Electronic energy mismatch among symmetry equivalent states

[Vahid]

Dear EPW Users,

In order to try and reduce the computational demand for the problem we are looking at, we're testing to see if we can use only the irreducible k-points in the BZ (using symmetry to identify the equivalent k-points from the full dense k-grid (but still using the full dense q-grid)). To test this we've started by looking at two examples: B-doped diamond, and pure silicon. From our tests we noticed an unexpected behavior. We have two questions we're hoping someone can help us with:

QUESTION 1:
When comparing the electron eigenenergy for a single k-point when using the irreducible k-grid vs. the full k-grid, it appears equivalent k-points have slightly different energies. This doesn't happen for diamond, but does happen for silicon (energies differ by as much as 80 meV, which can affect the self-energy calculation). Below are sets of eigenenergies at equivalent kpoints: the first two sets are for diamond (the first set with the irreducible k-grid, the second set with the full k-grid), and the next two sets are for silicon. We used both a 20x20x20 k-grid (reduced down to 256 irreducible k-points) and 20x20x20 q-grid to calculate the electron self energies for B-doped diamond and pure silicon.

Is there any reason why for diamond the energies at equivalent k-points would be identical, but would be different in the case of silicon? Note we tried varying the Gaussian broadening with the
values of 0.1, 0.03 and 0.01 eV, but the issue persists with silicon.

QUESTION 2:
In principle, will using the irreducible k-grid instead of the full k-grid in EPW give identical results for the electron self-energy calculations (still using the full q-grid)?

Code: Select all

----- LIST OF EIGENENERGIES AT EQUIVALENT K-POINTS -----
DIAMOND:
ik=178 in the irreducible k-grid:
  ik =     178 coord.:    0.0500000   0.4000000  -0.4500000
  E(   1 )= -13.4270 eV   Re[Sigma]=     -30.933681 meV       Im[Sigma]=      24.869431 meV
  E(   2 )= -10.3825 eV   Re[Sigma]=      -4.229391 meV       Im[Sigma]=      61.216478 meV
  E(   3 )=  -7.4753 eV   Re[Sigma]=     -13.901492 meV       Im[Sigma]=      66.070361 meV
  E(   4 )=  -5.7127 eV   Re[Sigma]=     -15.327615 meV       Im[Sigma]=      97.407976 meV

ik=6968 in the full k-grid (equivalent to the one above):
  ik =    6968 coord.:    0.8500000   0.4000000   0.3500000
  E(   1 )= -13.4270 eV   Re[Sigma]=     -30.952174 meV       Im[Sigma]=      24.860316 meV
  E(   2 )= -10.3825 eV   Re[Sigma]=      -4.215018 meV       Im[Sigma]=      61.172779 meV
  E(   3 )=  -7.4753 eV   Re[Sigma]=     -13.822378 meV       Im[Sigma]=      66.013742 meV
  E(   4 )=  -5.7127 eV   Re[Sigma]=     -15.250938 meV       Im[Sigma]=      97.573909 meV

SILICON:
ik=178 in the irreducible k-grid:
  ik =     178 coord.:    0.0500000   0.4000000  -0.4500000
  E(   2 )=  -7.5793 eV   Re[Sigma]=     -68.528328 meV       Im[Sigma]=      47.413559 meV
  E(   3 )=  -4.2495 eV   Re[Sigma]=     -40.153585 meV       Im[Sigma]=      35.669718 meV
  E(   4 )=  -3.3577 eV   Re[Sigma]=     -35.255624 meV       Im[Sigma]=      55.421246 meV
  E(   5 )=   0.5776 eV   Re[Sigma]=       0.617174 meV       Im[Sigma]=      14.450487 meV
  E(   6 )=   2.1666 eV   Re[Sigma]=      -1.575815 meV       Im[Sigma]=      79.771434 meV
  E(   7 )=   7.5324 eV   Re[Sigma]=      80.555970 meV       Im[Sigma]=      23.258165 meV
  E(   8 )=   8.2987 eV   Re[Sigma]=      40.674666 meV       Im[Sigma]=       4.181535 meV

ik=6968 in the full k-grid (equivalent to the one above):
  ik =    6968 coord.:    0.8500000   0.4000000   0.3500000
  E(   2 )=  -7.5787 eV   Re[Sigma]=     -68.242478 meV       Im[Sigma]=      47.824544 meV
  E(   3 )=  -4.2153 eV   Re[Sigma]=     -37.682325 meV       Im[Sigma]=      34.277042 meV
  E(   4 )=  -3.2997 eV   Re[Sigma]=     -32.718397 meV       Im[Sigma]=      49.295667 meV
  E(   5 )=   0.5991 eV   Re[Sigma]=      -1.460534 meV       Im[Sigma]=      16.483447 meV
  E(   6 )=   2.1402 eV   Re[Sigma]=      -7.812829 meV       Im[Sigma]=      75.284735 meV
  E(   7 )=   7.4590 eV   Re[Sigma]=      84.012895 meV       Im[Sigma]=      50.947981 meV
  E(   8 )=   8.2740 eV   Re[Sigma]=      39.402005 meV       Im[Sigma]=       2.551329 meV

