Re: [Wien] Metal or semimetal

2019-01-30 Thread E.A.Moore 
The Wikipedia article on metalloids says it's the overlaping valence and 
conduction bands that define semimetals in physics.


Following textbooks in chemistry could lead to confusion as many chemists still 
use the term for elements that cannot be classified definitively as metals or 
non-metals based on their chemical and physical properties. I assume you use 
metalloids for these.


Incidentally the IUPAC gold book does not define semimetal, semi-metal or 
metalloid.

Elaine A. Moore

From: Wien  on behalf of Fecher, 
Gerhard 
Sent: 30 January 2019 09:17
To: A Mailing list for WIEN2k users
Subject: Re: [Wien] Metal or semimetal

You did not read up to the end (Graphite with a strong directional dependence 
of the conductivity was an example, maybe it was not lucky to place it in front)
a little later you find:
"In some cases you have to check whether there might be an overlapp of the 
valence and and conduction bands at different k-points, resulting in a 
semimetallic or zero-bandgap type behavior."
and that is (for a semi-metal) the situation you find as C in Wikipedia, isn't 
it ?

By the way, the definition of semi-metals changed with time one just need to 
follow the textbooks in chemistry and physics
see also about metalloids (https://en.wikipedia.org/wiki/Metalloid) .
Metalloid - Wikipedia
en.wikipedia.org
A metalloid is a type of chemical element which has properties in between, or 
that are a mixture of, those of metals and nonmetals.There is neither a 
standard definition of a metalloid nor complete agreement on the elements 
appropriately classified as such. Despite the lack of specificity, the term 
remains in use in the literature of chemistry.. The six commonly recognised 
metalloids are boron ...




Ciao
Gerhard

DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy:
"I think the problem, to be quite honest with you,
is that you have never actually known what the question is."


Dr. Gerhard H. Fecher
Institut of Inorganic and Analytical Chemistry
Johannes Gutenberg - University
55099 Mainz
and
Max Planck Institute for Chemical Physics of Solids
01187 Dresden

Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von Karel 
Vyborny [vybor...@fzu.cz]
Gesendet: Mittwoch, 30. Januar 2019 09:08
An: A Mailing list for WIEN2k users
Betreff: Re: [Wien] Metal or semimetal

A quick question to add my bit into this conversation: where does the
definition of semimetal as "insulating (semiconducting) in one direction
and conducting in another one" come from? I have never heard of this; it
may the lack of my knowledge but I hold that it's the overlaping valence
and conduction bands that define semimetals (and indeed, bismuth has
always been the prime example for me). The strong anisotropy in
conductivity may be a concomitant feature, however, it is not necessary.

Cheers,

Karel


--- x ---
dr. Karel Vyborny
Fyzikalni ustav AV CR, v.v.i.
Cukrovarnicka 10
Praha 6, CZ-16253
tel: +420220318459


On Tue, 29 Jan 2019, Fecher, Gerhard wrote:

> Thank you for the Link, but I don't understand your remarks
> Wikipedia tells:
> - Schematic
> C) a semimetal (like tin (Sn) or graphite and the alkaline earth metals).
> and further
> - Classic semimetals
> The classic semimetallic elements are arsenic, antimony, bismuth, ?-tin (gray 
> tin) and graphite, an allotrope of carbon.
>
> on the page https://en.wikipedia.org/wiki/Graphite one finds
> Graphite has a layered, planar structure. The individual layers are called 
> graphene. ...
> Atoms in the plane are bonded covalently, with only three of the four 
> potential bonding sites satisfied. The fourth electron is free to migrate in 
> the plane, making graphite electrically conductive.
> However, it does not conduct in a direction at right angles to the plane.
>
> Maybe check the band structures of Graphite and Bi to find out what is common 
> and what is different.
>
> You did not understand the remark on the integration of the density of 
> states, please read it correctly.
> The initial question was on the Fermi energy beeing slightly below the top of 
> the valence band
> and this might be caused by a bad integration which depends on the number of 
> k-points (indeed among others)
> and is used to find the Fermi energy.
>
> PS.: Please check the definition of the density of states and you see why 
> flat (say better narrow) bands result in a high density of states and steep 
> ones in a low density of states.
> (that a band is horizontal at a certain point of the Brillouin zone does not 
> mean that the complete band is flat, this situation you have always at the 
> bottom of a parabola) .
>
> Ciao
> Gerhard
>
> DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy:
> "I think the problem, to be quite honest with you,
> is that you have never actually known what the question is."
>
>

