Re: symmetry questions

On Sat, 16 Dec 2006, John Chang wrote:
>  1) You used a rhombus shape unit cell which uses the centers of four 
> nearby holes as the unit cell's four corners. This unit cell doesn't 
> have any rotational symmetry. Will the result loss any symmetry 
> property, for example, degenerate modes?

The choice of a rhombus unit cell doesn't inherently break the symmetry.

However, numerically, there is always some broken symmetry due to the 
discretization and due to floating-point rounding error.  So, in a 
numerical eigensolver code you will essentially never see exactly 
degenerate modes.

>  I would think that using different unit cells should yield the same 
> result.

That's true, but this is a small numerical effect.  It's important to 
distinguish numerical effects from the exact solution (at infinite 
resolution and infinite precision), which is insensitive to the choice of 
unit cell (i.e. the choice of primitive lattice vectors).

> with only one hole in the center. Could you please explain why your 
> rhombus shape unit cell won't loss any symmetry property? Thanks.

Defining a unit cell corresponds to a chioce of primitive lattice vectors. 
The choice of primitive lattice vectors does not break the symmetry (in 
the exact solution) because they still define the same lattice once you 
include their linear combinations, which has the same rotational symmetry

> I read page 67 in "photonic crystals" by Joannopoulos. The book said the 
> mirror-reflection symmetry is still intact for kz=0, TE and TM modes 
> still decouple in in-plane propagation. Why will nonzero kz break 
> mirror-reflection symmetry?

Because nonzero kz makes makes z > 0 different from z < 0.  Read the book 
again, it explains why k must be preserved by the mirror plane in order to 
get pure TE/TM modes.

> In analogy to step index fiber, vector wave analysis does have non zero 
> kz for TE and TM modes, right? Is there any fundamental reason TE and TM 
> can't exist with nonzero kz?

See above.  If you are used to step-index fibers, they may mislead you: 
such fibers typically have very low index contrast, which allows them to 
be accurately described by a scalar-wave approximation where the modes are 
*approximately* purely polarized.  This approximation is not accurate in 
general.

(There is one important exception, where you can have a *different* 
definition of "TE" and "TM" modes that *strictly* holds even when k breaks 
all mirror symmetry: metallic waveguides with homogeneous, isotropic,
dielectric interiors.)

Steven