The name is absent

we can easily see the relation between FT and DFT

h(f_n) ≈ Ah_n.

The formula for the discrete inverse Fourier transform, which recovers the set of ∕ι⅛^,s ex-
actly from the h_n^,s is

(A. 12)

where we have used ɪ ∑_n e ^{2rri(j k,nlN} = δ_jk. Let’s check out those terms which involve

DFT, for simplicity, here we only consider an ID case. The derivative related terms are

= -4πV-' [∕²≠ω] ≈ -4√∆_z X         = -4√1 2
4π² J dff²∖φ{f)∖² ≈ 4π²∆y £ f²_n∖φ{f_n)∖² = 4π²- £ f²_n ∖φ_n ∣².

For dipole-dipole interaction energy, we need to find D(t) = ʃ dt'V(t - t')φ(t') which is
alternatively

!D<t_j) = ʃ dfe^2πiftiV(f)φ(f) ≈ ∆_z ∑ e^2πW(∕_nШ) = ɪ ∑ e^lmιl^Ql
this is exactly the inverse DFT. Since, in our case, we know the FT of V(t) analytically, we
use V(∕_n) from FT rather than V_n from DFT.

Forthermore, fast Fourier transform (FFT), instead of DFT, is used for the numerical
computation. FFT is slightly different from DFT in the way how it sorts the transformed
data. In FFT, f_n is ordered as

(A. 13)

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