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Can anyone show detailed proof of why radius of convergence of perturbation series in quantum electrodynamics is zero? And how is perturbation series constructed?

So, as this argument requires imaginary electric charge of bispinor field (negative coupling constant), how is this argument considered valid?

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By "detailed proof" you mean something more than Dyson's Argument (last paragraph in the wiki link) ? –  twistor59 Jan 17 '13 at 12:21
    
No, just Dyson's argument. some detailed explanation from step zero. –  solay Jan 17 '13 at 12:42
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2 Answers 2

There is no strict proof, but a qualitative reasoning. Dysons' argument is based on an invented (mental) situation with imaginary charge values. It is still not clear whether such a pathological situation is feasible. From my point of view, Dysons' understanding of QED physics is wrong and cannot be a guide to make a conclusion.

EDIT: Many think the perturbation parameter in QED is $\alpha\propto e^2$, but it is not the case. To obtain a meaningful calculation result, one has not only to perform renormalization, but also fulfill soft diagram summation. The latter is equivalent to taking into account $\alpha$ in another initial approximation. In other words, the meaningful perturbation expansion is different from the Dyson's one and may converge.

The subtlety is not immediately visible, but it is implied that each term of Dysons' expansion is different from zero, and it is not so! For example, a Compton scattering amplitude calculated in the first Born approximation (Klein-Nishina) represents an elastic process that never happens in reality and in the theory because soft photon radiation. Only inclusive cross sections are different from zero in QED. The Klein-Nishina formula is multiplied in QED by an elastic form-factor whose first term of the Taylor series is unity, but the rest, after summation, gives zero, like $e^{-x}=1-x+x^2/2-...$ when $x\to\infty$.

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The idea is the following: Consider a series expansion in elementary charge squared, $e^2$:

\begin{equation}S(e^2)=\alpha_0+\alpha_2 e^2+\alpha_4 e^4+\;\ldots\end{equation}

Assuming finite convergence radius, $S(e^2)$ is analytic at $e=0$. This leads to the analyticity of $S(-e^2)$ (i.e. $e\rightarrow ie$). A theory with imaginary charges possesses an unstable vacuum, leading to production of spatially separated pairs of electrons and positrons. One can think of this as an interaction potential which results in two electrons attracting each other, which contradicts known physics. The conclusion is that $S(e^2)$ is not analytic around $e=0$, i.e. its convergence radius cannot be finite, hence it is zero.

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