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  Principal value of 1/x and few questions about complex analysis in Peskin's QFT textbook

+ 6 like - 0 dislike

When I learn QFT, I am bothered by many problems in complex analysis.

1) $$\frac{1}{x-x_0+i\epsilon}=P\frac{1}{x-x_0}-i\pi\delta(x-x_0)$$

I can't understand why $1/x$ can have a principal value because it's not a multivalued function. I'm very confused. And when I learned the complex analysis, I've not watched this formula, can anybody tell me where I can find this formula's proof.

2) $$\frac{d}{dx}\ln(x+i\epsilon)=P\frac{1}{x}-i\pi\delta(x)$$

3) And I also find this formula. Seemingly $f(x)$ has a branch cut, then $$f(z)=\frac{1}{\pi}\int_Z^{\infty}dz^{\prime}\frac{{\rm Im} f(z^{\prime})}{z^{\prime}-z}$$ Can anyone can tell the whole theorem and its proof, and what it wants to express. enter image description here

Now I am very confused by these formula, because I haven't read it in any complex analysis book and never been taught how to handle an integral with branch cut. Can anyone give me the whole proof and where I can consult.

This post imported from StackExchange Physics at 2014-03-30 12:29 (UCT), posted by SE-user user34669
asked Mar 30, 2014 in Theoretical Physics by user34669 (205 points) [ no revision ]

2 Answers

+ 3 like - 0 dislike
Unfortunately, there are two completely unrelated meanings of the term "principal value". The kind referred to here is the Cauchy principal value, which assigns values to otherwise undefined improper integrals. This has nothing to do with the principal value you had in mind, which is for selecting single-valued branches of multi-valued functions. I know, it's stupid. You'd think someone would have fixed all these weird ambiguities by now, but alas math is not French.This post imported from StackExchange Physics at 2014-03-30 12:29 (UCT), posted by SE-user David H
answered Mar 30, 2014 by David H (90 points) [ no revision ]
@DavidH Thanks a lot! Then the last question, can you give me some clues?

This post imported from StackExchange Physics at 2014-03-30 12:29 (UCT), posted by SE-user user34669
@user34669 I think the last expression goes by the name of "Kramers-Kronig" relation, it is a way to express a complex function in its real or imaginary part. So with either the real or imaginary part, you can reconstruct the whole function. For a proof, see en.wikipedia.org/wiki/Kramers%E2%80%93Kronig_relations

This post imported from StackExchange Physics at 2014-03-30 12:29 (UCT), posted by SE-user Funzies
+ 2 like - 0 dislike

Identities like these are often used to perform calculations with Greens functions. In particular, given a Hamiltonian H, its resolvent 

\((ω-H)^{-1} = \sum_n \frac1{ω-E_n} |ψ_n\rangle\langleψ_n|\)

is an object of key interest. It's essentially the Fourier transform of the corresponding Greens function.

Now, the thing about the resolvent is that is analytic for all complex values ω that are not in the spectrum of the Hamiltonian H. Hence, we can apply identities from complex analysis to perform computations.

One of the key observation of complex analysis is that complex analytic functions are very rigid: knowing the values of a function \(f\)on a countable set of points with a limit point is enough to reconstruct the whole function \(f\). The most prominent example is Cauchy's integral formula, which states that if a function is complex analytic on and inside a disk, then the values inside can be reconstructed entirely from the values on the boundary, via the following integral formula:

\(f(ω) = \frac1{2πi} \oint_{\text{circle}} dz\frac{f(z)}{z-ω}\)

The identities you mention in your question are used for the same purpose, except that in this case, we want to reconstruct a function not from its values on a circle, but from its values on the real axis.

Formula 1 is to be understood in the sense of distributions: multiply with a test function \(g\)and integrate over the whole real axis. Then, the results are equal:

\(\int^∞_{-∞} dx \frac{g(x)}{x-x_0+iε} = \mathcal{P} \int_{-∞}^∞dx \frac{g(x)}{x-x_0}- iπg(x_0)\)

(We assume that g is any continuous, complex-valued test function that decays sufficiently fast at infinity.)

Formula 2 is actually formula 1 after you perform the differentiation on the left-hand side.

Formula 3 can be derived from formula 1 for the special case where the function g is related to the values of a function which is complex analytic in the upper and lower half plane. It is similar Cauchy's integral formula in the sense that you reconstruct the values of in the complex plane from just the imaginary part of its values on the real axis.

My favorite example for these kinds of calculations is the function \(f(z) = \sqrt{z}\). It is complex analytic everywhere, except for the negative real numbers, where this function has a branch cut. Plug this into the formula and try to calculate the result, doing so will hopefully help you understand what is going on. (Note that the branch cut is from -∞ to 0 here, i.e. mirrored compared to formula 3.)

answered Apr 4, 2014 by Greg Graviton (775 points) [ no revision ]

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