Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
    
(@Qmechnic applied the homework tag, that's fine with me --- but it's not a homework, I really want to understand that.) –  Ondřej Čertík Nov 26 '13 at 20:23
    
You don't need all this. Just replace, in your first equation, $f(x)$, by its value in function of $\tilde f(G)$ given by your second equation, and verify the coherence, that is : "$\tilde f(G) =\tilde f(G)$" –  Trimok Nov 27 '13 at 11:51
    
Thanks @Trimok for the suggestion. Yes, I know how to do that, but as I mentioned in the question, I am interested how to derive it directly from the 3D Fourier transform definition. There must be a way. –  Ondřej Čertík Nov 27 '13 at 19:41
1  
@Trimok: just to make it absolutely clear, I've added a new question and answered it myself, where I show in detail how to obtain the $f(\mathbf{x})=f(\mathbf{x})$ identity by substituting the first equation into the second: physics.stackexchange.com/q/88169. Here however I am interested in deriving it from the 3D Fourier transform. –  Ondřej Čertík Nov 28 '13 at 6:09

2 Answers 2

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.

share|improve this answer
    
Hi Volker, thanks a lot, I think you nailed it! I wrote it up in my answer below. Do you know how to prove ${N_\mathrm{cell}\over\Omega_\mathrm{BZ}}\tilde f(\mathbf{G}+\boldsymbol\omega) =\tilde f(\mathbf{G})\delta(\boldsymbol\omega)$ explicitly? Obviously it's true for $\boldsymbol\omega\ne0$, but I want to make sure all the factors are right for $\boldsymbol\omega=0$ as well. –  Ondřej Čertík Nov 29 '13 at 17:59

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.