[Wien] Bad formation energies for the charged vacancies

John Pask pask1 at llnl.gov
Sat Feb 27 00:30:40 CET 2010


Hi Peter,

> In the integrals below, \rho is just the electronic charge density  
> (without nuclei).
> Thus c \int{\rho] does NOT vanish and gives c * NE (number of  
> electrons).
> However, if rho comes from electronic states, each eigenvalue is  
> shifted by the constant c
> and thus the sum of eigenvalues cancels the  c * NE term
>
> However, when I add a "background charge" to neutralize the unit  
> cell, this does not come
> from any eigenvalue, so if I handle this in the "usual" way, \rho  
> will now integrate to
> NE + Q, and I get an extra c * Q term, which is not compensated by  
> an eigenvalue.

Actually, in the integrals below, \rho is the *total* (electronic +  
nuclear) charge, which must be net neutral to have a well-defined  
total energy (otherwise energy diverges).

With regard to the present question on charged-cell calculations, the  
point is just that the calculation must be performed on a neutralized  
cell in order to have well-defined total energy. So the Kohn-Sham  
calculation is performed on a neutral cell, whether or not the  
physical system is charged, and the corrections for non-neutrality, if  
any (e.g., Makov-Payne, Eq. (15)), are added after.

So as long as the neutralizing charge enters all potential and energy  
expressions along with the "physical charge", so that all expressions  
operate on a net-neutral total, the Kohn-Sham total energy must be  
invariant to arbitrary constants in V (because the total Coulomb  
energy is).

John

>
> John Pask schrieb:
>> Dear Peter,
>> Yes, the background charge must be taken into account as part of  
>> the net-neutral total charge in order to have well-defined total  
>> energy. Then as long as the compensation charge is then in exactly  
>> the same way as the remaining "physical" charge (i.e., enters all  
>> the same integrals), then the arbitrary constant in potential  
>> should not matter since:
>> \int{ \rho (V + c)}  = \int{ \rho V}  + c \int{ \rho} = \int {\rho  
>> V},
>> independent of arbitrary constant c.
>> John
>> On Feb 24, 2010, at 11:54 PM, Peter Blaha wrote:
>>>> Is the question regarding the computation of total energy per  
>>>> unit  cell in an infinite crystal with non-neutral unit cells? If  
>>>> so, then  the total energy diverges -- and so is not well- 
>>>> defined. (So  neutralizing backgrounds must be added in such  
>>>> cases to obtain  meaningful results, etc.)
>>>
>>> Yes, this is the question and yes, of course we add a positive or  
>>> negative background.
>>> We are quite confident that the resulting potential is ok, but the  
>>> question is if there
>>> is a correction to the total energy due to the background charge.
>>> I believe: yes (something like Q * V-col_average / 2), but my  
>>> problem is that V-coul
>>> is in an infinite crystal only known up to an arbitrary constant  
>>> and thus this correction
>>> is "arbitrary".
>>>
>>> -- 
>>> -----------------------------------------
>>> Peter Blaha
>>> Inst. Materials Chemistry, TU Vienna
>>> Getreidemarkt 9, A-1060 Vienna, Austria
>>> Tel: +43-1-5880115671
>>> Fax: +43-1-5880115698
>>> email: pblaha at theochem.tuwien.ac.at
>>> -----------------------------------------
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>
> -- 
> -----------------------------------------
> Peter Blaha
> Inst. Materials Chemistry, TU Vienna
> Getreidemarkt 9, A-1060 Vienna, Austria
> Tel: +43-1-5880115671
> Fax: +43-1-5880115698
> email: pblaha at theochem.tuwien.ac.at
> -----------------------------------------
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> Wien mailing list
> Wien at zeus.theochem.tuwien.ac.at
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