18.369 Problem Set 3 Solutions
Problem 1: (10+15 points)
In both parts of this problem, we need to prove that the variational quotient hH, Θ̂k Hi / hH, Hi < k2 for some
trial function H, or equivalently that
Z aZ ∞
−∞
0
(1 − ∆) |(∇ + ik) × Hk |2 dx dy − k2
Z aZ ∞
0
−∞
|Hk |2 dx dy < 0
for the trial Bloch envelope Hk = He−ikx , k = kx̂, and ε −1 = 1 − ∆.
(a) We will choose u(x, y) = e−|y|/L for some L > 0,
exactly asR in class—that is, it is the simplest conceivR
able periodic function of x, a constant. Thus, |u|2 = 2a 0∞ e−2y/L dy = aL over the unit cell. In this
case, the variational criterion above becomes, exactly as in class except for the factor of a:
Z aZ ∞
0
(1 − ∆) k2 + L−2 e−2|y|/L dxdy − k2 aL
< 0
−∞
=
a
−
L
Z aZ ∞
0
∆ · k2 + L−2 e−2|y|/L dx dy,
−∞
∞
which becomes negative
in the limit L → ∞ thanks to our assumption that 0a −∞
∆(x, y) dx dy > 0.
R
Note that the fact that |∆| < ∞ ensures that we can interchange the limits and integration, via the
dominated convergence theorem.
R R
(b) Let us assume that we can choose u(y) and v(y) to be functions of y only (i.e., again the trivial constantfunction periodicity in x). The fact that ∇ · H = 0 implies that (∇ + ik) · [u(y)x̂ + v(y)ŷ] = 0 = iku + v0 ,
and therefore u = iv0 /k. Therefore, it is convenient to choose v(y) to be a smooth function so that u is
differentiable. Let us choose
2
2
v(y) = e−y /2L
R ∞ −y2 /L2
R ∞ 2 −y2 /L2
√
2
2
in which case u(y) = − kLiy2 e−y /2L . Recall the Gaussian integrals −∞
e
dy = L π and −∞
y e
dy =
R
R
√
√
L3 π/2. So, |H|2 = a |u|2 + |v|2 = aL π[1 + k21L2 ]. Also, (∇ + ik) × [u(y)x̂ + v(y)ŷ] = (ikv − u0 )ẑ.
So,
2 2
y2
1
e−y /L .
|∇ × H|2 = |(∇ + ik) × Hk |2 = |u0 |2 + k2 |v|2 = k2 1 + 4 4 1 − 2
k L
L
Then, if we look at our variational criterion, we have two terms: |∇×H|2 and − ∆·|∇×H|2 . Again,
we can swap limits withRintegration in the latter by the dominated convergence theorem. Combining
the former with the −k2 |H|2 term in the variational criterion, we get:
R
Z
2
|∇ × H| − k
2
Z
2
|H|
R
√
1
y2
1
−y2 /L2
2
k 1+ 4 4 1− 2
= a
e
dy − k aL π 1 + 2 2
k L
L
k L
−∞
√
Z ∞
2
2
2
2
2
k
y
k aL π
= a
1 − 2 e−y /L dy − 2 2
4 4
L
k L
−∞ k L
√
2
√
a
L
a π
=
L π 1− 2 −
,
k 2 L4
2L
L
Z ∞
2
which goes to zero as L → ∞. Thus:
Z
(1 − ∆) |(∇ + ik) × Hk |2 − k2
Z
as L → ∞. Q.E.D.
1
|Hk |2 → −k2
Z aZ ∞
∆(x, y) dx dy < 0.
0
−∞
Problem 2: (10+20 points)
(a) Trivially from the given identity,
a
J=
2π
Z 2π/a
"
∑
0
#
∞
δ (x − na)δ (y)e
ikna−iωt
ẑ dk,
n=−∞
where the term [· · · ] is Bloch-periodic. Because irrep is conserved, from class and homework, the
resulting steady-state/time-harmonic E field from each Bloch-periodic term is also Bloch-periodic
(from periodicity) and TM-polarized (from z ↔ −z mirror symmetry), i.e. the field of the current [· · · ]
is:
Ek (x, y,t) = Ek (x, y)eikx−iωt ẑ,
where Ek (x + a, y) = Ek (x, y). By linearity, we can simply add up the solutions Ek from the integrand
of J to get the total field by superposition:
E=
a
2π
Z 2π/a
0
Ek dk = ẑ
a
2π
Z 2π/a
0
Ek (x, y)eikx−iωt dk.
