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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).

Connor Behan
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Jonathan Baxter
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    It sounds like "must diverge" -> "must have zero radius of convergence" would make you happy. – Connor Behan Aug 02 '22 at 15:09
  • No. It can have a non-zero radius of convergence. Just not around $e=0$. – Jonathan Baxter Aug 02 '22 at 15:32
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    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. – d_b Aug 02 '22 at 18:12
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    @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 + and −... – Jonathan Baxter Aug 02 '22 at 18:39
  • @hft surely then there must be a duplicate somewhere… – ZeroTheHero Aug 02 '22 at 20:05

2 Answers2

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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.

J. Murray
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  • You can use the same formula for the power series about zero for the positive part of my $F$ ($e^2>0$). You just can't continue it to negative $e^2$.

    So yes, the convergence of the Taylor series for $F(e^2) = \exp(e^2)$ for $e^2>0$ implies that the function defined by that Taylor series can be continued to $e^2<0$. But that defines a different function to my $F$ for $e^2<0$.

     – Jonathan Baxter Aug 02 '22 at 17:45
  • @JonathanBaxter No you can't. What would your $a_0$ be? Writing $F(z)= a_0 + a_1 z + \ldots$ obviously implies that $F(0)=a_0$, right? – J. Murray Aug 02 '22 at 17:52
  • No. I didn't give a formula for $F$ at zero. $F(0)$ is undefined. You can continue the Taylor series to zero if you want, but I am pretty sure the answer you get will be wrong. Dyson's argument shows that $F$ must be very badly behaved at $0$. – Jonathan Baxter Aug 02 '22 at 18:03
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    @JonathanBaxter See my edit - I think I understand your objection better. – J. Murray Aug 02 '22 at 18:48
  • Replying to your edited version. Assuming your characterization of Dyson's argument is correct (I don't agree that's what he had in mind, but I don't think it matters), then all Dyson has shown (by his physical argument) is that $f$ is not the correct function for both $e^2>0$ and $e^2<0$. Something has to give. But there's no reason to give up $e^2 > 0$ (i.e. claim the series must be everywhere divergent). Just jettison $e^2 < 0$. It's not clear to me that $e^2<0$ corresponds to an interesting physical theory at all, since all states apparently collapse to negative infinite energy. – Jonathan Baxter Aug 02 '22 at 19:07
  • @JonathanBaxter My characterization of Dyson's argument is that if we take the function $f$ defined in my answer, then either $U={0}$ or $U\supseteq (-\epsilon,\epsilon)$ for some $\epsilon>0$ (do we agree on this point?). If we assume the latter, then Dyson makes a physical argument that the power series should diverge for negative values, leading to a contradiction. This then implies that $U={0}$, and that the power series diverges for all nonzero values of its argument. – J. Murray Aug 02 '22 at 19:16
  • @JonathanBaxter If your issue is with Dyson's physical argument, then that's reasonable. But I am not sure what your objection to the preceding mathematical argument is. – J. Murray Aug 02 '22 at 19:18
  • @JMurray Ok, on rereading I think I agree with your characterization of Dyson's argument. But I still think my objection is correct. It comes down to what $F$ is describing. Dyson says only that $F$ is "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" So $F$ is a physical quantity. What does it mean to replace $e^2$ with $-e^2$ in that physical quantity? Let's be specific: what does it mean to replace $e^2$ with $-e^2$ for, say, the Lamb shift? I would claim it's meaningless. – Jonathan Baxter Aug 02 '22 at 19:37
  • By meaningless I mean, you'll get an answer, but what physical process does it correspond to? The upsidedown (I've been watching Stranger Things with my kids) where opposite charges repel almost certainly doesn't even have a Lamb shift. So what physical setup does $F(-e^2)$ correspond to?

    Ok, so maybe rather than starting with something as complex as the Lamb shift, we can just start with $F$ for the (positive-coupling) vacuum, and then flip the sign of $e^2$ on that. But again, I don't know if the right-side-up vacuum state even exists in the upsidedown...

     – Jonathan Baxter Aug 02 '22 at 19:44
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    @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 [...] – J. Murray Aug 02 '22 at 20:16
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    [...] 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. – J. Murray Aug 02 '22 at 20:23
  • Thanks. This clarifies the situation. Unlike Lamb shift (or even the vacuum), this $Z$ function corresponds to a physical process both in the upsidedown and the rightsideup. And in fact is essentially the same physical process in both. Yet we know it diverges in the upsidedown.

    Perhaps I misread Dyson, but this seems much stronger than Dyson's general argument (at least as it is ordinarily portrayed), because you have an identified physical quantity in both worlds that should be related by a change in sign of $e^2$, which is impossible.

