28 KiB
Formulation of the problem
Assume a system of compact EM scatterers in otherwise homogeneous and isotropic medium, and assume that the system, i.e. both the medium and the scatterers, have linear response. A scattering problem in such system can be written as
Aα = TαPα = Tα(∑βSα ← βAβ + P0α)
where Tα is the T-matrix for scatterer α, Aα is its vector of the scattered wave expansion coefficient (the multipole indices are not explicitely indicated here) and Pα is the local expansion of the incoming sources. Sα ← β is ... and ... is ...
...
∑β(δαβ − TαSα ← β)Aβ = TαP0α.
Now suppose that the scatterers constitute an infinite lattice
$$\sum_{\vect b\beta}(\delta_{\vect{ab}}\delta_{\alpha\beta}-T_{\vect a\alpha}S_{\vect a\alpha\leftarrow\vect b\beta})A_{\vect b\beta}=T_{\vect a\alpha}P_{0\vect a\alpha}.$$
Due to the periodicity, we can write $S_{\vect a\alpha\leftarrow\vect b\beta}=S_{\alpha\leftarrow\beta}(\vect b-\vect a)$ and $T_{\vect a\alpha}=T_{\alpha}$. In order to find lattice modes, we search for solutions with zero RHS
$$\sum_{\vect b\beta}(\delta_{\vect{ab}}\delta_{\alpha\beta}-T_{\alpha}S_{\vect a\alpha\leftarrow\vect b\beta})A_{\vect b\beta}=0$$
and we assume periodic solution $A_{\vect b\beta}(\vect k)=A_{\vect a\beta}e^{i\vect k\cdot\vect r_{\vect b-\vect a}}$, yielding
$$\begin{aligned}
\sum_{\vect b\beta}(\delta_{\vect{ab}}\delta_{\alpha\beta}-T_{\alpha}S_{\vect a\alpha\leftarrow\vect b\beta})A_{\vect a\beta}\left(\vect k\right)e^{i\vect k\cdot\vect r_{\vect b-\vect a}} & = & 0,\\
\sum_{\vect b\beta}(\delta_{\vect{0b}}\delta_{\alpha\beta}-T_{\alpha}S_{\vect 0\alpha\leftarrow\vect b\beta})A_{\vect 0\beta}\left(\vect k\right)e^{i\vect k\cdot\vect r_{\vect b}} & = & 0,\\
\sum_{\beta}(\delta_{\alpha\beta}-T_{\alpha}\underbrace{\sum_{\vect b}S_{\vect 0\alpha\leftarrow\vect b\beta}e^{i\vect k\cdot\vect r_{\vect b}}}_{W_{\alpha\beta}(\vect k)})A_{\vect 0\beta}\left(\vect k\right) & = & 0,\\
A_{\vect 0\alpha}\left(\vect k\right)-T_{\alpha}\sum_{\beta}W_{\alpha\beta}\left(\vect k\right)A_{\vect 0\beta}\left(\vect k\right) & = & 0.\end{aligned}$$
Therefore, in order to solve the modes, we need to compute the “lattice Fourier transform” of the translation operator,
$$W_{\alpha\beta}(\vect k)\equiv\sum_{\vect b}S_{\vect 0\alpha\leftarrow\vect b\beta}e^{i\vect k\cdot\vect r_{\vect b}}.\label{eq:W definition}$$
Computing the Fourier sum of the translation operator
The problem evaluating ([eq:W definition]) is the asymptotic behaviour of the translation operator, $S_{\vect 0\alpha\leftarrow\vect b\beta}\sim\left|\vect r_{\vect b}\right|^{-1}e^{ik_{0}\left|\vect r_{\vect b}\right|}$ that makes the convergence of the sum quite problematic for any d > 1-dimensional lattice. 1 In electrostatics, one can solve this problem with Ewald summation. Its basic idea is that if what asymptoticaly decays poorly in the direct space, will perhaps decay fast in the Fourier space. I use the same idea here, but everything will be somehow harder than in electrostatics.
