Why artificial lowering of symmetry changes the total energy? (from wien mailing list)

katrusiat Wed, 16 Apr 2008 13:37:36 +0400

 

Dear all,

I perform ferromagnetic (FM) (I make it FM by modifying the case.inst file 
accordingly) GGA+U calculations of the total energy of Cs2CuCl4, the compound 
with 4 equivalent magnetic Cu's in a unit cell. When I create an artificial 
unit cell of a lower symmetry with 2 inequivalent pairs of Cu's: Cu1 and Cu2, 
and again perform a FM GGA+U calculation, the total energy of the system is 
slightly different.

By lowering the symmetry I only add labels to Cu's making them inequivalent, 
and also the crystallographic axes are interchanged in the new unit cell. So, 
why after these seemingly harmless actions the total energy should change?

Thank you,
Kateryna Foyevtsova

 Answer of Prof Steefan Cotennier

Stefaan Cottenier Wed, 16 Apr 2008 12:17:43 +0200

 

1) How large is your energy change? Perhaps it is just numerical noise? 

2) If the change is large enough, and if it is a lowering of the energy upon symmetry reduction, you might have an onset of orbital ordering. Inspect the density matrices of the two inequivalent Cu-sites (dmat files), and see whether they are different from each other and/or different from the density matrix for the high-symmetry case. If there are differences, then this material apparently can lower its total energy by occupying the Cu-d orbitals in different ways in the two Cu-sites.

Stefaan

 

katrusiat Wed, 16 Apr 2008 17:24:14 +0400

Dear Stefaan,

To be exact, I have never calculated the high-symmetry case, but only the cases 
with a symmetry reduced in different ways, that is, picking up Cu pairs in a 
different way. 

For example, the high-symmetry (62_Pnma) unit cell can be reduced into:

{1} - (26 Pmc21)
or
{2} - (11 P21/m).

The total FM energy of {1} is -152655.214439 Ry
           and that of {2} is -152655.214842 Ry

The precision of both calculations was 0.000001 Ry since all the figures after 
the point are important for me.

I am including the case.dmatdn files of both cases. There are differences 
between the density matrices of {1} and {2}, but I do not know what they mean 
physically.

....

Stefaan Cottenier Thu, 17 Apr 2008 11:05:44 +0200

 

> {1} - (26 Pmc21)

 > or 

> {2} - (11 P21/m).

 > > The total FM energy of {1} is -152655.214439 Ry 

> and that of {2} is -152655.214842 Ry 

> > The precision of both calculations was 0.000001 Ry since all the figures 

> after the point are important for me. 

> > That looks like a difference that is more than numerical noise. But: we should distinguish between accuracy and self-consistency. The precision of 0.000001 Ry which you quote is probably coming from your convergence criterium, isn't it? That means that your solution is considered to be self-consistent if such (strong) criterium is met. But that isn't the completely identical to requiring high absolute accuracy : your result can be highly self-consistent but not very accurate. And then it is not guaranteed that two formally identical cases will yield exactly the same energy. You can try to repeat these calculations with a higher RKmax, and see whether the energy difference decreases.

> I am including the case.dmatdn files of both cases. There are differences 

 > between the density matrices of {1} and {2}, but I do not know what they mean > physically.

 > > These are 5x5 matrices, of which the diagonal elements correspond to the occupation of the different m-orbitals for d-electrons (roughly speaking...). At first sight, the diagonals look fairly identical in all cases. My bet is hence that you need a larger RKmax. 

 Stefaan

apost...@uni-osnabrueck.de Thu, 17 Apr 2008 13:14:13 +0200 (CEST)

 

Apart from checking the convergence in RKmax as emphasized by Stefaan,
the convergence in k-mesh is another issue to be checked, for two reasons
(both apply in the case of tetrahedron interation only):

 1) I think the assignment of tetrahedra, even over identical meshes of
k-points, may be done diferently for different symmetries (depending on
the heuristics of cutting k-microcells into tetrahedra) that may result
in numerical differences on integration. (However, I might be wrong on
whether this issue is indeed valid in WIEN2k).

 2) As you have one symmetry a subgroup of the other, the bands from one IBZ
will be folded into other (smaller) one. Now, even if all eigenvalues
are identical, the "connectivity" of bands might be different, giving rise
to a "wrong band crossings" error.

Best regards

Andrei Postnikov

katrusiat Thu, 17 Apr 2008 19:12:09 +0400

Dear Stefaan and Andrei,

To check whether not big enough RKM could cause the difference in the total 
energy, I made a test calculation of TiOCl system. In the high symmetry 
(59_Pmmn) unit cell, there are 2 equivalent Ti atoms, which I made inequivalent 
to generate a low symmetry unit cell (25 Pmm2). With -ec 0.00001 and RKM = 7, 
the energies were:

HS = -5562.790746
LS = -5562.790740

and with RKM = 6,

HS = -5562.726158
LS =  5562.726155

I think this is enough to say that in this case the energy does not depend on 
the space group at all.

This case is different from that of Cs2CuCl4 I quoted before in that now for HS 
and LS I had equal number of k-points in IBZ, whereas for {1} and {2} of 
Cs2CuCl4 the number of k-points in IBZ was 120 and 192.
I would conclude therefore that Andrei is right, and the results for Cs2CuCl4 
are not accurate because k-mesh is not dense enough.

Thank you very much for usefull suggestions,

Kateryna Foyevtsova

Reference: https://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/msg00388.html