Hi
I experimenting a bit with using the coulomb cutoff for a twodimensional system periodic in x and y. As I understand from Phys. Rev. B 73, 205119 (2006), one should use then use a cutoff in the z direction of length L/2 where L is the size of my supercell in the z direction, but I am wondering if this is always the best choice? I seem to get vey similar results for the QP gap if I use a cutoff of L instead of L/2 meaning that I can actually use a smaller supercell to obtain the converged result? However, I would think that setting the cutoff length equal to the supercell size means that the densities of periodic images can interact and I am puzzled to why I seem to get the same result?
In a related question (now that I am here:)), I would like to ask if the random integration method is something that need to be converged carefully (and how). I usually just use 1000000 random points and a single G vector (as recommended) and my results are then converged with respect to kpoint sampling. Does this mean that my RIM parameters are all right? Actually I do not fully understand the Gvector parameter, since I thought that only G=(0,0,0) is problematic and needs to be treated with this method?
BR
Thomas Olsen
Post Doc
Technical University of Denmark
Coulomb cutoff
Moderators: Davide Sangalli, andrea marini, Daniele Varsano, andrea.ferretti, myrta gruning
 Daniele Varsano
 Posts: 2963
 Joined: Tue Mar 17, 2009 2:23 pm
 Contact:
Re: Coulomb cutoff
Dear Thomas,
in order to set the z value, it depends on the cell length as well as on the spread of your density,
What one need, is t leave interact al the charge in the cell, and do not permit interactions between cells.
Now, I think that there is a factor two that may be have to be fixed. z=L it means that you will have a potential
1/r from L/2 to L/2, and zero otherwise. As what has meaning is from distance equal to zero to the
extreme of the cut, putting L in the input, it is considering an interaction distance of L/2. Sorry for that,
I will fix it soon. I do not know if it is clear, anyway you can activate the flag TestColCut, or something like that
(just put more verbosity in the creation of the input) in order to look at what your potential look like in real space.
Of course this is a back FFT, so the quality of the potential will depend on how many Gvectors you are including.
Remember , when constructing a box like potential, the random integration method have to be used mandatory,
because you need integrals over the Bz in order to construct the potential components.
About your second question, it is true the problematic one is the G(0,0,0), but the integral of 1/(q+G)^2 on your q point
grid could be badly evaluated if you have non tridimensional grid samplings, also for other G vector than the (0,0,0) anyway
in the report you have the error you are doing, comparing a Monte Carlo integral with just 1/q^2*Vq where Vq is the small
volume of qgrid around the point q. When considering G=(0,0,0) you have this comparison for all the values of q
in the Bz, looking at the edge you can estimate if you need more G or not. The more the function 1/(q+G)^2 will be flat, the
more the integrals evaluated as 1/(q+G)^2Vq is correct, usually at the edge of the Bz the error is small and considering
just one G is enough.
I hope, even if I doubt, it helps
Cheers,
Daniele
in order to set the z value, it depends on the cell length as well as on the spread of your density,
What one need, is t leave interact al the charge in the cell, and do not permit interactions between cells.
Now, I think that there is a factor two that may be have to be fixed. z=L it means that you will have a potential
1/r from L/2 to L/2, and zero otherwise. As what has meaning is from distance equal to zero to the
extreme of the cut, putting L in the input, it is considering an interaction distance of L/2. Sorry for that,
I will fix it soon. I do not know if it is clear, anyway you can activate the flag TestColCut, or something like that
(just put more verbosity in the creation of the input) in order to look at what your potential look like in real space.
Of course this is a back FFT, so the quality of the potential will depend on how many Gvectors you are including.
Remember , when constructing a box like potential, the random integration method have to be used mandatory,
because you need integrals over the Bz in order to construct the potential components.
About your second question, it is true the problematic one is the G(0,0,0), but the integral of 1/(q+G)^2 on your q point
grid could be badly evaluated if you have non tridimensional grid samplings, also for other G vector than the (0,0,0) anyway
in the report you have the error you are doing, comparing a Monte Carlo integral with just 1/q^2*Vq where Vq is the small
volume of qgrid around the point q. When considering G=(0,0,0) you have this comparison for all the values of q
in the Bz, looking at the edge you can estimate if you need more G or not. The more the function 1/(q+G)^2 will be flat, the
more the integrals evaluated as 1/(q+G)^2Vq is correct, usually at the edge of the Bz the error is small and considering
just one G is enough.
