# [Pw_forum] About the Frozen Phonon Method

Wed Aug 3 07:03:05 CEST 2011

```Dear Prof. Baroni,

Thank you so much.
It clarified a lot.

Yours,

On Tue, Aug 2, 2011 at 10:20 PM, Stefano Baroni <baroni at sissa.it> wrote:

>
>
> Dear Prof. Baroni
>
> Thanks again
>
>>  Dear Prof. Baroni
>>
>> 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
>
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> Why? See:  http://www.gnu.org/philosophy/no-word-attachments.html
>
>
>
>
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