# Correlation function which has branch cut in momentum space

When correlation function has branch cut in momentum space, how to find correlation in coordinate space? For example $$\tilde {G}(\omega) = \frac{2i}{\omega+(\omega^2-\nu^2)^{1/2}}$$ How to get the $G(t)$ usng Fourier transformation ?
t>0 HERE. This problem is from matrix model of Iizuka and Polchinski. They discuss the propagator in the model and find that the propagator $G(t)$ has power law decay behavior if there is branch cut in $\tilde{G}(\omega)$. If there is a pole in the lower half plane for $\tilde{G}(\omega)$, there is an exponential decay in $G(t)$.

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Context would help here. – Ron Maimon Jun 5 '12 at 5:55
en.wikipedia.org/wiki/… – genneth Jun 5 '12 at 13:50
Here we do not have poles. Does the method work? – Craig Thone Jun 6 '12 at 7:59
Following the suggestion of genneth, using formula Gradshteyn & Ryzhik, 8.464-1, the integral can be calculated giving the power law decay behavior $t^{-\frac{3}{2}} e^{\pm i \nu t}$ – Craig Thone Jun 6 '12 at 9:46
@CraigThone: Don't do it that way, the way I say is the right way to think about it. Besides it's not $\pm$, it's just minus. – Ron Maimon Jun 6 '12 at 17:11

### Cutology

The math books aren't good for this, you need seat-of-the-pants intuition. The quick and dirty physicist answer is that a branch cut is best thought of as a continuum of poles densely spread over a line. You reproduce branch cuts by integrating poles spread over an interval, for example with a constant "residue density" (this is not the standard term for it, see below)

$$\log({x-a\over x-b}) = \int_a^b {1\over x-u} du$$

And this is a continuous density of poles with unit residue between a and b, just as you can see from the right hand side, and a function with a cut between a and b on the left hand side. If you make the residue density between a and b some function $\rho(u)$

$$f(x) = \int_a^b {\rho(u)\over x-u} du$$

you get different functions, but always with a cut inside [a,b] wherever $\rho(u)$ is nonzero. The residue density is not usually called the residue density--- it's called the "cut discontinuity", because if you consider the value of the function f defined by integral just above and just below the real axis somewhere in the interval [a,b], and take the difference between the two values, you get 2\pi i times $\rho$ as the difference. This is because you can deform the two integrals in opposite sense into a small circle, or, if you like, because of the Cauchy-distribution representation of the delta function as:

$$\delta(x) = {1\over 2\pi i} ({1\over x-i\epsilon} - {1\over x+i\epsilon})$$

Which you can work out explicitly.

To show you how it works in detail, say you want reproduce the square-root function's branch cut, running along the negative real axis, you look for the jump-discontinuity in $\sqrt{x}$ along the negative axis. It goes from $i\sqrt{|x|}$ to $-i\sqrt{x}$, the jump discontinuity is the square root of the distance from the origin. So you write

$$f(x) = {1\over \pi} \int_{-\infty}^0 {\sqrt{|u|}\over x-u} du$$

And this should reproduce the square-root function. Except this is nonsense, since the integral is divergent! This is still morally true, however, by doing the following manipulation: change u to -u, and split the integral as follows:

$${2\over \pi} \int_0^\infty {1\over2\sqrt{u}} {u\over x+u} du$$

Then split ${u\over x+u}$ into $1 - {x\over x+u}$. Discard the 1, because this is an (infinite) x independent constant, and change variables to $\alpha=\sqrt{u}$, and you get

$${2\over \pi} \int_0^\infty {x\over x+\alpha^2} d\alpha$$

Now you can see that this evaluates to $\sqrt{x}$ by rescaling by x. So it works, but you have to watch out for divergences.

### Polology

When there is a single pole on the negative imaginary axis at position $-ai$, when you integrate with $e^{-i\omega t}$, for t positive, the integration contour can be moved down to the $\mathrm{Im}(\omega)=-ia$ line, just by sliding it down (there are no singularities along the way) and the integral along the new contour is a sum of contributions each with an exponential decay $e^{-at}$. So the dominant exponential decay at infinity is the pole closest to the real axis.

If the pole is the only singularity and with appropriate stuff at negative imaginary infinity, you can do the integral explicitly, but the contour moving is simple and doesn't depend on anything, and shows the exponential decay immediately. Usually you have many singularities, and you only want the leading behavior at infinity.

If there is a pole very close to the real axis with position $a-i\epsilon$, the fourier transform isn't exponentially decaying (or rather it is decaying with a rate $\epsilon$), but oscillating. There can't be any singularities in the positive imaginary $\omega$ halfplane, as this would lead to a blow up in the future. This means that you should imagine all singularities shifted infinitesimally to the negative imaginary half-plane, just like this.

When there is a continuous density of poles, you get a continuous density of decay rates, and if they accumulate arbitrarily close to the real axis, you get a continuous superposition of different decay rates that can reproduces a power law, if the pole-density near the real axis is the appropriate power.

In you case, you have a cut starting at $\omega=\nu$ on the real axis, and the cut runs off to infinity somehow along the negative imaginary axis. The discontinuity in the imaginary part along the cut can be found by writing the thing as
$${\omega - \sqrt{\omega^2 - \nu^2}}\over \nu^2$$
and you can see there is a square-root pole-density near $\omega=\nu$. Each pole gives a decy of $e^{-at}$ in real time, so you superpose the decays:
$$\int_0^\infty \sqrt{a} e^{-at} da \approx {1\over t^{3\over 2}}$$
This is all on top of a general oscillation of $e^{-i\nu t}$ which you get from the fact that this stuff is happening near $\omega=\nu$. This sort of stuff is very useful, and is secret lore for some reason.