Here is the input file: &control calculation = 'scf' restart_mode='from_scratch', pseudo_dir = './../../../Pseudo', prefix='arsenene', verbosity = 'high' / &system ibrav = 4, nat= 2, ntyp= 1, ibrav= 4, celldm(1) =7.8, celldm(3) =6.103648,
ecutwfc =30.0, occupations='smearing', smearing='gauss', degauss=0.02, lspinorb=.true noncolin=.true nbnd = 20 / &electrons !diagonalization='david' electron_maxstep = 300 !mixing_mode = 'plain' mixing_beta = 0.7 conv_thr = 1.0d-10 / ATOMIC_SPECIES As 74.9216 As.rel-pbe-n-rrkjus_psl.0.2.UPF ATOMIC_POSITIONS crystal As 0.333333333 0.666666666 0.524859275 As 0.666666666 0.333333333 0.475140725 K_POINTS crystal 1 0.5 0.0 0.0 1 Note that in the above highlighted portion I am using one of the TRIM points, i.e. M1. Where other TRIM points are M2, M3 and G with coordinates 0.0 0.5 0.0, -0.5 0.0 0.0 and 0.0 0.0 0.0 respectively. After SCF (pw.x) calculation, I run the bands,x calculation. The relevant portion of the output is : ************************************************************************** xk=( -0.50000, 0.28868, 0.00000 ) double point group C_2h (2/m) there are 8 classes and 4 irreducible representations the character table: E -E C2 -C2 i -i s_h -s_h G_3+ 1.00 -1.00 0.00 0.00 1.00 -1.00 0.00 0.00 G_4+ 1.00 -1.00 0.00 0.00 1.00 -1.00 0.00 0.00 G_3- 1.00 -1.00 0.00 0.00 -1.00 1.00 0.00 0.00 G_4- 1.00 -1.00 0.00 0.00 -1.00 1.00 0.00 0.00 imaginary part E -E C2 -C2 i -i s_h -s_h G_3+ 0.00 0.00 1.00 -1.00 0.00 0.00 1.00 -1.00 G_4+ 0.00 0.00 -1.00 1.00 0.00 0.00 -1.00 1.00 G_3- 0.00 0.00 1.00 -1.00 0.00 0.00 -1.00 1.00 G_4- 0.00 0.00 -1.00 1.00 0.00 0.00 1.00 -1.00 the symmetry operations in each class and the name of the first element: E 1 -E -1 C2 2 -C2 -2 i 3 -i -3 s_h 4 -s_h -4 Band symmetry, C_2h (2/m) double point group: e( 1 - 2) = -14.19192 eV 2 --> G_3- e( 1 - 2) = -14.19192 eV 2 --> G_4- e( 3 - 4) = -12.67503 eV 2 --> G_3+ e( 3 - 4) = -12.67503 eV 2 --> G_4+ e( 5 - 6) = -6.29105 eV 2 --> G_3+ e( 5 - 6) = -6.29105 eV 2 --> G_4+ e( 7 - 8) = -5.79866 eV 2 --> G_3- e( 7 - 8) = -5.79866 eV 2 --> G_4- e( 9 - 10) = -4.65984 eV 2 --> G_3- e( 9 - 10) = -4.65984 eV 2 --> G_4- e( 11 - 12) = -2.39948 eV 2 --> G_3+ e( 11 - 12) = -2.39948 eV 2 --> G_4+ e( 13 - 14) = -0.22470 eV 2 --> G_3- e( 13 - 14) = -0.22470 eV 2 --> G_4- e( 15 - 16) = 0.10174 eV 2 --> G_3+ e( 15 - 16) = 0.10174 eV 2 --> G_4+ e( 17 - 18) = 2.74250 eV 2 --> G_3+ e( 17 - 18) = 2.74250 eV 2 --> G_4+ e( 19 - 20) = 3.59948 eV 2 --> G_3- e( 19 - 20) = 3.59948 eV 2 --> G_4- Is highlighted K point in the output file okay?? since I used 0.5 0 0 but it is xk=( -0.50000, 0.28868, 0.00000 ). Moreover, I get the same result for M2 and M3 (which should not happen because then Z2 = 0 which contradicts the dedfinitions of TI) Please help. On Thu, Sep 27, 2018 at 4:19 PM Asad Mahmood <amahm...@phys.qau.edu.pk> wrote: > Hi everyone, > > I am working with 2D materials, applying biaxial strain. At some strain > value, the electronic structure exhibits the band diagram similar to that > of a Topological Insulator (as I could observe band inversion from partial > DOS too). The material I am working with has inversion symmetry which > implies that I can find parity eigen values at different bands(then I can > find Z2 Topological Invariant using parity eigen values). > My question is: > > How can we obtain parities (or directly Z2 values, if possible) using > Quantum Espresso for a given band diagram? > > Regards, > Asad Mahmood, > Physics Department, > Q.A.U, Islamabad, > Pakistan >
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