# Fourier Transforms Related to Green's Functions

I'm reading a text on field theory where there are a number of assertions made about Fourier transforms that I'm finding confusing. For example, let $G^R = -i \theta(t - t')e^{-i \omega_0 (t - t')}$. The text says that the FT of $G^R$ wtih respect to $t-t'$ is $(\epsilon - \omega_0 + i0)^{-1}$. What is the meaning of this $i0$ term?

Secondly, after inverting this expression the text takes the inverse transform. That is, the inverse FT of $\epsilon - \omega_0 + i0$ and obtains $\delta(t - t')(i\partial_{t'} - \omega_0 + i0)$. What is the meaning of this? How did the fourier transform of an expression become an operator?

Thanks!

• The Fourier transform of $G^R$ does not converge, unless you put in a regularization term $e^{-\eta t}$ where $\eta$ is taken to $0$ in the end of the calculation. Then you get $\epsilon-\omega_0+i\eta$. So $0$ here really means an infinitesimally small positive number. For the inverse transform, you need to do the complex integral using residue theorem and you will see $i0$ determines how you close the contour. – Meng Cheng Apr 2 '15 at 17:53
• Thanks. I see what is being done in the FT now. Can you explain the inverse FT in more depth? – Laplacian Apr 2 '15 at 18:00
• What I said was not entirely right. $i0$ means that the pole of the integrand is in the upper half plane. If $t>t'$, then you have to close your contour also in the upper half plane and then picks up a nonzero reside. If $t<t'$ the contour is closed in the lower half plane and no poles are enclosed, so you get zero. Therefore $i0$ gives you exactly the causal structure $\theta(t-t')$. – Meng Cheng Apr 2 '15 at 18:08

In essence, you are taking the Fourier transform of the Heaviside theta function, $\theta(t)$, to try and get $$\tilde\theta(\omega)=\int_{-\infty}^\infty e^{i\omega t}\theta(t)\text dt=\int_{0}^\infty e^{i\omega t}\text dt.$$ For nonzero $\omega$, this is perfectly fine and easily evaluates to ${1}/{i\omega}$ once you discard the term at infinity (which is also not easy!). For $\omega=0$, however, that result is obviously bogus, and the bare integral $\int_{0}^\infty 1\text dt$ also makes no sense.

To bring both of those problems into line, the usual approach is to employ a regularization procedure: introduce an exponential decay $e^{-\epsilon t}$ into the integrand, to force convergence at $\omega=0$ and to get rid of the $e^{i\omega·\infty}$ term, and then take the limit of $\epsilon\to 0$. This means that, when we talk about $\tilde\theta$, we're actually talking about $$\tilde\theta(\omega) =\lim_{\epsilon\to 0}\int_{0}^\infty e^{i\omega t-\epsilon t}\text dt =\lim_{\epsilon\to 0}\frac{-i}{\omega+i\epsilon}.$$ This notation is often contracted to the shorthand $$\tilde\theta(\omega)=\frac{-i}{\omega+i0},$$ where the term in $+i0$ stands for an infinitesimally small, but positive, imaginary part. This imaginary part comes in when integrating over $\omega$: it means that the ugly pole at $\omega=0$ should be displaced down before integration, which can be done quite easily in practice via the Sokhotski formula. This integration is important because often, in the end, what we actually want from $\tilde\theta$ is that it reproduce the time-domain function, $$\theta(t)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-i\omega t}\tilde\theta(\omega)\text d\omega,$$ so if it has non-integrable singularities it also needs to be equipped with instructions on how to deal with them.

If you want to get a bit more technical, both $\theta(t)$ and $\tilde\theta(\omega)$ are best thought of as distributions rather than functions, and the maths there can get a bit subtle. Essentially, a distribution is an object $D$ which (only?) makes sense when inside an integral, i.e. as $$\int D(x) f(x) \mathrm d x$$ for some unknown, suitably nice, function $f$. (Ring a bell?) More formally, $D$ is really shorthand notation for a functional which takes functions into numbers, i.e. for the mapping $$f\mapsto\int D(x) f(x) \mathrm d x.$$

One easy way to get distributions is to simply make $D$ a function. Even better, if $D$ is in $L^1$ then its distribution-sense Fourier transform is just its normal transform. If $D$ is not in $L^1$ (say it stays constant, or grows linearly, at large $x$), then it won't have a Fourier transform in the traditional sense, but it can still have a transform in a suitable distributional sense, which opens the way to using the Fourier transform in a much broader range of interesting cases, at the minor cost that the end result is again a distribution.

