[Pw_forum] Calculate the charge states? Fermi level?

Stefano Baroni baroni at sissa.it
Thu Dec 9 15:25:40 CET 2004

On Dec 9, 2004, at 12:45 PM, 胡树军 wrote:

> Dear Drs,
> When I calculate the formation energy of impurities which relate to 
> the charge
> states, how can i add or deprive an electron on the impurity atom, as 
> -1 or +1
> charged?

This is done by changing the total number of electrons in the 
simulation cell. With periodic boundary conditions (which are 
implicitly used in any plane-wave calculation) this would imply a 
macroscopically charged system, which requires proper neutralization, 
usually achieved by adding a unifrorm charge backgroung. A background 
with the appropriate charge state is implicitly assumed by the 
convetion that the G=0 Fourier component of the crystal electrostatic 
potential is finite (conventionally, zero)

> Should I construct the new pp,

definitely not. a "good" pp is supposed do describe different charge 
state of a same atom

> or set the value of variable "nelec" in
> the input file?


> ( I donnt think they are the proper methords above)
>     Another question, such message is extracted from the outfile of pw
> calculation:
>           k = 0.0000 0.0000 0.0000 (17221 PWs)   bands (ev):
>     -7.2587  -7.1101  -2.7191  -1.8977   2.7554   5.0637   6.8914   
> 6.8914
>      6.9326   6.9326   7.7895   9.2568   9.2569   9.3341   9.3341   
> 9.8081
>     15.2266  15.6076  15.6078  15.7622  15.7623  15.9133  16.9850  
> 17.0452
>      the Fermi energy is    12.4174 ev
> "real" Fermi level ,as an negative number indicating the ionization 
> energy of an
> electron on the Ef, is as one wants. How can i get it?

This a question whose answer requires some study of condensed-matter 
physics. Orbital energies in PWscf (as well as in any other code 
dealing with the electronic structure of infinite systems) are defined 
with respect to an *arbitrary* energy reference level. You should first 
understand well the fact that this arbitrariness is a necessary 
consequence of the fact that the system is infinite *and* of the 
long-range nature of the electrostatic interactions. A ionization 
potential can only be defined in a finite (or semi-infinite) system. In 
the latter case, the long-range nature of the Coulomb intercation 
determines a finite potential drop if a dipolar layer is super-imposed 
to the surface, yet this dipolar layer - not being associated with any 
electric field - has no observable effects deep in the bulk. To cut the 
long story short, the ionization potential of a solid is *not* a 
property of the bulk, but can only be obtained from a surface 
calculation ... Thisnk a bit about this and then revert to us if, after 
thinking, you still cannot find the answer to your legitimate question.

Take care -
Stefano Baroni

Stefano Baroni    ---  SISSA  &  DEMOCRITOS National Simulation Center
via Beirut 2-4 34014 Trieste Grignano / 
[+39] 040 3787 406 (tel) -528 

Please, if possible, don't  send me MS Word or PowerPoint attachments
Why? See:  http://www.gnu.org/philosophy/no-word-attachments.html
-------------- next part --------------
A non-text attachment was scrubbed...
Name: not available
Type: text/enriched
Size: 3181 bytes
Desc: not available
URL: <http://lists.quantum-espresso.org/pipermail/users/attachments/20041209/6ec64505/attachment.bin>

More information about the users mailing list