Fourier transform of a spectrum

Question

I have the following definitions of Fourier transform FT with its inverse IFT for the complex function $h$ such that $$ h(\omega)=\mathcal{F}\{h(t)\}=\int_{-\infty}^{\infty}h(t)e^{i\omega t}dt \\ h(t) = \mathcal{F}^{-1}\{h(\omega)\} = \frac{1}{2\pi}\int_{-\infty}^{\infty}h(\omega)e^{-i\omega t}d\omega $$ and the definition of the delta function as $$ \delta(\omega+\omega^{\prime}) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(\omega+\omega^{\prime})t}dt. $$ The spectrum of $h$ is hence defined as $$ S_{hh}(\omega) = \int_{-\infty}^{\infty}d\tau\,e^{i\omega\tau}\left<h^{\star}(t+\tau)h(t)\right>_{t=0} = \int_{-\infty}^{\infty}d\omega^{\prime}\left<h^{\star}(-\omega)h(\omega^{\prime})\right>. $$ My question is: Using the definition of the delta function and the FT, IFT of $h$, how do I show the second equality in the preceding equation?

Answer 1

Substitute the IFT expressions for $h$ and $h^*$ into the spectrum integral to get $$ \frac{1}{(2\pi)^2}\iiint_{-\infty}^\infty e^{i\omega \tau}\left\langle e^{i\omega''(t+\tau)}e^{-i\omega't}h^*(\omega'')h(\omega')\right\rangle_{t=0}d\omega''d\omega'd\tau $$ We now pull the functions of time out of the expectation value (I assume this is OK) and set $t= 0$ to get $$ \frac{1}{(2\pi)^2}\iiint_{-\infty}^\infty e^{i\omega \tau}e^{i\omega''\tau}\left\langle h^*(\omega'')h(\omega')\right\rangle d\omega''d\omega'd\tau = \frac{1}{2\pi}\iiint_{-\infty}^\infty \frac{e^{i(\omega+\omega'')\tau}}{2\pi}d\tau \left\langle h^*(\omega'')h(\omega')\right\rangle d\omega'' d\omega'\\=\frac{1}{2\pi}\iint_{-\infty}^\infty \delta(\omega + \omega'') \left\langle h^*(\omega'')h(\omega')\right\rangle d\omega'' d\omega' = \frac{1}{2\pi}\int_{-\infty}^\infty\left\langle h^*(-\omega)h(\omega')\right\rangle d\omega' $$ A factor of $2\pi$ seems to have escaped somewhere. Might want to check normalization conventions for the Fourier transform.