Three Dimensions

In three dimensions, in contrast with 2D, there exist localized solutions to the 3D wave-equation.For example, for a wavepacket propagating in the x-direction, of the form f (x, y, z, t) =

f1(x− ct) f2(y) f3(z), the wave-equation reduces to

for which non-zero solutions exist.However, we do not proceed with solving this equation, because it is not trivial, and it would only give us one of the fields.It is more convenient to simply extend the method of the previous section, and prepare a wavepacket by expanding a reference function in eigenmodes of the TDME of a cavity.

In three dimensions, the eigenmodes of a cavity with size Lx, Ly, Lz are the sets of coefficients ciand di.Omitting the nml indices, they yield

d1ω =klc2− kmc3, (A.34a) one of the fields, for example

Hx(x, y, z, t) = sin(k(x−x0−ct)) exp[−((x−x0−ct)/σx)¹⁰−((y−y0)/σy)²−((z−z0)/σz)²]. (A.35) The coefficients anml and bnml can be found by integration, and the other fields are recon-structed by carrying out the summations.When the reference function decomposes, for example f1(x− ct) f2(y) f3(z), and we choose Hx to equate this function, we would have to

A.2 Expansion in eigenmodes of a cavity 109

Figure A-3: Example of a wavepacket in three dimensions.System size is L= 12 × 12 × 12, with a mesh spacingδ = 0.1.Left: energy distribution density slice at y = 6 of the initial wavepacket, with widthsσ = (3, 0.75, 0.75)^T, centered at r0= (3.5, 6, 6)^T and wave-number k= 6.Right: energy distribution at t = 6.4, after integration with the Chebyshev algorithm (κ = 10⁻¹³).

The integrals in the non-propagating directions can be evaluated analytically by choosing f2(y) and f3(z) appropriately.The expressions for the summations in Eq.(A.32) do not simplify however, due to the presence of the ci(nml) and di(nml) coefficients, that couple all the modes.

A wavepacket, constructed by starting from Eq.(A.35) and reconstructing the other fields from the eigenmodes, is shown in figure A-3.As in the 2D case, the divergence of the wavepacket is clearly visible.

Appendix B

Density of states

The time-evolution of the EM fields yields the energy density distribution of each of the fields, allowing the measurement of various quantities, such as the scattering angle distri-bution.However, in some cases it is important to know the density of states (as a function of frequency) for a given system.In this appendix, we explain how the time evolution of the EM fields can be used to compute the density of states, or, eigenvalues of H, the discretized form ofH.In general, H is a (very) large matrix, usually too large to be stored in mem-ory.If only a few, well-separated eigenvalues of H are required, sparse-matrix techniques can be used to compute these eigenvalues [17, 26].However, if one is interested in global features of the distribution of eigenvalues, i.e. if we want to determine all eigenvalues, time-domain algorithms offer several advantages.In fact they are at the heart of so-called “fast”

algorithms to compute the density of states (DOS), defined as D(ω) =

n

δ(ω − ωn), (B.1)

or, using dispersion relationω = ω(k), D(ω) =L

2π 3

V

dkδ(ω − ω(k)), (B.2)

and other related quantities [95–100].

The basic idea of this time-domain approach was laid out by Alben et al.[95] who used it to compute the DOS of models for one electron moving in a disordered alloy.Applied to the case of the EM fields, the main idea is that a random initial condition contains all the possible eigenvalues, albeit with different amplitudes, some of which might be zero.These modes can be extracted by performing the Fourier transform of the function f (t), defined as

f (t)≡



V

dr (εE(0) · E(t) + µH(0) · H(t)) = Ψ(t)|Ψ(0) (B.3) Denoting the (unknown) eigenvalues and (unknown) eigenvectors of H by iωn and φn

respectively, and assumingΨ to be normalized, we can expand f (t) in eigenmodes

f (t)= Ψ(0)|Ψ(t) = Ψ(0)|e^tHΨ(0) =

where K (K = nx for 1D, K = 3nxny/4 for 2D, K = 3nxnynz/4 for 3D) is the dimension of the vector space on which H acts.From Eq.(B.4) it follows immediately that the Fourier transform of f (t) contains the information on all eigenvalues,

F (ω) =

This expression is very similar to the definition of the DOS.We use uncorrelated random numbers to initialize the elements of Ψ(0).It can be shown that for f (t), which is the average of f (t) over different realizations of the random initial state,F (ω) will correspond to the density of states

D(ω) =

 _∞

−∞dte^iωtf (t). (B.6)

It is often expedient to consider, the integrated density of states N(ω) =

 _ω

−∞D(u) du . (B.7)

The information on the eigenvalues of H, obtained through the use of a time-domain method is intimately related to the unconditional stability of the latter.As f (t) is band-limited, with frequencies ωn in the interval [− H , H ] (where H denotes the largest eigenvalue of H in absolute value), it follows from Nyquist’s sampling theorem that it is sufficient to sample f (t) at regular intervals ∆t = π/ H .If M denotes the number of data points used to sample f (t), frequenciesωn that differ less than ∆ω = π/M∆t are indistinguishable (although they will all contribute to the DOS).Extending the length of the time integration by a factor two increases the resolution in frequency by a factor two.This is a rather efficient and flexible procedure to trade accuracy for computational resources.

