The difference between the residuals and the modulus of the exciton transition dipole

[Jiale Shen]

Dear Daniele:
I am confused about the form of output in the o*exc*E file,I saw the previuos reply from the forum,and someone said that the output was the square modulus of the residuals.My understanding of the "residuals" is "BS_R=A_cv*rho_cv",the square modulus of the residuals is BS_R*BS_L,is it true?

In this Nano from the Palummo,https://pubs.acs.org/doi/full/10.1021/nl503799t,I see that the exciton decay rate can be obtained by using Equation(2).

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\gamma_{S}^{2D}(0)=\frac{8\pi e^2E_S(0)\mu_S^2}{A_{\mu c}\hbar^2c}

where the \mu_S^2 is defined as the square modulus of the BSE exciton transition dipole divided by N_k.

Below the seconded Equation(2) in this Nano from the Hsiao-Yi Chen,https://pubs.acs.org/doi/full/10.1021/a ... tt.8b01114,I get that

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\gamma_{S}^{2D}(0)=\frac{e^2 P_{S}^2}{\epsilon_0 m^2cAE_S(0)}

where the P_{S}^2 is also defined as the square modulus of the exciton transition dipole,and above the 11th equation in this article,

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P_S^2=\frac{m^2E_S^2(0)\mu_S^2}{\hbar^2}.

My question :

1. According to my understanding, \mu_S^2 and P_{S}^2 are actually a physical quantity, and the reason for the existence of the relationship between P_S^2=\frac{m^2E_S^2(0)\mu_S^2}{\hbar^2}.is actually due to different unit systems，is it true?
2. How do I get this \mu_S^2 from yambo,can it be directly obtained from the o*exc*E file?
3. How do I get this P_{S}^2 from yambo ,can it be directly obtained from the o*exc*E file?
4. I read the previous reply on exciton lifetime in the forum, and Professor Palumo once mentioned that viewtopic.php?t=1947

   you can use the o*sorted* file and yes , mu_s^2 is what is written inside o*exc* file beyond the normalization.

   (I imagine mu_s^2 is your Us^2 right?). But below this post, she mentioned again

   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:

   1. 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.
   2. From them to obtain the modulus square of excitonic dipoles mu_S^2 =residuals^2*Omega*N_k/(4*pi*HA2EV*n_spin).

   This seems contradictory,Is the output in the o * exc * E file residules^2 or u_ s^2? Because the formula obtained by expanding exciton dipole using dipole approximation is almost the same as that obtained by BS_ R*BS_ L .

In order to make better use of yambo, I really want to understand this kind of problem because it is highly prevalent in forums, but I haven't got a good understanding.
Thank you very much for your patient help!
Best wishes!
JiaLe

[Davide Sangalli]

Dear JiaLe,

\mu_S is the dipole operator in excitonic space, i.e.

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\mu_S = \sum_cvk A^S_{cvk} x_{cvk}

and corresponds to the value of BS_R

p_S is the momentum operator in excitonic space (it should be more properly called v_S, i.e. velocity operator)

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p_S =  \mu_S * E_S(0) 

As already pointed out, the ypp output contains |BS_R|^2 re-normalized to one.
Accordingly from the ypp output you can obtain |\mu_S|^2 following the indications from Maurizia.

Best,
D.

[Jiale Shen]

Dear Davide：
Thanks for your reply! I have read too many times about this article:Theory and Ab Initio Computation of the Anisotropic Light Emission
in Monolayer Transition Metal Dichalcogenides：https://pubs.acs.org/doi/pdf/10.1021/ac ... tt.8b01114
I think I got what you said.The P_S should be the <G|p|S>=-im<G|[x,H]|S>/\hbar=-im<G|x|S>*E_s/\hbar,the <G|x|S>should be the \mu_S,is that right?

But what confused me is that the units of the P_S^2 and the \mu_S^2,I think I have repeated a lot of derivation and read the code for ypp.