I hope, even if I doubt, it helps
Cheers,
Daniele
Dr. Daniele Varsano
S3CNR Institute of Nanoscience and MaX Center, Italy
MaX  Materials design at the Exascale
http://www.nano.cnr.it
http://www.maxcentre.eu/
S3CNR Institute of Nanoscience and MaX Center, Italy
MaX  Materials design at the Exascale
http://www.nano.cnr.it
http://www.maxcentre.eu/

 Posts: 32
 Joined: Thu Oct 28, 2010 9:45 am
Re: Coulomb cutoff
Dear Daniele
Thanks a lot for your reply.
I am sorry if I seem a bit slow, but just to get it completely straight. You say that setting the cutoff to L corresponds to the potential being zero outside the range L/2 to L/2. My slab is centered in the middle of the unit cell but that should not matter right? Since the coulomb potential is just a function of rr'. I am thinking that the potential is cut off when rr' is larger than L, but if r is in my slab then r' could be the periodic image if L is also the length of the unit cell? Is the correct or am I thinking about this in a wrong way?
I should say that my main problem presently is that when I use RIM only my gap seem to be very well converged with respect to kpoint sampling (for a fixed unit cell size). When I also use the cutoff method the, gap changes (as expected) but is easier to converge with respect to unit cell size. But now it changes when I increase the kpoint sampling from 15x15 to 30x30! (no change here when not using coulomb cutoff) Does one need a denser kpoint sampling when using the coulomb cutoff? Or am I doing something else wrong?
The output of a GW calculation with RIM and cutoff is pasted below.
BR
Thomas
[04] Coloumb potential Random Integration (RIM)
===============================================
[RD./SAVE//ndb.RIM]
Brillouin Zone Q/K grids (IBZ/BZ): 27 225 27 225
Coulombian RL components : 1
Coulombian diagonal components :yes
RIM random points : 1000000
RIM RL volume [a.u.]: 0.187592
Real RL volume [a.u.]: 0.187314
Eps^1 reference component :0
Eps^1 components : 0.000 0.000 0.000
RIM anysotropy factor : 0.000000
 S/N 005615  v.03.02.03 r.696 
Summary of Coulomb integrals for nonmetallic bands Q[au] RIM/Bare:
Q [1]:0.1000E40.9472 * Q [2]: 0.08177 0.89853
Q [9]: 0.141637 0.962746 * Q [3]: 0.163548 0.972193
Q [10]: 0.216353 0.984153 * Q [4]: 0.245322 0.988311
Q [16]: 0.283273 0.989550 * Q [11]: 0.294840 0.991820
Q [5]: 0.327096 0.995115 * Q [17]: 0.356444 0.995261
Q [12]: 0.374735 0.995251 * Q [6]: 0.408870 0.997731
Q [21]: 0.424910 0.998159 * Q [18]: 0.432707 0.997258
Q [13]: 0.455298 0.995599 * Q [7]: 0.490644 0.999148
Q [22]: 0.497411 0.999299 * Q [19]: 0.510678 0.998449
Q [14]: 0.536227 0.996669 * Q [25]: 0.566546 0.998411
Q [23]: 0.572417 1.000049 * Q [8]: 0.572417 1.000000
Q [20]: 0.589680 0.999207 * Q [15]: 0.617379 0.997348
Q [26]: 0.638675 0.998927 * Q [24]: 0.649060 1.000560
Q [27]: 0.708183 0.999285
[05] Coloumb potential CutOff :box z
====================================
[RD../SAVE//ndb.cutoff]
Brillouin Zone Q/K grids (IBZ/BZ): 27 225 27 225
CutOff Geometry :box z
Box sides length [au]: 0.00000 0.00000 21.80000
Sphere/Cylinder radius [au]: 0.000000
Cylinder length [au]: 0.000000
RL components : 6445
RL components used in the sum : 6445
RIM corrections included :yes
RIM RL components : 1
RIM random points : 1000000
 S/N 005615  v.03.02.03 r.696 
Thanks a lot for your reply.
I am sorry if I seem a bit slow, but just to get it completely straight. You say that setting the cutoff to L corresponds to the potential being zero outside the range L/2 to L/2. My slab is centered in the middle of the unit cell but that should not matter right? Since the coulomb potential is just a function of rr'. I am thinking that the potential is cut off when rr' is larger than L, but if r is in my slab then r' could be the periodic image if L is also the length of the unit cell? Is the correct or am I thinking about this in a wrong way?
