How to derive inverse Fourier transform for periodic functions (in crystal lattice)?

Question

I would like to derive the following two well-known formulas that work for crystal lattice [1]: $$ F[f(\mathbf{x})] \equiv \tilde f(\mathbf{G}) = {1\over\Omega_\mathrm{cell}} \int_{\Omega_\mathrm{cell}} f(\mathbf{x}) e^{-i\mathbf{G} \cdot \mathbf{x}}\,d^3 x $$ $$ F^{-1}[\tilde f(\mathbf{G})] = f(\mathbf{x}) = \sum_{\mathbf{G}} \tilde f(\mathbf{G}) e^{+i\mathbf{G} \cdot \mathbf{x}} $$ Specifically, I want to derive them from the general 3D Fourier transform (the other way is to simply plug the second formula into the first, one obtains a delta function and obtains an identity --- see this question where I have worked this out in details, but here I don't want to use this approach). How do I derive the second formula?

Following [1], here is how to derive the first formula from the basic definition of a 3D Fourier transform, divided by the volume of the crystal $\Omega_\mathrm{crystal}$ (to make it finite):

$$ F[f(\mathbf{x})] \equiv \tilde f(\boldsymbol\omega) = {1\over\Omega_\mathrm{crystal}}\int_{\Omega_\mathrm{crystal}} f(\mathbf{x}) e^{-i\boldsymbol\omega \cdot \mathbf{x}}\,d^3 x = $$ $$ = {1\over\Omega_\mathrm{crystal}} \sum_\mathbf{n} \int_{\Omega_\mathrm{cell}} f(\mathbf{x}+\mathbf{T}(\mathbf{n})) e^{-i\boldsymbol\omega \cdot (\mathbf{x}+\mathbf{T}(\mathbf{n}))}\,d^3 x = $$ $$ = {1\over\Omega_\mathrm{crystal}} \sum_\mathbf{n} \int_{\Omega_\mathrm{cell}} f(\mathbf{x}) e^{-i\boldsymbol\omega \cdot (\mathbf{x}+\mathbf{T}(\mathbf{n}))}\,d^3 x = $$ $$ = {1\over\Omega_\mathrm{crystal}} \sum_\mathbf{n} e^{-i\boldsymbol\omega \cdot \mathbf{T}(\mathbf{n})} \int_{\Omega_\mathrm{cell}} f(\mathbf{x}) e^{-i\boldsymbol\omega \cdot \mathbf{x}}\,d^3 x = $$ $$ = {1\over\Omega_\mathrm{crystal}} N_\mathrm{cell} \int_{\Omega_\mathrm{cell}} f(\mathbf{x}) e^{-i\boldsymbol\omega \cdot \mathbf{x}}\,d^3 x = $$ $$ = {1\over\Omega_\mathrm{cell}} \int_{\Omega_\mathrm{cell}} f(\mathbf{x}) e^{-i\boldsymbol\omega \cdot \mathbf{x}}\,d^3 x $$ In here, the function $f(\mathbf{x})$ is periodic: $f(\mathbf{x}+\mathbf{T}(n_1, n_2, n_3)) = f(\mathbf{x})$ and the sum $\sum_\mathbf{n} e^{-i\boldsymbol\omega \cdot \mathbf{T}(\mathbf{n})} = \sum_\mathbf{n} 1 = N_\mathrm{cell}$ for $\boldsymbol\omega=\mathbf{G}$, where $\mathbf{G}$ are reciprocal space vectors (defined by $e^{i\mathbf{G} \cdot \mathbf{T}(\mathbf{n})} = 1$). For $\boldsymbol\omega\neq\mathbf{G}$, the sum is bounded, and so in the limit $\Omega_\mathrm{crystal}\to\infty$ the factor before the integral sign above goes to zero.

For the second formula, there is no hint in [1] how to proceed. Here is my best effort so far: $$ F^{-1}[\tilde f(\boldsymbol\omega)] = f(\mathbf{x}) = {\Omega_\mathrm{crystal}\over(2\pi)^3}\int_{-\infty}^{\infty} \tilde f(\boldsymbol\omega) e^{+i\boldsymbol\omega \cdot \mathbf{x}}\,d^3 \omega = $$ $$ = {\Omega_\mathrm{cell}N_\mathrm{cell}\over(2\pi)^3}\int_{-\infty}^{\infty} \tilde f(\boldsymbol\omega) e^{+i\boldsymbol\omega \cdot \mathbf{x}}\,d^3 \omega = $$ $$ = {N_\mathrm{cell}\over\Omega_\mathrm{BZ}} \sum_{\mathbf{G}} \int_{\Omega_\mathrm{BZ}} \tilde f(\mathbf{G}+\boldsymbol\omega) e^{+i(\mathbf{G}+\boldsymbol\omega) \cdot \mathbf{x}}\,d^3 \omega = $$ $$ = {N_\mathrm{cell}\over\Omega_\mathrm{BZ}} \sum_{\mathbf{G}} e^{+i\mathbf{G} \cdot \mathbf{x}} \int_{\Omega_\mathrm{BZ}} \tilde f(\mathbf{G}+\boldsymbol\omega) e^{+i\boldsymbol\omega \cdot \mathbf{x}}\,d^3 \omega = \cdots $$ Here $\Omega_\mathrm{BZ} = {(2\pi)^3 \over \Omega_\mathrm{cell}}$ is the volume of the Brillouin zone. I have moved the integration over $\boldsymbol\omega$ to the Brillouin zone. As you can see, it's quite close, but I can't figure out how to finish it. Any ideas?

[1] Martin, R. M. (2004). Electronic Structure -- Basic Theory and Practical Methods (p. 642). Cambridge University Press.

Answer 1

The Fourier transform of a periodic function has discrete support, so your $\tilde{f}(G+\omega)$ is zero unless $\omega=0$ in your fundamental domain.

The regulator needs some care, the crystal volume and the (related) number of cells are infinite. Its probably easier to think of the combination $\tilde{f}(G+\omega) \cdot N/\Omega_{BZ} = \tilde{f}(G)\cdot \delta(\omega)$ as a delta-function. If you insist on doing the integral separately then it would be infinitely small, but multiplied by $N$ to give a finite value.

Answer 2

I think Volker (@vbraun) nailed it in his answer. Continuing where I left off: $$ \cdots = {N_\mathrm{cell}\over\Omega_\mathrm{BZ}} \sum_{\mathbf{G}} e^{+i\mathbf{G} \cdot \mathbf{x}} \int_{\Omega_\mathrm{BZ}} \tilde f(\mathbf{G}+\boldsymbol\omega) e^{+i\boldsymbol\omega \cdot \mathbf{x}}\,d^3 \omega = $$ $$ = \sum_{\mathbf{G}} \tilde f(\mathbf{G}) e^{+i\mathbf{G} \cdot \mathbf{x}} \int_{\Omega_\mathrm{BZ}} \delta(\boldsymbol\omega) e^{+i\boldsymbol\omega \cdot \mathbf{x}}\,d^3 \omega = $$ $$ = \sum_{\mathbf{G}} \tilde f(\mathbf{G}) e^{+i\mathbf{G} \cdot \mathbf{x}} $$ where we used the fact that: $$ {N_\mathrm{cell}\over\Omega_\mathrm{BZ}}\tilde f(\mathbf{G}+\boldsymbol\omega) =\tilde f(\mathbf{G})\delta(\boldsymbol\omega) $$ I still have to figure out how to prove this last fact explicitly, but it might follow from using the first equation for $\tilde f(\mathbf{G})$ in it somehow.