Re: [Wien] Transition-State/NEB Calculations

2018-06-15 Thread E.A.Moore 
It may be worth looking at the CRYSTAL17 tutorial on this. 
http://tutorials.crystalsolutions.eu/tutorial.html?td=ts=tutorial-ts_crystal#energytol
 CRYSTAL does not use NEB but does suggest you use a high SCF tolerance 10-11H. 
It may also be worth looking at their method.

The only place I have personally encountered NEB is In AMBER using classical 
force fields.

Elaine A. Moore



From: Wien [mailto:wien-boun...@zeus.theochem.tuwien.ac.at] On Behalf Of 
Laurence Marks
Sent: 15 June 2018 14:20
To: A Mailing list for WIEN2k users
Subject: Re: [Wien] Transition-State/NEB Calculations

Dear Victor,

Thanks for the email. The very specific question I am asking is whether it is 
"well known" with Vasp, Ab-init, Gaussian etc that one needs "good" parameters 
such as k-mesh etc for the transition state. I am hoping that there are people 
who have done NEB (for instance) calculations on this mailing list.

On Fri, Jun 15, 2018 at 8:14 AM, Víctor Luaña Cabal 
mailto:vic...@fluor.quimica.uniovi.es>> wrote:
* Laurence Marks mailto:l-ma...@northwestern.edu>> 
[2018-06-15 07:49:51 -0500]:
> Dear All,
>
> I am part way through developing new code for transition-state (TS)
> calculations. I observed one thing which may be "well-known" with other
> codes; since currently WIen2k cannot do such calculations I have no
> experience. The observation is that at/near the TS one needs a fairly good
> k-mesh, just using gamma fails. (This may be a Wien2k phenomenon.) Has
> anyone heard of this?

Prof. Marks,

I have no experience in solid state calculations but, regarding
molecules, it is kind of normal that TS become problematic to find.
In a minimum all gradient paths arrive to the critical point you are
looking for. In a TS you have gradient paths coming to but also a
gradient line escaping from the critical point.

Under "morse theory gradient flow" you will find scores of material.

Good luck developing the TS code! It will be a much appreciated addition
to wien2k, imho.

Best regards,
 Víctor Luaña
--
.  ."Never let your sense of morals prevent you from
   / `' \   doing what is right."
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/   '`'`   \ "Freedom!, freedom!, freedom! After that put whatever
|  \'`'`/  | term you like"
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 \/`'`'`'\/   shortly before dying)
==(((==)))===+ A person is slave of his words
! Dr.Víctor Luaña, in silico chemist & prof. ! and owner of his silences.
! Departamento de Química Física y Analítica !
! Universidad de Oviedo, 33006-Oviedo, Spain ! The collective intelligence of
! e-mail:  
mailto:vic...@fluor.quimica.uniovi.es>>  ! a 
comitee equals the CI of its
! phone: +34-984080927fax: +34-985103125 ! worst divided by the number
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 Articles:  

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--
Professor Laurence Marks
"Research is to see what everybody else has seen, and to think what nobody else 
has thought", Albert Szent-Gyorgi
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4D: MURI4D.numis.northwestern.edu
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Re: [Wien] Unit of magnétic suscptibility

2017-09-22 Thread E.A.Moore 
cm3/mol is the unit of the molar susceptibility in the cgs system. Did you want 
to convert it to the mass susceptibility in cm3/g? If so you need to divide by 
the relative molecular mass.

