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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