.41675827 0.085292632
Si 0.499782156 0.49955788 0.000861393
Si 0.582608779 0.582796164 0.08492405
Si 0.498891265 0.334010796 0.1706163
Si 0.571385857 0.419481169 0.262134574
Si 0.334633728 0.499504498 0.170620451
Si 0.41675 0.584843238 0.253519495
Si 0.000861911 0.333959699 0.333739196
Si 0.085292632 0.41675827 0.41676146
Si 0.170620451 0.499504498 0.334633728
Si 0.253519495 0.584843238 0.41675
Si 0.1706163 0.334010796 0.498891265
Si 0.262134574 0.419481169 0.571385857
Si 0.000861393 0.49955788 0.499782156
Si 0.08492405 0.582796164 0.582608779
Si 0.333252803 -0.000634212 0.333252803
Si 0.416757556
[QE-users] Van der Waals with LDA functional
Dear community, Is it possible to know if I need to take Van der Waals interactions into account by only looking at the scf output done with LDA functional? If no, what is the fastest way to know if I need them? My system is implantation of Sb in silicon. Thank you very much, Regards, Antoine Jay ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users
Re: [QE-users] ?==?utf-8?q? SCF convergeance without smearing when HO and LU are closed
The nscf was the sollution, Thank you very much Stefano Antoine Jay On Wednesday, August 15, 2018 21:09 CEST, Stefano Baroni wrote: On 15 Aug 2018, at 19:26, JAY Antoine wrote: Dear all, I'm performing a charge +2 supercell calculation for a silicon divacancy. The highest occupied and lowest unoccupied electronic states are very close: 6.1233 and 6.1405 so that I have to use a smearing for the scf calculation. Now I need to obtain the wfc and electronic states without smearing. What do you suggest to me? I already try all the possible combinations of the following parameters: mixing_mode, mixing_beta, startingwfc. Why not doing a simple non-selfconsistent calculation? SB — Stefano Baroni - Trieste — http://stefano.baroni.me ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users
[QE-users] SCF convergeance without smearing when HO and LU are closed
Dear all, I'm performing a charge +2 supercell calculation for a silicon divacancy. The highest occupied and lowest unoccupied electronic states are very close: 6.1233 and 6.1405 so that I have to use a smearing for the scf calculation. Now I need to obtain the wfc and electronic states without smearing. What do you suggest to me? I already try all the possible combinations of the following parameters: mixing_mode, mixing_beta, startingwfc. Regards, Antoine Jay ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users
Re: [QE-users] ?==?utf-8?q? Recover phonons frequencies from the eigenmodes
Dear Lorenzo, As X I convert the atomic displacement to meters: (R0+i*0.01*U)*alat*au2meters where alat is the unit cell parameter (in a.u.) au2meters convert a.u to meters. R0/i*0.01U is in alat units (cubic cell) as Y I used the enery obtained in Ry ploted in Joules so d^2E/dXdX is in kg/s^2. I think that the difficulty of obtaining someting comparable is in the divison by the masses to obtain a result homogeneous to 1/s^2 (omega^2) In fact the dynamical matrix is filled by 1/sqrt(M_ati*M_atj) d^2E/dRatidRatj So the equivalent mass for the eigenmode obtained by diagonalising the matrix must be more complicated than just the reduce mass? If all my atoms are the same I just have to divide by one mass, but if not Antoine On Saturday, July 14, 2018 10:51 CEST, Lorenzo Paulatto wrote: Hello Antoine, Your procedure does not look obviously wrong to me, but you did not say what X is. -- Lorenzo Paulatto Written on a virtual keyboard with real fingers On Sat, 14 Jul 2018, 10:43 JAY Antoine, wrote:Dear all, I would like to (re)obtain the phonons frequencies that I first obtained using DFPT but from finite difference. Lets be R0 the ground state atomic positions and U the normalised atomic displacement of a normal mode obtained from DFPT. I have calculated the total