The field Ek satisfies (∇ × ∇ × − ω 2 ε)Ek = iωJ, and since both Ek and J are Bloch-periodic we
can trivially reduce the domain to the unit cell (x, y) ∈ [0, a] × (−∞, ∞) with Bloch-periodic boundary
conditions.
(b) To get the total power, we also need
H=
1
a
∇×E =
iω
2π
Z 2π/a
0
Hk (x, y)eikx−iωt dk,
where Hk = ∇k × (Ek ẑ) is a periodic function. Hence, H, like E, is a superposition of Bloch-periodic
R
0
functions. Because partners of different irreps are orthogonal, the dx of ŷ · (Ek ẑeikx )∗ × (Hk0 eik x )
must necessarily be zero unless k = k0 , hence the total power P will be a superposition of terms Pk that
are integrals of ŷ · (Ek ẑeikx )∗ × (Hk eikx ): the Poynting flux of one k at a time.
However, getting the exact form of this superposition, including the normalization, is a bit tricky
because of the infinite bounds of the x integral. One approach is to try and use the Fourier identity
R ∞ i(k−k0 )x
dx = 2πδ (k −k0 ). However, applying this requires a little bit of care because the envelopes
−∞ e
Ek and Hk are periodic in x, not constants. In order to perform the x integral and apply the Fourier
identity, we can expand the periodic function ŷ · (Ek ẑ)∗ × (Hk0 ) in a Fourier series:
∞
p(k, k0 , x) = ŷ · Ek (x)∗ ẑ × Hk0 (x) =
∑
p̂n (k, k0 )ei
2πn x
a
n=−∞
(where we have evaluated everything at y = y0 ). Plugging this in, we obtain:
P=
1
2
Z ∞
−∞
ŷ · ℜ [E∗ (x, y0 ) × H(x, y0 )] dx
1
a
= ŷ · ℜ
2
2π
Z 2π/a
1
a
= ŷ · ℜ
2
2π
Z 2π/a
0
0
a
dk
2π
Z 2π/a
a
dk
2π
Z 2π/a
dk0
Z ∞
−∞
0
dk
0
Z ∞
−∞
0
"
∞
∑
p̂n (k, k0 )e
i(k0 −k+ 2πn
a )x
#
dx
n=−∞
"
#
2πn
∑ p̂n (k, k )2πδ (k − k + a ) dx.
n=−∞
∞
0
0
Now, the delta function is only nonzero for n = 0 in the sum, since |k0 − k| < 2π/a for k, k0 ∈ [0, 2π/a).
Thus, we obtain:
Z 2π/a
a2 1
P=
ŷ · ℜ
p̂0 (k, k)dk,
2π 2
0
2
but since p̂0 (k, k0 ) =
1 Ra
∗
a 0 ŷ · Ek (x) ẑ × Hk (x)dx
P=
by the usual Fourier-series formula, we find
a
2π
Z 2π/a
0
Pk dk
where
1
Pk = ŷ · ℜ
2
Z a
0
[ẑEk (x, y0 )]∗ × [Hk (x, y0 )] dx
is the Poynting flux in a single unit cell from the Bloch solution for a single k.
Alternatively, another approach is to adopt the same trick that we used to deriving the per-period
LDOS in class: instead of an infinite system in the x direction, we consider a supercell of M periods,
with periodic boundaries x ↔ x + Ma, and take the M → ∞ limit in the end. As in the notes, for such
2π
m for integers m = 0, . . . , P − 1, or
a supercell we get a subset of the Bloch solutions, only km = Ma
equivalently one can easily show:
#
"
1 M−1 M−1
J=
∑ ∑ δ (x − na)δ (y)eikm na−iωt ẑ ,
M m=0
n=0
E = ẑ
1 M−1
∑ Ekm (x, y)eikm x−iωt
M m=0
for the same Ek solutions as above. The Poynting flux is then
1
2
Z Ma
ŷ · ℜ [E∗ (x, y0 ) × H(x, y0 )] dx
#∗ "
#
"
Z Ma
1 M−1
1
1 M−1
ikm x−iωt
ikm0 x−iωt
×
Hkm0 (x, y0 )e
dx.