     – Jonathan Baxter Aug 02 '22 at 20:41
  • @JonathanBaxter It may also be worth noting that this vacuum polarization is ultimately the physical origin of the Lamb shift, in which an electron in a bound state experiences a distorted electric potential from the nucleus. In the “upside down” it’s true that the bound state itself wouldn’t exist (since the electron and nucleus would repel) but in some sense that’s somewhat downstream of the issue being raised here. – J. Murray Aug 02 '22 at 20:50
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    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. – Jonathan Baxter Aug 02 '22 at 21:01
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    @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 :) – J. Murray Aug 02 '22 at 21:35
  • On that note, I found this. The history of corrections to the Lamb shift makes me wonder... Or maybe this guy's just a crank. – Jonathan Baxter Aug 03 '22 at 01:00
  • @JonathanBaxter That author has no co-authors, is affiliated to an online polytechnic institute (as far as I can tell, not in a faculty capacity), and has only published work in The General Science Journal and Progress in Physics - a quick glance at which does not inspire great confidence in their respective review processes. While it is possible that they have uncovered a grand conspiracy in high energy physics ... I would not personally treat their review as authoritative. – J. Murray Aug 03 '22 at 01:23
  • I'm certainly not treating it as authoritative. And I don't think there's some vast conspiracy. I was more intrigued by the history of the corrections, and the possibility that we have on occasion judiciously truncated or stopped looking for an error as soon as we achieved agreement. Given we're potentially summing a divergent series,how unlikely is it that by summing only the first few terms we get great agreement with experiment, only to eventually lose that agreement if we keep adding terms? Or is the consensus that agreement between experiment and theory indicates the series is convergent? – Jonathan Baxter Aug 03 '22 at 02:14
  • @JonathanBaxter If Dyson's argument is to be believed, then we are certainly summing a divergent series. The general consensus is that non-perturbative effects (which manifest as terms $\sim e^{-1/g}$ with $g$ the coupling constant of the theory) spoil the perturbative expansion at order $1/g$ ($\approx 137$ for QED), past which the addition of more terms will make the perturbative result worse. AFAIK QED computations haven't made it past order 10 or so. The point is that there is enough information in the coefficients of the asymptotic series to yield fantastic agreement with experiment. – J. Murray Aug 03 '22 at 02:31
  • "enough information in the coefficients of the asymptotic series"... there's a lot of information in the words of that phrase - took me down another rabbit hole but I think I understand it now. But.... if non-perturbative effects spoil the expansion, do we expect the same non-perturbative effects in the upsidedown? I'm guessing we don't (upsidedown doesn't even have bound states). So there's that loophole in Dyson's argument again. Feel free to ignore me. I'm just musing out loud. – Jonathan Baxter Aug 03 '22 at 03:49
  • @JonathanBaxter It seems your doubt could be phrased as follows: "Yes, if $F(e^2)$ converges for fixed $e^2>0$, then $F(x^2)$ defines an analytic function of $x$ over some neighborhood of zero. Yes, this means $F(-e^2)$ converges and $F(-x^2)$ defines an analytic function similarly. But who cares? $F(-x^2)$ corresponds to a bad physical theory, but the real world uses $F(x^2)$, and the fictitious world either can't exist, or if it exists, must have a different $F$. Who says we need to be able to analytically continue $F$ to other worlds?" – WillG Aug 01 '23 at 07:19
  • At least, that's how I'm reading it, and I find it a very sensible complaint. – WillG Aug 01 '23 at 07:20
  • @WillG The point is that if $F(e^2)$ (the power series) converges, then $F(-e^2)$ must converge - however, $F(-e^2)$ shouldn't converge on physical grounds. But if that's the case, then $F(e^2)$ cannot converge either. – J. Murray Aug 01 '23 at 12:59
  • I guess the step I'm not understanding is: "$F(-e^2)$ defines a theory that could not exist" $\implies$ "$F(-e^2)$ should not converge." – WillG Aug 01 '23 at 13:18
  • @WillG That’s the physical part of the argument. Think about what interacting QED would look like if you switched the sign of $e^2$ everywhere. I gave the example of vacuum polarization in the comments. The vacuum becomes unstable to the formation of particle-antiparticle pairs, which means that one would expect physical quantities (e.g. scattering cross sections, decay rates, etc) to diverge (for example, now there are an unbounded infinity of energetically accessible states to decay into). This is the physical argument that $F(-e^2)$ should diverge. – J. Murray Aug 01 '23 at 13:53
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    @WillG You can take issue with that argument (though you should read Dyson and other far more eloquent physicists than me before you make up your mind), but that’s the essential character of the argument being made. – J. Murray Aug 01 '23 at 13:54
  • I see why changing the sign of $e^2$ leads to a bad theory, I just don't see why this should force $F(-e^2)$ to diverge. Taking a step back, the general argument is roughly of the form, "Consider mathematical object $X$, and use it to define a physical theory $T$. $T$ turns out to be pathological. Therefore, $X$, as a mathematical object, must also be pathological in some sense." It's the last sentence that feels murky to me. – WillG Aug 02 '23 at 02:40
  • I'm sure one can write down plenty of convergent series and other well-behaved mathematical objects, that, when used to define a physical theory in some way, still lead to an impossible theory. – WillG Aug 02 '23 at 02:41
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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 $e=0$.

hft
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mike stone
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  • 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. – Jonathan Baxter Aug 02 '22 at 16:52
  • @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... – hft Aug 02 '22 at 17:04
  • @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 $e^2>0$.

     – Jonathan Baxter Aug 02 '22 at 17:25
  • 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. – hft Aug 02 '22 at 17:29
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    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.

     – Jonathan Baxter Aug 02 '22 at 17:38