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Dyson's argument that the perturbative expansion for QED must diverge:

[...] 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[...]

He then argues that for negative coupling $-e^2$ (like-charges attract) the vacuum of positive-coupling QED will be unstable and will continuously generate positron-electron pairs of a lower energy state (that subsequently repel). That part of the argument is fine.

The issue I have is with the argument that $F(e^2)$ must be analytic at $e^2=0$. What's to prevent $F(e^2)$ being well-behaved on $e^2\gt 0$ and poorly-behaved on $e^2\lt 0$. E.g: $$ F(e^2) = \exp(e^2), e^2\gt 0\\ F(e^2) = \frac{1}{e^2}, e^2\lt 0 $$ and we make no claim for $F(0)$?

The Taylor series for the exponential converges everywhere, so the perturbative expansion of $F(e^2)$ is convergent for $e^2\gt 0$, as we would like to be true of QED. But this function does not have a convergent perturbative expansion in an open set around zero.

But nothing about the lack of a convergent perturbative expansion of $F(e^2)$ around $e^2=0$ implies that there is not a convergent perturbative expansion of $F(e^2)$ for $e^2\gt 0$. What am I missing?

(It seems to me that Dyson's argument is simply an argument that $F(e^2)$ must have such behavior around $e^2=0$. Switching from opposite-charges attract to opposite-charges repel while keeping matter/anti-matter pair creation the same (i.e. anti-matter has the opposite charge to matter) is not remotely a "continuous" change, no matter how small the coupling constant).

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    $\begingroup$ It sounds like "must diverge" -> "must have zero radius of convergence" would make you happy. $\endgroup$ Commented Aug 2, 2022 at 15:09
  • $\begingroup$ No. It can have a non-zero radius of convergence. Just not around $e=0$. $\endgroup$ Commented Aug 2, 2022 at 15:32
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    $\begingroup$ I don't know the answer, but I suspect that Dyson is making some unstated assumptions about the niceness of $F$ that make his argument go through. One could presumably argue for those assumptions on physical grounds. I think this is a good question, and I also think that some other users have approached it (and very oddly, you personally) rather uncharitably. $\endgroup$
    – d_b
    Commented Aug 2, 2022 at 18:12
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    $\begingroup$ @d_b (edited) Thanks for the encouragement. Yes, that thought had occurred to me, but there's a real tension between whatever physical arguments he may have had in mind to motivate smoothness of $F$, and his (again, physical) argument implying that $F$ is not at all well-behaved at zero (and in fact is likely not defined at zero). Infinitesimal positive $e^2$ has a stable vacuum that collapses for infinitesimal negative $e^2$. That's a lot of work between +𝜖 and −𝜖... $\endgroup$ Commented Aug 2, 2022 at 18:39
  • $\begingroup$ @hft surely then there must be a duplicate somewhere $\endgroup$ Commented Aug 2, 2022 at 20:05

2 Answers 2

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$F(z) = a_0 + a_1 z + \ldots$ is a power series in $z$ centered at $z=0$. If this series converges for some $z_0>0$, then its radius of convergence is at least $z_0$ and so it is guaranteed to converge on the open ball $|z|<z_0$.

If I understand your objection, you're saying that it's possible to have a function which is non-analytic at $0$ but which is equal to $F(z)=a_0 + a_1 z + \ldots\ $ for some (or possibly all) $z>0$. That is of course true, as your example demonstrates, but you're getting the argument twisted. The function which Dyson is referring to is the power series in question, defined on the domain on which it converges. In other words, letting $F$ be a formal power series, we are interested in the function $f:U\rightarrow \mathbb R$ where $$U:= \{z\in \mathbb R \ | \ F(z) \text{ converges}\}$$ $$f :U\ni z\mapsto F(z)$$

It is this $f$ which Dyson is referring to as being analytic at $0$ if it converges for some $z_0>0$. In your objection, you are saying that one could construct some $\hat f$ which coincides with $f$ for $z>0$ but is non-analytic at $0$, which is beside the point.