This is actually what's happening in both of the examples you mention. Here $\theta(t)$ and $1/\tilde\theta(\omega)=-(\omega+i0)$ stay constant, and grow linearly, at large arguments, so they wouldn't normally be given Fourier transforms. In the distributional sense, however, this is indeed valid, and their transforms $$\tilde\theta(\omega)=\frac{-i}{\omega+i0}$$ and $$\widetilde{1/\tilde\theta}(t)=\frac{i}{2\pi}\int_{-\infty}^\infty\omega e^{-i\omega t}\mathrm d\omega=\delta(t)\partial_t$$ are perfectly valid distributions - i.e. they make perfect sense when applied on some function $f$ and integrated over.

1. $i0$ is meant as $$\frac{1}{\epsilon - \omega_0 + i 0} = \lim_{x\to 0^+} \frac{1}{\epsilon - \omega_0 + i x}\, .$$ The idea here is that the Fourier transform of something containing a theta function is not well defined. This comes from the fact that the integral $$G(\epsilon) = -i \int_{-\infty}^{\infty} \theta(\tau) \text{e}^{-i \tau (\omega_0-\epsilon)} = -i \int_{0}^{\infty} \text{e}^{-i \tau (\omega_0-\epsilon)} \, ,$$ does not converge. However, if you add any positive imaginary part to $\epsilon \rightarrow \epsilon + i x$, it converges for all $x>0$. As long as $x>0$ all our integrals converge, $$G(\epsilon+ix) = -i \int_{0}^{\infty} \text{e}^{-i \tau (\omega_0-\epsilon)} \, \text{e}^{-\tau x} = -\frac{1}{\omega_0-\epsilon-ix}\, .$$ At the end of the calculation, we can therefore take $x$ as small as we want and it drops out of the final result. The sign of $x$ however is very important. If you take a negative $x$ and Fourier transform back, you will find the same $G(\tau)$ as before with the only difference that $\theta(\tau)$ is now $\theta(-\tau)$. This means that your response function $G(\tau)$ now responds to something that will happen in the future. Choosing $x>0$ is a means to enforce causality to response functions.

2. As before, the Fourier transform of $\epsilon$ leads to an integral that does not converge, $$G(\tau) = \int_{-\infty}^{\infty} d \epsilon \, \text{e}^{-i\tau\epsilon} \epsilon \, .$$ You can however write it formally as $$G(\tau) = i\frac{d}{d\tau} \int_{-\infty}^{\infty} d\epsilon \, \text{e}^{-i\tau\epsilon} = i\frac{d}{d\tau} \delta(\tau)\, .$$ $\delta(\tau)$ is the delta distribution, $\delta(\tau\neq 0) = 0$, $\delta(0) = \infty$, $\int \delta(\tau) = 1$. You cannot really interpret it as a function, but it makes a perfect operator acting on functions, $$(\delta f)(x) \equiv \int d y \, \delta(y-x) f(y) = f(0) \, .$$ The same happens with its derivative and you recover the expression from your book.

The $\mathrm{i}0$ is short for $$\lim_{\epsilon\rightarrow 0^+}\mathrm{i}\epsilon$$ and is used in the so-called $\mathrm{i}\epsilon$-prescription. It is necessary for some integrals so that the contour does not hit a pole and render the integral ill-defined. In this case, you must insert $\lim_{x\rightarrow 0}\mathrm{e}^{-x\epsilon}$ into your Fourier transform so that the integral converges. Note that the notation of your book is confusing -- it uses $\epsilon$ for the transform variable while $\epsilon$ is standard notation for the infinitesimal in $\mathrm{i}0$.

I think that you misread the text for the inverse transform. It should be $(\mathrm{i}\partial_{t'}-\omega_0+\mathrm{i}0)\delta(t-t')$, i.e. the derivative is acting on the delta function. Using the integral representation of the delta function, $$\delta(t)=\frac{1}{2\pi}\int_\mathbb{R}\mathrm{d}x\,\mathrm{e}^{\mathrm{i}tx}$$ we find that the inverse Fourier transform of a constant (note that $\omega_0$ and $\mathrm{i}0$ are just constants) is proportional to a delta function. You may also verify that the derivative of the integral representation of the delta function is proportional to the inverse Fourier transform of the time variable.