I should say that my main problem presently is that when I use RIM only my gap seem to be very well converged with respect to kpoint sampling (for a fixed unit cell size). When I also use the cutoff method the, gap changes (as expected) but is easier to converge with respect to unit cell size. But now it changes when I increase the kpoint sampling from 15x15 to 30x30! (no change here when not using coulomb cutoff) Does one need a denser kpoint sampling when using the coulomb cutoff? Or am I doing something else wrong?
The output of a GW calculation with RIM and cutoff is pasted below.
BR
Thomas
[04] Coloumb potential Random Integration (RIM)
===============================================
[RD./SAVE//ndb.RIM]
Brillouin Zone Q/K grids (IBZ/BZ): 27 225 27 225
Coulombian RL components : 1
Coulombian diagonal components :yes
RIM random points : 1000000
RIM RL volume [a.u.]: 0.187592
Real RL volume [a.u.]: 0.187314
Eps^1 reference component :0
Eps^1 components : 0.000 0.000 0.000
RIM anysotropy factor : 0.000000
 S/N 005615  v.03.02.03 r.696 
Summary of Coulomb integrals for nonmetallic bands Q[au] RIM/Bare:
Q [1]:0.1000E40.9472 * Q [2]: 0.08177 0.89853
Q [9]: 0.141637 0.962746 * Q [3]: 0.163548 0.972193
Q [10]: 0.216353 0.984153 * Q [4]: 0.245322 0.988311
Q [16]: 0.283273 0.989550 * Q [11]: 0.294840 0.991820
Q [5]: 0.327096 0.995115 * Q [17]: 0.356444 0.995261
Q [12]: 0.374735 0.995251 * Q [6]: 0.408870 0.997731
Q [21]: 0.424910 0.998159 * Q [18]: 0.432707 0.997258
Q [13]: 0.455298 0.995599 * Q [7]: 0.490644 0.999148
Q [22]: 0.497411 0.999299 * Q [19]: 0.510678 0.998449
Q [14]: 0.536227 0.996669 * Q [25]: 0.566546 0.998411
Q [23]: 0.572417 1.000049 * Q [8]: 0.572417 1.000000
Q [20]: 0.589680 0.999207 * Q [15]: 0.617379 0.997348
Q [26]: 0.638675 0.998927 * Q [24]: 0.649060 1.000560
Q [27]: 0.708183 0.999285
[05] Coloumb potential CutOff :box z
====================================
[RD../SAVE//ndb.cutoff]
Brillouin Zone Q/K grids (IBZ/BZ): 27 225 27 225
CutOff Geometry :box z
Box sides length [au]: 0.00000 0.00000 21.80000
Sphere/Cylinder radius [au]: 0.000000
Cylinder length [au]: 0.000000
RL components : 6445
RL components used in the sum : 6445
RIM corrections included :yes
RIM RL components : 1
RIM random points : 1000000
 S/N 005615  v.03.02.03 r.696 
 Daniele Varsano
 Posts: 2963
 Joined: Tue Mar 17, 2009 2:23 pm
 Contact:
Re: Coulomb cutoff
Dear Thomas,
don't worry, the argument here is a little bit tricky:
I try to plot a sketch:
==========
D
()
L
D is the dimension of your system, and it will decay to zero exponentially outside D.
L is the size of your supercell.
the truncated coulomb function will be 1/r if z < zcut
what we want here is that all the density of your system interact in the unit cell, but does not
interact with the replica. Now just sit down on one of the edge of your system
thie condition implies that
zcut > D
and pratically in the input file you should put that zcut/2 > D for the reason I explained in the previous mail,
the origin of your potential will be on the edge nad is defined it will be 1/r from zcut/2 to zcut/2.
Next in order to avoid it interact with the replica
zcut < LD
for the same reason explained before zcut in reality will be zcut/2. In this case, you will have interaction with the replica,
but in a range with zero density.
so as you can see, it is not independent from the cell size, the cell size have to be fixed in order to find a zcut that
fulfill the two conditions. This condition are not so straight, as the density decay exponentially.
Let me remember that each component of this shape of potential it obtained by integration in the Bz, unfortunately this
is not documented. For this reason, before calculating the cutoff, you have first to run a RIM calculation (or doing them in
the same input file).
Next:
from here you can see the ratio between the RIM and a naif calculation done as 1/q^2*Vq. The first component q=0 is treated
analytically (integral in a sphere around the origin, so the error is small (0.94). Next the two methods starts to deviate q(2) 0.89853,
then the function to integrate is more and more flat and at the edge of the box they are quite indential (1.0005). This is a fingerprint
that considering just G=0 is safe!!
to converge it with the volume of the supercell. Of course the number of kpoint increase as power of dimensions, so for 2D system it could be a problem.
