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[求助]
wien2k计算磁性 已有1人参与
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| 我在进行FeSe的自洽计算时，初始化Se原子为无磁性，最终收敛到Se有0.06782的剩余磁矩，这是为什么呢？ |
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 FS P LATTICE,NONEQUIV.ATOMS: 4 53 Pmna MODE OF CALC=RELA unit=bohr 20.145136 10.433182 10.072568 90.000000 90.000000 90.000000 ATOM -1: X=0.87500000 Y=0.00000000 Z=0.50000000 MULT= 4 ISPLIT= 8 -1: X=0.62500000 Y=0.00000000 Z=0.00000000 -1: X=0.12500000 Y=0.00000000 Z=0.50000000 -1: X=0.37500000 Y=0.00000000 Z=0.00000000 Fe1 NPT= 781 R0=0.00005000 RMT= 2.31 Z: 26.0 LOCAL ROT MATRIX: 0.0000000 0.0000000 1.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 ATOM -2: X=0.87500000 Y=0.00000000 Z=0.00000000 MULT= 4 ISPLIT= 8 -2: X=0.62500000 Y=0.00000000 Z=0.50000000 -2: X=0.12500000 Y=0.00000000 Z=0.00000000 -2: X=0.37500000 Y=0.00000000 Z=0.50000000 Fe2 NPT= 781 R0=0.00005000 RMT= 2.31 Z: 26.0 LOCAL ROT MATRIX: 0.0000000 0.0000000 1.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 ATOM -3: X=0.00000000 Y=0.26880000 Z=0.75000000 MULT= 4 ISPLIT= 8 -3: X=0.50000000 Y=0.73120000 Z=0.25000000 -3: X=0.50000000 Y=0.26880000 Z=0.75000000 -3: X=0.00000000 Y=0.73120000 Z=0.25000000 Se1 NPT= 781 R0=0.00005000 RMT= 2.20 Z: 34.0 LOCAL ROT MATRIX: 0.0000000 0.0000000 1.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 ATOM -4: X=0.75000000 Y=0.26880000 Z=0.25000000 MULT= 4 ISPLIT= 8 -4: X=0.75000000 Y=0.73120000 Z=0.75000000 -4: X=0.25000000 Y=0.73120000 Z=0.75000000 -4: X=0.25000000 Y=0.26880000 Z=0.25000000 Se2 NPT= 781 R0=0.00005000 RMT= 2.20 Z: 34.0 LOCAL ROT MATRIX: 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 1.0000000 0.0000000 0.0000000 以上就是struct文件，不知道怎么贴图。。。 |
4楼2024-07-25 14:22:25