# Loop corrections to the propagator

LAHLH
Hi,

I have a couple of questions on this (mostly based on Srednicki ch14).

If I understand things correctly the difference between the exact propagator (always shown in bold in Srednicki) and the "usual propagator" from earlier chapter (i.e. the one corresponding to lines in the Feynman graphs of Chapter 9 say) is that propagator was for the free theory and this exact propagator is for the interacting theory.

I understand that by the definition $$\frac{1}{i} {\bf \Delta}(x_1-x_2) =\delta_1 \delta_2 iW(J) \mid_{J=0}$$, that this means sum over all connected diagrams which originally had two sources (2 external lines), with the sources removed(by the $$\delta_i$$'s), since all other diagrams are either killed by the differentiation (i.e. those diagrams with less than two J's) or are killed when we put J=0 at the end, because they still contain J's. But J=2 to start with survives.

It is obvious that the simplest and lowest order in g diagram is just the barbell type diagram. Since this has two external lines to start with, and has no vertices (so it is of order ~1), after the $$\delta_i$$'s act, we are left with just the free propagator.

To get the next order of g, which will be O(g^2), since to have a connected diagram with two external sources, this means at least 2 propagators hanging around with the end not connected to the source, connected to a vertex. Considering if we just had one vertex it would have the two source propagators connected but then just a loose line if we didn't have another vertex, so next contributing term must be of order O(g^2).

So far so good, however I don't understand now why my 2nd, on left, diagram isn't included by Srednicki as a O(g^2) correction? Is this something to do with the fact we don't need to include diagrams with tadpoles (diagrams that when a single line is cut falls into two diagrams one of which has no sources)?

Now we also have counterterms at this order, from chapter 9 we have:

$$Z(J)=exp[-\frac{i}{2} \int d^4x (\frac{1}{i}\frac{\delta}{\delta J(x)}) (-A\partial^{2}_x+Bm^2)(\frac{\delta}{\delta J(x)})] Z_Y(J)$$

This obviously acts to create a new vertex by acting on the $$\int J \Delta J$$$$\int J \Delta J$$ to label both ends of the propagator by x, i.e. attach both end of the propagator to this new type of 2 valent vertex, that we symbolise with an X. Since A, B are O(g^2) the first counterm is that shown on the right side of my picture.

But then why does this diagram also appear as the rightmost figure of Srednicki's 14.3 as an O(g^2) contributor, how can it possibly be both?

#### Attachments

• Feynmandiag.jpg
15.2 KB · Views: 408

## Answers and Replies

LAHLH
My next question concerns out Srednicki gets the momentum representation of these diagrams, if we were working in spacetime coords, my 3rd LHS diagram (or Srednicki's first of fig 14.1), would be given by $$\frac{1}{2}(ig)^2 \left(\frac{1}{i}\right) ^2 \int d^4x \int d^4y \Delta(x_1-x_2)\Delta(x_2-x_3)\Delta(x_3-x_2)\Delta(x_3-x_4)$$ (after expanding the $$Z_g$$ and just keeping the term in contributing to a term of this order, namely the 1)

How do you get from this to what he has, in momentum space, I mean I know it somehow involves Fourier transforms, but I can't quite make it work.

$$\Delta(x_1-x_2)=\int \frac{d^4k}{(2\pi)^4} e^{ik(x_1-x_2)} \tilde{\Delta(k^2)}$$

So now if I plug all these into what I had:

$$\frac{1}{2}(ig)^2 \left(\frac{1}{i}\right) ^2 \int d^4x_2 \int d^4x_3 \int \frac{d^4k_1}{(2\pi)^4} e^{ik_1(x_1-x_2)} \tilde{\Delta(k^{2}_1)} \int \frac{d^4k_2}{(2\pi)^4} e^{ik_2(x_2-x_3)} \tilde{\Delta(k^{2}_2)} \int \frac{d^4k_3}{(2\pi)^4} e^{ik_3(x_3-x_2)} \tilde{\Delta(k^{2}_3)}\int \frac{d^4k_4}{(2\pi)^4} e^{ik_4(x_3-x_4)} \tilde{\Delta(k^{2}_4)}$$

$$\frac{1}{2}(ig)^2 \left(\frac{1}{i}\right) ^2 \int d^4x_2 \int d^4x_3 \int \frac{d^4k_1}{(2\pi)^4} \frac{d^4k_2}{(2\pi)^4} \frac{d^4k_3}{(2\pi)^4} \frac{d^4k_4}{(2\pi)^4} e^{ik_1 x_1} \tilde{\Delta(k^{2}_1)} e^{ix_2(k_2-k_3-k_1)} \tilde{\Delta(k^{2}_2)} e^{ix_3(k_3+k_4-k_2)} \tilde{\Delta(k^{2}_3)} e^{-ik_4x_4)} \tilde{\Delta(k^{2}_4)}$$

Performing the x integrals:

$$\frac{1}{2}(ig)^2 \left(\frac{1}{i}\right) ^2 \int \frac{d^4k_1}{(2\pi)^4} \frac{d^4k_2}{(2\pi)^4} d^4k_3 d^4k_4 e^{ik_1 x_1} \tilde{\Delta(k^{2}_1)} \delta^4(k_2-k_3-k_1) \tilde{\Delta(k^{2}_2)} \delta^4((k_3+k_4-k_2) \tilde{\Delta(k^{2}_3)} e^{-ik_4(x_4)} \tilde{\Delta(k^{2}_4)}$$

Now doing the $$k_1, k_2$$ integrals:

$$\frac{1}{2}(ig)^2 \left(\frac{1}{i}\right) ^2 \int \frac{d^4k_2}{(2\pi)^4} \frac{d^4k_3}{(2\pi)^4} e^{i(k_2-k_3)x_1} \tilde{\Delta((k_2-k_3)^{2})} \tilde{\Delta(k^{2}_2)} \tilde{\Delta(k^{2}_3)} e^{-i(k_2-k_3)(x_4)} \tilde{\Delta((k_2-k_3)^{2})}$$

Now I'm kinda stuck to get what he has as the first term in 14.4

Cheers for any help on either question, sorry about the long post.

