In QED, when dealing with the vacuum polarization and the photon propagator, some authors like Peskin & Schroeder introduce the so-called "1-particle irreducible" diagrams. These are defined as:

Let us define a one-particle irreducible (1PI) diagram to be any diagram that cannot be split in two by removing a single line.

So this is a "graphical definition", so that given a diagram we determine whether it is a 1PI or not by looking at whether or not a line can be removed leaving two diagrams that make sense by themselves.

That much I understand. What I don't understand is that Peskin & Schroeder then does the following: consider the 1-loop correction to the photon propagator. That would be the vacuum polarization diagram.

The authors denote its value by $i\Pi_2^{\mu\nu}(p)$. They then define $i\Pi^{\mu\nu}(p)$ to be "the sum of all 1PI insertions into the photon propagator". This is ilustrated by eq. (7.72)

enter image description here

Then they say on the bottom of p. 245 that the exact two point function is

enter image description here

Now I don't understand what he is doing here. For example, he claims that for $\Pi^{\mu\nu}(q)$ the Ward identity holds $q_\mu \Pi^{\mu\nu}(q)=0$.

My question is:

  1. What is the motivation for defining this $\Pi^{\mu\nu}$, namely to consider that "1-particle-irreducible insertions"?

  2. How do we deal mathematically with that? Because I just have one "pictorial" definition of what a single 1PI is, I have no idea what it actually means to consider "all possible 1PI insertions", and this confuses me.

  3. Why the full dressed propagator which is defined as the Fourier transform of $\langle \Omega |T\{A^{\mu}(x)A^{\nu}(y)\}|\Omega\rangle$ is expanded as that sum? The author doesn't seems to prove that.

Edit: Based on the answers I was thinking and I believe the point is that in the last equation the second term on the RHS is the sum over all 1PI's, the second is the sum over all diagrams with two 1PI pieces and so forth.

But it seems the author implies that: "the sum over all diagrams with two 1PI pieces is the same as the product of two sums of all 1PI's". Namely, I think the author is trying to write down the following (writing $G_0^{\mu\nu}$ for the bare propagator).


now I've tried to understand why the "sum over all diagrams with two 1PI pieces" is actually that, but I think I don't get it.

Let two diagrams be given that are decomposed in two 1PI pieces each. The first diagram has 1PI pieces with values $I_{\alpha\beta}$ and $II_{\alpha\beta}$ while the second has values $I'_{\alpha\beta}$ and $II'_{\alpha\beta}$. Summing them we have


now I can't reduce this to something with $I+I'$ and $II+II'$ which I think is what I need. What is wrong in my reasoning?


2 Answers 2

  1. There isn't really much of a motivation except that it's useful.

  2. There's really not much to deal with: You have the set of all Feynman diagrams. You define that a 1PI diagram is one that cannot become two non-trivial separate diagrams by cutting a single line. So each diagram that is not 1PI has such a line. The two pieces you get after cutting the single line are either 1PI or not, if not, you repeat the process. This decomposes every diagram as a string of 1PI diagrams, so the set of all diagrams is the union of "1PI diagrams", "2 1PI diagrams connected by a line", "3 1PI diagrams connected by a line", and so on. Saying one considers "all possible 1PI insertions" in the propagator simply means one considers the sum over all 1PI diagrams with two external legs.

  3. There's nothing to prove, really. You start from knowing that the dressed propagator is the sum over all diagrams, and since the strings of 1PI diagrams exhaust all diagrams, you can write the sum over all diagrams as the sum over 1PI diagrams plus the sum over all 2 1PI diagrams plus the sum over all 3 1PI diagrams and so on. This expansion is useful (as you will probably see shortly in the text you're reading) because it yields a geometric series in the 1PI contributions, which then lets us conclude that the 1PI contributions are precisely the mass shift between the bare and the dressed particles.

