I am using quantum espresso for band structure calculation and I am beginner in using this software.
How the nat is decided? As I read in manual that nat defines "no of atoms in unit cell". When I look at the examples, I found different nat(s) for different atoms as the type of crystal lattice remains same.?
Que2. When the value of ibrave is same then how nat is differing? Let say ibrav=2(fcc) for which we have read in books the no of atoms per unit cell is 4. How come nat=2 or 1 is written in the example files.
Que 3. How Xcrysden is using this info for making unit cell of crystal lattice?
Que4. What is the basis of choosing any particular "kpoints"?
1.no. of atoms (nat)-total number of atoms you define in atomic positions
For Example: Silicon (diamond structure) with Ibrav=0 ,nat = 8 because i m defining the positions of each atom in bulk
Si 0.0 0.0 0.0 //will generate the cubic crystal
Si 0.5 0.5 0.0 //will genetare a fcc atom in x-y plane
Si 0.25 0.25 0.25 //will genetare a one atom of diamond structure in xyz
Si 0.75 0.75 0.25 //will generate second atom of diamond structure in xyz
Si 0.0 0.5 0.5 // will genetare a fcc atom in y-z plane
Si 0.5 0.0 0.5 //will genetare a fcc atom in x-z plane
Si 0.25 0.75 0.75 //will generate third atom of diamond structure in xyz
Si 0.75 0.25 0.75 //will generate fourth atom of diamond structure in xyz
but with ibrav = 2 nat =2 beacuse i m defining only two positions for atoms.
For example :Silicon diamond structure
Si 0.0 .0.0 0.0 //will generate whole fcc structure of silicon
Si 0.25 0.25 0.25 // will generate the extra atoms of daimond structure of silicon
2.ibrav according to me is not that we read...acctually here ibrav means how we are defing the crystal
For Example:as i have given above..there with ibrav =0 i was making my own structure of silicon(diamond structure) but with i brav =2 ,software has already defined a fcc structure of silicon and nat = 2 means there 0 0 0 was fr fcc ibrav 2 and 0.25 0.25 0.25 was fr extra atoms of diamond structure
4.
It does not depend on the number of atoms.
You have to test the convergence at fixed smearing. E.g. you fix a value
smearing that's small enough for you, than you increase the number of
k-points until total energy converges. You may then try again for a
smaller/larger smearing and see if the results are consistent.
You have to do a convergence test for the right kpoint to use for your system under study.
@Razia Shabnam Thanks for making me to understand about the atomic positions . I`ll run the jobs and let you know..! Thanks
@Omamuyovwi Jolayemi Thanks for you kind response. But trying this approach on bigger systems will take huge amount of time for deciding the suitable kpoints. If there is any possibility that we can reduce the level of theory. So that it gets converge in 5-10 min ?
@Prateez Ahuja, if running calculations on your personal computer takes a longer time than expected, you can check the blocks on your input file. If everything seem fine, you can run the job on a supercomputer.
in quantum espresso kpoints consist of 6 inputs ( nx ny nz dx dy dz ) in first three you determine to which Brillion zone do you want to take or to make the calculation ( to first Brillion zone or second or ...) in each direction so if you just work on isolated atom or single molecule so you do not have a repetition or crystal structure so only make it 1 1 1 0 0 0
otherwise in bulk material you have an 3diminsion crystal so can chose larger numbers ex 4 4 4 0 0 0i so f you have 2dimintion material such as graphene
so, you have repetition only in 2 direction for example in y and z direction so you make your kpoints such as 1 4 4 0 0 0 so here we made an limitation to x direction
for more advanced level you make use of dx dy dz to take in between zones distance
another advanced case when you in bulk material has antisymmetric in certain direction or another problem so you reduce certain diction calculation
such as 3 5 5 0 0 0
at end i consist on to understand it will you must understand well what is (k-vector) and (reciprocal lattice) and (Brillion zones)
and you can look on this YouTube video
https://www.youtube.com/watch?v=9jD1KY426EM&t=948s
for more deeply Scientific concepts of k-points
K-point grid
k_grid k_grid_shift
0 0 0 @@@ 0 0 0
reciprocal lattice vector units If you have a large unit cell, you need fewer k-points because large distance in real space it is the smallest step in reciprocal space
So, if you have small unit cell so you will need to take large kpoints because you if you remember that smallest step in real space it is the biggest on in reciprocal space
. . . . In real
000 001 002 003
. . . . In reciprocal
000 003 002 001
So, if in real space has a unit cell with side dim =1
So must take in reciprocal space large k vale to take whole unit cell
And I am not sure about what is grid shift
But I thought it is use when you want to take calculation for the in between bands
Or in most probable think it is represent the difference between calculated k value and input one for more instruction look down on how we calculate the kpoints value (n of kx, n of ky , n of kz )
For example, si unit cell has a=10.41 angstrom
So, the rule of k=n*pi / a
So, for k value which cover the whole unit cell in another form the k value which represent 001 plane get it at (n*pi) = a so n=a/pi =10.41/pi=3.31 so if so we take n=3 for two reasons first n for bands start take only integer second tol sure our calculation will cover all cell but not preferer to take any large value because large value in reciprocal space will take use more in side unit cell so not cover the whole cell
So, in our si unit cell the most preferer kpoints are
(3 3 3 0 0 0 )
Imp note unit cell of isolate and semi it is large so have low density of states
But in metal unit cell it is small so has high density
So, in semi take k points value low compare to in metal need large k point
Omamuyovwi Rita Jolayemi The kpoint that I used for my unit cell? Do I have to use the same kpoint for scf and bandstructure for supercell?
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