[Pw_forum] Lanczos in Phonons to test the dynamical stability

Lorenzo Paulatto paulatz at gmail.com
Mon Oct 23 11:47:43 CEST 2017

The algorithm used by matdyn is not done for speed, it can be sped up 
hundred of times with some little tweaking. You can try using the codes 
from D3Q which are a bit more optimized: d3_q2r.x (same input as q2r) 
and d3_r2q.x (check the manual, requires some code editing as phonon dos 
is not implemented, but is trivial) Or you can try to profile and 
optimize matdyn.


P.S. I initially misread you question and gave you a very long and 
completely pointless answer, I leave it here for the records. The point 
still stand to some extent, because what takes time in matdyn is 
probably not the matrix diagonalization, but its interpolation.

No, because the Lanczos algorithm (and any other similar algorithm, like 
Davidson) allows you to compute the lower eigenvalues of a matrix M by 
applying it repeatedly, instead of diagonalizing it. But you still have 
to be able to apply the matrix.

This kind of algorithms are useful when the matrix is huge, for example 
in the electronic problem, the matrix has the dimension of the number of 
plane waves. By using Davidson, you can reduce it to (twice) the number 
of bands. Note that the scaling is still N^3, it is just the prefactor 
that is smaller.

But in the phonon case, the matrix is tiny, only 3 x number of atoms, it 
only takes a nanosecond to diagonalize it even for hundreds of atoms.

On 23/10/17 08:21, JAY Antoine wrote:
> Dear all,
> Is there a way to perform a phonon calculation only for the lowest 
> phonon frequency for exemple by using the Lanczos algorithm?
> I mean to test the dynamical stability of a structure, one only need 
> to know that "all the frequencies are positives over the full BZ",
> which is the same as "no frequencies are negatives" and in this case, 
> one only need to know the lowest value of all the phonons frequencies.
> For the biggest structures with a too big dynamical matrix, the phonon 
> DOS have a too high computationnal cost and this trick should be very 
> usefull...
> Best regards,
> Antoine Jay
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Lorenzo Paulatto - Paris

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