PS

Could I just do the FT over say k3:

$$\frac{1}{2}(ig)^2 \left(\frac{1}{i}\right) ^2 \int \frac{d^4k_2}{(2\pi)^4} \frac{d^4k_3}{(2\pi)^4} e^{ik_2(x_1-x_4)} \tilde{\Delta((k_2-k_3)^{2})} \tilde{\Delta(k^{2}_2)} \tilde{\Delta(k^{2}_3)} e^{-ik_3(x_4-x_1)} \tilde{\Delta((k_2-k_3)^{2})}$$

hmm but then I'm left with strange spacetime deltas with nowhere to go.....

Last edited:
DrFaustus
LAHLH -> Regarding the second question. You should plug your first formula from the second post into the LSZ formula, basically getting two additional spacetime integrations and two more delta functions because of the Klein-Gordon operators acting on the Feynman propagators connected to the external points. Then you get a lot of simplifications and should be able to extract a momentum conservation delta and obtain with a momentum integral convolution of the two internal (loop) propagators.

Regarding the first question -> You don't include your second diagram exactly because you don't include tadpoles. And you don't include tadpoles because you fix the counter term Y (in Srednicki's notation) precisely in such a way that the tadpole diagrams do not appear.

The A and B are of O(g^2), but you don't know that! At least not a priori. You essentially fix A and B so that they cancel your infinites arising through the divergent loop integrals, and you eventually find out that they are of O(g^2). And being of O(g^2) does NOT mean A = k g^2, B = h g^2 for some k and h. A and B are power series in g, that is $$A = \sum_n A_n g^n, \qquad B = \sum_n B_n g^n$$. And you fix them at every order by requiring that the divergences at some corresponding order get removed.
(I suspect that your question above has a spelling mistake and that you don't understand how can it be bot O(g^2) and O^(g^4), as in fig 14.3 he has O(g^4) contributions.)

Hope that helps.

LAHLH
LAHLH -> Regarding the second question. You should plug your first formula from the second post into the LSZ formula, basically getting two additional spacetime integrations and two more delta functions because of the Klein-Gordon operators acting on the Feynman propagators connected to the external points. Then you get a lot of simplifications and should be able to extract a momentum conservation delta and obtain with a momentum integral convolution of the two internal (loop) propagators.

I don't quite understand the rationale behind this (I have every confidence it will work to give me the desired result when I do it explicitly later, after reading what you have said), but I don't understand why you'd want to put this into the LSZ now. I know the LSZ does contain $$\langle 0\mid T \phi(x_1)\phi(x_{1'}) \mid 0 \rangle$$ to give the amplitude $$\langle f\mid i\rangle$$ for 1 particle in 1 particle out (i.e. no scattering just propagating). But we're not trying to calculate the amplitude, $$\langle f\mid i\rangle$$, we're trying to calculate the propagator, just in momentum space. [/QUOTE]

Regarding the first question -> You don't include your second diagram exactly because you don't include tadpoles. And you don't include tadpoles because you fix the counter term Y (in Srednicki's notation) precisely in such a way that the tadpole diagrams do not appear.

Ah, I suspected as much, thanks alot.

The A and B are of O(g^2), but you don't know that! At least not a priori. You essentially fix A and B so that they cancel your infinites arising through the divergent loop integrals, and you eventually find out that they are of O(g^2). And being of O(g^2) does NOT mean A = k g^2, B = h g^2 for some k and h. A and B are power series in g, that is $$A = \sum_n A_n g^n, \qquad B = \sum_n B_n g^n$$. And you fix them at every order by requiring that the divergences at some corresponding order get removed.

I see that now, my mistake was thinking they were just proportional somehow to g^2, and not knowing they contained higher order contributions too, so of course the counterm I had drawn does appear at O(g^4) and indeed at all higher orders too.

On a related note, does that mean the first diagram in 14.1 (the bubble with two arms, not the counterm), also appears at all orders, because we expanded $$Z_g=1+O(g^2)+...$$. So this exact same diagram should appear in the O(g^4) contributions to $$i\Pi(k^2)$$. Srednicki doesn't show it in the diagrams he does show for O(g^4) in 14.3 however, maybe he is just showing a sample of diagram contributing at this order?

(I suspect that your question above has a spelling mistake and that you don't understand how can it be bot O(g^2) and O^(g^4), as in fig 14.3 he has O(g^4) contributions.)

That is what I didn't understand, not sure what spelling mistake you're refering too however? I was confused as why the ------X------ counterm could be both in fig 14.1 (O(g^2)) and fig 14.3 (O(g^2)). But I think thanks to your post I now understand my error.

Thanks alot I appreciate it

Geometrick
Just a quick question, is the 2nd diagram in Fig 14.1 of order g? I don't see why there are two powers of g in this diagram. We have 2 propagators and one counterterm, giving us one power of g is Y = O(g). Can anyone clear up this quick question? Basically, I'm asking, is it ~ g and hence obviously O(g^2)?

LAHLH
This is a counter term to do with A and B not Y (see ch9, below eq(9.9)) and these are O(g^2)

Geometrick
Thank you so much!