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What does 'Absolute Chemical Shift tensors' mean in NMR calculation

Posted: Tue Nov 24, 2020 1:01 pm
by xhbao502
Dear vasp users,

I'm trying to deal with chemical shift anisotropy using vasp 5.4.4. There's a part called 'Absolute Chemical Shift tensors' in OUTCAR, which contains 'UNSYMMETRIZED TENSORS' and 'SYMMETRIZED TENSORS'. However, I find that they are exactly the same when ISYM=0, while different when ISYM=2. For instance, for the example of SiO2 in wiki, the results are as follow (just list the first Si atom as an example):

ISYM=2:

Code: Select all

  UNSYMMETRIZED TENSORS 
 ion    1
        420.703998         -8.599732         -2.590616
         -6.545519        416.487888         -8.715890
         -0.948753         14.027251        430.167267
  SYMMETRIZED TENSORS 
 ion    1
        420.703998          0.000000          0.000000
          0.000000        416.487888         -8.715890
          0.000000         14.027251        430.167267
ISYM=0:

Code: Select all

  UNSYMMETRIZED TENSORS 
 ion    1
        420.704774          0.001908          0.004644
          0.001119        416.488022         -8.715877
          0.005821         14.027311        430.167785
  SYMMETRIZED TENSORS 
 ion    1
        420.704774          0.001908          0.004644
          0.001119        416.488022         -8.715877
          0.005821         14.027311        430.167785
I know that the chemical shift tensor (shielding tensor) should be decomposed into a symmetric and antisymmetric component. But none of the 4 tensors(ISYM=0 & 2, UNSYMMETRIZED & SYMMETRIZED) is symmetric or antisymmetric. I cannot figure out whether the tensor gave by vasp is the original one, symmetric one, antisymmetric one, or something else. Also, I'v no idea about what 'SYMMETRIZED' and 'UNSYMMETRIZED' mean and how ISYM influence them. I've searched it with google and no information was found. Do you have any idea about it? Thanks.

My input files:

KPOINTS:
G4*4*4

INCAR(exactly the same as vasp wiki):

Code: Select all

SYSTEM = Si O2
 
  GGA    = PE
  ISTART = 1
  ICHARG = 0
  INIWAV = 1
  ISYM   = 2
  ISPIN  = 1
  
Ionic minimisation
  NSW    = 0
  ISIF   = 2
  IBRION = 2
#  EDIFFG = -2E-2 
  POTIM  = 0.1
    
Electronic minimisation 
  IALGO = 38
  
LWAVE  = .TRUE.

 EMIN =   -20.0  
 EMAX =   10.0
 NEDOS = 1601

EFG Calculation
LEFG = .TRUE.
QUAD_EFG = 0.0 25.5

Chemical Shift
PREC = Normal      # nice
ENCUT = 400.0      # typically higher cutoffs than usual are needed
ISMEAR = 0; SIGMA= 0.1 # no fancy smearings, SIGMA sufficiently small
EDIFF = 1E-9       # you'd need much smaller EDIFFs.

LCHIMAG = .TRUE.   # to switch on linear response for chemical shifts
DQ = 0.001         # often the default is sufficient
ICHIBARE = 1       # often the default is sufficient
LNMR_SYM_RED = .TRUE. # be on the safe side
NSLPLINE = .TRUE.  # only needed if LREAL is NOT set.
LREAL = A          # helps for speed for large systems, not needed
NBANDS = 30        # to safe memory, ??? = NELECT/2

Re: What does 'Absolute Chemical Shift tensors' mean in NMR calculation

Posted: Mon Aug 23, 2021 6:04 am
by marie-therese.huebsch
Hi xhbao502,

Sorry, your post has remained unanswered for so long! Probably you have moved on by now, but let me answer as it might be helpful for other users.

I think your question rather concerns the ISYM tag and the general use of symmetry in VASP. The symmetry of the whole system is determined based on the structure defined in the POSCAR file (and if present the magnetic moments set by the MAGMOM tag). The symmetry analysis is written rather at the beginning of the OUTCAR file. The ground state may observe the same symmetry as the structure, so you can save a lot of computational time by exploiting the symmetry of the system. In other words, the Kohn-Sham orbitals, charge density, forces, and many other quantities are symmetrized to obey the same symmetry operations as the structure. The extent to which the symmetry is exploited is controlled by the ISYM tag.

In that sense, the symmetrized chemical shift tensor is "the original one" to put it in your words. When you want to have the symmetric or antisymmetric part of that tensor, you need to decompose it as a post-processing step.

Best regards,
Marie-Therese