Pretty much everywhere I look it is stated that the full two point Green function (let's say for the Klein-Gordon field) is a geometric series in the one particle irreducible diagrams, ie. in momentum space,

$$G(k) = G_0(k)+G_0(k)\Sigma(k)G_0(k)+G_0(k)\Sigma(k)G_0(k)\Sigma(k)G_0(k) + \dots $$ $$ = G_0(k)\big(1+\Sigma(k)G_0(k)+\Sigma(k)G_0(k)\Sigma(k)G_0(k)+\dots\big)$$

and that the sum of this is, using the geometric series formula, $$G(k)=\frac{G_0(k)}{1-\Sigma(k)G_0(k)}=\frac{1}{k^2+m^2-\Sigma(k)-i\epsilon}$$

(for instance on the first page on the top right hand corner of here, or the bottom/top of pages 56/57 here).

However, what is the justification for summing this geometric series in that way? It never seems to be justified and it doesn't seem that $|\Sigma(k)G_0(k)|<1\,.$ Even if I do renormalize the mass $m$ so that $m_R^2 := m^2-\Sigma(k)$ is finite, if the geometric series summation isn't justified the infinities won't cancel out and everything will still diverge. Is there a resummation going on implicitly? Is this step completely nonperturbative?

$\mathbf{Edit\;1}$: Basically as I see it, the situation is this: I ask what the amplitude for the propagator is (in momentum space), and you say $$\frac{G_0(k)}{1-\Sigma(k)G_0(k)}=\frac{1}{k^2+m_R^2-i\epsilon}$$ where $m_R^2=m^2-\Sigma(k)$ is the (finite) mass. Then I notice that you can calculate it approximately using the perturbation series given by the theory, which is $$G_0(k)\big(1+\Sigma(k)G_0(k)+\Sigma(k)G_0(k)\Sigma(k)G_0(k)+\dots\big)\,.$$ However I notice that $\Sigma(k)G_0(k)>1$ and I know that for $x>1$ the perturbation series given by $$1+x+x^2+\cdots$$ is not a good approximation to $\frac{1}{1-x}\,,$ at any order of the perturbation series. So it would seem that perturbation theory fails because it doesn't give a good approximation, at any order, for the propagator. So then I ask what is the justification for this whole thing in the first place? Are they basically just seeing that the perturbation series for the propagator is the same as the perturbation series for $\frac{1}{1-x}\,,$ but realizing that $|x|>1$ and then assuming that the actual propagator must be $\frac{1}{1-x}\,?$ Because this would seem to be a completely nonperturbative step.

$\mathbf{Edit\;2}$: I am going to do the actual computation which is confusing me, and if someone can point out where it is wrong (if it is) that would be a great help. By the way I did a Wick rotation previously, but I am not going to do that this time:

We have that $$\frac{G_0(k)}{1-\Sigma(k)G_0(K)}=\frac{1}{k^2-m_R^2}$$

where $m_R$ is the physical (and finite) mass, and where $G_0(k)=\frac{1}{k^2-m^2}$ where $m$ is the cut-off dependent bare mass. Rearranging, I get $$\Sigma(k)G_0(k)=1-G_0(k)(k^2-m_R^2)=1-\frac{k^2-m_R^2}{k^2-m^2}\,.$$ Let's look at the region where $m_R^2<k^2<m^2\,.$ $m^2\to\infty$ as the cut-off is removed, so this is a very large region. Now it looks to me like that in the above expression for $\Sigma(k)G_0(k)\,,$ that the right hand side is ALWAYS greater than one, and even possibly near infinity for certain values of $k$ (or maybe it doesn't obtain values near infinity because $k$ has to be restricted to values below the cut-off, but this isn't really important). This would seem to make the whole geometric summation suspicious. Is there a mistake?

If I have a basic misunderstanding of how this works then I'd like to know, this is the most basic case of renormalization in QFT but I don't understand it.