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    $\begingroup$ @JonathanBaxter See my edit - I think I understand your objection better. $\endgroup$
    – J. Murray
    Commented Aug 2, 2022 at 18:48
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    $\begingroup$ @JonathanBaxter As an example, we might consider the physical (i.e. renormalized, as opposed to "bare") charge of an electron. The intuitive picture is that when you couple a bare electron with charge $q_0$ to the electromagnetic field, the EM field in the immediate vicinity becomes distorted in a way which effectively screens the bare charge so that other particles observe a renormalized charge $q=Zq_0$ with $Z<1$. If we flip the sign of $e^2$, then rather than being screened, the bare charge would be enhanced by the distortion in the EM field, which would cause further distortion and [...] $\endgroup$
    – J. Murray
    Commented Aug 2, 2022 at 20:16
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    $\begingroup$ [...] lead to a positive feedback loop and therefore a divergence in the effective charge of the electron $(Z\rightarrow \infty)$. This can be framed in terms of the polarization of a cloud of virtual particle/antiparticle pairs, if one were so inclined. In any case, this is the essence of Dyson's physical argument; if $Z$ (which is defined by a perturbative expansion and which plays the role of the function $f$) is well-defined at $e^2>0$ then that implies that it must be well-defined for some $\hat e^2<0$, but the latter should diverge on physical grounds. $\endgroup$
    – J. Murray
    Commented Aug 2, 2022 at 20:23
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    $\begingroup$ Very helpful. I really appreciate it. I've accepted your answer. I need to read a lot more to understand how to do these vacuum polarization calculations. I am still uneasy about Dyson's argument: QFT seems so precariously defined that concluding something about rightsideup QFT based on upsidedown QFT begs the question (to me) whether you omitted an important detail that really matters in the upsidedown but is irrelevant in the rightsideup. However, that's not the nature of my original objection. $\endgroup$ Commented Aug 2, 2022 at 21:01
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    $\begingroup$ @JonathanBaxter I think that's very reasonable - much work has to be done to put many aspects of interacting QFT on a mathematically firm footing. Indeed even the space of states in an interacting QFT is difficult to talk about rigorously. One would expect that ignoring these foundational issues and simply calculating via some formally divergent perturbative expansion would lead to a mess, but the results are in extraordinarily (and mysteriously) good agreement with experiment. I look forward to learning why someday :) $\endgroup$
    – J. Murray
    Commented Aug 2, 2022 at 21:35
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By being analytic at $e=0$ means that it has a perturbation expansion with a non-zero radius of convergence when expanded about $e=0$. As the radius of convergence is determined by the distance to the nearest singularity, means that there can be no nearby singularity for either positive or negative $e$. But there is is such a singularity (a branch point exactly at $e=0$) so the radius of convergence is zero, and by definition the function is not analytic at $e=0$.

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  • $\begingroup$ I am not arguing that the function $F(e^2)$ is not analytic at $e=0$. I agree that it is not. But Dyson argues the convergence of the perturbative expansion for $e^2>0$ implies that $F(e^2)$ is analytic at $e^2=0$. And then he uses a physical argument to imply a contradiction and therefore that the perturbative expansion for $e^2>0$ diverges. I think that's backwards. As my simple example shows, the perturbative expansion of $F(e^2)$ for $e^2>0$ can converge without $F(e^2)$ being analytic at $e^2=0$. So no contradiction. $\endgroup$ Commented Aug 2, 2022 at 16:52
  • $\begingroup$ @JonathanBaxter He already explained that in the answer. Your example is not a counterexample. You just picked out two different functions and pasted them together at $e^2=0$. One of them has an expansion about $e^2=0$ and the other doesn't... $\endgroup$
    – hft
    Commented Aug 2, 2022 at 17:04
  • $\begingroup$ @hft It's not "two different functions". It's one function, defined as stated. It's not meant to be a counterexample to analyticity at $e^2=0$. It's meant to show why Dyson's argument that the perturbative expansion of $F(e^2), e^2>0$ necessarily diverges is wrong. Dyson's physical argument suggests we expect some really bad behavior from $F$ as we pass from $e^2>0$ to $e^2<0$. Even for infinitesimal negative $e^2$ we expect the vacuum corresponding to infinitesimal positive $e^2$ to collapse. So maybe there's no perturbative expansion for $e^2<0$. But there still can be for $e^2>0$. $\endgroup$ Commented Aug 2, 2022 at 17:25
  • $\begingroup$ I think you need to understand (or remember) what it means for for a perturbative expansion about $z=0$ to converge. There is another answer by J. Murray that might help jog your memory. $\endgroup$
    – hft
    Commented Aug 2, 2022 at 17:29
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    $\begingroup$ I have a PhD in mathematics. I understand. Just take my $F(e^2)$. Write out the Taylor series for the exponential. Plug in any value of $e^2>0$. It converges. So we have a convergent perturbative expansion of $F$ for $e^2>0$. It's the same formula as a different function: $G(e^2) := \exp(e^2)$ for both positive and negative $e^2$, which is analytic at zero and hence the expansion works for all values of $e^2$ for $G$. The fact that the expansion only works for positive values of $e^2$ for $F$ is fine. That's all we need. $\endgroup$ Commented Aug 2, 2022 at 17:38

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