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On QFT, dimensionality and regularization

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Insaneworld82
#1
Aug31-13, 03:22 PM
P: 2
Hello everyone !!!


First of all, Quantum Field Theory is not my field of research. However, I have to investigate on some problems in QFT and I'm trying to get familiar with it again.

I'm basically working with scalar fields and I encounter some problems in dealing with renormalization, counterterms and so on.

To make my point clear, I will cite the two loop correction to the propagator in phi 4 theory, the so called "sunset diagram" (see for example, Peskin and Schroeder problem 10.3). To solve it, we proceed as usual, we make use of the Feynman trick to join the denominators, we integrate over momenta and we are left with an integral in the parameter $x$ which runs from 0 to 1 and have a momentum dependance of (p^2)^(n-3) (where $n$ denotes the number of dimensions including time).

So, in $n=4$, we get a correction proportional to $p^2$. The divergent part of it can be removed by means of the field strength renormalization counterterm. What happens in $n=3$ and $n=2$ ???

Explicitely, in $n=2$, we get a momentum dependence of $(p^2)^{-1}$. What does this correction represent? If we do not use counterterm renormalization, what is the meaning of such a contribution?

I'm aware that the divergence of a diagram depends strongly on the number of dimensions we are working in, but I simply don't understand what the resulting corrections to the propagator mean - if they renormalize the field strength, mass, etc.

I would really like to hear your opinions and suggestions.

Kindest regards !!!
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andrien
#2
Sep1-13, 04:17 AM
P: 1,020
You should be careful,A theory which is renormalizable in n=4 may not be renormalizable for other values.A famous example is Fermi four theory which is non-renormalizable in n=4 but it is renormalizable in n=2.This is actually the beginning point of effective field theories.When you go to any other dimensions,the coupling strength is multiplied with a mass parameter(with some power depending on the number of dimensions),it changes the mass dimension of coupling strength which is a somewhat a crucial point to check the renormalizability.
Insaneworld82
#3
Sep1-13, 09:22 AM
P: 2
Thanks for your reply Andrien. In fact I put as an example phi 4 scalar theory as it is super-renormalizable in 2 dimensions (1+1). I am aware that for example, in the massless case, when we go from n=4 to n=2, the above mentioned diagram divergence changes from ultraviolet to infrared. I just want to understand what does the diagram accounts for in n=2. Thanks !!!

andrien
#4
Sep1-13, 11:33 AM
P: 1,020
On QFT, dimensionality and regularization

Quote Quote by Insaneworld82 View Post
In fact I put as an example phi 4 scalar theory as it is super-renormalizable in 2 dimensions (1+1).
This is why it is not useful to analyze it in d=2.It acquires a mass dimension +2,at least in d=4 it has mass dimension zero so you need to be sure for renormalizability.
DarMM
#5
Sep2-13, 03:28 AM
Sci Advisor
P: 303
Quote Quote by Insaneworld82 View Post
Thanks for your reply Andrien. In fact I put as an example phi 4 scalar theory as it is super-renormalizable in 2 dimensions (1+1). I am aware that for example, in the massless case, when we go from n=4 to n=2, the above mentioned diagram divergence changes from ultraviolet to infrared. I just want to understand what does the diagram accounts for in n=2. Thanks !!!
It's a correction to the propagator. When you write out the expansion for the two-point function (propogator) the first term is just the free propagator (diagrammatically a single line). Every term after that is a correction. If you could sum the entire series, you would have the exact propagator for the interacting theory. That is its meaning in all dimensions.


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