LEFG: Difference between revisions

Revision as of 14:50, 27 February 2025

LEFG = .TRUE. | .FALSE.
Default: LEFG = .FALSE.

Description: The LEFG computes the Electric Field Gradient at positions of the atomic nuclei.

For LEFG=.TRUE., the electric field gradient tensors at the positions of the atomic nuclei are calculated using the method of Petrilli et al. ^[1].

The EFG tensors are symmetric. The principal components V_ii and asymmetry parameter η are printed for each atom. Following convention the principal components V_ii are ordered such that:

{\displaystyle |V_{zz}| > |V_{xx}| > |V_{yy}|. }

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

{{{2}}}

To convert the V_zz values into the C_q often encountered in NMR literature, one has to specify the nuclear quadrupole moment by means of the QUAD_EFG-tag.

The output of $C_q$ is in MHz. See references ^[2] and Ref. ^[3] for a compilation of nuclear quadrupole moments.

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

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

QUAD_EFG = 146.6 33.27 0.0

$^{27}\mathrm{Al}$ 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.

for heavy nuclei inaccuracies are to be expected because of an incomplete treatment of relativistic effects.

References

[petrilli:prb:1998-1] H. M. Petrilli, P. E. Blöchl, P. Blaha, and K. Schwarz, Electric-field-gradient calculations using the projector augmented wave method, Phys. Rev. B 57, 14690 (1998).

[pyykko:molphys:2008-2] P. Pyykkö, Year-2008 nuclear quadrupole moments, Mol. Phys. 106, 1965-1974 (2008).

[pyykko:molphys:2017-3] P. Pyykkö, Year-2017 nuclear quadrupole moments, Mol. Phys. 116, 1328-1338 (2018).

[1]

[2]

[3]

@@ Line 13: / Line 13: @@
 For so-called "quadrupolar nuclei", ''i.e.'', nuclei with nuclear spin I>1/2, NMR experiments can
 access ''V''<sub>zz</sub> and η.
+{{NB|Tip|Attaining convergence can require somewhat smaller {{TAG|EDIFF}} than the default of <tt>1.e-4</tt>
-'''Beware''': Attaining convergence can require somewhat smaller {{TAG|EDIFF}} than the default of <tt>1.e-4</tt>
 and somewhat larger cutoff {{TAG|ENCUT}} than default with {{TAG|PREC}}=A. Moreover, the calculation of
 EFGs typically requires high quality PAW data sets. Semi-core electrons can be important (check with
-<tt>*_pv</tt> or <tt>*_sv</tt> POTCARs) as well as explicit inclusion of augmentation channel(s) with ''d''-projectors.
+<tt>*_pv</tt> or <tt>*_sv</tt> POTCARs) as well as explicit inclusion of augmentation channel(s) with ''d''-projectors.}}
 To convert the ''V''<sub>zz</sub> values into the ''C''<sub>q</sub> often encountered in NMR literature, one has to specify the nuclear quadrupole moment by means of the {{TAG|QUAD_EFG}}-tag.
@@ Line 31: / Line 30: @@
 <math>^{27}\mathrm{Al}</math> is the stable isotope of Al with a natural abundance of 100% and <math>Q = 146.6</math>. The stable isotopes <math>^{12}\mathrm{C}</math> and <math>^{13}\mathrm{C}</math> are not quadrupolar nuclei, however, the radioactive <math>^{11}\mathrm{C}</math> is. It has <math>Q = 33.27</math>. For Si it is pointless to calculate a <math>C_q</math> since all stable isotopes have <math>I \le 1/2</math>. No moments are known for the other isotopes.
-'''Beware''': for heavy nuclei inaccuracies are to be expected because of an incomplete treatment of relativistic effects.
+{{NB|Important|for heavy nuclei inaccuracies are to be expected because of an incomplete treatment of relativistic effects.}}
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