Concern about Fermi energy position: in course of studying spin-spiral energetics in 4f metal

[Yo Pierre Mizuta]

Dear developers and users of Elk,

I am a newcomer to Elk, who wish to take advantage of the code for extracting the Heisenberg exchange parameters through the frozen-magnon approach.

As a starting point of such studies, I am trying to confirm the consistency of results I obtain via Elk and those reported for bcc europium by Kunes and Laskowski PRB 70, 174415 (2004) via Wien2k.

However, I have not been successful in obtaining even the consistent sign of the energy difference ΔE = E(H) - E(0) between two selected magnetic configurations of the uniform ferromagnet (q=0) and a spin spiral with q=(H point).
Since I assumed a cone angle of θ=30°, what is expected according to Fig.2 of the paper is ΔE ≒ (sin30°)^2×(-0.8 mHartree)= -0.2 mHartree, whereas my result is ΔE ≒ +0.1 mHartree.

Looking at the band dispersion for q=0 (attached), I suspect the large difference in the position of their Fermi energy E_F (blue dashed line ~ +0.8 eV; visually extracted from Fig.1 of the paper) and mine (0.0 eV) to be a fundamental origin of this discrepancy.

Possible factors that could largely modify the band structure in the present context are the parameters U & J, but the inconsistent E_F is not remedied by their adjustment (as is expected since the dispersive bands crossing E_F should have little f-character), nor is it by choosing to interpolate around-mean-field and fully-localized-limit treatments in double-counting correction.

The input file used for q=0 calculation is

! bccEu in FM structure with LDA+U and fully-localised-limit (FLL) double
! counting (dftu=1). 

tasks
 0 
 20 

xctype
3

autolinengy
true

mixtype
3

maxscl
200

vhighq
 .true.

dft+u
  1 1                             : dftu,inpdftu
  1 3 0.257 0.0276    : is, l, u, j 

spinpol
  .true.

avec
   4.303851250       4.303851250      -4.303851250    
  -4.303851250       4.303851250       4.303851250    
   4.303851250      -4.303851250       4.303851250    

atoms
   1                                    : nspecies
'Eu.in'                                 : spfname
   1                                    : natoms; atposl, bfcmt below
    0.00000000    0.00000000    0.00000000    0.00000000  0.00000000  1.00000000

reducebf
0.5

sppath
  '/home/mizuta/elk-5.2.14/species/'

ngridk
  10 10 10

plot1d
5 300
0.000  0.000  0.000
0.500 -0.500  0.500 
0.000  0.000  0.500
0.000  0.000  0.000
0.250  0.250  0.250 

and a typical input for the subsequent run for q=H is

! bccEu in spin-spiral(q=H) structure with LDA+U and fully-localised-limit (FLL) double
! counting (dftu=1). 

tasks
 1 

xctype
3

autolinengy
true

mixtype
3

maxscl
200

vhighq
 .true.

dft+u
  1 1                             : dftu,inpdftu
  1 3 0.257 0.0276    : is, l, u, j 

spinpol
  .true.

spinsprl
  .true.

vqlss
0.500 -0.500  0.500 

fsmtype
-2

mommtfix
1 1 0.500 0.000 0.866025403 

avec
   4.303851250       4.303851250      -4.303851250    
  -4.303851250       4.303851250       4.303851250    
   4.303851250      -4.303851250       4.303851250    

atoms
   1                                    : nspecies
'Eu.in'                                 : spfname
   1                                    : natoms; atposl, bfcmt below
    0.00000000    0.00000000    0.00000000    -0.500 0.000 -0.866025403

reducebf
0.5

sppath
  '/home/mizuta/elk-5.2.14/species/'

ngridk
  10 10 10

I also tried some variations: with a finer k-mesh, or with a larger muffin-tin radius, or fixing total moment instead of that inside muffin-tin, or starting from scratch for q=H instead of reading STATE.OUT from run for q=0, all of which gave only minor change.

Any comments or instructions would be very much appreciated.

Yo P. Mizuta
Postdoctoral Researcher,
Osaka University

[Lars Nordström]

Dear Yo,

Is your main concern that you get E(H)>E(0)? Well, it should be.
Probably their Eq 3 lacks a minus sign ... It is clearly written that the energy minimum corresponds to the maximum in their J(q) curve. and hence E(H)-E(0)>0 also for them. The factor two I cannot explain, but I suggest you do the full q-range with cone angle 90 degrees, which is more stable numerically and compare the curves ...

Good luck,
Lars

[Yo Pierre Mizuta]

Dear Professor Nordstrom,

Thank you so much for your quick help!
Ah...! Yes, that was my main concern. I should have been more careful...
I have the impression that there have been some confusion of signs in such formula in the literature. You gave me a really important lesson!
After trying what you suggest, I would like to report the quality of agreement.

Yours sincerely,
Yo P. Mizuta

[Anonymous]

The mentioned above factor of two can result from different systems of units, Hartree or Rydberg (it is just a guess though).