energy from DFT of 11 structures R0+i*0.01*U with i variing from -5 to 5. The so obtained curve is fitted with a second order polynom a0+a1*X+a2*X^2, so that I obtain the second order derivative of the total energy with respect to the atomic displacements of the studied mode: 2*a2. I then divided by the atomic mass (one type of mass) and I should obtain the omega^2, but my resulting value is 3 or four times to big. I use a cubic supercell with one type of atom. Did someone already performed this kind of work? How should I do with differents atomic masses? Thank you very much for your help, Antoine Jay ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users
[QE-users] Recover phonons frequencies from the eigenmodes
Dear all, I would like to (re)obtain the phonons frequencies that I first obtained using DFPT but from finite difference. Lets be R0 the ground state atomic positions and U the normalised atomic displacement of a normal mode obtained from DFPT. I have calculated the total energy from DFT of 11 structures R0+i*0.01*U with i variing from -5 to 5. The so obtained curve is fitted with a second order polynom a0+a1*X+a2*X^2, so that I obtain the second order derivative of the total energy with respect to the atomic displacements of the studied mode: 2*a2. I then divided by the atomic mass (one type of mass) and I should obtain the omega^2, but my resulting value is 3 or four times to big. I use a cubic supercell with one type of atom. Did someone already performed this kind of work? How should I do with differents atomic masses? Thank you very much for your help, Antoine Jay ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users
Re: [QE-users] ?==?utf-8?q? phonon convergence problem of supercells
Dear Christoph, Try to decrease alph_mix to 0.1. Is it enough to converge? Best regards, Antoine Jay On Wednesday, June 27, 2018 14:16 CEST, Christoph Wolf wrote: Dear all, I have recently encountered an "interesting phenomenon". Let's say I have determined the (zone boundary phonon frequency) convergence parameter for a unit cell to be a k-grid of 6x6x6 and a ecutwfc=120 Ry. If I now take a 2x2x2 supercell of the material (using the relaxed cell as reference) I (naively) assumed that that would translate to a 3x3x3 k-grid, however I can never manage to converge phonons with that kind of setting as during the initial electric fields calculation the minimization will start to oscillate like this: iter # 17 total cpu time : 8249.5 secs av.it.: 1.8 thresh= 2.804E-05 alpha_mix = 0.700 |ddv_scf|^2 = 1.396E-07 iter # 18 total cpu time : 8353.6 secs av.it.: 1.0 thresh= 3.737E-05 alpha_mix = 0.700 |ddv_scf|^2 = 1.506E-08 iter # 19 total cpu time : 8876.9 secs av.it.: 15.4 thresh= 1.227E-05 alpha_mix = 0.700 |ddv_scf|^2 = 6.969E-08 iter # 20 total cpu time : 9303.8 secs av.it.: 12.1 thresh= 2.640E-05 alpha_mix = 0.700 |ddv_scf|^2 = 6.076E-08 iter # 21 total cpu time : 9751.4 secs av.it.: 12.9 thresh= 2.465E-05 alpha_mix = 0.700 |ddv_scf|^2 = 3.246E-08 iter # 22 total cpu time : 10145.8 secs av.it.: 11.1 thresh= 1.802E-05 alpha_mix = 0.700 |ddv_scf|^2 = 3.519E-08 iter # 23 total cpu time : 10542.9 secs av.it.: 11.2 thresh= 1.876E-05 alpha_mix = 0.700 |ddv_scf|^2 = 3.168E-09 Is this a common problem or am I doing something completely wrong...? Your help is very much appreciated! Best,Chris --Postdoctoral Researcher Center for Quantum Nanoscience, Institute for Basic Science Ewha Womans University, Seoul, South Korea ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users
[QE-users] Phonons: link between normalised eigenmodes and real atomic displacement at a given temperature