Should you not use SI rather than cgs units?


Elaine A. Moore

From: Wien [mailto:wien-boun...@zeus.theochem.tuwien.ac.at] On Behalf Of karima 
Physique
Sent: 22 September 2017 16:12
To: A Mailing list for WIEN2k users
Subject: Re: [Wien] Unit of magnétic suscptibility

Thank you very much Dr. Martin Pieper

In this table there is only the conversion between cm^3/mol to m^3/mol or 
Hּm2/mol and for me i want to convert cm^3/mol to emu/g.

from my reading I think I have to divide cm3/mol on the molar mass only but I 
want a confirmation.

2017-09-22 17:04 GMT+02:00 pieper 
>:
Hi Karima,

see e.g.

http://www.ieeemagnetics.org/index.php?option=com_content=article=118=107

Best regards,



---
Dr. Martin Pieper
Karl-Franzens University
Institute of Physics
Universitätsplatz 
5
A-8010 Graz
Austria
Tel.: +43-(0)316-380-8564


Am 22.09.2017 15:16, schrieb karima Physique:
Dear Wien2k users:

how to convert the magnetic susceptibility from cm^3/mol to emu/g.
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Re: [Wien] About the magnetic moment of vanadium in vanadium sulphide

2017-09-07 Thread E.A.Moore 
Your calculations are probably fine.

Apologies for my previous posting but my point was partly that I think you said 
the  experimentalists claimed it was Pauli paramagnetic.

Pauli paramagnetism is not the type of paramagnetism that arises from unpaired 
electrons on metal ions in, for example, transition metal complexes. It is a 
property of metals. It arises if there is an unfilled conduction band. In an 
external magnetic field, one type of spin (up or down) acquires a different 
energy to the other type resulting in an excess of one type over the other and 
hence paramagnetism.

Both your result and the nonmagnetic finding are consistent with the existence 
of Pauli paramagnetism.

You results suggest that the 3 unpaired electrons on V are not localised on V 
but are in a delocalised band. Do you predict VS has a partially full 
conduction band?

Elaine A. Moore

From: Wien [mailto:wien-boun...@zeus.theochem.tuwien.ac.at] On Behalf Of 
Abderrahmane Reggad
Sent: 07 September 2017 16:11
To: wien@zeus.theochem.tuwien.ac.at
Subject: Re: [Wien] About the magnetic moment of vanadium in vanadium sulphide

Hi All

I have used the PBE+EECC calculation for 3 configurations: nm, fm and afm I and 
I found that the afm I is the most stable.

The energy criterion and charge are 0.001 Ry and 0.001 e respectively.

I don't worry about if the material is really antiferromagnetic or paramagnetic 
because of:

1- I found only one experimental study that they found the compound to be pauli 
magnetic and one theoritical study which they found the compound to be non 
magnetic and these two studies are not sufficient to judge the compound to be 
in a such state. The theoritical study used the GGA method which is not good 
for correlated systems.

2- In the anfiferromagnetic state afm I in the NiAs structure for vanadium 
sulphide I found the following results:

MMI for V1: 0.05 MB
MMI for V2 :- 0.05 MB
MMI for S:0 MB

My questions are now:

what's the definition of non magnetic compound ?

I think we can talk about non magnetic calculation and not about non magnetic 
compounds.

As Blaha said we can't silulate the paramagnetic state or at at least it's 
difficult to do it because we can't orientate the spins randomly ang maintain 
the total magnetic moment equals to zero.

Because of the Hind's prediction and because the impaired number of the V2+ ion 
to equal 3 I believe the atomic magnetic moment to be different from zero.

