EPW Forum

Dear all,

I have recently been using EPW to study polarons in transition metal oxides. When I increase nkf and nqf to 13 or above, the EPW program encounters the following error:
===================================================================
Use the band from 1 to 24 total 24
Including bands: 1 2 3 4 5 6 7 8 9 10
Including bands: 11 12 13 14 15 16 17 18 19 20
Including bands: 21 22 23 24
Use the band from 1 to 24 total 24 in saving g
Gathering eigenvalues of 24 bands and 2197 k points
The band extremes are at band 24
Fermi Energy is 10.4835306 (eV) located at kpoint 1 0.000 0.000 0.000 band 24
Allocating arrays and open files.

===================================================================================
= BAD TERMINATION OF ONE OF YOUR APPLICATION PROCESSES
= RANK 0 PID 24374 RUNNING AT cu74
= KILLED BY SIGNAL: 9 (Killed)
===================================================================================

should I use etf_mem = 2? But I don't know how to use this parameter, like how to extract formation energy from result(I tried this when I was learning to use EPW in the examples, but the epw2.out just has some basic information, I don't know the workflow).

And should I use efermi_read? I don't know what impact this parameter has on the results. When should I enable it?

Thanks a lot!

Regards
Snake

Attached is the input file.
epw2.13.in
&inputepw
prefix = 'm-HfO2'
outdir = './'

elph = .true.
epwwrite = .false.
epwread = .true.
lpolar = .true.
etf_mem = 1

nbndsub = 24
dvscf_dir = './save/'

wannierize = .false.
num_iter = 200
iprint = 2
proj(1) = 'O:p'
bands_skipped = 'exclude_bands = 1-24'
dis_win_max = 10.5
dis_froz_max = 10.5
dis_froz_min = 5
wannier_plot = .true.
wannier_plot_supercell = 3 3 3

! efermi_read = .true.
! fermi_energy = 10.4835

! lifc = .true.
! asr_typ = 'simple'

plrn = .true.
restart_plrn = .false.
type_plrn = 1
init_plrn = 1
init_sigma_plrn = 2.0
niter_plrn = 500
conv_thr_plrn = 1E-4
ethrdg_plrn = 1E-4
! adapt_ethrdg_plrn = .true.

nk1 = 6
nk2 = 6
nk3 = 6
nq1 = 3
nq2 = 3
nq3 = 3

nkf1 = 13
nkf2 = 13
nkf3 = 13
nqf1 = 13
nqf2 = 13
nqf3 = 13
/

Hi,

Can you explain a bit more regarding the calculation? How many cores you were using etc.?
Also, will be helpful if you can attach the traceback.

Best regards,
Sabya.