# Tag Info

1

When beginning by calculate transition amplitudes in position space, and taking the Fourier transform of these amplitudes, to get the transition amplitude in momentum space, you get terms (for instance in a $2 \to 2$ interaction) in $\int d^4v e ^{-i(p_1+p_2-p_3-p_4)v}$, and this is equals to $(2\pi)^4 \delta^4(p_1+p_2-p_3-p_4)$ An example of such an ...

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I suppose you talk about the standard $2\pi$ that appears in the rules for Fourier transform. The factor of $2\pi$ or $1/2\pi$ or two factors of $1/\sqrt{2\pi}$ have to appear "somewhere" in the Fourier transform rules because this is what the mathematics implies. At any rate, if this is your question, it is a mathematical question and you may learn it in ...

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The impression I got form the equation is that you model your system as being composed of a collection of point particles. The index $i$ indexes this set of point particles (as opposed to indexing the components of a vector). One could create a spatial velocity distribution by saying $\vec{v}(\vec{r})=\sum_i \vec{v}_i \delta(\vec{r} - \vec{r}_i)$, where ...

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I think Volker (@vbraun) nailed it in his answer. Continuing where I left off: $$\cdots = {N_\mathrm{cell}\over\Omega_\mathrm{BZ}} \sum_{\mathbf{G}} e^{+i\mathbf{G} \cdot \mathbf{x}} \int_{\Omega_\mathrm{BZ}} \tilde f(\mathbf{G}+\boldsymbol\omega) e^{+i\boldsymbol\omega \cdot \mathbf{x}}\,d^3 \omega =$$ ...

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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 ... 0 The simplest derivation is probably to take the first equation and substitute into the second: $$F^{-1}[\tilde f(\mathbf{G})] = f(\mathbf{x}) = \sum_{\mathbf{G}} \tilde f(\mathbf{G}) e^{+i\mathbf{G} \cdot \mathbf{x}} =$$ $$= \sum_{\mathbf{G}} \left({1\over\Omega_\mathrm{cell}} \int_{\Omega_\mathrm{cell}} f(\mathbf{x'}) ... 1 The energy you seem to refer to is the electric part of the Poynting energy expression for some volume V:$$ E_{\text{Poynting}}(t) = \int_V \frac{1}{2}\epsilon_0 \left|\mathbf E(\mathbf x, t)\right|^2 + \frac{1}{2\mu_0}\left|\mathbf B(\mathbf x, t)\right|^2 \,d^3\mathbf x.$$The vector$\mathbf E(\mathbf x, t)\$ in this expression is the electric vector ...

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