1.The output in o * exc * Esorted should be（8*\pi/q^2*\Omega*N_q ）*|\sum_cvk A^S_cvk*rho_cvk|^2/Rmax，but I need to ensure that,the |\mu_S|^2 is the |\sum_cvk A^S_cvk*rho_cvk|^2,

so I could get that from the data which printed in the o*exc*Esorted{（8*\pi/q^2*\Omega*N_q ）*|\sum_cvk A^S_cvk*rho_cvk|^2},the units of the data is eV,but I could change that units to Hartree,then I could use the atomic units to calculate the |\mu_S|^2,the units of the |\mu_S|^2 should be |bohr|^2,is that true?

2.But in that article:https://pubs.acs.org/doi/pdf/10.1021/ac ... tt.8b01114
I see that you use the |P_S|^2 to calculate the exciton lifetime,and the units is SI,|P_S|^2=(m^2*\|mu_S|^2*E_S^2)/\hbar^2,

so I confuse is that should the units of the |P_S|^2 is |bohr|^2?And if I use the SI units,I should use the \mu_S^2 (the units is (bohr)^2),multiply the e^2=(1.602*10^(-19)C)^2,then continue multiply the (m^2**E_S^2)/\hbar^2 to get the correct SI units value of the |P_S|^2?

3.I also repeated the Toluene in that article: https://journals.aps.org/prb/abstract/1 ... 100.075135, P_S^2=0.59*10^6(atomic units),I think the units of the P_S^2 is |bohr|^2,is it right?

I really want to understand this series of calculations combined with yambo, and I think it will be very helpful for me to learn yambo.
This greatly stimulated my interest in learning and using yambo.
Thanks for your help again!
Best wishes!
Jiale

[chenhsiaoyi]

Dear Jiale:

This is Hsiao-Yi Chen.
Thank you for being interested in my works.
Maybe I can help you with some questions.

1. Since I am not the Yambo developer, I can not be sure about the relation among output quantities.
But what I can tell you is that the quantity \sum_{cvk} A^S_{cvk}*rho_{cvk} is the P_S not the \mu_S.
And you'd need to know that in the Yambo code, it calculates some different dipoles. Here, rho_{cvk} corresponds to the "dipole_iR" in the source code.

2. I think since dipole_iR is already in unit of Bohr, so as for P_S.

3. Yes.

Best,
Hsiao-Yi

[Jiale Shen]

Dear Hsiao-Yi Chen:
Thanks a lot for your reply! I read the code find that,when I use the Gauge=length,that will be calculate the <G|r|S> as the exciton transition,if I use the Gauge= velocity,that will be the <G|p|S>.I don't know if my understanding is correct.
As you said,there are some different dipoles in yambo code.I see that from Daniele:viewtopic.php?p=11499&hilit=three+ndb#p11499.

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266            !
267           ! DIP_iR(c,v) = i <v|r|c> = i <v|[r,H]|c>/(Ec-Ev) =
268            !             = i <v|i p|c>/(Ec-Ev) + i<v|[x,Vnl]|c>/(Ec-Ev) =
269            !             = - <v|-i grad|c>/(Ec-Ev) +i<v|[x,Vnl]|c>/(Ec-Ev) =
270            !             =   <v|-i grad|c>/(Ev-Ec) -i<v|[x,Vnl]|c>/(Ev-Ec) =
271            !             =   P_vc/(Ev-Ec) -i<v|[x,Vnl]|c>/(Ev-Ec)

Combined with the formular:P_S(Q)=-im/\hbar*<G|[x,H]|SQ> and the P_S^2=(m^2*E_S^2*\mu_S^2)/\hbar^2.If you said the P_S=the DIP_iR(c,v),so what is the \mu_S?
They should only differ by one energy difference in atomic unit mass.Compared to the formula, you might want to say：the P_S= i <v|[r,H]|c>,the \mu_S= i <v|[r,H]|c>/(Ec-Ev).
Of course, I know all of the above are single electron dipoles, P_ S and mu_ S are exciton dipoles that can be unfolded by them.

Also, thank you very much for your reply. I am very interested in your work and have a strong desire to communicate with you. Thank you again for your help！
Best wishes！
Jiale