Dear all, After a phonon calculation, and a Bose-Einstein distribution we know the number n_i of phonons with frequency w_i at a given temperature. But how can we have access to the real atomic displacement induced by this phonon? I mean, the eingenmodes (X1,Y1,Z1,X2,Y2,Z2,...,XN,YN,ZN)_i in the dynfile are normalised, so doing just n_i*(X1,Y1,Z1,X2,Y2,Z2,...,XN,YN,ZN)_i is non sence... Thank you for your help, Antoine Jay ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users
[QE-users] Relevance of electron-phonon at Gamma only for defects in bulk
Dear community, I need to calculate the electron-phonons matrix elements of a simple defect in a bulk materials. I use a super cell that is large enough to study the system at gamma only: - For the scf: k=gamma and k=2^3 give the same total energies (+/- 1 meV/at). - For the phonons: q=gamma and q=2^3 give the same ph-DOS (after interpolations). However, reading the el-ph documentation and examples, it is suggested that these two steps must be done on a dense k-points and q-points grids. Has this suggestion been done only for simple cells? (and is then not necessary for huge super cells...) Any comment would be apreciated, Best regards, Antoine Jay ___ users mailing list users@lists.quantum-espresso.org https://lists.quantum-espresso.org/mailman/listinfo/users
[Pw_forum] custum-definitions in Xcrysden: complete list of key words
Dear community, Xcrysden users,
I'am trying to impose my drawing style through the file:
~/.xcrysden/custom-definitions
but the only key words I have found in the examples are:
set atmRad(14) 1.255
set atmCol(9) {1.0 0.0 0.0}
Does anyone have a complete example with all the possible key words?
In my case I only need to impose:
atomic covalent radius (11)=0.5
dispay Crystal cell =false
background color =white
but a complete review would be apreciated.
Thank for all
Antoine Jay
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Re: [Pw_forum] Physical meaning of two imaginary phonon frequencies at NEB saddle point and Transition State Theory
Dear Pascal, Dear Stefano, dear all
In a symetric system where two reactions A1->B1 and A2->B2 go trough the same
saddle point, following the eigen vector responssible in each NEB is quite easy.
For example in NEB 1, here I plot the number of "negative" frequencies as a
function of the path/15 steps
1 0 state A1
2 0
3 0
4 -1 inflexion point on the E({Ri}) curve
5 -1 This "negative" frequency w1 with eigen vector V1 is the one
responsible of the transition A1->B1
6 -1
7 -2 here appear the frequency w2 responsible of the transition A2->B2
8 -2 saddle point that is a 2 dimension saddle in a 3Nat dimension space.
9 -2
10 -1
11 -1
12 -1 inflexion point on the E({Ri}) curve
13 0
14 0
15 0 state B1
the same for NEB2 by exchanging w1 and w2.
The two paths are separeted by high energy barriers that is always higher than
the DeltaE of the saddle point except at the saddle point where this barrier is
null.
Not that in this example I give 4 potential wells link together through the
same saddle point (the reaction A1->A2 is also possible),
but it is mathematically possible to have even more potential wells as we are
in a 3Nat dimension space.
For Stefano, the state 8 mathematically exists and is physically due to a high
symmetry point, so I do not see why I should look for another saddle point.
NEB raffinement is un precise in this "plateau region".
For Pascal, following eigen vector V1 from the saddle point give me B1 (or A1
as a function of the sign of the imposed displacement) and following V2 give me
B2 (or A2),
but the problem is to know the transition rate from the Delta E and the
omega_i, I already know which eigen vector correspond to which path...
Antoine Jay
On Monday, November 13, 2017 20:27 CET, degironc <degir...@sissa.it> wrote:
I think that if there are two unstable directions at the saddle point this
means there is another saddle point with a lower Activation energy nearby. Just
slide the neb downhill stefano Sent from my Samsung Galaxy smartphone.