Let us re-express the sum in ([eq:W definition]) in terms of integral with a delta comb
$$W_{\alpha\beta}(\vect k)=\int\ud^{d}\vect r\dc{\basis u}(\vect r)S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})e^{i\vect k\cdot\vect r}.\label{eq:W integral}$$
The translation operator S is now a function defined in the whole 3d space; $\vect r_{\alpha},\vect r_{\beta}$ are the displacements of scatterers α and β in a unit cell. The arrow notation $S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})$ means “translation operator for spherical waves originating in $\vect r+\vect r_{\beta}$ evaluated in $\vect r_{\alpha}$” and obviously S is in fact a function of a single 3d argument, $S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})=S(\vect 0\leftarrow\vect r+\vect r_{\beta}-\vect r_{\alpha})=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha})$. Expression ([eq:W integral]) can be rewritten as
$$W_{\alpha\beta}(\vect k)=\left(2\pi\right)^{\frac{d}{2}}\uaft{(\dc{\basis u}S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0))\left(\vect k\right)}$$
where changed the sign of $\vect r/\vect{\bullet}$ has been swapped under integration, utilising evenness of $\dc{\basis u}$. Fourier transform of product is convolution of Fourier transforms, so (using formula ([eq:Dirac comb uaFt]) for the Fourier transform of Dirac comb)
$$\begin{aligned}
W_{\alpha\beta}(\vect k) & = & \left(\left(\uaft{\dc{\basis u}}\right)\ast\left(\uaft{S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\right)(\vect k)\nonumber \\
& = & \frac{\left|\det\recb{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\left(\dc{\recb{\basis u}}^{(d)}\ast\left(\uaft{S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\right)\left(\vect k\right)\nonumber \\
& = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\sum_{\vect K\in\recb{\basis u}\ints^{d}}\left(\uaft{S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\left(\vect k-\vect K\right).\label{eq:W sum in reciprocal space}\end{aligned}$$
As such, this is not extremely helpful because the the whole translation operator S has singularities in origin, hence its Fourier transform $\uaft S$ will decay poorly.
However, Fourier transform is linear, so we can in principle separate S in two parts, $S=S^{\textup{L}}+S^{\textup{S}}$. $S^{\textup{S}}$ is a short-range part that decays sufficiently fast with distance so that its direct-space lattice sum converges well; $S^{\textup{S}}$ must as well contain all the singularities of S in the origin. The other part, $S^{\textup{L}}$, will retain all the slowly decaying terms of S but it also has to be smooth enough in the origin, so that its Fourier transform $\uaft{S^{\textup{L}}}$ decays fast enough. (The same idea lies behind the Ewald summation in electrostatics.) Using the linearity of Fourier transform and formulae ([eq:W definition]) and ([eq:W sum in reciprocal space]), the operator Wαβ can then be re-expressed as
$$\begin{aligned}
W_{\alpha\beta}\left(\vect k\right) & = & W_{\alpha\beta}^{\textup{S}}\left(\vect k\right)+W_{\alpha\beta}^{\textup{L}}\left(\vect k\right)\nonumber \\
W_{\alpha\beta}^{\textup{S}}\left(\vect k\right) & = & \sum_{\vect R\in\basis u\ints^{d}}S^{\textup{S}}(\vect 0\leftarrow\vect R+\vect r_{\beta}-\vect r_{\alpha})e^{i\vect k\cdot\vect R}\label{eq:W Short definition}\\
W_{\alpha\beta}^{\textup{L}}\left(\vect k\right) & = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\sum_{\vect K\in\recb{\basis u}\ints^{d}}\left(\uaft{S^{\textup{L}}(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\left(\vect k-\vect K\right)\label{eq:W Long definition}\end{aligned}$$
where both sums should converge nicely.
Finding a good decomposition
The remaining challenge is therefore finding a suitable decomposition $S^{\textup{L}}+S^{\textup{S}}$ such that both $S^{\textup{S}}$ and $\uaft{S^{\textup{L}}}$ decay fast enough with distance and are expressable analytically. With these requirements, I do not expect to find gaussian asymptotics as in the electrostatic Ewald formula—having ∼ x − t, t > d asymptotics would be nice, making the sums in ([eq:W Short definition]), ([eq:W Long definition]) absolutely convergent.
The translation operator S for compact scatterers in 3d can be expressed as
$$S_{l',m',t'\leftarrow l,m,t}\left(\vect r\leftarrow\vect 0\right)=\sum_{p}c_{p}^{l',m',t'\leftarrow l,m,t}\ush p{m'-m}\left(\theta_{\vect r},\phi_{\vect r}\right)z_{p}^{(J)}\left(k_{0}\left|\vect r\right|\right)$$
where Yl, m(θ,ϕ) are the spherical harmonics, zp(J)(r) some of the Bessel or Hankel functions (probably hp(1) in all meaningful cases; TODO) and cpl, m, t ← l′, m′, t′ are some ugly but known coefficients (REF Xu 1996, eqs. 76,77).