One may object that by integrating the TDME over longer and longer times (larger and larger M) the error on Ψ(t) will increase, possibly leading to no gain in accuracy at all.

However, it has been shown rigorously [101] that the error on the eigenvalues of H vanishes as τ²/M if one uses unconditionally stable algorithms based on the second-order Suzuki product-formula.The proof given in Ref.[101] applies to the fourth-order algorithm T2S4 as well, the exponent ofτ being four instead of two.Furthermore, the statistical error on

f (t) vanishes as 1/√

S K where S is the number of statistically independent samples of Ψ(0) [101].The fact that the statistical error decreases with the number of lattice points K/2 gives a tremendous boost to the efficiency of the method.

B.1 Scaling the density of states

Using definition (B.1), the total number of eigenvalues is K:

N(∞) =

But, if we use Eq.(B.6), then we obtain N(∞) =

B.1 Scaling the density of states 113

Hence, in the Fourier transform of f (t), the total number of eigenvalues is lost and must be manually recovered to assign physical significance to the amplitude of the DOS.Note that if we only need to know the location of the eigenvalues, for example to determine whether or not a system has a PBG, this procedure is not necessary.

It is instructive to see how accurately the computed DOS agrees with the known analyt-ical expression for vacuum.Using definition (B.2) for the DOS, and the dispersion relation for vacuum, i.e.ω(k) = c|k|, we obtain the following expressions for the DOS

D1D(ω) = 2 × L

The factor 2 in front of each equation accounts for the two different polarization modes (TM and TE) each field has.

From a computational point of view, the dispersion relation deviates from vacuum, simply due to the fact that the fields are defined on a grid.For the simplest dicretization scheme, the dispersion relation on the grid in d dimensions is given by (cf.Eq.(3.24))

ω(k) = 2c

Using this dispersion relation, it is possible, in one dimension, to obtain an exact expression for the density of states on the grid.In higher dimensions we need to resort to numerical methods. Inserting the dispersion relation (B.11) into definition (B.2), we obtain

D1D(ω) = L The total integrated density of states becomes

N(∞) = where Nx is the number of modes of one of the fields for one polarization using periodic boundary conditions.For the k-vector we have explicitly

k= 0, ±2π L , ±4π

L, . . . , ±(N_x− 2)π L ,N_xπ

L . (B.14)

Taking into account the different TM and TE mode polarizations, the total number of modes is indeed 2Nx.For higher dimensions, the total number of modes (for each of the

Figure B-1: Density of states per unit volume as a function of frequency.

Solid line: DOS obtained by timestepping integration with computational parameters (method,τ,∆t,M,S )=(T2S2,0.01,0.1,512,100), for a system measuring L = 100.Dashed-dotted line: vacuum value:D(ω)/L = 2/πc.Dotted line : analytical result, cf.Eq.(B.12).

vector fields separately, and using periodic boundary conditions) is 2NxNy and 2NxNyNz

respectively.The exact shape of the DOS on the grid in two and three dimensions can be obtained by simply counting the number of modes in a given interval dω by using dispersion relation (B.11). We can now compare these results with the DOS obtained by direct measurement (cf.Eq.(B.6)).The procedure to obtain the latter is as follows: first we measure a DOS, using an initial condition in which the fields do not have a vanishing divergence, we then remove the trivial contribution¹ atω = 0 and finally scale such that

&

dωD(ω) = N, where N = 2,d

i=1Ni.The resulting DOS is usually expressed in DOS per unit volume.Note that if the divergence of the fields is initially zero, for example by taking H(r)= ∇ × f (r), where f is a random vector field, the resulting DOS is quite different.

The DOS for vacuum is thus obtained by measuring it using the two initial conditions (divergence equal to zero or not equal to zero, in both two and three dimensions) for the time integration algorithm.The results are compared with the analytical result (on the grid) in one dimension, the histogram method in two dimensions, and the known analytical result for vacuum (Eq.(B.10)) in continuum.The resulting DOS are shown in figures B-1, B-2 and B-3.Notice the singularities at ω = 2/δ for one dimension, ω2√

2/δ for two dimensions, andω2√

3/δ for three dimensions.On the one hand, the agreement between the results obtained by using Fourier analysis and an initial condition for which the divergence of the fields is not zero, compared with the histogram/analytical results is quite good.

For low frequencies, the results of both methods converge to the continuum value.On the other hand, if the divergence of the initial fields is zero (as the Maxwell Equations demand), then we see a profoundly different DOS.One explanation could be that the set

1ω = 0 modes are valid solutions of the TDME, when one does not require that the divergence of the fields must vanish.