# What is the significance of the difference in the eigenvalue equations of Bloch functions for electrons vs photons?

any text on photonic crystals will highlight the almost perfect analogy between electrons in a periodic potential and photons in a periodic dielectric. The analogies are:

$$V(\vec r + \vec R) = V(\vec r) \Leftrightarrow \epsilon(\vec r + \vec R) = \epsilon(\vec r)$$

$$\psi(\vec r)=e^{i\vec k\cdot \vec r}u_n(\vec r) \Leftrightarrow \vec H(\vec r)=e^{i\vec k\cdot \vec r}\vec u_n(\vec r)$$

But they have a slight difference in their respective operators in their eigenvalue equations:

$$(\frac{-\hbar^2}{2m}\nabla^2 + V(\vec r))\psi = E\psi \Leftrightarrow \nabla^2\vec H= \epsilon(\vec r) (\frac{\omega}{c})^2\vec H$$

That is, the QM version adds the periodic part, whereas the EM version multiplies it. Why is this, and is it significant?

I don't know enough about differential equations, so maybe it's very insignificant and you could represent either in either way. If I had to guess, I'd say it's related to the fact that electrons and photons have different $k$ dependencies in their dispersion relations ($k^2$ and $k$), but I don't see how the two are related.

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The master equation used in metamaterial calculations is usually phrased in its most general form as the eigenvalue equation $$\Theta \mathbf{H}=\left(\frac{\omega}{c}\right)^2\mathbf{H}$$ where $$\Theta=\nabla\times\frac{1}{\epsilon(\mathbf{r})}\nabla\times$$ is a linear operator which typically has some translational symmetry property due to some sort of periodicity in $\epsilon(\mathbf{r})$.

Likewise, the Schrodinger equation is of the form $$H\psi=E\psi$$ where $$H=T+V(\mathbf{r})$$ is a linear operator which typically has some translational symmetry property due to some sort of periodicity in $V(\mathbf{r})$.

In that sense, both situations involve computing the eigenstates of a translation-invariant operator, either $\Theta$ or $H$ respectively. You mentioned that

the QM version adds the periodic part, whereas the EM version multiplies it

but if you look at it closely, the QM version does actually multiply the periodic part once you distribute out the parenthesis, in the form of $V(\mathbf{r})\psi$. So the situations are nearly exactly the same, it's just that the shift-invariant operators are different.

As for why $\Theta$ and $H$ take the specific forms they do, the master equation is derived straight from Maxwell's equations, and Schrodinger's equation is of course obtainable from a number of approaches.

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Hi, the potential obviously multiplies the wave function in the QM case, but the difference is that its modification to the eigenvalue is subtractive, rather than multiplicative. –  YungHummmma Feb 26 '14 at 3:03
@YungHummmma: I'm not totally sure what you mean by the phrase "modification to the eigenvalue". By the change of expression $V_E(\mathbf{r})= E-V(\mathbf{r})$, the Schrodinger equation can be rewritten in the form $T\psi=V_E\psi$, where $V_E$ has the translation symmetry of the system. Rewritten in this form, the Schrodinger equation has the exact same form as your expression for the master equation, so you could argue that it's "modification to the eigenvalue" is also multiplicitive, once written in that form. –  DumpsterDoofus Feb 26 '14 at 3:42
Actually, maybe my previous comment didn't make sense. But in that case, I don't really know how to answer the question other than to simply say "the electron and photon equations are different because Schrodinger's equation and Maxwell's equations are different". They come about from a description of two different phenomena, one describing the movement of charges and one describing wavefunctions, so there's no reason to expect them to behave exactly the same. Maybe there's some deeper mathematical significance, but I don't know it; maybe ask Lubos? –  DumpsterDoofus Feb 26 '14 at 3:51