Re: calculation of kmag

Oliver Willekens <[email protected]>
Newsgroups gmane.comp.science.photonic-bands
Message-ID <1260925754.14472.109.camel@cinnamoon>
Thank you very much. I had indeed lost track of the orientation of the
reciprocal lattice vectors (switched them with the real lattice
vectors). For the current simulations, then indeed kmag gives me a nice
cone when plotted versus k1 and k2 (transformed to an orthogonal basis
of course). 
Unfortunately it means I will have to rerun the simulations.

Thank you very much! :-)

On Tue, 2009-12-15 at 17:56 -0500, Steven G. Johnson wrote:
> On Dec 15, 2009, at 3:52 PM, Oliver Willekens wrote:
> > I wanted to plot the lightcone by plotting kmag versus k1 and k2 in a
> > triangular lattice. However, it does not look like a cone(segment).
> > Plotting it versus sqrt(k1^2+k2^2) works of course.
> 
> kmag is not sqrt(k1^2 + k2^2), except in a square lattice.  The reason  
> is that k1 and k2 are in the reciprocal basis which is non-Cartesian  
> in a non-square lattice, and hence the Pythagorean theorem does not  
> apply.

> In MPB, kmag is (vector3-norm (reciprocal->cartesian k))
> 
> > k1	k2	k3	kmag/2pi	(comment)
> > 0.3333	0.3333 	0	0.3849		(K-point)
> > 0.5000	0	0	0.5774		(M-point)
> >
> > However, in this basis, the magnitude of the vector Gamma-K should be
> > 1/sqrt(3), no?
> 
> No.  The correct answer is 2/3 / sqrt(3) = 0.384900179459751.
> 
> The reciprocal lattice vectors for your lattice are 2pi/a (1/sqrt(3),  
> +/- 1).  Hence the K point, in Cartesian coordinates, is the sum of  
> these two vectors times 1/3, or 2pi/a (2/sqrt(3), 0) / 3.  Hence the  
> length is 2/3 / sqrt(3) in units of 2pi/a.
> 
> See appendix B of our book if you need more explanation about  
> reciprocal lattices (http://ab-initio.mit.edu/book).
> 
> In general, when you plot kmag over the irreducible Brillouin zone  
> boundaries, the only portions that will be straight lines ("look like  
> a cone") will be the portions that correspond to straight lines from  
> the origin (Gamma).
> 
> Steven
> 
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