  • 1
    $\begingroup$ This things are hardly ever justified in the literature. I guess, the best a-posteriori justification is that it simply works: perturbative QFT gives sensible results on the $n$-loop level, but I hope somebody will answer this question to give a better one. $\endgroup$ Sep 22, 2016 at 18:32
  • $\begingroup$ Theoretical physicists love to talk about asymptotic expansion in these cases. But I also hope that someone in the field answers with more details and clarifies the issue. $\endgroup$
    – DelCrosB
    Sep 22, 2016 at 18:39
  • $\begingroup$ I am not sure you are aware that the two propagators $\frac{1}{k^2+m^2}$ and $\frac{1}{k^2-m^2}$ are not wick rotated part but actually are propagators in two different metric convention used in literature. They are written such that $k_0$ has poles $\sqrt {|k|^2+m^2}$ on real axis. The problem you are describing is just a result of the failed analysis you have done and would also appear with $\frac{1}{k^2+m^2}$. I will address this in further comments. $\endgroup$
    – ved
    Sep 29, 2016 at 6:20
  • $\begingroup$ Using two different metric convention is not going to change any physics and as such $\Sigma(k)$ will have the same behaviour. In the first case, you had $m_R^2=m^2-\Sigma(k)$ as the physical mass as a result bare mass has to go to $+\infty$ to cancel the $+\infty$ coming from $\Sigma(k)$ to get a finite result. In the second case, bare mass has to go to $-\infty$ to cancel the $+\infty$ coming from $\Sigma(k)$ in $-(m^2+\Sigma(k))$ as changing the metric convention won't change $\Sigma(k)$ behaviour. $\endgroup$
    – ved
    Sep 29, 2016 at 6:32
  • $\begingroup$ This would basically lead to $m_R^2<k^2<-\infty$, which does not make sense and you might want to reverse it to $-\infty<k^2<m_R^2$. If $k^2$ is always than physical mass then it would never appear as a pole in the propagator and you will never be able to generate physical mass by radiative correction as a result your whole calculation is for nothing and that's why I was treating k in high-energy limit in the answer. $\endgroup$
    – ved
    Sep 29, 2016 at 6:39

4 Answers 4


You are asking a legitimate question. We can turn the whole situation on its head. Rather than beginning with the geometric series $$ \begin{equation} G(k) = G_0(k)+G_0(k)\Sigma(k)G_0(k)+G_0(k)\Sigma(k)G_0(k)\Sigma(k)G_0(k) + \dots, \quad (1) \end{equation} $$ and arriving at $$ G(k)=\frac{G_0(k)}{1-\Sigma(k)G_0(k)}=\frac{1}{k^2+m^2-\Sigma(k)-i\epsilon} ,\quad (2) $$ we can go the other way around. Instead, we shall assert that eq. (2) is valid and call the dubious expansion of eq. (1) into question, and rightfully so (OP has called out the folly in assuming $|\Sigma(k)G_0(k)|<1$).

If you do your homework by poking around the QFT literature, you will uncover plenty of non-perturbative ways of deriving eq. (2) without resort to eq. (1). The old school perturbative QFT makes intuitive sense, but is a mathematical train wreck.

  • 2
    $\begingroup$ Can you give a reference for a book that derives this in a non perturbative way? $\endgroup$
    – Suriya
    Dec 20, 2021 at 11:47
  • 1
    $\begingroup$ @Suriya, try any book detailing Schwinger-Dyson Equation and its first order approximate solution. $\endgroup$
    – MadMax
    Mar 9, 2022 at 21:56

Perturbative expansions are typically divergent in field theory so I would venture that the full resumation of the 1 particle irreducible diagrams is actually divergent.

But this is no trouble if you just want to renormalize. Keep in mind that in practice you always renormalize at finite order in your expansion parameter so there really is no need to worry about the infinite resummation. The sum of the truncated geometric series is well defined, so in order to renormalize you define the counterterms such that your propagator is finite at the order on which you are doing your computations.

  • 1
    $\begingroup$ OP's question was: how can be taking the sum of the complete geometrical series (which diverges, so the expression for the sum is not correct) justified in actual computations. I have no doubt that what you wrote is true, but it does not answer OP's question. $\endgroup$ Sep 23, 2016 at 6:30
  • $\begingroup$ @SolenodonParadoxus I answered it in the sense that I have said that the complete series is (most likely) divergent so rigorously it is not justified to do it. When have you seen the resummation of the geometric series of all 1PI diagrams in an actual computation? $\endgroup$
    – Yossarian
    Sep 23, 2016 at 9:57
  • $\begingroup$ @AnarchistBirdsWorshipFungus Are you suggesting that if I set $m^2 = m_R^2+\Sigma(k)$ for some fixed $m_R\,,$ then the perturbation series, at some order, is a good approximation to $\frac{1}{k^2+m_R^2-i\epsilon}$? Because it doesn't seem to be and if not how is the summation justified? $\endgroup$
    – JLA
    Sep 23, 2016 at 17:50
  • $\begingroup$ JLA are you aware that even after renormalization, when you get finite contributions at every order in perturbation theory, and even when you are considering the coupling much smaller than one the radius of convergence of perturbative expansions is zero? this is the key of my point. The full summation of the perturbative expansion is (believed) to be divergent. Chech this physics.stackexchange.com/q/70411 . Therefore most likely it is never legitimate to sum all the diagrams, so don't worry about doing it $\endgroup$
    – Yossarian
    Sep 30, 2016 at 13:47
  • $\begingroup$ The full $\Sigma$ after summing all diagrams even adding all counterterms is (most likely) divergent. $\endgroup$
    – Yossarian
    Sep 30, 2016 at 13:48

