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[Fabien Bruneval <fabien.bruneval@polytechnique.fr>]

Dear Deyu,

Sorry for continuing an old discussion, but there is a point I can't get 
in your reasoning.

If I consider your example of the scissor operator, the Hamiltonian reads
H' = H_{KS} + \sum_c \Delta | c > < c |

This Hamiltonian has the effect of shifting up all the conduction 
energies, but it has the same eigenvectors as the KS Hamiltonian.

When using the commutator trick with H', you'll end up with
< v | r | c > = < v | [ H' , r ] | c > / (E'_v - E'_c)

The commutator of the scissor operator with the operator r is not zero, 
since the scissor is a non-diagonal operator in r-representation:
\sum_c  < r | c > < c | r' > .ne. \delta( r - r').

This means that if you want to introduce H' instead of H_{KS}, you'll 
have to calculate explicitly this commutator in addition.

But in the end, introducing the commutator with H_{KS} or H' should give 
exactly the same result for the matrix element < v | r | c > (it is only 
a mathematical trick for commodity).

I hope I understood correctly your point.
Regards,

Fabien



deyulu@yahoo.com wrote:
> Fabien:
>       Thank you for your explanation, but I don't agree with your argument. 
> In doing scissor+G0W0, self-consistent GW on the energies-only, one should use 
> H', i.e., H_{KS}+scissor or H_{KS}+Delta sigma, to evaluate quasi-particle energies.
>       It applies also to the commutator trick. If we takes the 1st order app.
> on H' to assume it is diagonal, one has:
> < v | r | c > = < v | [ H' , r ] | c > / (E'_v - E'_c),
> where E'_v and E'_c are quasi-particle energies instead of KS energies.
>       For a wide range of materials, it has been shown in Fiorentitni and 
> Balderesch's paper (PRB, 51:17916, 1995) that the effect of Delta sigma can
> be largely attributed to a static screened exchange term, which commutes with r. Then it 
> leads to < v | r | c > = < v | [ H_{KS} , r ] | c > / (E'_v - E'_c).
>       In my opinion, this expression is consistent with the formula for G!=0
> terms.
>
>      Best
>      
> Deyu***************************************************************************
> Deyu Lu (Ph.D)
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