I was working through a lecture on quantum optics, in which we calculate the spectrum of electric field correlations of fields produced by two level emitters. Now, the part where I got stuck was actually rather elementary I'm afraid, but I simply don't get it. In the end we find an expression for the correlation function $g(\tau)$ (The exact form probably doesn't matter, but adding it here for completeness) enter image description here

Up to this point everything was fine, I understood the derivation. My teacher then went on to say that the implications are best understood by looking at the spectrum, which he defines as enter image description here So to be clear, this is twice the real part of the integral (from 0 to infinity) of e^{i omega tau} times the correlation function. I have to admit that I don't really understand this form. The way I see it, it is the inverse fourier transform (due to no minus sign), and it is being taken twice from 0 to infinity, implying that it is even. And we only take the real part, for a reason I also don't understand.

Could someone explain to me why this exact form gives us a resonance spectrum? Should it not be a normal fourier transform (minus sign)? I don't see why the function gives above is even (it doesn't seem to be to me), so why take it from 0 to inf and multiply by two? And why are we only interested in the real part? I thought that in order to get the spectrum one should take the absolute value. However, I have to admit that I no longer remember exactly what the real part and the imaginary part of a fourier transform contain, information wise. I can't find it online either, I'm afraid.


It turns out that this form is just a fancy way of writing a normal Fourier transform. Use the definition of the intensity correlation function $$g^{(2)}(\tau) = \langle n(\tau) n(0) \rangle, $$ where $n(t)$ denotes the (Hermitian) intensity at the position of the detector, evaluated at time $t$ in the Heisenberg picture. The angle brackets $\langle \bullet\rangle$ denote an expectation value with respect to the quantum state. For a stationary process, one has $$\langle A(t) B(s) \rangle = \langle A(t-s)B(0)\rangle. $$ This condition actually defines a stationary process. For example, if your dynamics is generated by a time-independent Liouvillian (e.g. a Lindblad equation), then the dynamics is stationary. Therefore we are dealing with a stationary process. Using these relations, along with the fact that $\langle A\rangle^{\ast} = \langle A^{\dagger}\rangle$, you should be able to show that $$2 \mathrm{Re} \int_0^{\infty} e^{i\omega t} g^{(2)}(\tau) = \int_{-\infty}^{\infty} e^{i\omega t} g^{(2)}(\tau), $$ which is just a standard Fourier transform (up to a choice of sign convention for the complex phase).

  • $\begingroup$ Thank you, that actually makes a lot of sense. We did indeed start with a Lindblad equation, so that follows quite well. The sign convention is stupid I guess, that I didn't think of that. One small comment (this is unclear from the handwritten stuff above), these are actually $^{1}$ correlations, not intensity. The one looks like a two, I agree. $\endgroup$ – user129412 Nov 22 '14 at 17:25
  • $\begingroup$ @user129412 Okay, well then if the operators you are dealing with are not Hermitian it's probably the case that you have the Fourier transform of an anti-commutator. $\endgroup$ – Mark Mitchison Nov 22 '14 at 17:45

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