= ŷ · ℜ
ẑ
∑ Ekm (x, y0 )e
2
M m=0
M m∑
0
0 =0
P=
0
As above, the m 6= m0 cross terms must integrate to zero (since they are partner functions of different
irreps of the symmetry group: translations by na for n = 0, . . . , M − 1). What remains is
1
1 M−1
P = ŷ · ℜ 2 ∑
2
M m=0
1
1 M−1
= ŷ · ℜ
∑
2
M m=0
=
Z Ma
0
Z a
0
[ẑEkm (x, y0 )]∗ × [Hkm (x, y0 )] dx
[ẑEkm (x, y0 )]∗ × [Hkm (x, y0 )] dx
1 M−1
∑ Pkm ,
M m=0
where we have used the periodicity of Ek and Hk and defined Pk as above. Finally, by multiplying and
2π
a R 2π/a
dividing by ∆k = Ma
as in class, we can take the M → ∞ limit to recover the integral P = 2π
Pk dk
0
as above.
Problem 3: (5+5+10+10 points)
(a) The 2dwaveguide.ctl file by default is already for a large enough k (0 to 2 · 2π/a) to get five even
modes (in fact, there are more, as we would see if we increased num-bands, and the result is shown
in figure 1 (left). If we double the size of the computational cell (from Y = 10 to Y = 20), then the
change is insignificant—we can’t even see on the regular graph. If we zoom in on the crossover point
for band 4, as in figure 1 (right), then we can see a slight change, but only to the point of the band that
3
2
0.91
1.8
Y=10
Y=20
light cone
0.909
1.6
1.4
0.908
ω (2πc/a)
ω (2πc/a)
1.2
1
0.907
0.906
0.8
0.6
0.905
0.4
0.904
0.2
0
0
0.2
0.4
0.6
0.8
1
k (2π/a)
1.2
1.4
1.6
1.8
2
0.904
0.906
0.908
0.91
kx (2π/a)
0.912
0.914
Figure 1: Band diagram (ω vs. k) for five even TM bands of εhi = 12 waveguide structure with thickness
h = a. Right: zoom in on point where band 4 enters the light cone, showing effect of increasing cell size
from Y = 10 (blue) to Y = 20 (red).
is above the light line. To get the crossover points, I increased the number of k points to k-interp=100,
and then interpolated the intersection points. The first band, of course, is guided starting at k = 0,
just as we predicted. The next three bands intersect the light line at ka/2π of 0.2979, 0.5954, and
0.8926,
p even mode whenever
p respectively. Our analytical prediction was that we would get a new
kh 1/ f − 1/2π was an integer, i.e. for kh/2π an integer multiple of 1/ 1/ f − 1. In this case,
h = a = 1, and f = 1/12, so we should get modes starting at ka/2π of 0.3015, 0.6030, and 0.9045.
This matches our numerical calculation to an accuracy of better than 2%, which is as good as we can
expect without increasing the number of k points, etc.
(b) If we plot the fields at ka/2π = 1 on a log scale in figure 2(left), we see that the amplitude decays as
a straigth line (thus, exponentially) at first, but then just becomes noisy. What is going on here? The
answer is twofold. First, MPB solves for the modes by an iterative process, optimizing the Rayleigh
quotient until some tolerance (by default, 10−7 ) in the eigenvalue is achieved. However, because very
small values of the field have little effect on the eigenvalue, MPB does not try to converge them, and
thus we see random tiny values once the field decays beyond a certain point. We can improve the
accuracy of the small field values by reducing this tolerance...if we run MPB with tolerance=1e-14,
we see the field shown in figure 2(right), which decays linearly (exponentially) for a much larger range
of y. It still has some noise at the boundaries, however, because the finite precision of floating point
arithmetic does not let us get more accurate than this in our computation. If we reduce the cell size,
however, so that it did not decay so much before reaching the boundary, however, we would see the
field flatten out to a minimum value at the boundary for even modes (or go to a node for odd modes),
due to the periodic boundary conditions.