The spherical Hankel functions can be expressed analytically as (REF DLMF 10.49.6, 10.49.1)
$$h_{n}^{(1)}(r)=e^{ir}\sum_{k=0}^{n}\frac{i^{k-n-1}}{r^{k+1}}\frac{\left(n+k\right)!}{2^{k}k!\left(n-k\right)!},\label{eq:spherical Hankel function series}$$
so if we find a way to deal with the radial functions sk0, q(r) = eik0r(k0r) − q, q = 1, 2 in 2d case or q = 1, 2, 3 in 3d case, we get absolutely convergent summations in the direct space.
2d
Assume that all scatterers are placed in the plane $\vect z=0$, so that the 2d Fourier transform of the long-range part of the translation operator in terms of Hankel transforms, according to ([eq:Fourier v. Hankel tf 2d]), reads
$$\begin{gathered}
\uaft{S_{l',m',t'\leftarrow l,m,t}^{\textup{L}}\left(\vect{\bullet}\leftarrow\vect 0\right)}(\vect k)=\\
\sum_{p}c_{p}^{l',m',t'\leftarrow l,m,t}\ush p{m'-m}\left(\frac{\pi}{2},0\right)e^{i(m'-m)\phi}i^{m'-m}\pht{m'-m}{h_{p}^{(1)\textup{L}}\left(k_{0}\vect{\bullet}\right)}\left(\left|\vect k\right|\right)\end{gathered}$$
Here $h_{p}^{(1)\textup{L}}=h_{p}^{(1)}-h_{p}^{(1)\textup{S}}$ is a long range part of a given spherical Hankel function which has to be found and which contains all the terms with far-field (r → ∞) asymptotics proportional to ∼ eik0r(k0r) − q, q ≤ Q where Q is at least two in order to achieve absolute convergence of the direct-space sum, but might be higher in order to speed the convergence up.
Obviously, all the terms ∝ sk0, q(r) = eik0r(k0r) − q, q > Q of the spherical Hankel function ([eq:spherical Hankel function series]) can be kept untouched as part of $h_{p}^{(1)\textup{S}}$, as they decay fast enough.
The remaining task is therefore to find a suitable decomposition of sk0, q(r) = eik0r(k0r) − q, q ≤ Q into short-range and long-range parts, $s_{k_{0},q}(r)=s_{k_{0},q}^{\textup{S}}(r)+s_{k_{0},q}^{\textup{L}}(r)$, such that $s_{k_{0},q}^{\textup{L}}(r)$ contains all the slowly decaying asymptotics and its Hankel transforms decay desirably fast as well, $\pht n{s_{k_{0},q}^{\textup{L}}}\left(k\right)=o(z^{-Q})$, z → ∞. The latter requirement calls for suitable regularisation functions—$s_{q}^{\textup{L}}$ must be sufficiently smooth in the origin, so that
$$\pht n{s_{k_{0},q}^{\textup{L}}}\left(k\right)=\int_{0}^{\infty}s_{k_{0},q}^{\textup{L}}\left(r\right)rJ_{n}\left(kr\right)\ud r=\int_{0}^{\infty}s_{k_{0},q}\left(r\right)\rho\left(r\right)rJ_{n}\left(kr\right)\ud r\label{eq:2d long range regularisation problem statement}$$
exists and decays fast enough. Jν(r) ∼ (r/2)ν/Γ(ν+1) (REF DLMF 10.7.3) near the origin, so the regularisation function should be ρ(r) = o(rq − n − 1) only to make $\pht n{s_{q}^{\textup{L}}}$ converge. The additional decay speed requirement calls for at least ρ(r) = o(rq − n + Q − 1), I guess. At the same time, ρ(r) must converge fast enough to one for r → ∞.
The electrostatic Ewald summation uses regularisation with 1 − e − cr2. However, such choice does not seem to lead to an analytical solution (really? could not something be dug out of DLMF 10.22.54?) for the current problem ([eq:2d long range regularisation problem statement]). But it turns out that the family of functions
$$\rho_{\kappa,c}(r)\equiv\left(1-e^{-cr}\right)^{\text{\ensuremath{\kappa}}},\quad c>0,\kappa\in\nats\label{eq:binom regularisation function}$$
might lead to satisfactory results; see below.
