2
$\begingroup$

I am reading "Photonic crystals - molding the flow of light" by Joannopoulos et al. (available on-line). The figures below are reproduced from there.

This is a diagram of a triangular lattice of air columns drilled in a dielectric, and a corresponding band diagram: Triangular columns in dielectric Triangular lattice band diagram

The reciprocal lattice vectors are, in terms of real unit vectors:

$$\vec{b}_1=(2\pi / a)(\hat{x}+\hat{y}/ \sqrt{3}), \vec{b}_2=(2\pi / a)(\hat{x}-\hat{y}/\sqrt{3}),$$ as depicted below:

Real and reciprocal lattices

According to MIT Photonic Bands tutorial, symmetry points M and K, as seen in the band diagram, can be constructed as follows:

$$ M = \frac{1}{2}\vec{b}_2, K = \frac{1}{3} (-\vec{b}_1+\vec{b}_2) $$

The above formulation of the symmetry points in the irreducible Brillouin zone is not obvious to me, so bonus points for explaining them in the answer. However, my main question is:

Does it make physical sense to convert between the symmetry points in the reciprocal lattice and the directions in the real lattice? If so, how to make this conversion?

In the concrete example above, $M=\frac{1}{2}\vec{b}_2=\frac{2\pi}{a}\vec{a}_2$. If an optical mode propagates in the above structure with a frequency of $0.51\frac{\omega a}{2 \pi c}$ (i.e. on the TE band edge, see band diagram above), does it mean that it can only propagate along $\vec{a}_2$? And if I want to find out what symmetry point corresponds to $\hat{x}$, should I look at $\vec{b}_1+\vec{b}_2=4 \pi a \hat{x}$ and an equivalent point in the irreducible zone, which would be K in this case?

$\endgroup$
3
$\begingroup$

Does it make physical sense to convert between the symmetry points in the reciprocal lattice and the directions in the real lattice? If so, how to make this conversion?

No. You need to convert from reciprocal vectors back to real-space vectors. You've already provided the formulae required.

In your specific example, you state that $M=\frac{1}{2}\vec{b}_2=\frac{2\pi}{a}\vec{a}_2$, but in the second image, you can pictorially see that this is not true. Notice the $\hat{x}$ and $\hat{y}$ directions are defined by the Cartesian axes in the second picture. It applies to each of the 3 sub-plots. If you were to translate $\vec{b_2}$ from the second to the first sub-plot, you'd see they do not point in the same direction. Mathematically you would see this if you expand the lattice vector expressions and $\bf{M}$ in terms of the Cartesian coordinates: $$\vec{b_2} = \frac{2 \pi}{a}\hat{x} - \frac{2 \pi}{\sqrt{3}a}\hat{y}$$ $$\vec{a_2} = \frac{a}{2}\hat{x} - \frac{a\sqrt{3}}{2}\hat{y}$$

The $\bf{LHS}$ of given equation for $\textbf{M}$:

$$\textbf{M} = \frac{1}{2}\vec{b}_2 = \frac{1}{2} (\frac{2 \pi}{a}\hat{x} - \frac{2 \pi}{\sqrt{3}a}\hat{y}) = \frac{\pi}{a}\hat{x} - \frac{\pi}{\sqrt{3}a}\hat{y}$$

And the $\textbf{RHS}$ of given equation for $\textbf{M}$:

$$\textbf{M} = \frac{2\pi}{a}\vec{a}_2 = \frac{2\pi}{a} (\frac{a}{2}\hat{x} - \frac{a\sqrt{3}}{2}\hat{y}) = \pi\hat{x} - \pi\sqrt{3}\hat{y}$$

You can see $\textbf{LHS}\neq\textbf{RHS}$.

Wave direction is given by its wave vector direction. (Note: The wave vector is also the wave's momentum vector, which is the sum of one or more reciprocal unit vectors multiplied by their scalars). The wave vector direction is defined in terms of the reciprocal space vectors, here $\vec{b_1}$ and $\vec{b_2}$. So, if a wave has a momentum vector only in $\vec{b_2}$, the wave is traveling in that direction in real space too. You can use the Cartesian vector expression of $\vec{b_2}$ to get its direction components in real space.


If an optical mode propagates in the above structure with a frequency of $0.51\frac{\omega a}{2 \pi c}$... does it mean that it can only propagate along ...

Actually no. Waves with momentum exactly equal to the band edge are not traveling waves, but are known as standing waves. Read on to see why. If you had chosen another frequency above the gap, the direction would not necessarily be the same because this is a folded band diagram. You'd need to add $0.5$ to whatever value was indicated on the abscissa to account for the extra momentum from a neighboring Brillouin zone, which give you the total momentum at that frequency. The higher order bands actually are extensions from neighboring Brillouin zones. E.g., a frequency of $0.55\frac{\omega a}{2 \pi c}$ indicates $~\frac{2}{3} \Gamma M = \frac{1}{3}\vec{b_2}$, which can't be the total momentum, because that is momentum at $~0.28\frac{\omega a}{2 \pi c}$ also, and these are not the same modes. There is an extra component of either $-\frac{1}{3}\vec{b_2}$ from the neighboring $\Gamma M$ zone, or there is a component in the $M K$ zone, which would mean a direction off axis from the reciprocal space unit vectors.

So, right at the edge of the first Brillouin zone, a wave vector with momentum exactly equal to $\Gamma M$ is equivalent to a wave vector with 0 momentum, because you can subtract 0.5 and end up with the same frequency solution.

Generally, if a horizontal line defined by some frequency intersects multiple bands at multiple points, that is how many momentum vectors solve system's eigen-problem to give that eigenvalue. On this diagram, for TE bands, the momentum (and thus direction with a factor of $\pm1$ due to negative and positive folding) are only exactly known at the frequency you chose, from about $~0.34-0.34\frac{\omega a}{2 \pi c}$, and around $~0.76\frac{\omega a}{2 \pi c}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.