[Pw_forum] About the Frozen Phonon Method

[Stefano Baroni]

On Aug 2, 2011, at 8:18 AM, mohaddeseh abbasnejad wrote:

> Dear Prof. Baroni
>  
> Thanks again
>> Dear Prof. Baroni
>>  
>> Thanks for your reply.
>> I wonder shouldn't the plot show the same frequency for a range of displacements?
> 
> what range? why should it?
> 
> 
> -- Because if we are in the neighberhood of a minimum "where a quadratic could be fitted", any displacement in that range should result the same frequency. By displacement, I mean the same pattern but different in amplitude.

who said this? around a minimum, the only thing you can say is that the Taylor expansion of the energy in powers of the atomic displacements lacks any linear terms. In general, it has both quadratic terms (which are the one you are interested in), as well as higher order terms ... No reason why the Taylor expansion should be *exactly* quadratic (which is what would give rise to a "same frequency"). Note however, that it is your definition of frequency that is probably flawed. I suspect that you define the frequency as the ratio between the energy variation and the square of the displacement, which obviously depends on the magnitude of the displacement, if the energy variation is not exactly quadratic. The correct definition, instead, is the *limit* of that ration when the displacement goes to zero. That limit is obviously well defined, independently of the magnitude of any particular distortion that you may consider.

>  
>> And the same freq. for the range to be the harmonic frequency? or the zero displacement limit, is the harmonic frequency?
> 
> yes, by definition

See the above discussion

> 
>> And when I test the displacements with the same amplitude, but different in sign (regarding the inward and outward atomic movements in the basis) I do not get the same results.
> 
> why should you?
> 
>> Shouldn't the plot have symmetry vs. the vertical axis?
> 
> why?
> 
> -- Consider for example a linear chain of atoms, with 2 atoms per basis. The optical mode is the result of the outward movement of the atoms (The one in +a/2 moves towards +x and the one in -a/2 moves towards -x). I expect that if I displace the atoms in an inward movement with the same amplitude as the outward, I should see the same frequency (The one in +a/2 moves towards -x and the one in -a/2 moves towards +x) because the 2 patterns present the same mode, right? That is what I've done in the case of Diamond. Is my expectation wrong?

Yes. If the Taylor expansion of the energy difference vs displacement has cubic terms (as it usually has, but in very special cases dictated by symmetry), there is no reason why the energy difference should be even with respect to energy displacement. Actually, it usually isn't ...

Hope this clarifies a bit the situation ...

Take care
SB

---
Stefano Baroni - SISSA  &  DEMOCRITOS National Simulation Center - Trieste
http://stefano.baroni.me [+39] 040 3787 406 (tel) -528 (fax) / stefanobaroni (skype)

La morale est une logique de l'action comme la logique est une morale de la pensée - Jean Piaget

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 --------------
An HTML attachment was scrubbed...
URL: <http://lists.quantum-espresso.org/pipermail/users/attachments/20110802/0838a7ae/attachment.html>