[DFTB-Plus-User] R: electron phonon interaction and matrix derivative

[Alessandro Pecchia]

Dear Haimen,

 

you are right! The factor ½ always created some headache to me… Back at the time when we did most of the el-ph work in molecules with Gemma Solomon and J. Reimers, we derived the coupling starting from the eigenproblem Hc=ESc, so as a perturbation to the eigenvalues and got the factor ½. I should dig out the derivations, if you need. 

More recently I re-derived the couplings starting from a second-quantized formula H=Sum_ij H_ij b_i* b_j, where b_i*, b_j are the creation operators but in the dual representation (bi-orthogonal), so the b’s include S^1/2. If you make a first order expansion of H you now get the couplings without 1/2.   

I practice I now think the correct formula is without ½. As Gabriele pointed out it would be interesting to see the difference between the two expressions. I always though it is small and since we are in any case making several approximations it was not a fundamental problem. 

 

I hope this helps.

 

Alessandro Pecchia

 

 

Da: DFTB-Plus-User [mailto:dftb-plus-user-bounces at mailman.zfn.uni-bremen.de] Per conto di Haimen Mu via DFTB-Plus-User
Inviato: mercoledì 28 ottobre 2020 05:01
A: dftb-plus-user at mailman.zfn.uni-bremen.de
Cc: Haimen Mu
Oggetto: Re: [DFTB-Plus-User] electron phonon interaction and matrix derivative

 

Hi Gabriele,
 
I am also interested in the electron-phonon coupling related to dftb+. I read your python code ‘elcoupl.py’which gave me much inspiration,but I still have a small confusion. In your formulas ‘coupling = coupling - 0.5 * work2’ and ‘coupling = (coupling - 0.5 * work2) * prefactor’, there is a coefficient ‘0.5’that did not occour in some literatures. I saw formular in papers is like what Dirk Ziemann written in previous mail. That is ‘ gamma_ab^q   \approx  d H_ab / d_q  -  \sum_mn  d S_am / d_q * S^-1_mn * H_nb - \sum_mn  H_am * S^-1_mn * d S_nb / d_q ’. Would you explain the coefficient 0.5 here is for what?  
 

Looking forward to your reply.

 

Thank you .

 

Haimen Mu

 

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