Your basic misunderstanding is that $|\Sigma(k)G_0(k)|>1$, which is not true and basically three principles are used at least to make sure that $|\Sigma(k)G_0(k)|<1$. First is very basic which is the renormalizability of the theory and second is local nature of counterterms. A third ingredient is not needed except in case of scalar particles which is termed as "naturalness" in the standard model.

The mass correction term $\Sigma(k)$ is a sum of infinitely many graphs contributing to the self energy of the particle which in general will have a linear divergence (like self-energy of electron) or quadratic divergence (like self energy of photon) in high-energy limit. In case of particles other than spin zero different regularization schemes can be used to make this behaviour only logarithmically divergent and a redefinition of mass parameter ( renormalization) absorbs this divergence.

This basically tells you that $\Sigma(k)$ behaves as $ln(k/m)$ with k can be replaced by a scale $\Lambda$ and you will have $|\Sigma(k)G_0(k)|$ behaviour as $ln(k)/k^2$ as $$G_0(k)=\frac{1}{k^2+m^2}$$

Now $ln(k)/k^2$ is always less than $1$ (as $ln(k)-k^2$ is always negative).

In case of scalar particles however you have $\Sigma(k)$ behaving as badly as quadratically divergent (not enough symmetries in the theory), for which you need to have a lot of cancellation involving $\Lambda^2$ term to smoothen out the resulting integrals to not diverge as badly as quadratic. There is however no reason to expect that such cancellations should "naturally" occur as a result you need to fine-tune the theory to obtain a proper cancellation in which you can have $m^2_{physical}<<\Lambda$. As a result of these cancellations, you will have a soften up behaviour of $\Sigma(k)$ which will not diverge as badly as $\Lambda^2$ and you will have $|\Sigma(k)G_0(k)|<1$ as required.

Fine-tuning problem however is not a physical solution and hence must be supplemented with extra symmetry of theory. A case of which is supersymmetry which supplies the symmetry behaviour of spin 1/2 superpartner to make the divergence logarithamic but that is another question.

I hope this explain all your concerns and try to get a grab of good qft book like Srednicki's qft.

  • 1
    $\begingroup$ Apart from domain related issues, note that the rhs of your first equation is a well defined operator precisely if the geometric series with $Q=BA^{-1}$ converges. (take, for instance, $A=1$, $B=2$). $\endgroup$
    – pppqqq
    Sep 23, 2016 at 6:39
  • $\begingroup$ All right, see above. $\endgroup$
    – ved
    Sep 23, 2016 at 9:51
  • $\begingroup$ @ved I appreciate your response, can you see my edit to my question? Hopefully it's more clear now. $\endgroup$
    – JLA
    Sep 23, 2016 at 18:11
  • $\begingroup$ @JLA I have removed the unnecessary portions and made it as clear as I can. $\endgroup$
    – ved
    Sep 24, 2016 at 7:40
  • $\begingroup$ @ved Thank you for your response. Can you see the "Edit 2" in my post? I wrote it largely in response to your post. $\endgroup$
    – JLA
    Sep 24, 2016 at 19:42

You may consider this as a selective diagramm summation and a definition of eventually divergent series as a finite function. This was known even in early Euler times how to sum such series when the formal expanstion parameter is not actually small. Read Hardy's book on this subject.

The true problem is not here. In Classical Electrodynamics one can calculate exactly the back reaction, and elas, it is a divergent addendum to the particle mass. It is because the back reaction (self-action) is a mainly a self-induction of a point-like charge. No external force can accelerate a charge due to this infinite selft-induction. Note, the self-action idea is an ansatz, not something unquestionable. That is why we have to discard the mass corrections in practically any theory: even finite, they are unnecessary because we had the observed mass in the original equations. Thus, discarding (mass renormalization).

In addition, the electron propagator is not gauge invariant.

Worse, in QED there always a practically unity probability to emit soft photons because the electron is permanently coupled to the photon oscillators. Your exercise does not take it into account; in QED it is taken into account later on. An electron is actually an infra-particle (google it).


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