Original message From: Pascal Boulet
<pascal.bou...@univ-amu.fr>Date: 13/11/2017 19:45 (GMT+01:00)To: PWSCF Forum
<pw_forum@pwscf.org>Subject: *spam*Re: [Pw_forum] Physical meaning of two
imaginary phonon frequencies at NEB saddle point and Transition State Theory
Hello, My guess: you have to remove one of the imaginary frequencies (most
probably the smallest one) by following the corresponding eigenvector (=moving
the atoms accordingly). Then, you will have to make sure that the transition
state connects the reactants and products you are interested in. HTH,Pascal
Pascal Boulet—Professor in computational chemistry - DEPARTEMENT OF
CHEMISTRYAix-Marseille University - ST JEROME - Avenue Escadrille Normandie
Niemen - F-13013 Marseille - FRANCETél: +33(0)4 13 55 18 10 - Fax : +33(0)4 13
55 18 50Site : http://madirel.univ-amu.fr/pages_web_BOULET_PASCAL/infos - Email
: pascal.bou...@univ-amu.fr Le 13 nov. 2017 à 12:44, JAY Antoine
<antoine@isae-supaero.fr> a écrit : Dear all,
I have been using NEB to obtain the Delta E and phonon DOS at
begining/end/saddle point to properly evaluate the vibrationnal entropy
contribution in the transition state theory.
Commonly, one frequency in "negative" at the saddle point, the one that is
"responssible" of the transition and the following of its eigen vector from the
initial state give the minimum energy path.
However, when two phonon frequencies are "negatives" at the saddle point, that
means that at least 3 energy minima are link together to the same saddle point.
The transition state theory cannot be applied there as the preexponential term
is the ratio of the 3N-3 phonon frequencies of initial state over the 3N-4
frequencies of the saddle point, giving the dimension of a time (state life
time)
Using 3N-5 frequencies without the two negatives ones would not have a physical
meaning for the initial partition function.
Any idea on how to treat such a problem to evaluate the transition rate would
be appreciated.
Best regards.
Antoine Jay ___
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[Pw_forum] Physical meaning of two imaginary phonon frequencies at NEB saddle point and Transition State Theory
Dear all, I have been using NEB to obtain the Delta E and phonon DOS at begining/end/saddle point to properly evaluate the vibrationnal entropy contribution in the transition state theory. Commonly, one frequency in "negative" at the saddle point, the one that is "responssible" of the transition and the following of its eigen vector from the initial state give the minimum energy path. However, when two phonon frequencies are "negatives" at the saddle point, that means that at least 3 energy minima are link together to the same saddle point. The transition state theory cannot be applied there as the preexponential term is the ratio of the 3N-3 phonon frequencies of initial state over the 3N-4 frequencies of the saddle point, giving the dimension of a time (state life time) Using 3N-5 frequencies without the two negatives ones would not have a physical meaning for the initial partition function. Any idea on how to treat such a problem to evaluate the transition rate would be appreciated. Best regards. Antoine Jay ___ Pw_forum mailing list Pw_forum@pwscf.org http://pwscf.org/mailman/listinfo/pw_forum
Re: [Pw_forum] Lanczos in Phonons to test the dynamical stability
Dear Lorenzo, Maybe I misunderstand your answer because I misunderstand what is done during the long computationnal time steps of matdyn that begin with plotting "representation #i" in the ph.out file. Corrige me if I'am wrong, the goal of the step "convergeance for one representation" is to fill the 3*Nat terms of the dynamical matrix d^2E/(dx_i*dy_j) using DFPT and interpolation (sometimes 2or 3*3*Nat terms if the representation is degenerated). Using Lanczos, we can know the lowest eigen values without the knowledge of all the (3*Nat)^2 terms of the dynamical matrix, So by choosing correctlly this terms is could be possible to drastically decrease the computationnal time. For exemple with 400 atoms, if someone wants to know the 4 lowest eigen values (3 acoustics + eventually one with imaginary omega^2), only 10 to 20 representations are needed instead of 1200. (time/~100: whouaou!) So I do not understand why you say that I can not perform this algorithm here... Best regards, Antoine Jay On Monday, October 23, 2017 11:47 CEST, Lorenzo Paulatto <paul...