# Photonic crystal fundamentals

1. Oct 11, 2009

### krindik

Hi,
I have this quite basic question regarding photonic crystals.
As I understand, a periodic dielectric structure (photonic crystal) has modes (spatial profiles) at discrete frequencies from:
$$\nabla \times \Big( \frac{1}{\epsilon(r)}\nabla \times H(r) \Big) = \frac{\omega^2}{c^2} H(r)$$

However, dispersion diagrams ($$\omega$$ vs. $$k$$) are continuous plots.

Doesn't each $$k$$ characterize a different mode?
I'm refering "Photonic Crystals Molding the Flow of Light" and believe it's the best book to learn about photonic crystals.

Really appreciate some guidance.

Thanks

2. Oct 11, 2009

### Manchot

Yes, each k represents a different mode (or more accurately, each k represents a single mode in each band). The dispersion relation is technically only continuous for infinite crystals, and if you consider finite crystals, the allowed k's become discretized. For example, if you were to apply periodic boundary conditions on a 1D crystal with a total of N unit cells of lattice constant a, you'd find that the k's are separated by a distance of $\frac{2 \pi}{N a}$.

So, you're right: a continuous dispersion relation makes no sense in the context of a finite crystal. To get the continuous dispersion relation, what you're really doing is trying to find solutions of the form $\mathbf{H}(r) = e^{i k \cdot r} \mathbf{u}_k(r)$ (where k can be continuous). If you plug that back into your eigenequation and solve for $\mathbf{u}_k$ under periodic boundary conditions on the unit cell, you get a Hermitian eigenproblem that is also k-dependent. Its solutions are countable, and correspond to different bands.

Practically speaking, the reason people solve for the continuous dispersion relation is that the boundaries of the crystal are usually far away, and that means that the discretization of k is usually negligible.

Last edited: Oct 11, 2009
3. Oct 11, 2009

### krindik

Thanks for the words of wisdom !
I was a bit confused in going from chapter 2 in this excellent book as nowhere did I read about the discrete nature of k: $\frac{2 \pi}{N a}$ in an non-infinite crystal.

I can understand the periodic nature of k in: $\mathbf{H}(r) = e^{i k \cdot r} \mathbf{u}_k(r)$ where the period number is called 'band'. (guess this is correct).
But does this imply the continuity of k within a period?
Or why does k has to be continuous even in an infinite crystal?
Does it mean any wavelength (excluding the band gap) can freely exist in the crystal?
Then again, the derivation of discrete frequencies was not based on infinite extent of the crystal.

Really appreciate if u could give some advice.

Thanks again.