Hankel transforms of the long-range parts, „binomial“ regularisation[sub:Hankel-transforms-binom-reg]
Let
$$\begin{aligned}
\pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & \equiv & \int_{0}^{\infty}\frac{e^{ik_{0}r}}{\left(k_{0}r\right)^{q}}J_{n}\left(kr\right)\left(1-e^{-cr}\right)^{\kappa}r\,\ud r\nonumber \\
& = & k_{0}^{-q}\int_{0}^{\infty}r^{1-q}J_{n}\left(kr\right)\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}e^{-(\sigma c-ik_{0})r}\ud r\nonumber \\
& \underset{\equiv}{\textup{form.}} & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\pht n{s_{q,k_{0}}^{\textup{L}1,\sigma c}}\left(k\right).\label{eq:2D Hankel transform of regularized outgoing wave, decomposition}\end{aligned}$$
From [REF DLMF 10.22.49] one digs
$$\begin{gathered}
\pht n{s_{q,k_{0}}^{\textup{L}1,c}}\left(k\right)=\frac{k^{n}\Gamma\left(2-q+n\right)}{2^{n}k_{0}^{q}\left(c-ik_{0}\right)^{2-q+n}}\hgfr\left(\frac{2-q+n}{2},\frac{3-q+n}{2};1+n;\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right),\\
\Re\left(2-q+n\right)>0,\Re(c-ik_{0}\pm k)\ge0,\label{eq:2D Hankel transform of exponentially suppressed outgoing wave as 2F1}\end{gathered}$$
and using [REF DLMF 15.9.17] and [REF DLMF 14.9.5]
$$\begin{gathered}
\pht n{s_{q,k_{0}}^{\textup{L}1,c}}\left(k\right)=\frac{k^{n}\Gamma\left(2-q+n\right)}{k_{0}^{q}\left(c-ik_{0}\right)^{2-q+n}}\left(\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right)^{-\frac{n}{2}}\left(1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}\right)^{\frac{q}{2}-1}P_{q}^{-n}\left(\frac{1}{\sqrt{1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}}}\right),\\
k>0\wedge k_{0}>0\wedge c\ge0\wedge\lnot\left(c=0\wedge k_{0}=k\right)\label{eq:2D Hankel transform of exponentially suppressed outgoing wave expanded}\end{gathered}$$
with principal branches of the hypergeometric functions, associated Legendre functions, and fractional powers. The conditions from ([eq:2D Hankel transform of exponentially suppressed outgoing wave as 2F1]) should hold, but we will use ([eq:2D Hankel transform of exponentially suppressed outgoing wave expanded]) formally even if they are violated, with the hope that the divergences eventually cancel in ([eq:2D Hankel transform of regularized outgoing wave, decomposition]).
One problematic element here is the gamma function Γ(2−q+n) which is singular if the arguments are negative integers, i.e. if q − n ≥ 3; but at least the necessary minimum of q = 1, 2 would be covered this way. The associated Legendre function can be expressed as a finite “polynomial” if q ≥ n. In other cases, different expressions can be obtained from [eq:2D Hankel transform of exponentially suppressed outgoing wave as 2F1] using various transformation formulae from either DLMF or
Прудников
.
In fact, Mathematica is usually able to calculate the transforms for specific values of κ, q, n, but it did not find any general formula for me. The resulting expressions are finite sums of algebraic functions, Table 3 shows how fast they decay with growing k for some parameters. The only case where Mathematica did not help at all is q = 2, n = 0, which is unfortunately important. But if I have not made some mistake, the expression ([eq:2D Hankel transform of exponentially suppressed outgoing wave expanded]) is applicable for this case.
| n | ||||
| 0 | 1 | 2 | ||
| 1 | 2 | 1 | 1 | |
| 2 | x | w | 0 |
| 0 | 1 | 2 | 3 | 4 | ||
| 1 | w | 3 | 2 | 2 | 2 | |
| 2 | x | 1 | w | 1 | 1 |
| 0 | 1 | 2 | 3 | 4 | ||
| 1 | 0/w | 3 | 4 | 3 | 3 | |
| 2 | x | 3 | 2 | 2 | 1 |
3d (TODO)
$$\begin{gathered}
\uaft{S_{l',m',t'\leftarrow l,m,t}\left(\vect{\bullet}\leftarrow\vect 0\right)}(\vect k)=\\
\sum_{p}c_{p}^{l',m',t'\leftarrow l,m,t}\ush p{m'-m}\left(\theta_{\vect k},\phi_{\vect k}\right)\left(-i\right)^{p}\usht p{z_{p}^{(J)}}\left(\left|\vect k\right|\right)\end{gathered}$$
Major TODOs and open questions
Check if ([eq:2D Hankel transform of exponentially suppressed outgoing wave expanded]) gives a satisfactory result for the case q = 2, n = 0.
