[Pw_forum] FFT - Help

[Paolo Giannozzi]

On Tue, 2015-02-24 at 19:08 +0100, Aritz Leonardo Liceranzu wrote:

> So if I understood correctly once the Ecutoff is set, the file
> "gvectors.dat" contains the complete list of G vectors inside the
> sphere with a radius 4*Ecut where magnitudes such like density can be
> safely represented.
> 
> Starting from here, how is the real-space grid generated? I ask this
> because for my particular calculation there are around 3000 G
> different vectors for a real space grid that has nr1=nr2=nr3=24
> points.
> 
> according to the definition:
> r= (i-1)/nr1+(j-1)/nr2+(k-1)/nr3

r= (i-1)*a_1/nr1+(j-1)*a_2/nr2+(k-1)*a_3/nr3, where a_1, a_2, a_3 are
the three vectors that generate the lattice

> The real space grid is denser than the reciprocal grid, so there has
> to be some kind of mapping from one to each other that I am missing.
> According to the above transparencies both grids should have equal
> amount of points. 

G = h*b_1 + k*b_2 + k*b_3, where b_1, b_2, b_3 are the three vectors 
that generate the reciprocal lattice. Negative indices (h,k,l) are
refolded to positive one atthe other end of the FFT box. The
correspondence between G vectors and (j,k,l) indices is provided 
by array "nl"

Paolo

> I could still do (i think) brute force transformations using the
> forward and inverse transformations defined in transparency 5 but I if
> wanted to use fftw in order to be more efficient, shouldn't they be
> the same in size?
> 
> As I said, I would appreciate if somebody could address me a reference
> or notes to read where these issues are explained. Thanks! 
> 
> 
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-- 
 Paolo Giannozzi, Dept. Chemistry&Physics&Environment, 
 Univ. Udine, via delle Scienze 208, 33100 Udine, Italy
 Phone +39-0432-558216, fax +39-0432-558222