Best regards
-- The Open University is incorporated by Royal Charter (RC 000391), an exempt 
charity in England & Wales and a charity registered in Scotland (SC 038302). 
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Re: [Wien] About the magnetic moment of vanadium in vanadium sulphide

2017-09-07 Thread E.A.Moore 
 I am used to setting up calculations with spins set to be parallel or 
antiparallel to each other and I would probably class those set up with 
antiparallel spins as AFM. You can also set up calculations that don't include 
spin and so will not have any magnetic moments. I tend to regard these as 
nonmagnetic calculations. However I take the point that something set up as AFM 
could turn out to have a 0 magnetic moment. 

Sorry you are right it should be the V magnetic moments that balance out. 

A Pauli paramagnet should have 0 magnetic moment from these calculations (it 
could be antiferromagnetic or nonmagnetic) as the magnetic moment is only 
produced when a magnetic field is applied.

-Original Message-
From: Wien [mailto:wien-boun...@zeus.theochem.tuwien.ac.at] On Behalf Of 
Fecher, Gerhard
Sent: 07 September 2017 10:39
To: A Mailing list for WIEN2k users
Subject: Re: [Wien] About the magnetic moment of vanadium in vanadium sulphide

Sorry but there is obviously a lot of nonsense in the comments where you should 
first think about:

Please explain why a spin polarized calculation will always result in a 
ferromagnetic (or antiferromagnetic) state ? How do you define a ferromagnet 
(or antiferromagnet) ?  
What happens when the magnetic part of the Hamilton becomes Zero in a spin 
polarized calculation ? What is a ferromagnet or antiferromagnet without 
magnetic moments at the atoms ?
Why should a paramagnet become an antiferromagnet in the calculation ?

Why is VS an antiferromagnet when V has a magnetic moment of 0.05 muB and S one 
of -0.05 muB ? Don't you think it is possible that the magnetisation of two V 
atoms may have to cancel ?

Ciao
Gerhard

DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy:
"I think the problem, to be quite honest with you, is that you have never 
actually known what the question is."


Dr. Gerhard H. Fecher
Institut of Inorganic and Analytical Chemistry Johannes Gutenberg - University
55099 Mainz
and
Max Planck Institute for Chemical Physics of Solids
01187 Dresden

Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von E.A.Moore 
[e.a.mo...@open.ac.uk]
Gesendet: Donnerstag, 7. September 2017 11:02
An: A Mailing list for WIEN2k users
Betreff: Re: [Wien] About the magnetic moment of vanadium in vanadium sulphide

I have been following this thread and I think there is some confusion.

1. On the thread it said that the experiment showed it was Pauli paramagnetic. 
This is the type of magnetism displayed by some metals e.g. sodium which is 
only apparent if you apply a magnetic field.

2. If you include spin in your calculation (GGA or GGA + U) you can only get 
ferromagnetic, antiferromagnetic or ferromagnetic states. (An earlier thread 
deals with how to get paramagnetic states). I think you can only get a 
nonmagnetic state if you do not include spin? A material with Pauli 
paramagnetism will be antiferromagnetic in straight forward spin-including 
calculations.

3. I assume the 0.05 muB refers to the magnetic moment on V. If vanadium 
sulphide is antiferromagnetic and the magnetic moment on Vanadium is 0.05 muB, 
then there must be a balancing magnetic moment on the sulphur.

4. I suspect this compound might be alloy-like. Is there considerable mixing of 
V and S in the valence bands?

5. Assuming your formula is VS, it might be worth noting that VO shows some 
metallic physical properties.

Elaine A. Moore
Reader in theoretical chemistry
The Open University



-Original Message-
From: Wien [mailto:wien-boun...@zeus.theochem.tuwien.ac.at] On Behalf Of 
Fecher, Gerhard
Sent: 07 September 2017 08:12
To: A Mailing list for WIEN2k users
Subject: Re: [Wien] About the magnetic moment of vanadium in vanadium sulphide

0.05 muB does not mean that it is antiferromagnetic ! what was your charge 
convergence criterion ?

You did never answer my question whether you started the EECE calculation from 
a converged GGA calculation.

Why do you like to have an afm state when the experiment tells it is not ?

Ciao
Gerhard

DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy:
"I think the problem, to be quite honest with you, is that you have never 
actually known what the question is."