(c) To put ε = 2.25 instead of air on the y < −h/2 side, we simply modify the geometry list to:
> (set! geometry
>
(list (make block (center 0 (/ Y -4) 0)
>
(size infinity (/ Y 2) infinity)
>
(material (make dielectric (epsilon 2.25))))
>
(make block (center 0 0 0)
>
(size infinity h infinity)
>
(material (make dielectric (epsilon eps-hi))))))
4
0
0
10
10
band 1
band 2
band 3
−1
10
band 1
band 2
band 3
−2
10
−2
10
−3
−4
10
|Ez| (arbitrary units)
|Ez| (arbitrary units)
10
−4
10
−5
10
−6
−6
10
−8
10
10
−7
10
−10
10
−8
10
−9
10
−5
−12
−4
−3
−2
−1
0
y (a)
1
2
3
4
5
10
−5
−4
−3
−2
−1
0
y (a)
1
2
3
4
5
Figure 2: Fields |Ez |for first three even TM modes of εhi = 12 waveguide, for Y = 10 cell at ka/2π = 1,
showing exponential decay (straight line on log scale) until “noise floor” caused by finite numerical error in
the fields is reached. Left: default (10−7 ) tolerance in MPB. Right: decreased (10−14 ) tolerance in MPB; we
can’t decrease the tolerance much further because of floating-point errors.
That is, we added another block, of ε = 2.25, before the waveguide block. The new block has width
Y /2, but where it overlaps with the waveguide the waveguide takes precedence (because it comes after
in the geometry list). Now: BE CAREFUL – the original 2dwaveguide.ctl file computed the y-even
and y-odd modes separately, but now there is no y = 0 mirror plane. We must just use (run-te) and
(run-tm).
Now, if we plot the TM and TE modes, it looks at first as if there is no cutoff for the fundamental
TM mode! This isn’t the case, however. The problem is that, for modes very near the light cone,
they become delocalized and our computational cell needs to be larger. If we increase the size to
Y = 40, and zoom in on the origin, we see that the first TM mode does indeed have a cutoff at around
ωa/2πc = 0.02 (whereas the TE cutoff is at a frequency around 0.1).
R
(d) RThis waveguide should have a TM (and a TE) guided mode for all values of k, because ∆(y) dy =
(1−1/ε) dy = (h/2)·(0.5−0.25) > 0, applying our variational proof from class and from problem 2.
To show this numerically, we look at the first TM guided mode, for small values of k: we change
num-bands to 1, kmax to 0.1, and k-interp to 200. (Similar to (c), above, we modify the geometry to
contain two blocks of thickness h/2.) Since we are looking at small ω (large λ ), we don’t need such
a high resolution and reduce resolution to 10; this will allow us to look at much larger computational
cells. Then, in figure 4, we plot ck − ω, how far we are above the light cone—this should be positive
for guided modes and should go to zero for k → 0. This is precisely what we see, plotting on a log-log
scale to see the power-law dependence.
However, we have to be careful: as k → 0, we must increase the computational cell size so that
the guided mode does not “see” the boundary. In particular, as we increase Y from 10 to 20 to ... to
1280, we see that ck − ω is indeed converging to a positive, decreasing function (at the right side of
the plot), whereas small k values (at the left side of the plot) are not yet converged, but the trend is
clear: it is going towards a steeper power-law decay (whereas if there were a cutoff ω the curve would
diverge towards −∞).
