general question on calculating metals using cp2k

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Jun

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Nov 2, 2011, 1:35:32 PM11/2/11
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Hi,

I intend to run some calculations on metals. I am wondering if experts
could give me some ideas on the performance of cp2k. For example, to
model a metal surface, say Pt(111), what size of the system is big
enough for compensating the lack of k-point sampling? Usually, how
many iterations are needed to reach SCF convergence? How long it
roughly takes for each SCF step? How slow is the matrix
diagonalisation compared to OT? ...

It would be nice if someone shows me an input example for metals. I
know there was one given by Prof. Hutter about two years ago. Any up-
to-date one?

Thanks a lot,

Jun

carlo antonio pignedoli

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Nov 3, 2011, 4:53:28 AM11/3/11
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Dear Jun,
regarding the size of the cell, it depends on which property of the
system you would like to reproduce.
There is a simple rule:
if your property is well converged with a k-point cacluation for a
unit cell on a MxM regular grid not shifted (I mean
including Gamma point..)
then a supercell as big as MxM the unit cell will give you "exactly"
the same results.

Of course overriding symmetries could bring the MxM system to a
geomoetry not compatible
with the unit cell (this does not mean that the result is wrong...)
that's in part why I was saying "exactly".


If you are interested in slab calculations, no k-points in the
direction perpendicular to the slab surface,
then the thickness of the slab, again, depends "solely" on the
particular properties of the system that you
would like to reproduce.

For the behavior of the optimization algorithms i let somebody more
expert to reply.

Ciao

Carlo

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Jun

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Nov 3, 2011, 6:34:53 AM11/3/11
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Thanks, Carlo.

Yes, I want to do slab calculations. The properties I am interested in
are, for example, adsorption energies of some atoms or molecules, work
function, DOS et al. For start, I guess the thickness of four layers
should be enough, I mean for Pt(111). What I am not so sure is the xy
dimensions. Would MxM, say M=5, be enough? I read some discussions on
the magnetic state of Ni that Ni slab of M=6 and 6 layers is necessary
to get a reasonable spin state. I hope it is less demanding for non-
magnetic metals like Pt.

Cheers,
Jun

On Nov 3, 8:53 am, carlo antonio pignedoli <c.pigned...@gmail.com>
wrote:

carlo antonio pignedoli

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Nov 3, 2011, 6:43:33 AM11/3/11
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Dear Jun form the experience I have for systems I worked on,
a slab 4 layer thick could be the minimum thickness for a first investigation,
be careful that you would have, in such a thin slab, a spurious interaction
among surface states.
To reduce this, a "empirical" way is hydrogenation of one side of the slab.

Dont' forget to consider the relevance of van der Waals corrections.


Usually the smallest slab I consider in xy is in the order of 6x6
compared to a primitve cell
but this is just an indication and the result depend really on the
particular system you are dealing with.

Magnetic systems are always complex.. I have not enough experience on
this to give
you suggestions.

Ciao

Carlo

Jun

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Nov 3, 2011, 7:04:37 AM11/3/11
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Thanks very much again for the prompt reply, Carlo.

I agree a thickness of 4 layers is a minimum. I hope the screening of
metals is strong enough to eliminate the interaction between two
surfaces. I know termination with H atoms to saturate surface dangling
bonds is often used for covalent solids with a band gap. Is this
necessary for metals?

I will definitely do some tests on the size of slab before I really
run the calculations.

Cheers,
Jun

On Nov 3, 10:43 am, carlo antonio pignedoli <c.pigned...@gmail.com>
wrote:

carlo antonio pignedoli

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Nov 3, 2011, 7:12:02 AM11/3/11
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Dear Jun,
yes you are right H saturation is standard for semiconducting and
insulating systems,
there is in the literature (I will have to find it) a discussion on the ability
of H saturation in Au(111) (if I remember well) to reduce spurious interactions
among surface states at the sides of the slab.

Ciao

Carlo

Jun

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Nov 3, 2011, 8:27:21 AM11/3/11
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Could anyone comment on the efficiency of cp2k on metals, for example,
a 5x5 Pt(111) slab of 4 layers (100 atoms)?

Thanks in advance!

Jun

On Nov 3, 11:12 am, carlo antonio pignedoli <c.pigned...@gmail.com>
wrote:
> Dear Jun,
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