[Wien] Density matrix and QTL

[Peter Blaha]

> I am trying to understand how QTL works in performing its spectral
> decomposition of the density matrix. I could not find QTL - Technical
> report on the web anywhere so I hope you can help me.
>
> If we have a d-orbital density matrix, then the matrix is a 5x5
> corresponding to matrix elements:
> n_mm' = SUM_E<Ef < KS_k | phi_m> < phi_m' | KS_k>.
>
> Here, KS_k is a Kohn-Sham orbital and phi_m is a predefined orbital.
> Here with angular momentum m. I presume the quantization axis for m is
> set by the local rotation matrix? The summation is over energies less
> than the Fermi energy.

Yes, the quantization axis is defined by the crystal lattice and the 
local rotation matrices (with respect to the crystal axis).

However, we do NOT perform the next step indicated below. The density 
matrix is NOT rotated into diagonal form, but the "real" symmetries 
(like d-xy,...) are obtained by the well-known linear combinations from 
the complex density matrix components.

Using the program QTL you can now also define your "own" coordinate 
system and describe the partial charges with respect to this rotated 
frame. Usually, this is done to get the qtl's in an "approximate" 
symmetry frame (like a distorted octahedron,...) instead of the strict 
"highest symmetry" frame, which is the default.

Peter Blaha

>
> Then if we do a spectral decomposition for the QTL partial density of
> states we rewrite this matrix as:
> n_mm' = ( V * D * V^-1 )_mm'
> Here, D is a diagonal matrix and V is another square matrix whose
> columns should be the eigenvectors. We are now in a Hilbert space in
> which the occupations are diagonal.
> What I don't understand then is how these new eigenvectors are assigned
> to be dxy, dxz etc by QTL? Is it via a following projection? With
> respect to what spatial axis is the z-axis defined (still by the local
> rotation matrix in the struct file)?
> Also, if we set our own z-axis (and maybe x) for QTL does this simply
> alter the following projection?
>
> Finally, if it is easy to get the occupation matrix in a diagonal basis,
> why do codes use the rotationally invariant form?
>
> Many thanks,
> David.
>
>
>
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-- 
Peter Blaha
Inst.Materials Chemistry
TU Vienna
Getreidemarkt 9
A-1060 Vienna
Austria
+43-1-5880115671