@gmail.com> wrote: The algorithm used by matdyn is not done for speed, it can be sped up hundred of times with some little tweaking. You can try using the codes from D3Q which are a bit more optimized: d3_q2r.x (same input as q2r) and d3_r2q.x (check the manual, requires some code editing as phonon dos is not implemented, but is trivial) Or you can try to profile and optimize matdyn. hth P.S. I initially misread you question and gave you a very long and completely pointless answer, I leave it here for the records. The point still stand to some extent, because what takes time in matdyn is probably not the matrix diagonalization, but its interpolation. No, because the Lanczos algorithm (and any other similar algorithm, like Davidson) allows you to compute the lower eigenvalues of a matrix M by applying it repeatedly, instead of diagonalizing it. But you still have to be able to apply the matrix. This kind of algorithms are useful when the matrix is huge, for example in the electronic problem, the matrix has the dimension of the number of plane waves. By using Davidson, you can reduce it to (twice) the number of bands. Note that the scaling is still N^3, it is just the prefactor that is smaller. But in the phonon case, the matrix is tiny, only 3 x number of atoms, it only takes a nanosecond to diagonalize it even for hundreds of atoms.On 23/10/17 08:21, JAY Antoine wrote:Dear all, Is there a way to perform a phonon calculation only for the lowest phonon frequency for exemple by using the Lanczos algorithm? I mean to test the dynamical stability of a structure, one only need to know that "all the frequencies are positives over the full BZ", which is the same as "no frequencies are negatives" and in this case, one only need to know the lowest value of all the phonons frequencies. For the biggest structures with a too big dynamical matrix, the phonon DOS have a too high computationnal cost and this trick should be very usefull... Best regards, Antoine Jay ___ Pw_forum mailing list Pw_forum@pwscf.org http://pwscf.org/mailman/listinfo/pw_forum -- Lorenzo Paulatto - Paris ___ Pw_forum mailing list Pw_forum@pwscf.org http://pwscf.org/mailman/listinfo/pw_forum
[Pw_forum] Lanczos in Phonons to test the dynamical stability
Dear all, Is there a way to perform a phonon calculation only for the lowest phonon frequency for exemple by using the Lanczos algorithm? I mean to test the dynamical stability of a structure, one only need to know that "all the frequencies are positives over the full BZ", which is the same as "no frequencies are negatives" and in this case, one only need to know the lowest value of all the phonons frequencies. For the biggest structures with a too big dynamical matrix, the phonon DOS have a too high computationnal cost and this trick should be very usefull... Best regards, Antoine Jay ___ Pw_forum mailing list Pw_forum@pwscf.org http://pwscf.org/mailman/listinfo/pw_forum
Re: [Pw_forum] Phonons: Parallelisation over the representations
Ok I will do it 'manually' as proposed in these examples. Thanks for all. On Saturday, October 07, 2017 10:25 CEST, Lorenzo Paulatto <paul...@gmail.com> wrote: Great idea! Now you only need to spend 10k man/hours to actually do it. Oh, wait, someone did it already, but it requires some thinking to us it. Check PHonon/examples/GIRD_example Also the thermopw package should do this <http://people.sissa.it/~dalcorso/thermo_pw_help.html> although I have never used it personally. cheers On 07/10/17 10:19, JAY Antoine wrote:Dear devellopers, dear all, ?? Is there a way to do a phonon parallelisation over the representations of a same q-point ?? I think they are all calculated separatelly and communicate only at the end of each convergeance to fill the dynamical matrix, didn't it? This would be a much more efficent parallelisation than the nimage one: no processors would 'sleep' because of the high number of representations, and this allow to complete phonon calculation for much larger cells. Best regards, Antoine Jay ___ Pw_forum mailing list Pw_forum@pwscf.org http://pwscf.org/mailman/listinfo/pw_forum -- Lorenzo Paulatto - Paris ___ Pw_forum mailing list Pw_forum@pwscf.org http://pwscf.org/mailman/listinfo/pw_forum
[Pw_forum] Phonons: Parallelisation over the representations
Dear devellopers, dear all, ?? Is there a way to do a phonon parallelisation over the representations of a same q-point ?? I think they are all calculated separatelly and communicate only at the end of each convergeance to fill the dynamical matrix, didn't it? This would be a much more efficent parallelisation than the nimage one: no processors would 'sleep' because of the high number of representations, and this allow to complete phonon calculation for much larger cells. Best regards, Antoine Jay Save the date le 14 octobre 2017 : Venez d��couvrir l'ISAE-SUPAERO de 10h �� 18h pour les portes ouvertes du leader mondial de la formation en ing��nierie a��rospatiale. Rencontre avec Thomas Pesquet, dipl��m�� Supa��ro 2001, au programme. Plus d'infos sur isae-supaero.fr ___ Pw_forum mailing list Pw_forum@pwscf.org http://pwscf.org/mailman/listinfo/pw_forum
Re: [Pw_forum] symmetries during a vc-relax--> Non zero values in celldm !!
