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THG
Posted: Mon Dec 09, 2024 5:11 pm
by sanoj
Dear All,
I am trying to calculate the THG (field direction 100) for few crystals. As a benchmark, I have used LBO crystals as a benchmark and I am able to reproduce the results from the literature. In all the calculations, I am using correct ibrav for the calculations.
I have two crystals which doesn't give me reasonable result when I do the calculations with correct ibrav. The ibrav for those crystals are 7 and -13. When I do the calculations with ibrav=0 for those crystals, I get reasonable result. With correct ibrav, I get value in the order of 1E-6 which is wrong and with ibrav=0, I get the value in the order of 1E-12 which is a reasonable number. I want to do all the calculations with correct ibrav as the computation time is faster. I also want to point out that I have another crystal whose ibrav is -13 and I get correct number for THG value with or without ibrav.
Here are things, I have tried so far
1. I have tested LBO crystal with correct ibrav=5 and ibrav=0, I get similar number for THG which is correct.
2. Some ibrav have off diagonal elements for the lattice parameter, so I have tried NoDiagSc.
3. I have tried to converge the calculation with kpoints, NLbands and HARRLvcs.
Any suggestion would be appreciated. I can also upload the scf, nscf and other files if necessary. Thank you in advance!
Best,
Sanoj
Re: THG
Posted: Tue Dec 10, 2024 12:43 am
by lyzhao
Dear Sanoj,
1. I have tested LBO crystal with correct ibrav=5 and ibrav=0, I get similar number for THG which is correct.
Did you test LBO or BBO? LBO should be ibrav=8, as I remember. Also BBO generally should be ibrav=-5
And could you tell me the reference for THG of BBO or LBO, which you used as a benchmark?
Best regards.
Re: THG
Posted: Tue Dec 10, 2024 10:05 am
by claudio
Do you have the same symmetries with ibrav=0 and ibrav/=0?
this is already a simple check to see if everything works fine.
Moreover you can put the cell generate with ibrav/=0 in the calculation with ibrav=0
and see if you get the same results
let us know
best
Claudio
Re: THG
Posted: Tue Dec 10, 2024 8:48 pm
by sanoj
Thank you for your reply! I think I do have the same symmetry operation. I have added a part of output where it prints the symmetry operation as the file I have is bigger than 256 KB. I can also upload other output files in which the other crystal have space group 82 (ibrav=7)
scf_withsymm.txt - ibrav ≠ 0
In this case, it prints the symmetry operation but the point group
2 Sym. Ops. (no inversion) found
s frac. trans.
isym = 1 identity
cryst. s( 1) = ( 1 0 0 )
( 0 1 0 )
( 0 0 1 )
cart. s( 1) = ( 1.0000000 0.0000000 0.0000000 )
( 0.0000000 1.0000000 0.0000000 )
( 0.0000000 0.0000000 1.0000000 )
isym = 2 180 deg rotation - cart. axis [0,1,0]
cryst. s( 2) = ( 0 1 0 )
( 1 0 0 )
( 0 0 -1 )
cart. s( 2) = ( -1.0000000 0.0000000 -0.0000000 )
( 0.0000000 1.0000000 -0.0000000 )
( 0.0000000 0.0000000 -1.0000000 )
point group C_2 (2)
there are 2 classes
the character table:
E C2
A 1.00 1.00
B 1.00 -1.00
the symmetry operations in each class and the name of the first element:
E 1
identity
C2 2
180 deg rotation - cart. axis [0,1,0]
Cartesian axes
scf.txt - ibrav = 0 - This file size is big so
In this case, it prints the symmetry operation but not the point group
2 Sym. Ops. (no inversion) found
s frac. trans.
isym = 1 identity
cryst. s( 1) = ( 1 0 0 )
( 0 1 0 )
( 0 0 1 )
cart. s( 1) = ( 1.0000000 0.0000000 -0.0000000 )
( 0.0000000 1.0000000 0.0000000 )
( 0.0000000 0.0000000 1.0000000 )
isym = 2 180 deg rotation - cart. axis [0,1,0]
cryst. s( 2) = ( -1 0 0 )
( 0 1 0 )
( 0 0 -1 )
cart. s( 2) = ( -1.0000000 0.0000000 0.0000000 )
( 0.0000000 1.0000000 0.0000000 )
( 0.0000000 0.0000000 -1.0000000 )
___________________________________
Moreover you can put the cell generate with ibrav/=0 in the calculation with ibrav=0
and see if you get the same results
Could you potentially elaborate more on this? I do not think I know how to use the keyword in the QE input.
Re: THG
Posted: Tue Dec 10, 2024 9:34 pm
by sanoj
lyzhao wrote: ↑Tue Dec 10, 2024 12:43 am
Dear Sanoj,
1. I have tested LBO crystal with correct ibrav=5 and ibrav=0, I get similar number for THG which is correct.
Did you test LBO or BBO? LBO should be ibrav=8, as I remember. Also BBO generally should be ibrav=-5
And could you tell me the reference for THG of BBO or LBO, which you used as a benchmark?
Best regards.
