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Quxiao
Posts: 92
Joined: Fri Mar 26, 2021 11:27 am

Dear palful:
I want to know the physical expression of the "Dipole Oscillator Strengths" and the residual,which printed in the file o-2D_WR_WC.exc_qpt1_E_sorted (o-2D_WR_WC.exc_qpt1_I_sorted ) .And I want to know what is the relationship between residual and this Dipole Oscillator Strengths.

In the "Supplementary Information for:Exciton Radiative Lifetimes in Layered Transition Metal Dichalcogenides",in the Equation 8，

[\mu _s^2 = \frac{{{\hbar ^2}}}{{{m^2}E_S^2(0)}}\frac{{{{\left| {\left\langle {G|{p_\parallel }|{\psi _S}(0)} \right\rangle } \right|}^2}}}{{{N_k}}}\].

In the excitons_sort_and_report.F，I see that code，which may means the Residuals is not only the BS_R*BS_L,that may print the \mu_s^2,which calculated by the mu_ s^2 = residuals^2*Omega*N_ K/(4 * pi * HA2EV * n_spin),is it ?

Code: Select all

 126Residuals(:) = real(BS_R(:)*conjg(BS_R(:)),SP)
127    if (write_widths)  Residuals(:) = abs(BS_R(:))
128    if(iq==1) q_norm=q0_def_norm**2
129    if(iq> 1) q_norm=iku_v_norm(BSqpts(:,iq))**2
130    Residuals(:) = Residuals(:)*real(spin_occ,SP)/(2._SP*pi)**3*d3k_factor*4._SP*pi/q_norm*HA2EV


These make me very confused, I can see this process in that. F file,but I could not sure which printed in the o*exc*file,residuals^2 or \mu_s^2, Can you help me solve the confusion?

I have seen similar posts from you before, but the images cannot be loaded，and the topic is locked.Thank you very much for your help!

Best wishes!
Quxiao
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Quxiao in BIT,calculate the exciton

palful
Posts: 45
Joined: Tue Jan 26, 2016 11:23 am
Location: Modena and Milan

Dear Quxiao,

The equation missing from my previous post is:
eps2.png
(note that here we are considering the exciton Q-momentum equal to zero).
The discussion of the terms is the same as in my previous post, but let me add something more related to your question.

First, what we usually call "residual" for exciton state \lambda is the term: \sum_{cvk}A^\lambda_{cvk} d^{cvk}. This corresponds to the "BS_R" variable in the yambo code and is the quantity saved in the ndb.BS_diago database. The other factors multiplied at line 130 of the code you posted are dimensional factors lumped into the "C" term in the equation.

Using the equations of the reference you mention, the dipoles d^{cvk} are Eq. (3), and the residuals are Eq. (4).

The values printed in the o-* output of "ypp -e s" are a list of BS_R*BS_L for each exciton (i.e., what is called "Residuals" in the code but renormalised in such a way that the largest value is 1), which with standard approximations in the BSE kernel is equal to BS_R*conj(BS_R).

Cheers,
Fulvio
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Dr. Fulvio Paleari
S3-CNR Institute of Nanoscience and MaX Center
Modena, Italy

Quxiao
Posts: 92
Joined: Fri Mar 26, 2021 11:27 am

Dear palful：
So exciting recived your reply! The above issues have been bothering me for a long time,Thanks for your kindness again!
I also want to confirm that according to what you said, "residuals" refers to Equation 4 in Article" Supplementary Information for:Exciton Radiative Lifetimes in Layered Transition Metal Dichalcogenides".
I have also seen the Equation (9) in the "Theory and Ab Initio Calculation of Radiative Lifetime of Excitons in Semiconducting Carbon Nanotubes" https://journals.aps.org/prl/abstract/1 ... .95.247402 and the formula below the Equation (2) in the "Theory and Ab Initio Computation of the Anisotropic Light Emission in Monolayer Transition Metal Dichalcogenides"https://pubs.acs.org/doi/full/10.1021/a ... tt.8b01114.
Therefore, what they refer to is actually exciton transiton dipole:P_s(Q),which also called the residuals?Because the P_s(Q)=BS_R,so the P_s(Q)^2=residuals^2,which printed in the o*exc*Esorted,and the units of that residual^2 is eV? Is it true?
Best wishes！
Quxiao
Quxiao in BIT,calculate the exciton

Quxiao
Posts: 92
Joined: Fri Mar 26, 2021 11:27 am

Dear palful:
Thank you very much for your previous help. I need to clarify a concept.

The exciton transiton dipole:P_S^2(Q) defined in the the "Theory and Ab Initio Computation of the Anisotropic Light Emission in Monolayer Transition Metal Dichalcogenides"https://pubs.acs.org/doi/full/10.1021/a ... tt.8b01114. and the \mu _S^2 defined in the https://pubs.acs.org/doi/full/10.1021/nl503799[/url] are both related to the residuals (|A_cv*rho_cv|^2)which is printed from the o*exc*Esorted,the are called the exciton transition dipole.

This exciton transition dipole can be expanded by transition dipoles(<c|p|v> ）with different expansion coefficients A_cv，the|A_cv|^2 could be printed in the o*exc_weights_at_1.