Dr. Gerhard H. Fecher
Institut of Inorganic and Analytical Chemistry Johannes Gutenberg - University
55099 Mainz
and
Max Planck Institute for Chemical Physics of Solids
01187 Dresden

Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von Abderrahmane 
Reggad [jazai...@gmail.com]
Gesendet: Donnerstag, 7. September 2017 00:26
An: wien@zeus.theochem.tuwien.ac.at
Betreff: Re: [Wien] About the magnetic moment of vanadium in vanadium sulphide

Hi Martin

The problem is that I want to know if it's possible to get a such value of 0.05 
MB for atomic magnetic moment for the AFM state

Re: [Wien] About the magnetic moment of vanadium in vanadium sulphide

2017-09-07 Thread E.A.Moore 
I have been following this thread and I think there is some confusion.

1. On the thread it said that the experiment showed it was Pauli paramagnetic. 
This is the type of magnetism displayed by some metals e.g. sodium which is 
only apparent if you apply a magnetic field.

2. If you include spin in your calculation (GGA or GGA + U) you can only get 
ferromagnetic, antiferromagnetic or ferromagnetic states. (An earlier thread 
deals with how to get paramagnetic states). I think you can only get a 
nonmagnetic state if you do not include spin? A material with Pauli 
paramagnetism will be antiferromagnetic in straight forward spin-including 
calculations.

3. I assume the 0.05 muB refers to the magnetic moment on V. If vanadium 
sulphide is antiferromagnetic and the magnetic moment on Vanadium is 0.05 muB, 
then there must be a balancing magnetic moment on the sulphur.

4. I suspect this compound might be alloy-like. Is there considerable mixing of 
V and S in the valence bands?

5. Assuming your formula is VS, it might be worth noting that VO shows some 
metallic physical properties.

Elaine A. Moore
Reader in theoretical chemistry
The Open University



-Original Message-
From: Wien [mailto:wien-boun...@zeus.theochem.tuwien.ac.at] On Behalf Of 
Fecher, Gerhard
Sent: 07 September 2017 08:12
To: A Mailing list for WIEN2k users
Subject: Re: [Wien] About the magnetic moment of vanadium in vanadium sulphide

0.05 muB does not mean that it is antiferromagnetic ! what was your charge 
convergence criterion ?

You did never answer my question whether you started the EECE calculation from 
a converged GGA calculation.

Why do you like to have an afm state when the experiment tells it is not ?

Ciao
Gerhard

DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy:
"I think the problem, to be quite honest with you, is that you have never 
actually known what the question is."


Dr. Gerhard H. Fecher
Institut of Inorganic and Analytical Chemistry Johannes Gutenberg - University
55099 Mainz
and
Max Planck Institute for Chemical Physics of Solids
01187 Dresden

Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von Abderrahmane 
Reggad [jazai...@gmail.com]
Gesendet: Donnerstag, 7. September 2017 00:26
An: wien@zeus.theochem.tuwien.ac.at
Betreff: Re: [Wien] About the magnetic moment of vanadium in vanadium sulphide

Hi Martin

The problem is that I want to know if it's possible to get a such value of 0.05 
MB for atomic magnetic moment for the AFM state of vanadium sulphide in NiAs 
structure.

Hafner and Hobbs have found all the calculations converged to the non magnetic 
state because they have used the GGA method. To get the AFM state they have to 
use either the EECE or GGA+U methods.

I hope you touch the problem


Best regards
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Re: [Wien] Discrepancy in the simulation of the paramagnetic state

2016-11-28 Thread E.A.Moore 
There is some confusion here about types of paramagnetism.


If the spin-polarised and non-spin polarised results are the same, it merely 
means that the spin up and spin down bands are at equal energies.  Pt has no 
unpaired spins so no magnetic moment. It could from the calculation be 
diamagnetic or Pauli paramagnetic. As it is a metallic conductor, the latter is 
likely, so the non-magnetic form is the Pauli paramagnetic ground state. The 
spin up and spin down bands will acquire different energies if you apply a 
magnetic field.


