calculate the QP properties (yambo vs. exp)

Deals with issues related to computation of optical spectra in reciprocal space: RPA, TDDFT, local field effects.

Moderators: Davide Sangalli, andrea.ferretti, myrta gruning, andrea marini, Daniele Varsano, Conor Hogan

Post Reply
chwolf
Posts: 44
Joined: Mon Jul 11, 2016 2:11 pm
Location: South Korea
Contact:

calculate the QP properties (yambo vs. exp)

Post by chwolf » Tue Oct 18, 2016 4:38 am

Dear all!

I am a bit confused by the definition of transferred momenta ("QpntsRXd") and I cannot achieve reasonable agreement between calculated and measured optical properties without scissoring.

My material (cubic CsPbBr3) has a direct gap at the R-point of a cubic unit cell (i.e. k=(0.5,0.5,0.5)); According to the GS calculation that point is:

QE: k(56)= (-0.5000 -0.5000 -0.5000)
YAMBO: *X* K [56] -0.5000 -0.5000 -0.5000 (iku)

So I don't need to add additional grid points. I have 0-22 filled bands and 23-60 empty bands, as reflected in r_setup. r_setup also gives me a direct and indirect gap

Indirect Gaps [ev]: 1.751380 3.439574
Direct Gaps [ev]: 1.751380 5.588216

The experimental gap is ~2.3 eV and from the DOS calculation I usually get 1.7 eV (good enough, considering GGA, see also [1]).

I follow the LiF exciton tutorial (what I want to calculate eventually is the "exciton binding energy") and I find the following

QpntsRXd = 1 | 1 give interestingly the best agreement with the data (but this must be physically meaningless...)
QpntsRXd= 56 | 56 does not give reasonable agreement in RPA and the shown LRC approach (the latter I just use the value given for LRC_alpha=-8.7, so no surprise)

Now, finally, to make this a question and not a monologue: which direction should I follow with the calculation to get better agreement between yambo and experiment?

The expected value for the binding energy is somewhere between 40...60 meV according to experiments;

[1] https://arxiv.org/pdf/1405.1706.pdf

I am new to yambo and any hint is very much appreciated!

Yours,
Chris

PS: k=12x12x12 (converges around 4x4x4 for the GS) and cutoff for the wavefunctions is 60 Ry
You do not have the required permissions to view the files attached to this post.
Christoph Wolf

Postech university, South Korea
chwolf@postech.ac.kr

User avatar
Daniele Varsano
Posts: 4198
Joined: Tue Mar 17, 2009 2:23 pm
Contact:

Re: calculate the QP properties (yambo vs. exp)

Post by Daniele Varsano » Tue Oct 18, 2016 7:15 am

Dear Christoph,

The variable QpntsRXd indicates the transferred momentum.
You are calculating spectra in linear response, and if you are interested in optical absorption you will need to consider the optical limit, i.e. q->0.
The momentum carried by the photon is extremely small compared with the crystal momenta.
This is exactly QpntsRXd = 1. If you look at your report you will see that the q index=1 is alway q=0.
Changing the index of QpntsRXd different from 1, you calculate the response for finite transferred momenta, which is related to electron energy loss spectra (EELS), but not to absorption.

You can see difference between EELS and Absorption in the documentation pages or in the lecture notes.

Consider then that if your goak is to compute exciton binding energies, simple linear response is not enough. You nee do resort to GW/BSE scheme, or alternative using the LRC_alpha kernel as you did, but the alpha value is an empiric parameter.

Hope it helps,

Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/

chwolf
Posts: 44
Joined: Mon Jul 11, 2016 2:11 pm
Location: South Korea
Contact:

Re: calculate the QP properties (yambo vs. exp)

Post by chwolf » Tue Oct 18, 2016 7:24 am

Dear Daniele!

Thank you very much for your reply (mille grazie if my high-school Italian doesn't betray me.. ;) ) and for clarifying the difference between q and k. So, if I understood this, QpntsRXd with index 1 corresponds to q(1)=0, i.e. vertical transitions in the band-structure image, whereas q(not equal 1) corresponds to some other momentum, more precisely to all transitions together (cumulative) that have the same q*=k-k'.

So irrespective of the k-point where the transition occurs, as long as it is vertical q(1)=0 is always right (actually this is crystal clear from reading the docs again.. sorry I asked!).

I will give the links a look!

Thanks a lot!

Chris
Christoph Wolf

Postech university, South Korea
chwolf@postech.ac.kr

User avatar
Daniele Varsano
Posts: 4198
Joined: Tue Mar 17, 2009 2:23 pm
Contact:

Re: calculate the QP properties (yambo vs. exp)

Post by Daniele Varsano » Tue Oct 18, 2016 7:32 am

Dear Christoph,
mille grazie
Non c'e' di che!!
If I understood this, QpntsRXd with index 1 corresponds to q(1)=0, i.e. vertical transitions in the band-structure image, whereas q(not equal 1) corresponds to some other momentum, more precisely to all transitions together (cumulative) that have the same q*=k-k'.
Totally correct.
So irrespective of the k-point where the transition occurs, as long as it is vertical q(1)=0 is always right
Yes, as it is a Fermi Golden Rule, all the vertical transition are summed up: and many transition can contribute to the spectrum, not only the one at the direct gap.


Best,
Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/

Post Reply