Analyse the behaviour k → k0.
Find a general algorithm for generating the expressions of the Hankel transforms.
Three-dimensional case.
(Appendix) Fourier vs. Hankel transform
Three dimensions
Given a nice enough function f of a real 3d variable, assume its factorisation into radial and angular parts
$$f(\vect r)=\sum_{l,m}f_{l,m}(\left|\vect r\right|)\ush lm\left(\theta_{\vect r},\phi_{\vect r}\right).$$
Acording to (REF Baddour 2010, eqs. 13, 16), its Fourier transform can then be expressed in terms of Hankel transforms (CHECK normalisation of jn, REF Baddour (1))
$$\uaft f(\vect k)=\frac{4\pi}{\left(2\pi\right)^{\frac{3}{2}}}\sum_{l,m}\left(-i\right)^{l}\left(\bsht{f_{l,m}}{}\right)\left(\left|\vect k\right|\right)\ush lm\left(\theta_{\vect k},\phi_{\vect k}\right)$$
where the spherical Hankel transform $\bsht l{}$ of degree l is defined as (REF Baddour eq. 2)
$$\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).$$
Using this convention, the inverse spherical Hankel transform is given by (REF Baddour eq. 3)
$$g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\, k^{2}\bsht lg(k)j_{l}(k),$$
so it is not unitary.
An unitary convention would look like this:
$$\usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition}$$
Then $\usht l{}^{-1}=\usht l{}$ and the unitary, angular-momentum Fourier transform reads
$$\begin{aligned}
\uaft f(\vect k) & = & \frac{4\pi}{\left(2\pi\right)^{\frac{3}{2}}}\sqrt{\frac{\pi}{2}}\sum_{l,m}\left(-i\right)^{l}\left(\usht l{f_{l,m}}\right)\left(\left|\vect k\right|\right)\ush lm\left(\theta_{\vect k},\phi_{\vect k}\right)\nonumber \\
& = & \sum_{l,m}\left(-i\right)^{l}\left(\usht l{f_{l,m}}\right)\left(\left|\vect k\right|\right)\ush lm\left(\theta_{\vect k},\phi_{\vect k}\right).\label{eq:Fourier v. Hankel tf 3d}\end{aligned}$$
Cool.
Two dimensions
Similarly in 2d, let the expansion of f be
$$f\left(\vect r\right)=\sum_{m}f_{m}\left(\left|\vect r\right|\right)e^{im\phi_{\vect r}},$$
its Fourier transform is then (CHECK this, it is taken from the Wikipedia article on Hankel transform)
$$\uaft f\left(\vect k\right)=\sum_{m}i^{m}e^{im\phi_{\vect k}}\pht mf_{m}\left(\left|\vect k\right|\right)\label{eq:Fourier v. Hankel tf 2d}$$
where the Hankel transform of order m is defined as
$$\pht mg\left(k\right)=\int_{0}^{\infty}\ud r\, g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}$$
which is already self-inverse, $\pht m{}^{-1}=\pht m{}$ (hence also unitary).
(Appendix) Multidimensional Dirac comb
1D
This is all from Wikipedia
Definitions
$$\begin{aligned}
Ш(t) & \equiv & \sum_{k=-\infty}^{\infty}\delta(t-k)\\
Ш_{T}(t) & \equiv & \sum_{k=-\infty}^{\infty}\delta(t-kT)=\frac{1}{T}Ш\left(\frac{t}{T}\right)\end{aligned}$$
Fourier series representation
$$Ш_{T}(t)=\sum_{n=-\infty}^{\infty}e^{2\pi int/T}\label{eq:1D Dirac comb Fourier series}$$
Fourier transform
With unitary ordinary frequency Ft., i.e.