The original query was concerned with Gd which has unpaired f electron spins 
and it is this type of system that becomes paramagnetic as you raise the 
temperature.


NiS which was also mentioned I assume contains Ni 2+ ions. In square planar 
environments these have no unpaired spins and so no magnetic moment and the 
compounds will be diamagnetic. In tetrahedral environments the ion has unpaired 
spins and so a magnetic moment. The change to no magnetic moment coincided with 
a first iorder phase transition so it is most likely linked to a change in  
structure and hence the local environment of Ni.


Elaine A. Moore


The Open University, UK




From: Wien  on behalf of Fecher, 
Gerhard 
Sent: 28 November 2016 07:33
To: A Mailing list for WIEN2k users
Subject: Re: [Wien] Discrepancy in the simulation of the paramagnetic state

I hope you agree that Pt is paramagnetic
I did two calculations for Pt, one was  spin polarized the other not.
The results are identical, no resulting magnetic moment (indeed, I started with 
one in the spin polarized case), did I play a trick or did Wien2k play a trick ?
but may be Wien2k can not be used to calculate the electronic structure of Pt, 
because it is paramagnetic (Pt, not Wien2k !).

I hope you agree that Pt is paramagnetic even at Zero temperature.
why do I need to include temperature effects to calculate the ground state of 
Pt (at 0 K, where else) ?
... and what should MtC calculations tell me about it ?

Remark 1:
Calculations may be  "spin polarized" (LSDA) or not (LDA) or they may be even 
more sophisticated "non-colinear spin polarized" or they may be for "disordred 
local moments"
or for "spin spirals", or ???,  just to name some.

Remark 2:
Materials may be diamagnetic, paramagnetic (Langevin, Pauli, van Vleck), 
ferromagnetic (localised moments, itinerant), ferrimagnetic (collinear, 
non-collinear), etc..

Therefore, I repeat my question:   How do you distinguish diamagnetic, 
paramagnetic, ferromagnetic, and ... states ?

The answer is for you, not for me.

I tried to calculate for Pt using Hohenberg Kohn DFT, but I could not find the 
functional, all I found was some approximation using wave functions.
Don't worry I will not ask a question about it ;-)

Ciao
Gerhard

DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy:
"I think the problem, to be quite honest with you,
is that you have never actually known what the question is."


Dr. Gerhard H. Fecher
Institut of Inorganic and Analytical Chemistry
Johannes Gutenberg - University
55099 Mainz
and
Max Planck Institute for Chemical Physics of Solids
01187 Dresden

Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von Xavier 
Rocquefelte [xavier.rocquefe...@univ-rennes1.fr]
Gesendet: Sonntag, 27. November 2016 12:46
An: wien@zeus.theochem.tuwien.ac.at
Betreff: Re: [Wien] Discrepancy in the simulation of the paramagnetic state

Just to add one more point to this funny discussion, the term
"paramagnetic" is sometimes used in the DFT litterature in an improper way.

It could clearly lead to misunderstanding for researchers who do not
know so much on how magnetic properties could evolve with temperature
and applied magnetic field. When you see in a paper "paramagnetic state"
simulated using DFT ... it is NOT paramagnetic at all, it is simply a
trick which must be considered with care as previously mentionned by
Peter, Eliane and Martin.

If you want to simulate a paramagnetic state you need to include the
temperature effects, i.e. you should consider the spin dynamics and the
competition between magnetic exchange interactions and thermal
fluctuations. This could be done, at least, using Monte-Carlo
calculations based on an effective hamiltonian constructed on top of DFT
parameters (including magnetic exchange and anisotropy at least).