If symmetries are conserved, why do I obtained non zero terms in the
CELL_PARAMETERS where there should be zero terms?
Here is a full example in which i asked for relaxation of a monoclinic
base-center system with cell parameters
v1 = ( a/2, 0, -c/2),
v2 = (b*cos(gamma), b*sin(gamma), 0),
v3 = ( a/2, 0, c/2)
but in the output v1(2) and v3(2) are not exactelly null!!
To obtain the celldm, if I use the reversed equations
celldm(2)=b/a
celldm(3)=c/a,
celldm(4)=cos(ab)
I do not take into account these non zero values, which then gives the wrong
atomic positions.
Fortunatelly, the loose of symmetry is small, and if use these "wrong" celldm
in a second vc-relax, then third and fourth,
until the relaxation becomes useless, I finally obtain v1(2) and v3(2)
exactelly null so the reversed equations gives the right celldm.
But this is not the good way to do it as one vc-relax that concerve the
symmetries the should be the good way.
My input/output example:
as input:
ibrav = 13,
celldm(1) = 16.288976472
celldm(2) = 0.544519937
celldm(3) = 0.668172534
celldm(4) = 0.501142811
nat= 13,
ntyp= 2,
ecutwfc =80,
occupations='smearing',
smearing='mp',
degauss=0.0015,
/
diagonalization='david'
mixing_beta = 0.7
conv_thr = 4.0d-13
electron_maxstep=800
/
ion_dynamics='damp'
pot_extrapolation='second_order'
wfc_extrapolation='second_order'
/
cell_dynamics='bfgs',
press = 0,
press_conv_thr=0.1D0
cell_factor=2
ATOMIC_SPECIES
B 10.810 B-EB.pw
C 12.011 C-EB.pw
ATOMIC_POSITIONS {crystal}
B -0.188793817 -0.218234974 0.307727082
B -0.192719364 0.340793252 -0.192719364
B 0.307727082 -0.218234974 -0.188793817
B 0.190169938 0.219331860 -0.303953219
B 0.194974825 -0.339438271 0.194974825
B -0.303953219 0.219331860 0.190169938
B 0.007880732 -0.005471582 0.338682645
B 0.010098557 0.335888236 0.010098557
B 0.338682645 -0.005471582 0.007880732
B -0.005624351 0.003265079 -0.333219342
C -0.003954045 -0.319654369 -0.003954045
B -0.333219342 0.003265079 -0.005624351
C 0.425826764 0.426219771 0.425826764
K_POINTS {automatic}
12 12 12 1 1 1
as output:
CELL_PARAMETERS (alat= 16.28897647)
0.498490550 0.013603023 -0.308425490
0.284663797 0.492181426 -0.0
0.498490550 0.013603023 0.308425490
ATOMIC_POSITIONS (crystal)
B -0.202019099 -0.212621756 0.346315439
B -0.205370945 0.354119566 -0.205370945
B 0.346315439 -0.212621756 -0.202019099
B 0.206012998 0.209180405 -0.314139284
B 0.204538774 -0.320989968 0.204538774
B -0.314139284 0.209180405 0.206012998
B 0.005743670 0.002927514 0.339636302
B 0.004144354 0.351749983 0.004144354
B 0.339636302 0.002927514 0.005743670
B 0.006794941 -0.000361453 -0.329132341
C 0.002164917 -0.321786041 0.002164917
B -0.329132341 -0.000361453 0.006794941
C 0.382406681 0.380246423 0.382406681
Best regards,
Antoine Jay
On Monday, October 02, 2017 21:43 CEST, Paolo Giannozzi <p.gianno...@gmail.com>
wrote:
On Mon, Oct 2, 2017 at 4:25 PM, JAY Antoine <antoine@isae-supaero.fr>
wrote: I would like to perform a vc-relax that conserves the point group of
each atom and the symmetry of the cell. vc-relax conserves the starting
symmetry of the crystal. There is no way to conserve a symmetry that is not
there at the beginning of the structural optimization.
