[Wien] Question on EFG and eigenvectors and sugestion for the output

[Martin Pieper]

Sorry for seing this somewhat late and not beeing able to stop myself from 
probably reintroducing  some confusion :-)

Personally, I am not too fond of the identification of the EFG eigensystem 
with the principal axis of an ellipsoid because the length of a principal 
axis is a positive number. This picture is convenient at first sight because 
(standard) NMR cannot determine the sign of V_zz (Moessbauer can in 
principal). But the picture does not describe a traceless tensor and I feel 
thats important here. 

I think one can see the ambiguity in the choice of directions of the 
eigenvectors if one considers the equation defining them: 

A*e_i=a_i*e_i

with some (square) matrix A, its eigenvalues a_i, and its eigenvectors e_i. 

If e_i satisfies this equation, so does any multiple s*e_i of it, and it does 
so with the same eigenvalue a_i. 

To be useful as a basis set one wants a normalized set (length 1), which 
leaves the options s=+/-- 1 (baring complex numbers). 

In addition one might want a right-handed eigenbasis.  Change of sign of all 
eigenvectors of a right-handed set gives a right-handed set and everything is 
ok. However, inverting just one e_i, or, equivalently, all except one (that 
is 2 for the EFG-tensor) results in a left-handed system. 

Stefaan is right, there is no stupid question in this field :-) I was unable 
to convince myself that the convention |V_z|>=|V_y|>=|V_x| necessary leads to 
a right handed system. 

If it does, or if one chooses to stay in a left-handed one if this convention 
implies it, I see only two possible choices for the orientation of the 
eigenbasis. The four choices mentioned by Florence Boucher would then be due 
to the ambiguity of the sign of V_zz in NMR - did I get this right?

Best regards,

Martin Pieper



Am Freitag, 7. Dezember 2007 08:48 schrieb Florent Boucher:
> Thank you Stephan for your kind help.
>
> Stefaan Cottenier a écrit :
> >> Now what about the eigenvectors or principle axis of the tensor.
> >> Is the direction of the  principle axis defined without ambiguity ?
> >> For instance, if I have an atom on a site with a local symmetry -1. I
> >> was assuming that the principle axis of the EFG tensor can be either in
> >> one direction or the opposite ?
> >> Am I true ?
> >> If not, why ?
> >
> > You can change the direction of the axis, yes. The problem of finding
> > the principle axis system for an EFG is mathematically equivalent to
> > finding the principle axes of a 3D ellipsoid. Obviously you can have
> > the freedom there as well to invert your axes.
> >
> > You can convince yourself about this by plotting a 1/r potential in
> > one dimension, as well as the corresponding field (=the first
> > derivative) and the gradient of the field (the second derivative of
> > the potential). The latter will turn out to be symmetrical with
> > respect to an inversion of the direction of your axis.
>
> I was arrived at the same conclusion as your (looking the EFG tensor
> like an ellipsoïd), I just wanted to be sure.
> It means that in such case there are at least four equivalent choices of
> eulerian angles to orient the EFG tensor in the cartesiane set of axis.
> The same is true for the NMR CSA tensor (that can be calculated with the
> GIPAW method).
> So, it can be conclude that they are 16 equivalent choices of Eulerian
> angles if you want to orient the EFG tensor with respect to the CSA one.
> I am not sure that people that use NMR and want to make comparison doing
> DFT calculations are aware of this specific point.
>
> Stefaan Cottenier a écrit :
> > Do you really think this is useful? It's a multiplication by
> > fundamental constants only (e/h), and the quantity that really
> > determines the value is Q, which you have to specify for any
> > individual case anyway...?
>
> On the last point for the output of lapw0, it is just a way to help
> users. If you don't do very often EFG calculations, you will have to
> remember every time how to convert the Vzz into a Cq quantity.
> I know that PWSCF, Paratec and Castep do the Cq calculation for you if
> you give Q in barn. I was just finding this option very convenient for
> the user (and not too complicated to implement...)
> Regards
> Florent

-- 
Dr. Martin W. Pieper
Institut f. Physik, 
Karl-Franzens Universität Graz
Universitätsplatz 5, 
A - 8010 Graz, AUSTRIA
Tel.: +43-316-380-8564, Fax: +43-316-380-9816,
email: martin.pieper at ifp.tuwien.ac.at