5
0.1
0.5
0.09
0.45
light cone of ε=2.25
0.08
0.4
0.07
0.35
0.06
ω (2πc/a)
ω (2πc/a)
0.3
0.25
0.05
0.04
0.2
0.03
0.15
0.1
0.02
0.05
0.01
0
0
0.1
0.2
0.3
0.4
0.5
k (2π/a)
0.6
0.7
0.8
0.9
0
1
0
0.01
0.02
0.03
0.04
0.05
k (2π/a)
0.06
0.07
0.08
0.09
0.1
Figure 3: TM (blue) and TE (red) bands of εhi = 12√waveguide with ε = 2.25 on one side, which causes a
cutoff in the modes. (The light cone here is ω ≥ ck/ 2.25.) At right is the same band diagram, but zoomed
in to show the cutoff for the first TM band, using a larger (Y = 40) cell.
−2
10
−3
10
Y=10
Y=20
Y=40
Y=80
Y=160
Y=320
Y=1280
−4
ck − ω (2πc/a)
10
−5
10
−6
10
−7
10
−3
−2
10
10
−1
10
k (2π/a)
Figure 4: The amount ck − ω by which the lowest TM mode of the εhi = {2, 0.8} system is below the
light line, as a function of k, on a log-log scale (straight line = power law). As we increase the size of
the computational cell Y from 10 to 1280, this curve decreases for small k (where the large wavelength is
strongly affected by a finite computational cell), and is clearly converging (at the right side of the plot) to a
steeper power-law decay.
6
80
70
60
gap size (%)
50
40
30
20
10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
thickness d1 / a
0.7
0.8
0.9
1
Figure 5: Gap size (% of mid-gap) for ε of 1 and 12 layers (period a) vs. thickness d1 /a of ε = 12 layer.
The maximum gap occurs at d1 ≈ 0.22a.
Problem 4: (10+5+5 points)
(a) To vary d1 from 0 to a, the easiest thing is to write a loop. One can write a loop in Scheme, modifying the bandgap1d.ctl file, but instead I sometimes find it easier to vary the parameters directly at the
Unix shell that modifies the parameters, using the mpb d1=0.3 command to modify the d1 parameter.
Using the bash shell (note that the default Athena shell is tcsh, however), one can do a sequence of
parameters and extract the results via:
(for d in ‘seq 0 0.01 1‘; do echo -n "$d, "; mpbi d1=$d bandgap1d.ctl |grep Gap
|grep "band 1" |cut -d, -f2 |tr -d ’%’; done) > p2a.dat
This results in the data shown in Figure 5, with the maximum gap
√ at about 0.22a, which is equal
(within our d1 resolution) to the quarter-wave thickness d1 /a = 1/( 12 + 1) ≈ 0.22401, as expected.
(b) At the quarter-wave thickness d1 , the mid-gap is at ωa/2πc = a/λ ≈ 0.32217 (as can be found either
from the MPB simulation or from the analytical formula discussed in class). Thus, a = 0.32217λ =
0.32217 · 1.55µm = 0.49936µm, and d1 = 0.22401a = 0.11186µm (d2 = a − d1 ).
(c) In figure 6 are shown the TM band diagrams for the quarter-wave d1 ≈ 0.22401a and the “random”
d1 = 0.12345a. (I increased num-bands to 10 to get this many bands, and increased k-interp to 19
to get smoother plots.) The quarter-wave bands have the following special features (besides having
the largest fundamental gap as already noted above). First, all of their gaps ∆ω are exactly the same
size (in absolute units, not as a percentage of mid-gap). Second, their mid-gap frequencies are spaced
exactly at the odd integer multiples of the first mid-gap! Third, the quarter-wave structure has no gaps
edges at k = 0, where the bands are accidentally degenerate. (The d1 = 0.12345a structure almost
has an accidental degeneracy at ka/2π = 0.5 between its third and fourth bands, but there is actually
a very small gap there.)
7
3
2.5
ω (2πc/a)
2
1.5
1
0.5
d1 = 0.12345a
quarter−wave
0
0
0.05
0.1
0.15
0.2
0.25
k (2π/a)
0.3
0.35
0.4
0.45
0.5
Figure 6: TM band diagram of ε = 12/1 multilayer film, with quarter-wave thickness d1 ≈ 0.22401a (blue,
solid) or thickness d1 = 0.12345a (red, dashed) of the ε = 12 layer. Quarter-wave gaps are shaded blue.
8