Explicit computation of mass counter term diagrams

[CAF123]

In most treatments of the mass renormalisation in dim reg, I see sources find the mass counter term by extracting the coefficient of e.g the $m^2$ term accompanied by a pole in epsilon. I know the mass counter term is found by placing an X on diagrams where there is usually a self energy correction and the counter term subtracts off the corresponding UV divergence, typically the ‘vertex’ Feynman rule is of the form $-i(Z_m-1)m$ etc..

If I write down the Feynman rules for the mass counter term diagram, given the above rule, it seems the diagrams are always the tree level multiplied by the above ‘vertex’ factor. Is there any additional structure to the above or is that correct?

There has to be some additional structure otherwise one could not separate the single line at tree level with the two obtained in the counter term diagram by placing a ‘vertex’ X.

 

[king vitamin]

If I write down the Feynman rules for the mass counter term diagram, given the above rule, it seems the diagrams are always the tree level multiplied by the above ‘vertex’ factor. Is there any additional structure to the above or is that correct?

This certainly isn't the only diagram involving the counterterm vertex. At higher orders, you must consider the insertion of the counterterm vertex into loops, and these will be needed to correctly renormalize the theory. But at every order in perturbation theory, generically the the highest-order divergence needs some cancellation by the tree-level counterterm vertex you're referring to, in addition to the insertion of counterms into loop diagrams.

In particular, the counterterm cannot be momentum-dependent. Generically, when you calculate a loop diagram in dim reg, you will find terms with $1/\epsilon$ poles which multiply momentum-dependent terms - but since these cannot be canceled by the simple $-i(Z_m - 1)m$ term you mentioned, they must instead be perfectly canceled by the insertions of $-i(Z_m - 1)m$ in loop diagrams between tree level and the order of perturbation theory you are working at. If it does not, then you must have made a mistake at some point in your calculation.

This is part of why I prefer dim reg - it tells you when you've made a mistake! If you've missed a diagram, your calculations will sometimes slap you in the face and tell you that you're wrong.

 

[CAF123]

Thanks @king vitamin Actually when I wrote my post I had a 1 loop order in mind so that all $1/\epsilon$ divergences in diagrams with a self energy correction to the tree level diagrams would be canceled by the corresponding mass counter term in place of the self energy.

What I'm wondering however is when one puts an X on the tree level diagram, this splits a single internal line into two internal lines, that is

--------------

is now

------X-------However, this is not reflected in a simple insertion of $-i(Z_m-1)m$ because it is multiplicative to the tree level diagram. So, at one loop, we would have schematically $$-i(Z_m-1)m \cdot \text{tree} + \text{1 loop} = \text{finite}?$$ From the above, I kind of expected to see additional structure in the Feynman rule which results in ------ to ----X-----, i.e. with a gamma matrix or something.

 

[vanhees71]

In a renormalizable theory the self-energy of a particle can be made finite by the wave-function renormalization and mass counterterms. For details, see my QFT manuscript:

https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

 

[CAF123]

@vanhees71 thanks, that is clear to me - any term with a pole proportional to pslash (mass) is accounted for in wave function (mass ) counter term. I’m more concerned with the expressions for the mass counter term diagrams after one uses the Feynman rules on them. In my last post I say it appears the mass counter term diagram at one loop is simply a mass counter term vertex multiplicative to the tree level graph but is that correct?

 

[vanhees71]

Maybe it's easier to understand, where your problem is, if you give a concrete example.

The usual proof of renormalizability goes by induction (usually the number of loops, i.e., powers of $\hbar$). To evaluate a proper one-particle-irreducible vertex diagram at $L$ loops you first need to renormalize all probably present subdivergences inside the diagram by adding the corresponding counter terms. The most clear way to understand this is to study BPHZ and Zimmermann's "forest formula" (see Sects. 5.8 ff in my manuscript).

 

[CAF123]

Maybe it's easier to understand, where your problem is, if you give a concrete example.

I was just thinking of any tree level process with a mass counter term insertion on one of the propagators accounting for the mass divergence in the corresponding one loop graph where the mass counter term is replaced with a self energy. Now I want to look at writing out the Feynman rules for the mass counter term diagram. Using the rule $-i(Z_m-1)$ for the mass counter term vertex means the Feynman rules amplitude for this diagram is simply the tree level graph multiplied by this factor. Yes?