Another physical quantity should be the transition dipole<c|p|v>,that could be printed by yambopyhttps://www.yambo-code.eu/wiki/index.ph ... _databases,so that is not the P_S^2 or the \mu_s^2.

My final question is how do I get the desired P_S^2and \mu _S^2 from o * exc * Esorted,the Pro.Palummo mentioned that:viewtopic.php?t=1947&start=10.

" In any case if you want to obtain mu_S^2 from the residuals^2 written by yambo in the o*exc*E_sorted file do in this way: take the second column of the file and multiply by the Maximum Residual Value you read at the beginning of the file these are the modulus square of the residuals now not normalized to 1.From them to obtain the modulus square of excitonic dipoles mu_S^2 = residuals^2*Omega*N_k/(4*pi*HA2EV*n_spin)",The residuals^2 here I understand as residual,which is the modulus square of the A_cv*rho_cv,?

What I don't understand is why I have to multiply by the coefficient Omega*N_k/(4*pi*HA2EV*n_spin)？In mathematical expressions, shouldn't the modulus square of the residuals(|A_cv*rho_cv|^2) be directly equal to the mu_S^2 or the P_S^2?if the units of the residual from the o*exc*Esorted is eV,let the Omega=1000bohr^3=148*1*10^(-30)(Angstrom）， In the SI system of units, there is a large order of magnitude difference between residuals and mu_S^2.This makes me particularly confused!

With the help of Pro Palomo, I previously assumed that the data obtained from o * exc * Esorted was mu_ S ^ 2, and then calculate the exciton lifetime of MoS2, obtaining a similar result.But now If I use the Omega*N_k/(4*pi*HA2EV*n_spin) multiply the residual，according to the above analysis, the order of magnitude will significantly change.

I don't know where I made a mistake. I really need your help to understand that.
I have been using yambo code for three years, and I particularly enjoy the process of using yambo for research. I hope to have a better understanding so that I can engage in future research work.
Best wishes
Quxiao
Quxiao in BIT,calculate the exciton

palful
Posts: 45
Joined: Tue Jan 26, 2016 11:23 am
Location: Modena and Milan

Dear Quxiao,

The quantity printed in the file o*exc*sorted is |P_S|^2/ MAX(|P|^2) = |\sum_cvk A^S_cvk*rho_cvk|^2/ MAX(|P|^2), with |P_S|^2 being in hartree squared units.

That being said, I am not sure if the quantity \mu_S should be exactly the same as P_S or should involve the dimensional factors that you mention, which come from discrete k-integration, spin summation, long-range Coulomb interaction squared, and Fourier transform of the delta function. These factors are surely needed if you are computing the imaginary part of the dielectric function and the absorption coefficient, in order to have the correct intensity values. But for the quantity \mu_S, maybe it is better to wait for the reply of an author of that paper.

Cheers,
Fulvio
Dr. Fulvio Paleari
S3-CNR Institute of Nanoscience and MaX Center
Modena, Italy

Quxiao
Posts: 92
Joined: Fri Mar 26, 2021 11:27 am

Dear palful:
I am very excited recived your reply! But regarding this o * exc * Esorted, I now have a different view from yours，I see that codes in the /work/home/acqt29ykyw/soft/yambo-self/yambo-
5.0.3/ypp/excitons/excitons_sort_and_report.F

Code: Select all

126Residuals(:) = real(BS_R(:)*conjg(BS_R(:)),SP)
127    if (write_widths)  Residuals(:) = abs(BS_R(:))
128    if(iq==1) q_norm=q0_def_norm**2
129    if(iq> 1) q_norm=iku_v_norm(BSqpts(:,iq))**2
130    Residuals(:) = Residuals(:)*real(spin_occ,SP)/(2._SP*pi)**3*d3k_factor*4._SP*pi/q_norm*HA2EV.

The code in line 130 indicates that the output in o * exc * Esorted should bet C*|\sum_cvk A^S_cvk*rho_cvk|^2，C=real(spin_occ,SP)/(2._SP*pi)**3*d3k_factor*4._SP*pi/q_norm*HA2EV.[/code]
The

Code: Select all

134      Residuals=Residuals/Rmax
show that C*|\sum_cvk A^S_cvk*rho_cvk|^2/Rmax.C=8*\pi/q^2*\Omega*N_q
My understanding is that the output quantity must be divided by C(8*\pi/q^2*\Omega*N_q )to obtain the exciton transition dipole :real(BS_R(:)*conjg(BS_R(:)),SP)=|<G|P|S>|^2.

My final question is that Daniele mentioned that the output unit is eV for the quantity in o * exc * Esorted. I also saw that the Hartee to eV conversion was indeed performed, but what puzzled me was that the unit of |<G | P | S>| ^ 2 seems to be bohr ^ 2,I also see that ，in some articles, the unit of this quantity has appeared in the form of Debye，but I could not know the units of the P_s^2 in that article,which in the page7:https://journals.aps.org/prb/abstract/1 ... 100.075135
And if I could print that from ypp change the code,what is the units of the real(BS_R(:)*conjg(BS_R(:)),SP) and the \Omega,Bohr^2,Bohr^3?
Best wishes
Quxiao
Quxiao in BIT,calculate the exciton