On Mon, 12 Nov 2007, Robert Rehammar wrote:
> What I do not understand is why the symmetry is odd. It is not the case
> that eps(x, y) = -eps(x, -y) (eps is an odd function in y). Further E is
> a vector, so this quantity does not behave as a pseudovector under
> reflection, or?

The symmetry is the symmetry of the fields, not of the dielectric 
structure.

Of course, the symmetry of the dielectric structure affects the symmetry 
of the fields---in this case, the dielectric structure is even in y, which 
means that the fields can be either even or odd.  Which symmetry the 
fields have (if any) is determined by the choice of source.

In this case, you have an "Ey" (dipole) point source at y=0.  This is odd. 
Imagine an arrow pointing in the y direction.  If you mirror flip y to -y, 
then the arrow flips direction, which means that E points in the opposite 
direction (because E is a vector).  That is, the mirror flip of the source 
(and field) is the original source (and field) multiplied by -1, and hence 
it is odd.

People are often confused if they look at the components as individual 
scalar functions.  If E is odd with respect to y, then Ey(y) is an even 
function and Ex(y) and Ez(y) are odd functions.  That way lies madness, 
however:  you have to look at how the vector field transforms as a whole. 
This is the whole meaning of having a "vector" field, as opposed to a 
collection of random scalar function.  A vector field is one in which the 
components transform in related ways under rotations (they transform in 
the same way as the coordinates).  (A pseudovector like the magnetic field 
picks up an extra sign flip under mirror flips.)

In order to define a consistent meaning of "even" and "odd" for vector 
(and pseudovector) functions like E (and H), one has no choice but to look 
at the proper transformation rules for the field as a while.

See also chapter 3 of "Photonic Crystals: Molding the Flow of Light" by 
Joannopoulos et al.

Regards,
Steven G. Johnson

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