# Understanding Feynman Diagrams in Loop Corrections to the propagator $\phi ^3$ theory [closed]

I found other posts talking about the same chapter in the same book, but none of them were exactly about what I am asking here.

In Srednicki's chapter 14 (Loop corrections to the propagator), we are presented the two diagrams that give the $$O(g^2)$$ corrections to the propagator.

Where in

$$\frac{1}{i}\tilde{\boldsymbol{\Delta }}(k^2 )=\frac{1}{i}\tilde{\Delta }(k^2 )+ \left( \frac{1}{i}\tilde{\Delta }(k^2 ) \right) \left[ i\Pi (k^2 ) \right] \left( \frac{1}{i}\tilde{\Delta }(k^2 ) \right) +O(g^4 ).\tag{14.2}$$

We have

$$i\Pi (k^2 )=\frac{1}{2}(ig)^2 \left( \frac{1}{i} \right) ^2 \left( \int \frac{d^dl}{(2\pi )^d}\tilde{\Delta }((l+k)^2)\tilde{\Delta }(l^2 ) \right) -i(Ak^2 +Bm^2 ) +O(g^4 ).\tag{14.4}$$

And to be precise:

$$\tilde{\Delta }(k^2 )=\frac{1}{k^2 +m^2 -i\epsilon }\tag{14.3}$$

The contribution from the second diagram is clear (I think) to me. (To be precise, this is given by rule 9, page 77, on chapter 10 on Scattering amplitudes)

Include diagrams with the counterterm vertex that connects two propagators, each with the same four-momentum $$k$$. The value of this vertex is $$-i(Ak^2 +Bm^2 )$$, where $$A=Z_{\phi }-1$$ and $$B=Z_m-1$$.

So the value of this diagram (the one with the counterterm vertex) is

$$\left( \frac{1}{i}\tilde{\Delta }(k^2 ) \right) \left( -i(Ak^2 +Bm^2 ) \right) \left( \frac{1}{i}\tilde{\Delta }(k^2 ) \right)$$

Where $$-i(Ak^2 +Bm^2 )$$ is what really enters in the definition of $$i\Pi (k^2 )$$.

Is that correct?

Then for the first diagram (with loop of momentum $$l$$) we again have:

• The two propagators (both of momentum $$k$$) that will sandwich $$i\Pi$$
• Factor of $$1/2$$ due to symmetry
• We have 2 vertices so a factor of $$(ig)^2$$
• Loop integral

My question then is, how do I write down the contribution of the loop? I know rule 7 states

A diagram with $$L$$ closed loops will have $$L$$ internal momenta that are not fixed by rule number 5. Integrate over each of these momenta $$l_i$$ with measure $$d^4 l_i /(2\pi )^4$$

Why is (as Srednicki shows):

$$\int \frac{d^dl}{(2\pi )^d}\tilde{\Delta }((l+k)^2)\tilde{\Delta }(l^2 )$$

the integrand? What are the factors of $$\tilde{\Delta }((l+k)^2)$$ and $$\tilde{\Delta }(l^2 )$$ representing there?

• They are propagators as you've already had defined in equation $(14.3)$ referring to the two parts of the loop with momenta $k$ and $k+l$ no? Commented Jul 8 at 18:25
• Yes, they are propagators. I am just not sure as to why there are two propagators for that loop, and with those momentums. Commented Jul 8 at 18:31
• Perhaps you're missing Srednicki's rule 6 which says for each internal line labeled with momentum $p$ you get a factor of the propagator with momentum $p$. Those particular momenta are dictated by momentum conservation, since $k + \ell - (k+\ell)=0$. But you could split up the momenta in different ways, e.g. $\ell+k/2$ on the upper leg and $\ell - k/2$ on the lower leg, as that still satisfies $k + (\ell - k/2) - (\ell + k/2)=0$. That's fine, these are equivalent by a change of variables of $\ell$ and will produce the same result. Commented Jul 8 at 21:48
• Thanks a lot @SethK ! Yeah, this is the kind of answer I was looking for. In the equation you wrote $k+l-(k+l)=0$, the first $k$ comes from the horizontal line, right? Further, why do we subtract the upper leg? I am aware we could flip the way we add things up, but what exactly are we trying to do here? Is it that we want the total momentum at that first vertex to be zero? So we have incoming $+k$ and from the lower leg an incoming $l$, but from the upper one it is directed away from the vertex so it is negative. Is this line of thought correct? Thanks again! Commented Jul 8 at 23:58
• @SethK you might want to extend that comment as an answer to the question. Commented Jul 9 at 0:30

I was asked to expand my comment to an answer. Maybe it would be useful to keep in mind a more physical notion of what the loop diagrams represent.

