Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Doubt in Dyson's argumet about the divergent nature of the perturbative expansion in QED

+ 2 like - 0 dislike
239 views

I am trying to understand Dyson's argument about the divergent nature of the perturbative expansion in QED. Quoting his own words [Source: hep-ph:0508017 (PDF)]

... let 

$$F(e^2)=a_0+a_1e^2+a_2e^4+\ldots$$

be a physical quantity which is calculated as a formal power series in $e^2$ by integrating the equations of motion of the theory over a finite or infinite time.Suppose, if possible, that the series...  converges for some positive
value of $e^2$; this implies that $F(e^2)$ is an analytic function
of $e$ at $e=0$. Then for sufficiently small value of $e$, $F(−e^2)$ will also be a well-behaved analytic function with a convergent power series expansion.

My question is, why does the convergence of the series for some positive value of $e^2$ imply that it must be analytic at $e=0$?

asked Dec 25, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ revision history ]
edited Dec 25, 2015 by dimension10

Your question in the text is purely mathematical. It concerns all functions.

As to real doubts about Dyson's argument, I have already written many times that the used in practice series are not in powers of $e^2$ or $\alpha$. For example, the meaningful inclusive cross section is a selective sum over all orders of $\alpha$: $F(\alpha)=G(\alpha) + b_0+b_1\cdot\alpha+...=G(1/137) + b_0+b_1\cdot\alpha+...$. First, G(1/137) is finite and the remaining series is IR finite too. Next, all non-analyticity may be contained in $G(\alpha)$ and the remaining series may represent an analytical function. This question has never been studied, so the Dyson's argument is just wrong as dealing with a wrong object - an IR-divergent series.

You can read a toy model about it: https://www.academia.edu/14945605/On_integrating_out_short-distance_physics

@VladimirKalitvianski: The Dyson series is being used in may places for finding the magnetic moment of the electron at first nontrivial order. Higher order computations of magnetic moments, lamb shifts, etc.  in QED are done in NRQED, which is a joint expansion in $\alpha$ and $Z\alpha$; see. e.g., this article by Kinoshita.

Your method, on the other hand, has never been used in QED, only in your toy models. 

@ArnoldNeumaier : The question is not about "my method", but about QED series. And "my method" is about possibilities of representation of a function in different ways. Factually, it is already written generally as $F(x)=G(x)$+another series in my comment above, without necessity to look into details of "my method".

@VladimirKalitvianski: The question is explicitly about the Dyson series in QED, not about resummation techniques that rearrange terms. So strictly speaking your comment was off-topic. 

@ArnoldNeumaier : "Resummation" is a necessity and banality in QED; it is given in all QED textbooks, but if you think that I am off-topic, you may delete my comments entirely.

1 Answer

+ 4 like - 0 dislike

Analytic at $x=0$ means convergent for $x$ in the interior of some disk around zero; its maximal radius $r$ is the radius of convergence of the series. The root test for the convergence of an infinite series implies that if a power series in $x$ converges for some nonzero $x$ then the radius of convergence is at least $|x|$. In particular, if the Dyson series converges for some $e^2=x>0$,  it also converges for $e^2=-x/2<0$.

answered Dec 25, 2015 by Arnold Neumaier (12,355 points) [ revision history ]
edited Dec 25, 2015 by Arnold Neumaier

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysi$\varnothing$sOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...