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In my previous question, Calculating the coherence length from a spectrum, it turned out that I can calculate the coherence length of my light source from the autocorrelation function, which can be obtained by taking the fourier transform of the spectrum, according to the Wiener-Khinchin theorem.

In one of the answers, the answerer takes two times the real part of a Fourier transform. I wonder what the physical meaning of this is. This is the equation in question, slightly rewritten for clarity, where $\mathscr{G}(f)$ is the spectrum and $\Gamma(t)$ the autocorrelation function:

$$\tilde{\Gamma}(t) = \int_{-\infty}^\infty e^{2\pi\,i\,f\,t} \mathscr{G}(f) \,df=2\,\mathrm{Re}\left(\int_{f_{min}}^{f_{max}} e^{2\pi\,i\,f\,t} \mathscr{G}(f) \,df\right)$$

I am also surprised that the answerer apparently claims that the part before the equals sign, without taking the real part is equal to the part after the equals sign, with taking the real part. Why is this?

In practice, to me it seems that in the context taking the absolute values of the transform is much more useful, as it yields the envelope of the real part. However, I am curious about the meaning of the answer.

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I am also surprised that the answerer apparently claims that the part before the equals sign, without taking the real part is equal to the part after the equals sign, with taking the real part. Why is this?

The statement $$\tilde{\Gamma}(t) = \int_{-\infty}^\infty e^{2\pi ift} G(f) \,df=2\,\mathrm{Re}\left(\int_{f_\text{min}}^{f_\text{max}} e^{2\pi ift}G(f) \,df\right)$$ is in fact correct, and follows from symmetry. The function $G(f)$ is symmetric with respect to inversion, ie $G(-f)=G(f)$, since the power contained in negative frequencies should be the same as the power in positive frequencies.

In practice it is implicit that your experimental data spectrum $G(F)$ is taken over a positive frequency range, ie $0<f_\text{min}<f_\text{max}$. So $$\tilde{\Gamma}(t) = \int_{-\infty}^\infty e^{2\pi ift} G(f) \,df =\int_{-f_\text{max}}^{-f_\text{min}}e^{2\pi ift} G(f) \,df+\int_{f_\text{min}}^{f_\text{max}}e^{2\pi ift} G(f) \,df \\ =\int_{f_\text{min}}^{f_\text{max}}e^{-2\pi ift} G(-f) \,df+\int_{f_\text{min}}^{f_\text{max}}e^{2\pi ift} G(f) \,df \\ =2\,\mathrm{Re}\left(\int_{f_\text{min}}^{f_\text{max}} e^{2\pi ift}G(f) \,df\right).$$

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  • $\begingroup$ It is far more common that $G(-f) = G(f)^*$, which is what happens when the time domain version of $G$ is purely real. $\endgroup$ – DanielSank Sep 26 '18 at 23:19

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