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\begin_layout Section*
18.369 Problem Set 3
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\begin_layout Standard
Due Wednesday, March 6.
\end_layout

\begin_layout Subsection*
Problem 1: Bloch-periodic eigenproblems
\end_layout

\begin_layout Standard
Suppose that we have a periodic system with period 
\begin_inset Formula $a$
\end_inset

 in the 
\begin_inset Formula $x$
\end_inset

 direction, and we look for Bloch-periodic eigenfunctions 
\begin_inset Formula $\vec{H}(x+a,y,z)=e^{ika}\vec{H}(x,y,z)$
\end_inset

 of the 
\begin_inset Formula $\hat{\Theta}=\nabla\times\varepsilon^{-1}\nabla\times{}$
\end_inset

 operator with these boundary conditions in 
\begin_inset Formula $x$
\end_inset

, acting on a unit cell 
\begin_inset Formula $x\in[0,a]$
\end_inset

 (with some other boundary conditions in 
\begin_inset Formula $y$
\end_inset

 and 
\begin_inset Formula $z$
\end_inset

).
 (That is, we 
\emph on
don't
\emph default
 rewrite in terms of the periodic Bloch envelope and use 
\begin_inset Formula $\hat{\Theta}_{k}$
\end_inset

.)
\end_layout

\begin_layout Enumerate
Explain why 
\begin_inset Formula $\hat{\Theta}$
\end_inset

 is still Hermitian with these boundary conditions: when we integrated by
 parts, we had some boundary terms that we needed to vanish, and explain
 why the boundary terms from 
\begin_inset Formula $x=0$
\end_inset

 and 
\begin_inset Formula $x=a$
\end_inset

 still vanish with Bloch-periodic boundary conditions.
 (You can assume that the 
\begin_inset Formula $y$
\end_inset

 and 
\begin_inset Formula $z$
\end_inset

 boundary conditions were chosen so that those boundaries vanished.)
\end_layout

\begin_layout Enumerate
Why do 
\begin_inset Formula $k$
\end_inset

 and 
\begin_inset Formula $k+\frac{2\pi}{a}$
\end_inset

 give the same solutions to this Bloch-periodic eigenproblem? (Yes, we already
 discussed the periodicity of 
\begin_inset Formula $k$
\end_inset

 from other perspectives in class, but you should be able to see it directly
 here without reference to any of our previous arguments.)
\end_layout

\begin_layout Subsection*
Problem 2: Periodic waveguide guidance proof
\end_layout

\begin_layout Standard
In class, we showed by a variational proof that any 
\begin_inset Formula $\varepsilon(y)$
\end_inset

, in two dimensions, gives rise to at least one guided mode whenever 
\begin_inset Formula $\varepsilon(y)^{-1}=\varepsilon_{\textrm{lo}}^{-1}-\Delta(y)$
\end_inset

 for 
\begin_inset Formula $\int\Delta>0$
\end_inset

 and 
\begin_inset Formula $\int|\Delta|<\infty$
\end_inset

.
\begin_inset Foot
status collapsed

\begin_layout Plain Layout
As in class, the latter condition on 
\begin_inset Formula $\Delta$
\end_inset

 will allow you to swap limits and integrals for any integrand whose magnitude
 is bounded above by some constant times 
\begin_inset Formula $|\Delta|$
\end_inset

 (by Lebesgue's dominated convergence theorem).
\end_layout

\end_inset

 At least, we showed it for the 
\begin_inset Formula $H_{z}$
\end_inset

 polarization (
\begin_inset Formula $\vec{H}$
\end_inset

 in the 
\begin_inset Formula $\hat{\vec{z}}$
\end_inset

 direction).
 Now, you will show the same thing much more generally, but using the same
 basic technique.
\end_layout

\begin_layout Enumerate
Let 
\begin_inset Formula $\varepsilon(x,y)^{-1}=1-\Delta(x,y)$
\end_inset

 be a periodic function 
\begin_inset Formula $\Delta(x,y)=\Delta(x+a,y)$
\end_inset

, with 
\begin_inset Formula $\int|\Delta|<\infty$
\end_inset

 and 
\begin_inset Formula $\int_{0}^{a}\int_{-\infty}^{\infty}\Delta(x,y)dxdy>0$
\end_inset

.
 Prove that at least one 
\begin_inset Formula $H_{z}$
\end_inset

-polarized guided mode exists, by choosing an appropriate (simple!) trial
 function of the form 
\begin_inset Formula $\vec{H}(x,y)=u(x,y)e^{ikx}\hat{\vec{z}}$
\end_inset

 .
 That is, show by the variational theorem that 
\begin_inset Formula $\omega^{2}<c^{2}k^{2}$
\end_inset

 for the lowest-frequency eigenmode.
 (It is sufficient to show it for 
\begin_inset Formula $|k|\leq\pi/a$
\end_inset

, by periodicity in 
\begin_inset Formula $k$
\end_inset

-space; for 
\begin_inset Formula $|k|>\pi/a$
\end_inset

, the light line is not 
\begin_inset Formula $\omega=c|k|$
\end_inset

.)
\end_layout

\begin_layout Enumerate
Prove the same thing as in (a), but for the 
\begin_inset Formula $E_{z}$
\end_inset

 polarization (
\begin_inset Formula $\vec{E}$
\end_inset

 in the 
\begin_inset Formula $\hat{\vec{z}}$
\end_inset

 direction).
 Hint: you will need to pick a trial function of the form 
\begin_inset Formula $\vec{H}(x,y)=[u(x,y)\hat{\vec{x}}+v(x,y)\hat{\vec{y}}]e^{ikx}$
\end_inset

 where 
\begin_inset Formula $u$
\end_inset

 and 
\begin_inset Formula $v$
\end_inset

 are some (simple!) functions such that 
\begin_inset Formula $\nabla\cdot\vec{H}=0$
\end_inset

.
\begin_inset Foot
status collapsed

\begin_layout Plain Layout
You might be tempted, for the TM polarization, to use the 
\begin_inset Formula $\vec{E}$
\end_inset

 form of the variational theorem that you derived in problem
\begin_inset space ~
\end_inset

1, since the proof in that case will be somewhat simpler: you can just choose
 
\begin_inset Formula $\vec{E}(x,y)=u(x,y)e^{ikx}\hat{\vec{z}}$
\end_inset

 and you will have 
\begin_inset Formula $\nabla\cdot\varepsilon\vec{E}=0$
\end_inset

 automatically.
 However, this will lead to an inequivalent condition 
\begin_inset Formula $\int(\varepsilon-1)>0$
\end_inset

 instead of 
\begin_inset Formula $\int\Delta=\int\frac{\varepsilon-1}{\varepsilon}>0$
\end_inset

.
\end_layout

\end_inset


\end_layout

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