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Normally, refractive index of a medium is defined as

$$n=\frac c{v_\text{p}},$$

where $c$ is the speed of light in vacuum, and $v_\text{p}$ is the phase speed of light in the medium. Phase speed is defined as

$$v_\text{p}=\frac{\omega}k,$$

where $\omega$ is the frequency, and $k$ is the wavenumber of light (which is the norm of the wavevector).

But in a photonic crystal we don't have a single well-defined wavevector: we have quasi-wavevector, which is defined up to a vector of the reciprocal lattice. So, how do we then determine the refractive index? Do we simply take the wavevector from the first Brillouin zone and calculate the phase velocity from it? (If yes, how can we justify this?) Or do we actually have multiple refractive indices so that a light beam splits into several beams as if by a diffraction grating?

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In photonic crystals, the dielectric function $\varepsilon(r)$ plays the role of confining potential for the light, so the refractive index $n = \sqrt{\varepsilon(r)}$ is not uniform at least along one dimension. And in general, the dispersion relation $\omega(k)$ will deviate from the linear dispersion relation, especially near the Brillouin zone boundary. But for a small wavevector, near the zone center, the dispersion relation will be linear, because the light does not "see" the potential variation and thus it only sees an effective average dielectric constant (see the discussion on page 56 on this book, which is freely available to public).

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  • $\begingroup$ This is all fine, but it still doesn't address the ambiguity of wavevector vs quasi-wavevector in relation to the phase speed. $\endgroup$ – Ruslan Jun 10 at 15:18
  • $\begingroup$ @Ruslan, phase velocity is hard to define exactly because of the ambiguity you mentioned, even when the dispersion relation is approximately linear. You can add reciprocal lattice vector to $k$ and the slope does not change. It's much more helpful to think about group velocity in photonic crystals. $\endgroup$ – wcc Jun 10 at 15:28
  • $\begingroup$ @Ruslan, your last sentence in OP also hints toward why there is no unique phase velocity in a photonic crystal. There is no unique wavefront to reference and measure the phase, as the multiply scattered waves coherently sum up to make a periodic wavefunction, just like that of electron in a crystalline solid. $\endgroup$ – wcc Jun 10 at 15:31
  • $\begingroup$ But normal crystals (e.g. diamond) can also be viewed as photonic crystals: they are also periodic, they sum up multiply scattered waves as you say, but still they do have a well-defined phase speed, don't they? $\endgroup$ – Ruslan Jun 10 at 15:36
  • $\begingroup$ @Ruslan, or do you mean diamonds have a well-defined dielectric constant in an average sense as I described? Any dielectric material has well-defined group velocity but you can't say the same for phase velocity. And I am not sure diamonds can be viewed as a photonic crystal - the carbon atoms are only angstroms apart, so the light mostly sees an average uniform dielectric material. $\endgroup$ – wcc Jun 10 at 16:05
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I would not say that you have ill-defined wave-vectors. As you mention, commonly, when doing an analysis in the First Brillouin Zone we obtain the wavenumber modulo $\pi/(2 a)$, being $a$ the lattice parameter.

For a bilayer material, we have the following dispersion relation

$$\cos(\kappa a) = \cos\left(\frac{\omega a}{2 c_1}\right) \cos\left(\frac{\omega a}{2 c_2}\right) - \frac{c_1^2 + c_2^2}{2 c_1 c_2}\sin\left(\frac{\omega a}{2 c_1}\right) \sin\left(\frac{\omega a}{2 c_2}\right)\, .$$

If you plot the solutions over the first Brillouin zone you would obtain the following figure.

enter image description here

If instead, you invert the relationship and consider the shift for each branch you end up with the following figure.

enter image description here

Now, if you ask me, the refractive index is a material parameter and you would need to average to obtain it. Thus, it is a function of frequency and, in general, is anisotropic what leads to a tensor rather than a single scalar:

$$n = [n_{ij}(\omega)]\, .$$

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