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.

\item {\verb+QUAD_EFG+}-tag (default: 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 ($Q$ is the element and isotope specific quadrupole moment): \[ C_q = \frac{e Q V_{zz}}{h} \] The \verb+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 $C_q$ is in MHz. See Ref.~\cite{pyykko} for a compilation of nuclear quadrupole moments.

Suppose a solid contains Al, C and Si, than the \verb+QUAD_EFG+-tag could read: \begin{verbatim} QUAD_EFG = 146.6 33.27 0 \end{verbatim} $^{27}$Al is the stable isotope of Al with a natural abundance of 100~\% and $Q = 146.6$. The stable isotopes $^{12}$C and $^{13}$C are not quadrupolar nuclei, however, the radioactive $^{11}$C is. It has $Q = 33.27$. For Si it is pointless to calculate a $C_q$: Again 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. \end{itemize}

Beware: for heavy nuclei inaccuracies are to be expected because of an incomplete treatement of relativistic effects.

References

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