Best Regards

Xavier




Le 27/11/2016 à 10:01, Fecher, Gerhard a écrit :
> How do you distinguish a diamagnetic, a paramagnetic, a ferromagnetic, and an 
> antiferromagnetic state.
>
> Think !
>
> This will answer your question, hopefully.
>
> Ciao
> Gerhard
>
> DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy:
> "I think the problem, to be quite honest with you,
> is that you have never actually known what the question is."
>
>

Re: [Wien] Discrepancy in the simulation of the paramagnetic state

2016-11-27 Thread E.A.Moore 
In the paramagnetic state, as Prof. Blaha says, the atoms still have magnetic 
moments but they are randomly oriented. This arises when the thermal energy is 
sufficient to overcome the spin-spin coupling. I would expect a calculation on 
Gd at 0K to give you a ferromagnetic state with very small spin-spin coupling. 
You can check the coupling by a run with one spin reversed.


I am not convinced you can model a paramagnetic state with a DFT calculation 
and zero moments is not a good model. Your second example reads as though it is 
reporting experimental results on the magnetisation and does not seem to 
provide a model for calculations.


I would also agree with Prof. Blaha about the factors influencing efg. 
Interatomic distance is very important in calculating this.

Elaine A. Moore
The Open University
UK


From: Wien  on behalf of Abderrahmane 
Reggad 
Sent: 26 November 2016 21:30
To: wien@zeus.theochem.tuwien.ac.at
Subject: Re: [Wien] Discrepancy in the simulation of the paramagnetic state


Thank you Prof Blaha for your quick answer.

The Ni atom is 3d transition metal . But my question is about the simulation of 
the paramagnetic state. There are many people that considere that the 
paramagnetic state is the non-spin polarierd one and the magnetic moment is 
zero, but you say no and the magnetic moments exist in arbitrary directions and 
my quoting is about that.

I have given 2 examples for that discrepancy with your statement.

Best regards
--
Mr: A.Reggad
Laboratoire de Génie Physique
Université Ibn Khaldoun - Tiaret
Algerie


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[Wien] calculations of monoclinic and triclinic systems

2010-07-13 Thread E.A.Moore 
There is no reason you should not do a calculation for monoclinic and triclinic 
systems. Look at the error message you have. Then check the User Guide and 
previous questions. Only after trying suggestions from there write to the 
mailing list with details of your files and error message.

Elaine A. Moore


From: Nirpendra Singh [mailto:nsingh@gmail.com]
Sent: 11 July 2010 12:58
To: wien-mailing
Subject: [Wien] calculations of monoclinic and triclinic systems

Dear all
  I am trying to do  the calculation for organic compounds having 
Monoclinic and triclic structures. Is it possible to do the calculation of 
these systems in wien2k. During init_lapw script, I am geeting error in x 
dstart.

suggestions are requested in this regard


with regards
nsingh

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[Wien] enthalpy of formation

2010-07-13 Thread E.A.Moore 
Try P21/c or space group 11, but be carefuk as it maybe a non-standard setting.

Elaine A. Moore
Open University


From: Sherif Yehia [mailto:wien542...@yahoo.com]
Sent: 09 July 2010 14:18
To: A Mailing list for WIEN2k users
Subject: Re: [Wien] enthalpy of formation

Dear  Wien users

I would  like to   set up  a case  RNiO3(R=Gd , Y , Sm etc)   I know the 
lattice parameters  and the angles from publications.

Structure is monoclinic(space group is P21/n)

My question is when I build case.structwhat would be my input for the 
space group.

Thank  you


--- On Thu, 7/1/10, Stefaan Cottenier Stefaan.Cottenier at UGent.be wrote:

From: Stefaan Cottenier stefaan.cotten...@ugent.be
Subject: Re: [Wien] enthalpy of formation
To: A Mailing list for WIEN2k users wien at zeus.theochem.tuwien.ac.at
Date: Thursday, July 1, 2010, 5:41 PM

 I woluld like to know how to calculate the enthalpy of formation for Fe2Ti???