Thank you,

Vahid

Vahid Askarpour
Department of Physics and Atmospheric Science
Dalhousie University,
Halifax, NS, Canada

[sponce]

Dear Vahid,

Silicon and diamond should indeed have the same Fd-3m space group (Space group number: 227).

Therefore equivalent k-points by symmetry should indeed give the same eigenenergies.

Can you check that Quantum Espresso (scf run) gives the same solution for diamond and silicon. You have to disable symmetry.
Then check that the output of the nscf run that you used for EPW also does gives the same results for equivalent points.

Now, I naturally also though about using symmetry for the homogeneous fine k/q grid.

However, I see no benefit in doing so. You should used random or quasi random (e.g. Sobol or Monte Carlo) k/q point sampling instead. This should converge faster or equal to an homogeneous grid with symmetries.

EPW provides random k/q point generations.

By using random k-point, you avoid over-representing high symmetry point (done with homogeneous sampling).
Off course you could randomly shift your homogeneous grid to avoid this and then use symmetry. But this should be as efficient as random sampling.
See for example Fig. 3 of Computational Materials Science 83, 341 (2014).
There I compare IBZ sampling from homogeneous mesh VS random points. The convergence is similar.

For those reasons I did not implement sym. in fine homogeneous grids.


Naturally, If you manage to implement it nicely and cleanly, I'm very happy to add this inside EPW.

Best,

Samuel

[Vahid]

Dear Samuel,

I am very grateful for your suggestions. I ran scf (with nosym=.true. and noinv=.true.) followed by nscf on both diamond and Si using a coarse 6x6x6 gamma-centered grid. For both systems, symmetry equivalent points have equal KS energies. In addition, the scf run gives identical energies to the nscf run for all the bands and k-points in both systems.

Could the energy mismatch we observe in the fine k mesh of Si be the result of Wannier interpolation from the coarse unto the fine grid? One difference between Si and diamond is the spread of the Wannier functions. Si has a wider spread (~2.24 and 2.6 A^2 for valence and conduction bands, respectively) relative to diamond (0.7A^2 for valence bands).

You mentioned that there is no benefit in using symmetry. I was hoping to replace the fine k-grid with the irreducible wedge. We noted that in diamond, the scattering rates for equivalent points are identical. This finding suggests that instead of running self-energy calculations on a 20x20x20 fine k grid, one can just run on a 256 list of irreducible points. This should speed up the calculations greatly. Since you do not see a benefit in doing so, then I must be missing a fundamental concept here and I would welcome your input as to what might be wrong with my approach.

Thank you and best wishes,

Vahid

[sponce]

Hi Vahid,

To check weather this is due to the Wannierization you can check the decay files. You can compare them. If the one for diamond decay much more quickly, it could be an explanation indeed.

As for the symm, as I was mentioning in my last post, off course you will take less time computing 256 points than 20**3 ...., I was just saying that this should be roughtly equivalent to running 256 random points.

Now, you do not really need to implement symmetries. You can just give an input file to EPW using "filkf" that contains the 256 points. You need to correctly put the weight off course. You can check that you get the same result as a full 20x20x20.

Best,

Samuel

[Vahid]

Dear Samuel,

The decay files are in fact showing different decay rates so I tried running Si with 8x8x8 coarse grids ( k and q). The mismatch still exists and may be related to the wannierization routine. I then tried 10x10x10 coarse grids but the run required 1TB of memory which is beyond our capability. I would appreciate any other suggestion you may have.

Thank you,

Vahid

[carla.verdi]

Dear Vahid

Maybe you could test the eigenenergies by running just wannier90?

Best
Carla

[roxana]

Dear All,

Symmetry can be used to generate the irreducible k-points in the BZ by using mp_mesh_k =.true. flag (this is done in loadkmesh.f90). However, this flag is currently used only when solving the anisotropic Eliashberg equations.

It could be indeed the wannierization that causes the mismatch between energies at equivalent points.

Best,
Roxana