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What is the simplest derivation of the following two well-known formulas that work for crystal lattice [1]: $$ F[f(\mathbf{x})] \equiv \tilde f(\mathbf{G}) = {1\over\Omega_\mathrm{cell}} \int_{\Omega_\mathrm{cell}} f(\mathbf{x}) e^{-i\mathbf{G} \cdot \mathbf{x}}\,d^3 x $$ $$ F^{-1}[\tilde f(\mathbf{G})] = f(\mathbf{x}) = \sum_{\mathbf{G}} \tilde f(\mathbf{G}) e^{+i\mathbf{G} \cdot \mathbf{x}} $$ See the question How to derive inverse Fourier transform for periodic functions (in crystal lattice)? for context. The difference from that question is that here we allow any derivation.

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

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2 Answers 2

The simplest derivation is probably to take the first equation and substitute into the second: $$ F^{-1}[\tilde f(\mathbf{G})] = f(\mathbf{x}) = \sum_{\mathbf{G}} \tilde f(\mathbf{G}) e^{+i\mathbf{G} \cdot \mathbf{x}} = $$ $$ = \sum_{\mathbf{G}} \left({1\over\Omega_\mathrm{cell}} \int_{\Omega_\mathrm{cell}} f(\mathbf{x'}) e^{-i\mathbf{G} \cdot \mathbf{x'}}\,d^3 x'\right) e^{+i\mathbf{G} \cdot \mathbf{x}} = $$ $$ = {1\over\Omega_\mathrm{cell}} \int_{\Omega_\mathrm{cell}} f(\mathbf{x'}) \sum_{\mathbf{G}} e^{i\mathbf{G} \cdot (\mathbf{x}-\mathbf{x'})}\,d^3 x' = $$ $$ = {1\over\Omega_\mathrm{cell}} \int_{\Omega_\mathrm{cell}} f(\mathbf{x'}) (2\pi)^3 \delta\left({(2\pi)^3\over\Omega_\mathrm{cell}}(\mathbf{x}-\mathbf{x'})\right) \,d^3 x' = $$ $$ = {1\over\Omega_\mathrm{cell}} \int_{\Omega_\mathrm{cell}} f(\mathbf{x'}) (2\pi)^3 {\Omega_\mathrm{cell}\over (2\pi)^3} \delta(\mathbf{x}-\mathbf{x'}) \,d^3 x' = $$ $$ =f(\mathbf{x}) $$ Thus we obtained an identity $f(\mathbf{x})=f(\mathbf{x})$. We have used: $$ \delta(\mathbf{x}) = {1\over (2\pi)^3} \sum_{\mathbf{n}} e^{i\mathbf{n}\cdot \mathbf{x}} $$ and $$ \mathbf{G_n} = {(2\pi)^3\over\Omega_\mathrm{cell}}\mathbf{n} $$ Note: the reciprocal space vector $\mathbf{G}$ is actually defined using the three reciprocal primitive vectors, see e.g. here for more info. I didn't want to complicate the notation, as it is obvious how to get these little details right and one gets the same result, just in a more complex way.

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I guess you are looking for the purely mathematical result. In that context, your elements $G$ are points of the crystal lattice, let's call it $\Lambda$, and $\Omega_{cell}$ is a fundamental domain of the lattice in $\mathbb R^3$ (or its volume), which is a parallelepiped that when translated by lattice elements exactly tesselates $\mathbb R^3$ (if I interpret your notation correctly).

Let $A:\mathbb R^3\rightarrow\mathbb R^3$ be a linear tranformation. Then by a straightforward application of the substitution theorem we get

$$\int f(x)g(Ax)dx = {1\over|\det A|}\int f(A^{-1}y)g(y)dy$$

Using this your result can be derived from the ordinary theory of Fourier series (going to 3D in orthogonal directions, which is trivial) by changing the lattice $\mathbb Z^3$ to the lattice $\Lambda$ through a linear transformation $A$ whose columns are generators of $\Lambda$, in other words, $A(\mathbb Z^3) = \Lambda$. Note that the (absolute value of the) determinant of $A$ is the volume of a fundamental domain. Some work still has to be done, but it is pretty straightforward.

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