My LBO has a space group is 161 and it will have ibrav=5. I know that different LBO has slightly different space group which might change the ibrav. You can find the reference paper here (
https://doi.org/10.1002/pssb.202200453)
What space group your LBO has?
Re: THG
Posted: Wed Dec 11, 2024 12:17 am
by lyzhao
Dear Sanoj,
My LBO has a space group is 161 and it will have ibrav=5.
What does your LBO refer to ? I mean LiBi3O5 whose space group is 33.
Best regards.
Re: THG
Posted: Wed Dec 11, 2024 1:07 am
by sanoj
lyzhao wrote: ↑Wed Dec 11, 2024 12:17 am
Dear Sanoj,
My LBO has a space group is 161 and it will have ibrav=5.
What does your LBO refer to ? I mean LiBi3O5 whose space group is 33.
Best regards.
Dear Youzhao,
I am referring to LiNbO3.
Thanks
Sanoj
Re: THG
Posted: Sat Dec 14, 2024 3:59 pm
by sanoj
sanoj wrote: ↑Tue Dec 10, 2024 8:48 pm
Thank you for your reply! I think I do have the same symmetry operation. I have added a part of output where it prints the symmetry operation as the file I have is bigger than 256 KB. I can also upload other output files in which the other crystal have space group 82 (ibrav=7)
scf_withsymm.txt - ibrav ≠ 0
In this case, it prints the symmetry operation but the point group
2 Sym. Ops. (no inversion) found
s frac. trans.
isym = 1 identity
cryst. s( 1) = ( 1 0 0 )
( 0 1 0 )
( 0 0 1 )
cart. s( 1) = ( 1.0000000 0.0000000 0.0000000 )
( 0.0000000 1.0000000 0.0000000 )
( 0.0000000 0.0000000 1.0000000 )
isym = 2 180 deg rotation - cart. axis [0,1,0]
cryst. s( 2) = ( 0 1 0 )
( 1 0 0 )
( 0 0 -1 )
cart. s( 2) = ( -1.0000000 0.0000000 -0.0000000 )
( 0.0000000 1.0000000 -0.0000000 )
( 0.0000000 0.0000000 -1.0000000 )
point group C_2 (2)
there are 2 classes
the character table:
E C2
A 1.00 1.00
B 1.00 -1.00
the symmetry operations in each class and the name of the first element:
E 1
identity
C2 2
180 deg rotation - cart. axis [0,1,0]
Cartesian axes
scf.txt - ibrav = 0 - This file size is big so
In this case, it prints the symmetry operation but not the point group
2 Sym. Ops. (no inversion) found
s frac. trans.
isym = 1 identity
cryst. s( 1) = ( 1 0 0 )
( 0 1 0 )
( 0 0 1 )
cart. s( 1) = ( 1.0000000 0.0000000 -0.0000000 )
( 0.0000000 1.0000000 0.0000000 )
( 0.0000000 0.0000000 1.0000000 )
isym = 2 180 deg rotation - cart. axis [0,1,0]
cryst. s( 2) = ( -1 0 0 )
( 0 1 0 )
( 0 0 -1 )
cart. s( 2) = ( -1.0000000 0.0000000 0.0000000 )
( 0.0000000 1.0000000 0.0000000 )
( 0.0000000 0.0000000 -1.0000000 )
___________________________________
Moreover you can put the cell generate with ibrav/=0 in the calculation with ibrav=0
and see if you get the same results
Could you potentially elaborate more on this? I do not think I know how to use the keyword in the QE input.
Dear All,
I am commenting again to get any thoughts or help if possible. Thank you for the help in advance!
Best,
Sanoj
Re: THG
Posted: Mon Dec 16, 2024 11:52 am
by claudio
Dear Sanoj
> Could you potentially elaborate more on this? I do not think I know how to use the keyword in the QE input.
when you use ibrav/=0 QE write the cell in the output file, you can take this cell
and copy in a new input file with ibrav=0, and check that results are the same.
let us know
best
Claudio
Re: THG
Posted: Mon Dec 16, 2024 1:52 pm
by sanoj
claudio wrote: ↑Mon Dec 16, 2024 11:52 am
Dear Sanoj
> Could you potentially elaborate more on this? I do not think I know how to use the keyword in the QE input.
when you use ibrav/=0 QE write the cell in the output file, you can take this cell
and copy in a new input file with ibrav=0, and check that results are the same.
let us know
best
Claudio
Dear Claudio,
Thank you for the response.
With ibrav/=0, QE does not output the full unit cell (It only prints lattice symmetry atomic position). So, I can't take the cell generated and run the new input file with ibrav=0. Unless I am missing something (like a keyword to print the atomic position in the full unit cell) in QE, I can't take the cell generated and set ibrav=0 and compare the result.
What I have done so far is that, I use the QE output to write down the cif file and then use this cif file to write down the POSACR file. I use this POSCAR file (has the lattice parameter and all the atoms in the unit cell) to make new QE input file with ibrav=0. The result (symmetry with and without ibrav) I shared before based on this approach.
Best,
Sanoj