$$\uoft f(\vect{\xi})\equiv\int_{\mathbb{R}^{n}}f(\vect x)e^{-2\pi i\vect x\cdot\vect{\xi}}\ud^{n}\vect x$$
we have
$$\uoft{Ш_{T}}(f)=\frac{1}{T}Ш_{\frac{1}{T}}(f)=\sum_{n=-\infty}^{\infty}e^{-i2\pi fnT}\label{eq:1D Dirac comb Ft ordinary freq}$$
and with unitary angular frequency Ft., i.e.
$$\uaft f(\vect k)\equiv\frac{1}{\left(2\pi\right)^{n/2}}\int_{\mathbb{R}^{n}}f(\vect x)e^{-i\vect x\cdot\vect k}\ud^{n}\vect x\label{eq:Ft unitary angular frequency}$$
we have (CHECK)
$$\uaft{Ш_{T}}(\omega)=\frac{\sqrt{2\pi}}{T}Ш_{\frac{2\pi}{T}}(\omega)=\frac{1}{\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-i\omega nT}$$
Dirac comb for multidimensional lattices
Definitions
Let d be the dimensionality of the real vector space in question, and let $\basis u\equiv\left\{ \vect u_{i}\right\} _{i=1}^{d}$ denote a basis for some lattice in that space. Let the corresponding lattice delta comb be
$$\dc{\basis u}\left(\vect x\right)\equiv\sum_{n_{1}=-\infty}^{\infty}\ldots\sum_{n_{d}=-\infty}^{\infty}\delta\left(\vect x-\sum_{i=1}^{d}n_{i}\vect u_{i}\right).$$
Furthemore, let $\rec{\basis u}\equiv\left\{ \rec{\vect u}_{i}\right\} _{i=1}^{d}$ be the reciprocal lattice basis, that is the basis satisfying $\vect u_{i}\cdot\rec{\vect u_{j}}=\delta_{ij}$. This slightly differs from the usual definition of a reciprocal basis, here denoted $\recb{\basis u}\equiv\left\{ \recb{\vect u_{i}}\right\} _{i=1}^{d}$, which satisfies $\vect u_{i}\cdot\recb{\vect u_{j}}=2\pi\delta_{ij}$ instead.
Factorisation of a multidimensional lattice delta comb
By simple drawing, it can be seen that
$$\dc{\basis u}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right)$$
where $c_{\basis u}$ is some numerical volume factor. In order to determine $c_{\basis u}$, let us consider only the “zero tooth” of the comb, leading to
$$\delta^{d}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\delta\left(\vect x\cdot\rec{\vect u_{i}}\right).$$
From the scaling property of delta function, δ(ax) = |a| − 1δ(x), we get
$$\delta^{d}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert ^{-1}\delta\left(\vect x\cdot\frac{\rec{\vect u_{i}}}{\left\Vert \rec{\vect u_{i}}\right\Vert }\right).$$
From the Osgood’s book (p. 375):
$$\dc A(\vect x)=\frac{1}{\left|\det A\right|}\dc{}^{(d)}\left(A^{-1}\vect x\right)$$
Fourier series representation
Fourier transform (OK)
From the Osgood’s book https://see.stanford.edu/materials/lsoftaee261/chap8.pdf, p. 379
$$\uoft{\dc{\basis u}}\left(\vect{\xi}\right)=\left|\det\rec{\basis u}\right|\dc{\rec{\basis u}}^{(d)}\left(\vect{\xi}\right).$$
And consequently, for unitary/angular frequency it is
$$\begin{aligned}
\uaft{\dc{\basis u}}\left(\vect k\right) & = & \frac{1}{\left(2\pi\right)^{\frac{d}{2}}}\uoft{\dc{\basis u}}\left(\frac{\vect k}{2\pi}\right)\nonumber \\
& = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{\rec{\basis u}}^{(d)}\left(\frac{\vect k}{2\pi}\right)\nonumber \\
& = & \left(2\pi\right)^{\frac{d}{2}}\left|\det\rec{\basis u}\right|\dc{\recb{\basis u}}\left(\vect k\right)\nonumber \\
& = & \frac{\left|\det\recb{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{\recb{\basis u}}\left(\vect k\right).\label{eq:Dirac comb uaFt}\end{aligned}$$
Convolution
$$\left(f\ast\dc{\basis u}\right)(\vect x)=\sum_{\vect t\in\basis u\ints^{d}}f(\vect x-\vect t)$$
Note that d here is dimensionality of the lattice, not the space it lies in, which I for certain reasons assume to be three. (TODO few notes on integration and reciprocal lattices in some appendix)↩︎