[Pw_forum] problem of Bi2Se3 with spin orbit coupling parity analysis at TRIM points
Andrea Dal Corso
dalcorso at sissa.it
Sun May 28 19:02:19 CEST 2017
Quoting balabi <balabi at qq.com>:
> Dear developers,
> previously, I send an email about symmetry analysis in calculating
> Bi2Se3 http://www.mail-archive.com/pw_forum@pwscf.org/msg31533.html
> But I pasted the wrong input file in that email. The symmetry info in
> that post is generated for Bi2Se3 without spin orbit coupling.
>
> Now, I calculated Bi2Se3 with spin orbit coupling. What I got is not
> right. There is no parity information labeled u and g.
> Below is the output of bands.x for gamma point
>
> -----------------------------
> xk=( 0.00000, 0.00000, 0.00000 )
>
> double point group C_1 (1)
> there are 2 classes and 1 irreducible representations
> the character table:
>
> E -E
>
> G_2 1.00 -1.00
>
> the symmetry operations in each class and the name of the first
> element:
>
> E 1
>
>
> -E -1
>
>
>
> Band symmetry, C_1 (1) double point group:
>
> e( 1 - 4) = -16.63750 eV 4 --> 4 G_2
>
> e( 5 - 6) = -16.62527 eV 2 --> 2 G_2
>
> e( 7 - 8) = -16.62265 eV 2 --> 2 G_2
>
> e( 9 - 10) = -13.66107 eV 2 --> 2 G_2
>
> e( 11 - 12) = -13.65905 eV 2 --> 2 G_2
>
> e( 13 - 14) = -13.64175 eV 2 --> 2 G_2
>
> e( 15 - 16) = -13.63940 eV 2 --> 2 G_2
>
> e( 17 - 18) = -13.63741 eV 2 --> 2 G_2
>
> e( 19 - 20) = -13.63672 eV 2 --> 2 G_2
>
> e( 21 - 22) = -6.40199 eV 2 --> 2 G_2
>
> e( 23 - 24) = -5.17683 eV 2 --> 2 G_2
>
> e( 25 - 26) = -5.15601 eV 2 --> 2 G_2
>
> e( 27 - 28) = -1.79777 eV 2 --> 2 G_2
>
> e( 29 - 30) = -1.02045 eV 2 --> 2 G_2
>
> e( 31 - 32) = 5.18513 eV 2 --> 2 G_2
>
> e( 33 - 34) = 5.40747 eV 2 --> 2 G_2
>
> e( 35 - 36) = 6.03590 eV 2 --> 2 G_2
>
> e( 37 - 38) = 6.14972 eV 2 --> 2 G_2
>
> e( 39 - 40) = 6.19631 eV 2 --> 2 G_2
>
> e( 41 - 42) = 6.51310 eV 2 --> 2 G_2
>
> e( 43 - 44) = 6.80580 eV 2 --> 2 G_2
>
> e( 45 - 46) = 7.20470 eV 2 --> 2 G_2
>
> e( 47 - 48) = 7.50637 eV 2 --> 2 G_2
>
> e( 49 - 50) = 8.05825 eV 2 --> 2 G_2
>
> e( 51 - 52) = 9.13333 eV 2 --> 2 G_2
>
> e( 53 - 54) = 9.27043 eV 2 --> 2 G_2
>
> e( 55 - 56) = 9.86223 eV 2 --> 2 G_2
>
> e( 57 - 58) = 11.09081 eV 2 --> 2 G_2
>
> e( 59 - 60) = 11.46791 eV 2 --> 2 G_2
>
> e( 61 - 62) = 13.77176 eV 2 --> 2 G_2
>
> e( 63 - 64) = 14.35855 eV 2 --> 2 G_2
>
> ----------------------
>
> Is this right? There are no u and g like Bi2Se3 without SOC. How can I tell
> the parity? And the symmetry seems too simple.
>
> ----------------------
>
> below is my Bi2Se3 with SOC bands.in
>
> &CONTROL
> prefix='Bi2Se3_SOC',
> calculation='bands',
> restart_mode='from_scratch',
> wf_collect=.true.,
> verbosity='high',
> outdir='./quantum_espresso/qe_tmpdir',
> pseudo_dir='./quantum_espresso/pseudo',
> /
> &SYSTEM
> ibrav = -5,celldm(1) =18.5969068371645,celldm(4)=0.9115970970052608,nat =
> 5,ntyp = 3,
> ecutwfc = 40,ecutrho = 500,
> noncolin=.true.,lspinorb=.true.,starting_magnetization=0.,
> nbnd=64,
> nosym=.true.,
nosym=.TRUE. means that no symmetry is used.
HTH,
Andrea
> /
> &ELECTRONS
> conv_thr = 1.0d-10, !default 1d-6
> diago_full_acc=.true.,!increase empty bands accuracy
> /
> ATOMIC_SPECIES
> Bi 208.98040 Bi.rel-pbe-dn-kjpaw_psl.0.2.2.UPF
> Se178.971Se.rel-pbe-n-kjpaw_psl.0.2.UPF
> Se278.971Se.rel-pbe-n-kjpaw_psl.0.2.UPF
> ATOMIC_POSITIONS crystal
> Bi0.39900.39900.3990
> Bi0.60100.60100.6010
> Se11.00001.00001.0000
> Se20.20600.20600.2060
> Se20.79400.79400.7940
> K_POINTS crystal_b
> 5
> 0.00000 0.00000 0.00000 1 !gG
> 0.50000 0.50000 0.50000 1 !Z
> 0.50000 0.50000 -0.00000 1 !F
> 0.00000 0.00000 0.00000 1 !gG
> 0.00000 0.00000 -0.50000 1 !L1
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