  • $\begingroup$ So let me see if I got this right. I pick one diagram, one incoming photon, something arbitrary in the middle and an outgoing photon. If it is not a 1PI there's a line in the middle that can be broken to yield two diagrams. So its form must be like in the second figure: photon-arbitrary-photon-arbitrary-photon. Now, I repeat this to this "arbitrary" thing until it can't be done anymore. Then any diagram with two external legs is a string of 1PI's and its value is the product of all such 1PI's. This breaks the set of diagrams as you've said right? $\endgroup$
    – Gold
    Jul 5, 2017 at 1:24
  • $\begingroup$ Now, when computing the dressed propagator, I first sum all 1PI's, this is the second term in the last figure with two external photons and a loop with 1PI inside. Then I pick all 2 1PI's connected by a line and sum them all. This is the third term in the last figure, symbolized by two external photons and two circles with 1PI's inside. Is that the point? $\endgroup$
    – Gold
    Jul 5, 2017 at 1:25
  • $\begingroup$ @user1620696 Yes, you understood correctly. $\endgroup$
    – ACuriousMind
    Jul 5, 2017 at 10:03
  • $\begingroup$ Just one thing, when writing down the geometric series you talk about, I'm getting something wrong. From the last figure I've posted the author seems to imply that "the sum of all 2 1PI diagrams connected by a line" is the same as the product of two sums of all 1PI diagrams together with a photon propagator in the middle. So the first diagram he writes as $G_0^{\mu\alpha}\Pi_{\alpha\beta}G_0^{\beta\nu}$ and the second $G_0^{\mu\alpha}\Pi_{\alpha\beta}G_0^{\beta\rho}\Pi_{\rho\sigma}G_0^{\sigma\nu}$ being $G_0$ the bare photon propagator. In my edit it doesn't seem to work out. What's wrong? $\endgroup$
    – Gold
    Jul 5, 2017 at 14:25
  • $\begingroup$ @user1620696 I don't really understand what you're doing in your edit, but the idea is that it's a variant of the Cauchy product. Just look at the two sums, on one hand the "sum of all 2 1PI diagrams connected by a line", on the other at "sum over all 1PI diagrams times the propagator times sum over all 1PI diagrams". You should be able to convince yourself that these are actually the same sets of diagrams, i.e. there's a bijection there. From every pair of 1PI diagrams you get one "2 1PI diagram with aline" and vice versa. That's it. $\endgroup$
    – ACuriousMind
    Jul 5, 2017 at 14:52

The point of defining the 1PI diagrams is that calculating all the diagrams (including the reducible ones) is redundant. Say you already calculated the lowest order for the self energy, the diagram with an electron loop. What if you have an electron loop and then another electron loop (so a bit like your second picture)? Is that any harder? No, it's just the value of one electron loop squared. So if we just calculate the 1PI diagrams we can get all the diagrams with very little extra work.

All possible 1PI insertions just means that, that inside of the shaded circle you can put anything as long as it connects with the external lines and it's 1PI. The pictorial intuition is a good intuition, because, again, the point of the 1PI diagrams is to simplify. If a diagram can be separated in half by a single cut, then it's just the product of two simpler diagrams.

As for the propagator, recall that earlier in the book you calculated the two point function for the scalar field and it turned out to be the sum of all diagrams with two external points at fixed positions $x$ and $y$. The Fourier transform is just the momentum space version of that. And the sum of all possible diagrams with two external photons is what P&S draw in the second picture, by definition of 1PI: if some diagram is not 1PI, it can be separated into two 1PI pieces.

In regards to your edit: what the book claims is only true if you include all diagrams at a given order. In your example, you've missed that you should also have a $I'-II$ and a $I-II'$ diagram. The sum of all four will give what you want.

  • $\begingroup$ I think I'm getting it. So in the last figure I've posted, the first term is the bare propagator, the second is the sum over all 1PI diagrams, the third is the sum over all 2 pieces of 1PI's connected by a line and so forth. An arbitrary diagram is a string of 1PI's, so it will be inside some of these sums and we have them all. Is this the idea? $\endgroup$
    – Gold
    Jul 5, 2017 at 1:48
  • $\begingroup$ Yes, that's correct. $\endgroup$
    – Javier
    Jul 5, 2017 at 3:03
  • $\begingroup$ @user1620696 I've answered the edited part of your question. $\endgroup$
    – Javier
    Jul 5, 2017 at 15:08
  • 1
    $\begingroup$ now I see. If I just pick two diagrams with 2 1PI pieces and sum them, this won't turn out to be a product of the two sums with a propagator. This will happen only when we include them all, which is what the author meant. Is that the point? $\endgroup$
    – Gold
    Jul 5, 2017 at 15:58

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