The convergence with the volume turns to be really slow, and I cannot say now what's the best procedure to follow. You can also imagine to calculate
you gap without cutoff for different values and extrapolate to infinite supercell.
Hope it helps,
Cheers,
Daniele
don't worry, the argument here is a little bit tricky:
I try to plot a sketch:
==========
D
()
L
D is the dimension of your system, and it will decay to zero exponentially outside D.
L is the size of your supercell.
the truncated coulomb function will be 1/r if z < zcut
what we want here is that all the density of your system interact in the unit cell, but does not
interact with the replica. Now just sit down on one of the edge of your system
thie condition implies that
zcut > D
and pratically in the input file you should put that zcut/2 > D for the reason I explained in the previous mail,
the origin of your potential will be on the edge nad is defined it will be 1/r from zcut/2 to zcut/2.
Next in order to avoid it interact with the replica
zcut < LD
for the same reason explained before zcut in reality will be zcut/2. In this case, you will have interaction with the replica,
but in a range with zero density.
so as you can see, it is not independent from the cell size, the cell size have to be fixed in order to find a zcut that
fulfill the two conditions. This condition are not so straight, as the density decay exponentially.
Let me remember that each component of this shape of potential it obtained by integration in the Bz, unfortunately this
is not documented. For this reason, before calculating the cutoff, you have first to run a RIM calculation (or doing them in
the same input file).
Right, should not matter!!My slab is centered in the middle of the unit cell but that should not matter right?
Next:
Code: Select all
Q [1]:0.1000E40.9472 * Q [2]: 0.08177 0.89853
Q [9]: 0.141637 0.962746 * Q [3]: 0.163548 0.972193
Q [10]: 0.216353 0.984153 * Q [4]: 0.245322 0.988311
Q [16]: 0.283273 0.989550 * Q [11]: 0.294840 0.991820
Q [5]: 0.327096 0.995115 * Q [17]: 0.356444 0.995261
Q [12]: 0.374735 0.995251 * Q [6]: 0.408870 0.997731
Q [21]: 0.424910 0.998159 * Q [18]: 0.432707 0.997258
Q [13]: 0.455298 0.995599 * Q [7]: 0.490644 0.999148
Q [22]: 0.497411 0.999299 * Q [19]: 0.510678 0.998449
Q [14]: 0.536227 0.996669 * Q [25]: 0.566546 0.998411
Q [23]: 0.572417 1.000049 * Q [8]: 0.572417 1.000000
Q [20]: 0.589680 0.999207 * Q [15]: 0.617379 0.997348
Q [26]: 0.638675 0.998927 * Q [24]: 0.649060 1.000560
Q [27]: 0.708183 0.999285
analytically (integral in a sphere around the origin, so the error is small (0.94). Next the two methods starts to deviate q(2) 0.89853,
then the function to integrate is more and more flat and at the edge of the box they are quite indential (1.0005). This is a fingerprint
that considering just G=0 is safe!!
Yes that's the reason we want to use it! usually it increase, but I have not a clear justification on that.When I also use the cutoff method the, gap changes (as expected) but is easier to converge with respect to unit cell size.
Yes it can happen!!, I studied it just for a wire, and I found that anyway it was easier and faster to converge it with a bigger kpoint sampling thanBut now it changes when I increase the kpoint sampling from 15x15 to 30x30! (no change here when not using coulomb cutoff) Does one need a denser kpoint sampling when using the coulomb cutoff? Or am I doing something else wrong?
to converge it with the volume of the supercell. Of course the number of kpoint increase as power of dimensions, so for 2D system it could be a problem.
The convergence with the volume turns to be really slow, and I cannot say now what's the best procedure to follow. You can also imagine to calculate
you gap without cutoff for different values and extrapolate to infinite supercell.
Hope it helps,
Cheers,
Daniele
Dr. Daniele Varsano
S3CNR Institute of Nanoscience and MaX Center, Italy
MaX  Materials design at the Exascale
http://www.nano.cnr.it
http://www.maxcentre.eu/
S3CNR Institute of Nanoscience and MaX Center, Italy
MaX  Materials design at the Exascale
http://www.nano.cnr.it
http://www.maxcentre.eu/

 Posts: 32
 Joined: Thu Oct 28, 2010 9:45 am
Re: Coulomb cutoff
Thanks a lot!
It was really the factor of 1/2 I did not really get before now. But it all makes sense now.
BR
Thomas
It was really the factor of 1/2 I did not really get before now. But it all makes sense now.
BR
Thomas