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<p>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.<br>
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<p>hth</p>
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<p>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.<br>
</p>
<p>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.</p>
<p>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.<br>
</p>
<p>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.</p>
<div class="moz-cite-prefix">On 23/10/17 08:21, JAY Antoine wrote:<br>
</div>
<blockquote type="cite"
cite="mid:4b6d-59ed8a80-2f-21c16f80@256466804">Dear all,<br>
Is there a way to perform a phonon calculation only for the lowest
phonon frequency for exemple by using the Lanczos algorithm?<br>
<br>
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",<br>
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.<br>
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...<br>
<br>
Best regards,<br>
<br>
Antoine Jay<br>
<br>
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<pre class="moz-signature" cols="72">--
Lorenzo Paulatto - Paris</pre>
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