Calculating $\frac{1}{x-i\epsilon}$ in Quantum Field Theory

[koolmodee]

\frac{1}{x-i\epsilon}=\frac{x}{x²+a²}+\frac{ia}{x²+a²}= P \frac{1}{x}+i pi \delta(x)

P means the principal value, a is possibly infinitesimal (?), i is the imaginary unit

Where does the pi, Dirac delta come from? What principal value?
It is from a quantum field theory book.

 

[Mute]

The expression contains a Dirac Delta, so that should be your clue to integrate against a test function, \varphi(x) that goes to zero as x \rightarrow \pm \infty. In particular, you want to consider a contour integral

\int_\gamma dz \frac{\varphi(z)}{z - i\epsilon}

about some contour \gamma. Note that the integrand has a pole at z = i\varepsilon. To evaluate, consider the closed contour in the upper half plane that consists of a large semicircle of radius R with a small semicircle of radius b around the pole in the integrand at z = i\varepsilon. Since the contour contains no poles, the contour integral is zero. Parameterize the contour:

\int_{-R}^{-b}dx \frac{\varphi(x)}{x - i\epsilon} + \int_{b}^{R}dx \frac{\varphi(x)}{x - i\epsilon} + \int_0^{\pi}d(Re^{i\theta}) \frac{\varphi(Re^{i\theta})}{Re^{i\theta}-i\varepsilon} + \int_\pi^{0}d(be^{i\theta}) \frac{\varphi(be^{i\theta})}{be^{i\theta}-i\varepsilon} = 0

Now, take the limits as R \rightarrow \infty and b \rightarrow 0, and use the definition of the principal part integral:

\mathcal{P}\int_{-\infty}^{\infty}dx~f(x) = \lim_{a \rightarrow 0} \int_{-\infty}^{-a}dx~f(x) + \int_{a}^{\infty}dx~f(x)

and the half-residue theorem, that states that for a semi-circular arc about a pole, the contribution from that arc about the pole as the radius goes to zero is 2\pi i times half the residue about that pole. To get the delta function you note that at the pole the test function will give a contribution \varphi(0), and so the action of the imaginary part of (x - i\epsilon)^{-1} is to pick out the value of the test function at the origin, which is just what a Dirac delta does. In the end you can set \epsilon = 0.

 

[koolmodee]

Mute, thanks so much!

 

[koolmodee]

WAIT!

In my OP there was a mistake! I'm very sorry.

It is x/(x²+a²)+ia/(x²+a²) = P 1/x + i pi delta(x) and a possibly infinitesimal.

So RHS are two innocent algebraic terms, the LHS has something to two with integration.

I still can't see how that works.

 

[Mute]

The \epsilon[/tex] is an infinitesimal. It&#039;s there to enforce causality. You can set it to zero after you do your integrals over the left hand side because after that causality has been enforced and you don&#039;t need the infinitesimal anymore.<br /> <br /> The right hand side of the expression only makes sense inside an integral. What it says is that when you integrate<br /> <br /> \int_{-\infty}^{\infty}dx \frac{\varphi(x)}{x - i\epsilon}<br /> <br /> it&#039;s equivalent to integrating<br /> <br /> \int_{-\infty}^{\infty}dx \left[\mathcal{P}\left(\frac{1}{x}\right) + i\pi \delta(x)\right]\varphi(x) = \mathcal{P}\left[\int_{-\infty}^{\infty}dx~\frac{\varphi(x)}{x}\right] + i\pi \varphi(0).

 

[koolmodee]

I really appreciate the effort Mute, but I can't see what you mean here.

 

[koolmodee]

I wrote the author and he was so kind answering me. If anybody interested here is what he wrote.

It's pretty easy to see how the delta function part works; for any
nonzero x, the function e/(x^2+e^2) goes
to zero as e goes to zero, but the area under it is pi for any e; so it
behaves like pi times delta(x) as e->0.

The principle-value part works similarly; one needs to show that the
integral of x/(x^2+e^2) times a smooth
test function f(x) equals, in the limit e->0, the principle value of
the integral of (1/x)f(x). (To get the principle value,
one replaces (1/x) with zero when -e<x<e, then takes the limit e->0
after doing the integral.)

After comparing that with what Mute said, I think that I understand it now.