[Pw_forum] easy question

[Varadharajan Srinivasan]

Dear Tomson,

This is a genuine confusion. The dependence on the nuclear positions comes
in through the Born Oppenheimer approximation. While the basis set are not
dependent on the nuclear positions the coefficient of these are
parametrically dependent on them. This is because we solve the Kohn Sham
equations for a given set of nuclear positions (for a given external
potential). So changes in density due to change in nuclear positions will
be effected through the change in the corresponding coefficients of plane
waves. n(r;R) = \sum_G C_G (R) e^i(G.r)

You should actually view the derivative as  d n(r;R)/d R. Besides, we only
treat valence electrons through the plane-wave expansion and the core
density from every atom is added directly in real-space. Since this
definitely depends on the nuclear positions it gives one dependence of the
density on the nuclear positions.

Hope that helps.

Best,
Vardha.


On Fri, Jul 5, 2013 at 7:20 AM, <kanta_chan25 at yahoo.co.jp> wrote:

> Dear all,
>
> It is begun these days to use QE and DFPT is studied.
> There is an easy question.
> Electron density(n(r)) is made from the plane wave set in this code?
> A plane wave cannot be dependent on nuclear coordinates.(exp(ikr))
> In DFPT, the differentiation in the nuclear coordinates of electron
> density is required.
> I will think that the differentiation in the nuclear coordinates of
> electron density will disappear.
> If surely a nucleus changes, density is also likely to change.
> However, a base is a plane wave.
> It is not dependent on a nuclear position.
>
> sum of plane wave set (sigma {exp(ikr)}) → density n(r) → dn(r)/dR is
> vanish ... ???
>
> Where do I mistake?
> please help me.
>
>
> Best,
> Tomson
>
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