For example, in a trigonal cell, ibrav 5 imposes the symmetries:
v1 = a(tx,-ty,tz)
v2 = a(0,2ty,tz)
v3 = a(-tx,-ty,tz)
but these symmetries seem to be broken by the relaxation, see above: symmetries
that were present in the starting structure are not broken. If a "symmetry" is
"broken by structural relaxation", it wasn't a symmetry.
so that at the end of the relaxation, v2(1) is not exactly null and v1(1) is
not exactly -v3(1),
or for the atomic positions, a coordinate (x, x, x) alat becomes (x, 0.99 x,
1.001x ).
?? Is it possible to impose that these constraints on the unit-cell and on the
atomic positions stay fix during the relaxation process ??
This is a major issue for the use of these relaxed values in a second step for
phonons for example, because it will then not detect the symmetries.
I think it could be a good idea to add into the vc-relax output file a lign
that gives the corresponding celldm(i) and not only the CELL_PARAMETERS. I also
think that it is a good idea, but it takes a bit of time and effort, that
becomes a byte, or even a word (32 or 64 bits) if one wants to write a piece of
code that works perfectly in all cases
PaoloIn fact, in the scf before the phonon use, we would like to have ibrav
=/0 and celldm in state of ibrav 0 and quasi symmetric CELL_PARAMETERS...
Best regards,
Antoine Jay
--Paolo Giannozz
[Pw_forum] symmetries during a vc-relax
Dear devellopers, dear all, I would like to perform a vc-relax that concerves the point group of each atom and the symmetry of the cell. I know it is possible to impose symmetries on the BZ's k-points or to fix some atoms coordinates, but this does not give the expected result. For example, in a trigonal cell, ibrav 5 imposes the symmetries: v1 = a(tx,-ty,tz) v2 = a(0,2ty,tz) v3 = a(-tx,-ty,tz) but these symmetries seem to be broken by the relaxation, so that at the end of the relaxation, v2(1) is not exactelly null and v1(1) is not exactelly -v3(1), or for the atomic positions, a coordinate (x, x, x) alat becomes (x, 0.99 x, 1.001x ). ?? Is it possible to impose that these constraints on the unit-cell and on the atomic positions stay fix during the relaxation process ?? This is a major issue for the use of these relaxed values in a second step for phonons for example, because it will then not detect the symmetries. I think it could be a good idea to add into the vc-relax output file a lign that gives the corresponding celldm(i) and not only the CELL_PARAMETERS. In fact, in the scf before the phonon use, we would like to have ibrav =/0 and celldm in state of ibrav 0 and quasi symmetric CELL_PARAMETERS... Best regards, Antoine Jay Save the date le 14 octobre 2017 : Venez d��couvrir l'ISAE-SUPAERO de 10h �� 18h pour les portes ouvertes du leader mondial de la formation en ing��nierie a��rospatiale. Rencontre avec Thomas Pesquet, dipl��m�� Supa��ro 2001, au programme. Plus d'infos sur isae-supaero.fr ___ Pw_forum mailing list Pw_forum@pwscf.org http://pwscf.org/mailman/listinfo/pw_forum
