[Wien] input conversion error in QTL calculation
明文美
iphyboy at hotmail.com
Sun Feb 10 14:53:53 CET 2008
Dear Wien2K users, after I successfully finished the spin-polarization scf calculation of NaCuPO4, I want to move onto its Dos calculation.but when I typed in commamds: x lapw2 -c qtl -up -p , some error as the followings appeared: [phymwm at alpha0 NaCuPO4_fm]$ forrtl: severe (64): input conversion error, unit 1003, file /pub/phymwm/opt_work/NaCuPO4_fm/./NaCuPO4_fm.helpup033Image PC Routine Line Source lapw2c 0000000000893E5B Unknown Unknown Unknownlapw2c 000000000089233A Unknown Unknown Unknownlapw2c 0000000000867D36 Unknown Unknown Unknownlapw2c 0000000000838D9C Unknown Unknown Unknownlapw2c 00000000008389EC Unknown Unknown Unknownlapw2c 000000000084E73D Unknown Unknown Unknownlapw2c 0000000000456BFF Unknown Unknown Un!
knownlapw2c 00000000004486FE Unknown Unknown Unknownlapw2c 0000000000451B95 Unknown Unknown Unknownlapw2c 000000000040996A Unknown Unknown Unknownlibc.so.6 0000003A9261C3FB Unknown Unknown Unknownlapw2c 00000000004098AA Unknown Unknown UnknownFurthermore, I checked that those NaCuPO4_fm.helpup$i files already existed.Also I did't find any item in previous mailing-list questions attached below is the corresponding NaCuPO4.struct file : NaCuPO4 P LATTICE,NONEQUIV.ATOMS: 719_P212121 MODE OF CALC=RELA unit=bohr 18.346413 9.080516 13.542539 90.000000 90.000000 90.000000 ATOM -1: X=0.68762000 Y=0.01855000 Z=0.23173000 MULT= 4 ISPLIT= 8 -1: !
X=0.81238000 Y=0.98145000 Z=0.73173000 -1: X=0.18762000 Y=0.48145
000 Z=0.76827000 -1: X=0.31238000 Y=0.51855000 Z=0.26827000Na1 NPT= 781 R0=0.00010000 RMT= 2.27 Z: 11.0 LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000ATOM -2: X=0.54163500 Y=0.49751000 Z=0.94537700 MULT= 4 ISPLIT= 8 -2: X=0.95836500 Y=0.50249000 Z=0.44537700 -2: X=0.04163500 Y=0.00249000 Z=0.05462300 -2: X=0.45836500 Y=0.99751000 Z=0.55462300Cu2 NPT= 781 R0=0.00010000 RMT= 1.90 Z: 29.0 LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000ATOM -3: X=0.63491000 Y=0.56252000 Z=0.52722000 MULT= 4 ISPLIT= 8 -3: X=0.86509000 Y=0.43748000 Z=0.02722000 -3: X=0.13491000 Y=0.93748000 Z=0.47278000 -3: X=0.36509000 Y=0.06252000 Z=0.97278000P 3 !
NPT= 781 R0=0.00010000 RMT= 1.43 Z: 15.0 LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000ATOM -4: X=0.77944000 Y=0.65778000 Z=0.47646000 MULT= 4 ISPLIT= 8 -4: X=0.72056000 Y=0.34222000 Z=0.97646000 -4: X=0.27944000 Y=0.84222000 Z=0.52354000 -4: X=0.22056000 Y=0.15778000 Z=0.02354000O 4 NPT= 781 R0=0.00010000 RMT= 1.43 Z: 8.0 LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000ATOM -5: X=0.53354000 Y=0.70607000 Z=0.39355000 MULT= 4 ISPLIT= 8 -5: X=0.96646000 Y=0.29393000 Z=0.89355000 -5: X=0.03354000 Y=0.79393000 Z=0.60645000 -5: X=0.46646000 Y=0.20607000 Z=0.10645000O 5 NPT= 781 R0=0.00010000 RMT= 1.43 Z: 8.0 LOCAL !
ROT MATRIX: 1.0000000 0.0000000 0.0000000 0.000
0000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000ATOM -6: X=0.60214000 Y=0.69187000 Z=0.72025000 MULT= 4 ISPLIT= 8 -6: X=0.89786000 Y=0.30813000 Z=0.22025000 -6: X=0.10214000 Y=0.80813000 Z=0.27975000 -6: X=0.39786000 Y=0.19187000 Z=0.77975000O 6 NPT= 781 R0=0.00010000 RMT= 1.43 Z: 8.0 LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000ATOM -7: X=0.62284000 Y=0.24448000 Z=0.52455000 MULT= 4 ISPLIT= 8 -7: X=0.87716000 Y=0.75552000 Z=0.02455000 -7: X=0.12284000 Y=0.25552000 Z=0.47545000 -7: X=0.37716000 Y=0.74448000 Z=0.97545000O 7 NPT= 781 R0=0.00010000 RMT= 1.43 Z: 8.0 LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 !
4 NUMBER OF SYMMETRY OPERATIONS 1 0 0 0.00000000 0 1 0 0.00000000 0 0 1 0.00000000 1-1 0 0 0.50000000 0-1 0 0.00000000 0 0 1 0.50000000 2-1 0 0 0.00000000 0 1 0 0.50000000 0 0-1 0.50000000 3 1 0 0 0.50000000 0-1 0 0.50000000 0 0-1 0.00000000 4 Any suggestion is greatly appreciated, and thanks for your help in advance! Best regards. Ming Wenmei
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