Recall we are trying to understand our interacting quantum field theory by using perturbation theory around a free field theory. The interactions are encoded in the Lagrangian of our theory where multiple field operators at a single point, like $$\phi^3(x)$$. These interactions give rise to vertices in the Feynman diagrams, where three $$\phi$$ particles interact. These interactions all conserve momentum, as they should in a Poincare-invariant theory. As Srednicki derives, the Feynman diagrams capture contributions from interactions to correlation functions of the fields, which we've seen in Sec 5 tell us about scattering amplitudes.

A given Feynman diagram is roughly capturing a scattering process with some ingoing particles of definite momenta and outgoing particles of definite momenta. These incoming and outgoing lines are labeled with those fixed momenta, using arrows to denote the direction of the labeled momentum through the diagram. Particles with a given momenta 'propagate' along the edges of the graph. In the interior of the diagram, these lines meet at vertices with momentum conserving interactions. If momenta $$k_1$$ and $$k_2$$ go into a vertex, the third line must have $$k_1 + k_2$$ going out of that vertex (or equivalently $$-k_1-k_2$$ going into that vertex).

In a tree-level diagram, the requirement of momentum conservation at each vertex fixes the momenta of all of the internal lines. You can find such examples in Srednicki's Fig 10.1-2. You can track the momenta through the diagram by following the arrows, and the sum of the ingoing momenta at any vertex minus the outgoing momenta at that vertex must vanish. Keep in mind the sum of ingoing momenta in external lines minus the outgoing momenta of external lines also must vanish. Keep also in mind that there is nothing sacred about the direction of the arrow. Whether you label something with an right-pointing arrow with momentum $$q$$ or a left-pointing arrow with momentum $$-q$$ it is exactly the same, but you must make a choice.

In the quantum theory we must sum over all possible interactions that can produce the given ingoing/outgoing states (all possible time evolution which could have occurred in between), which corresponds to summing over all Feynman diagrams with those external lines. In principle this is a sum over arbitrary numbers of interactions taking place in the interior---arbitrarily complex internal graphs---but for analytical calculations we are interested in the corrections with a fixed number of interaction vertices. At some order in perturbation theory this is going to include graphs which have internal closed loops. For these graphs, the requirement of momentum conservation at each vertex no longer fixes the momentum of all of the internal lines, as you will see by tracking the arrows through the diagram.

In the case of the two-point function at one loop, the incoming and outgoing external lines must have the same momentum $$k$$. In the one-loop diagram, for the internal lines let us fix the arrow direction as in Fig 14.1 and label the two lines with general momenta $$q_1$$ above and $$q_2$$ below. Now if we consider (incoming minus outgoing) momentum conservation at the left vertex we have $$k + q_2 - q_1 = 0$$. With this we can solve for $$q_1 = k + q_2$$. But now momentum conservation at the right vertex works automatically $$q_1 - q_2 - k = (k+q_2) - q_2 - k$$ is trivially satisfied and does not give us a constraint on q_2.

Recall in the quantum theory we must sum over every process with these external lines. So we must integrate over all of the possible internal momenta $$q_2$$ as $$\int \frac{dq_2}{(2\pi)^d}$$. If we 'change variables' to $$q_2 = \ell$$ we have the diagram labeled as in Fig 14.1. But we could also change variables to $$q_2 = \ell - k/2$$, which would give the upper line $$q_1 = k+q_2 = \ell + k/2$$. Sometimes in complicated calculations in can be useful to use this freedom cleverly to simplify algebra. But the answer doesn't change no matter what you choose.

...

And to be precise:

$$\tilde{\Delta }(k^2 )=\frac{1}{k^2 +m^2 -i\epsilon }\tag{14.3}$$

...

Why is (as Srednicki shows):

$$\int \frac{d^dl}{(2\pi )^d}\tilde{\Delta }((l+k)^2)\tilde{\Delta }(l^2 )$$

the integrand?

Look at the left side of your picture:

There are two internal propagators, the one on the top of the loop has momentum $$k+\ell$$, and so is associated with a factor of $$\Delta(k+\ell)$$. The one on the bottom of the loop has momentum $$\ell$$ and so is associated with a factor of $$\Delta(\ell)$$.

What are the factors of $$\tilde{\Delta }((l+k)^2)$$ and $$\tilde{\Delta }(l^2 )$$ representing there?

They are literally the same thing as you already wrote down in the above-quoted Eq. 14.3. The propagators are evaluated at arguments $$k+\ell$$ and $$\ell$$, respectively: $$\Delta(k+\ell) = \frac{1}{(k+\ell)^2 +m^2 - i\epsilon}$$ $$\Delta(\ell) = \frac{1}{\ell^2 +m^2 - i\epsilon}$$