[Wannier] degenerate bands in group velocity

[Jonathan Yates]

On 2 Jul 2008, at 08:16, nafise rezaei wrote:

Nafise,

  Firstly, I think that the best way to validate your implementation  
is to compare the analytic (ie via MLWF) gradients against numeric  
gradients (ie via pwscf and a spline fit) along some path in the BZ.  
Such as figure 2 in my paper (although this is for curvatures).  
Obviously you want to understand the math as well, but I found this  
helpful for making sure things were correct.

>
> I read your "Spectral and Fermi surface properties from Wannier
> interpolation" paper.
> but I do not know exactly, what should I do at the degenerate
> states.?elements off-diagonal HH_alpha(k) matrix are zero at the
> non-degenerate band.Would you please you explain about elements of
> this matrix at the degenerate states? HH_alpha(k) matrix is sum of two
> term: HHbar_alpha(k) term and multiplication of HH(k) and
> DH_alpha(k).
>
> at the degenerate states second term is zero.

So the second (and third) term of Eqn 26 is zero for both degenerate  
and non-degenerate states.

> Are elements of first
> matrix (HHbar_alpha(k)) for m/=n zero?

In general they won't be zero (deg or non-degen)

> I wrote program for calculation of group velocity.at  the degenerate
> states elements of HHbar_alpha(k) are zero for (m/=n) and non-zero for
> (m=n) . is this true?

Might be... it depends....

  If you are _exactly_ at a degenerate point (2 bands cross), then  
the 2x2 matrix
H^H_\alpha is ill-defined. You can take any combination of those (2)  
states at that k and they are still eigenvectors of the Hamiltonian.  
The "correct" choice are the vectors which diagonalise the 2x2  
submatrix of H^H_\alpha, and the eigenvalues of that (2x2) matrix are  
the band gradients.

  The question is "in a numeric implementation, how close to a  
degenerate point do you have to be to use the degenerate theory". In  
my experience you rarely need to worry about degenerate points for  
band gradients. Take Fig 2. You will hit the degenerate point at  
gamma (unless you shift your kpoint grid) - but the gradients are  
both zero. The crossing between Gamma and K has zero measure -  
however fine I sampled I could not find it to sufficient accuracy  
that I needed to use the degenerate form - maybe this is what is  
happening in your case.
  As above: benchmark this against the numeric gradients to give you  
a feel for how sensitive these things are at degenerate points.

  If you want to compute curvatures then you need to worry more about  
degeneracies

  Jonathan


--
Dr Jonathan Yates         |    Theory of Condensed Matter Group
Corpus Christi College    |    Cavendish Laboratory
Cambridge, CB2 1RH, UK    |    Cambridge, CB3 OHE, UK
email jry20 at cam.ac.uk     |    Tel +44 (0)1223 337461