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