See for instance http://dx.doi.org/10.1103/PhysRevB.73.144104 (Eq. 1)

Stefaan

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[Wien] Interstitial N-N pair in GaP supercell

2010-07-04 Thread E.A.Moore 
If you want to study a pair of interstitial N atoms, you must add two N atoms, 
not replace P. Or if you are replacing P by two interstitial N atoms, you 
remove one P and add two N at new positions. 
Interstitial means the atoms are not on lattice sites. 
If the reduced cell has 16 inequivalent atoms, it sounds as  though you have 64 
not 65 atoms, so possibly you have just replaced P by N on four sites. 

Are you trying to add an interstitial N2 molecule? If not why do you need two N 
atoms?

Elaine A. Moore


From: Kakhaber Jandieri [kakhaber.jandi...@physik.uni-marburg.de]
Sent: 04 July 2010 15:04
To: A Mailing list for WIEN2k users
Subject: Re: [Wien] Interstitial N-N pair in GaP supercell

Dear Swati,

Thank you for advice, but I still want to clarify some questions:

first of all I think you can't replace one atom by a pair of atoms.

My final goal is to obtain the relaxed atom positions in GaP supercell
where one of the P atoms is replaced by the interstitial pair of
Nitrogen-Nitrogen atoms.
To say truth, I am not sure in correctness of my approach but I did the
following:
Step1 - Using the program supercell I generated 2x2x2 supercell for
GaP containing 64 atoms
Step2 -  I construct a new supercell with 64  inequivalent atoms using
the positions obtained in Step1
Step3 -  I split one P atom in 2 atoms
Step4 -  I changed the atoms obtained in Step3 from P to N
Step5 -  I ran x sgroup. It converted my p supercell with 65
inequivalent atoms into the 111_p-42m supercell with 16 inequivalent atoms
Step6 - I ran the program mini.

Could you advice me the correct way for constructing the supercell with
interstitial pair of  N-N atoms in  GaP matrix?


Set 5% less RMT for all types of atoms instead of reduction of RMT of N
only and run mini

In interstitial N-N pair the interatomic distance between N atoms might be 
significantly smaller then the distance between Ga and P atoms.

If instead of 2.0 I take 5% less RMT, then how cam I prevent the overlapping of 
spheres between the N atoms in the interstitial pair?


Beforehand thankfull,
Kakha





swati chaudhury wrote:
 Hello,
 first of all I think you can't replace one atom by a pair of atoms.
 Set 5% less RMT for all types of atoms instead of reduction of RMT of N only 
 and run mini.
 best wishes.
 swati

 --- On Sun, 4/7/10, Kakhaber Jandieri Kakhaber.Jandieri at 
 physik.uni-marburg.de wrote:


 From: Kakhaber Jandieri Kakhaber.Jandieri at physik.uni-marburg.de
 Subject: [Wien] Interstitial N-N pair in GaP supercell
 To: wien at zeus.theochem.tuwien.ac.at
 Date: Sunday, 4 July, 2010, 5:17 PM
 In addition to my previous letter.

 I thought that may be the problem is in very large
 difference between the RMT(Ga,P)=2.0 and RMT (N)=1.0, but I
 cannot increase the RMT(N) because
 of overlapping spheres and, on the other hand, I cannot
 decrease the RMT(Ga,P) because of charge leakage.


 Could somebody advice the solution of this problem?

 I will be extremely thankful for any suggestion.

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[Wien] Mg in GaN

2010-03-15 Thread E.A.Moore 
You may want to study Mg-doped GaN as this would be a p-type semiconductor but 
you certainly would not want to study antiferromagnetic coupling between Mg2+ 
ions as Prof. Blaha has pointed out.
However if you were interested in Mg-doped GaN you would probably want a low 
level of dopant and hence a large supercell.
Or you may have been asked to look at manganese (Mn) rather than magnesium 
(Mg).  You would then need to decide on the oxidation state of the Mn ion, 
possibly Mn3+ on a Ga site.

Elaine A. Moore
Department of Chemistry and Analytical Science
The Open University
UK




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