 

[vanhees71]

Shouldn't it rather be a factor $-\mathrm{i}(Z_m-1)m^2$ for the counter-term vertex (with of course two legs) and then there should be also one (if the self-energy insertion is on a truncated external leg of the proper vertex function you evaluate) or two (if the self-energy insertion is in an inner propagator line) propagators. As I said, a concrete example may help more for a better understanding!

 

[CAF123]

Shouldn't it rather be a factor $-\mathrm{i}(Z_m-1)m^2$ for the counter-term vertex (with of course two legs) and then there should be also one (if the self-energy insertion is on a truncated external leg of the proper vertex function you evaluate) or two (if the self-energy insertion is in an inner propagator line) propagators. As I said, a concrete example may help more for a better understanding!

Oh yes, I missed the m^2 factor. The general procedure is clear to me but never in books do I see them write out the Feynman rules for the mass counter term diagram, they just write it out pictorially (Sum of mass counter term diagram plus one loop is finite) and say the divergence is put into the counter term order by order. They compute the one loop diagrams, extract the pslash and m factors then write out $Z_m$ and $Z_2$ by hand. The general process I’m considering is associated with my project but Just thought my questions above are such that they can be answered without reference to anything in particular.

 

[vanhees71]

Sure, you define your counter terms such that your renormalization conditions are fulfilled. If you have no massless particles in the theory you can choose the on-shell scheme. In other cases a mass-independent renormalization scheme like (modified) minimal subtraction defined by using dimensional regularization is better suited to avoid additional trouble with IR and collinear divergences if massless particles are present.

Indeed the counterterms are defined order by order in perturbation theory. You start with tree-level diagrams, which are finite and thus have no counter terms (using the finite renormalized parameters of the theory). Then you go to one loop and evaluate all the divergent proper vertex diagrams to determine the corresponding counter terms at order $\hbar$. That's the easy part, because there are by topology of the one-loop diagrams no subdivergences and particularly no overlapping divergences.

At order $\hbar^L$ ($L$=number of loops) you write down the naive diagram and then add all counterterms up to order $\hbar^{L-1}$ for divergent subdiagrams, treating also the overlaping divergences in this way. According to the BPHZ analysis (with Zimmermann's forest formula as the final result) then there are no more overlapping subdivergences left, and the maybe present overall divergence then defines together with the renormalization conditions the counter terms at order $\hbar^L$. It's an iterative process, becoming quite tedious quickly. In my QFT manuscript you find the treatment of the two-loop self-energy sunset diagram of $\phi^4$ theory. It's already pretty complicated at two-loop level!

https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf (sorry for giving the wrong link above, which I've corrected too).

 

[CAF123]

Maybe it's easier to understand, where your problem is, if you give a concrete example.

Say I have the 2->2 process, $(g q \rightarrow gq)$, illustrated in the attachment. I'd like to write out the Feynman rules for both diagrams. The second one I know how to do but what is the Feynman rules expression for the diagram with the mass counter term insertion (the 'X')?

 

[CAF123]

In the first diagram, without the counter term insertion, it's just a s channel type diagram with propagator momentum $p_1+p_2$. Now add the counter term. It splits the propagator line into two lines, each with momentum $p_1+p_2$. The Feynman rules is then $$\frac{(\not{p_1} + \not{p_2})_{\alpha \beta} + m \delta_{\alpha \beta}}{(p_1+p_2)^2-m^2} \cdot -i(Z_m-1)m \cdot \frac{(\not{p_1} + \not{p_2})_{\beta \eta} + m \delta_{\beta \eta}}{(p_1+p_2)^2-m^2}$$

Could someone tell me if that is correct feynman rules for the counter term?

 

[CAF123]

Is my question unclear?

 

[Reggid]

Could someone tell me if that is correct feynman rules for the counter term?

If the -i in the counter term is just a sloppy (-i), then I would agree (up to factors of i that I did not check).
Note that you will also need to include the wave function counter term.