Electric Field Gradient

Electric field gradients at the positions of the atomic nuclei can be calculated by VASP using the method described in reference .

The following flags control the behaviour of VASP (the given value is the default value):

• LEFG=.FALSE. This tag switches on the calculation of the electric field gradient tensors (LEFG=.TRUE.). The EFG tensors are symmetric. The principal components $V_{ii}$ and asymmetry parameter $\eta$ are printed out for each atom. Following convention is made for the principal components $V_{ii}$:

$|V_{zz}| > |V_{xx}| > |V_{yy}|$

The asymmetry parameter $\eta = {(V_{yy} - V_{xx})}/ V_{zz}$. For so-called "quadrupolar nuclei", i.e. nuclei with nuclear spin $I > 1/2$, NMR experiments can access $V_{zz}$ and $\eta$.

Beware: Attaining convergence can require somewhat smaller EDIFF parameters than the default of EDIFF=1.e-4 and a somewhat larger cutoff ENCUT than the default one with PREC=A. Moreover, the calculation of EFGs typically requires high quality PAW data sets. Semi-core electrons can be important (check the POSCAR files with *_pv or *_sv) as well as explicit inclusion of augmentation channels with $d$-projectors.

• QUAD_EFG=1 This tag allows the conversion by VASP of the $V_{zz}$ values into the $C_q$ often encountered in NMR literature. The conversion formula is given as follows($Q$ is the element and isotope specific quadrupole moment):

$C_q = \frac{e Q V_{zz}}{h}$

The QUAD_EFG tag consists of the nuclear quadrupole moment in millibarns for each atomic species, in the same order as in the POTCAR file. The output of $C_q$ is in MHz. See reference for a compilation of nuclear quadrupole moments.

Suppose a solid contains Al, C and Si, than the QUAD_EFG tag could read:

QUAD_EFG = 146.6 33.27 0

$^{27}\mathrm{Al}$ is the stable isotope of Al with a natural abundance of 100% and $Q = 146.6$. The stable isotopes $^{12}\mathrm{C}$ and $^{13}\mathrm{C}$ are not quadrupolar nuclei, however, the radioactive $^{11}\mathrm{C}$ is. It has $Q = 33.27$. For Si it is pointless to calculate a $C_q$ since all stable isotopes have $I \le 1/2$. No moments are known for the other isotopes.

Beware:

• Several definitions of $C_q$ are used in the NMR community.
• For heavy nuclei inaccuracies are to be expected because of an incomplete treatment of relativistic effects.

Related Tags and Sections

LEFG

Examples that use this tag

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