# Homework Help: Renormalization of a non-local QFT

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1. Jul 27, 2014

### rubbergnome

Hi guys. I'm working on a model described by a non-local QFT. I think I got the Feynman rules right, but I get divergences from $\delta(0)$-like factors.

1. The problem statement, all variables and given/known data

It's a QFT for a complex scalar field $\psi(x)=\psi(\mathbf{x},t)$ with action $$S= \int dx \dot{\bar{\psi}}\dot{\psi}-m^2\bar{\psi}\psi -\int dx dy \bar{\psi}(x)\psi(y) \Omega(x,y) \bar{\psi}(x)\psi(y)$$

where $\Omega$ is a known, positive function. The space-time of the theory is of the form $X \times \mathbb{R}$, where $X$ is a measure space. There is no a priori differentiable structure on $X$, therefore I can't use spatial derivatives of momentum space in this model (maybe just formally), just time derivatives. The problem is to renormalize the theory because loop diagrams are "divergent" in the sense that they are proportional to factors of the form $\delta(\mathbb{x},\mathbf{x})$, whatever those mean in a general measure space setting.

2. Relevant equations

It's not clear to me whether the Delta function $\delta(x,y)$ makes sense. Maybe just as a Dirac measure, not even as a distribution (there's no topology on $X$, unless you can build one from the measure). Formally one has the free propagator $$\Delta(x,y)=\int \frac{d\omega}{2 \pi}\frac{i\delta(\mathbf{x},\mathbf{y})}{\omega^2-m^2}e^{-i\omega(t-t')}$$

The Feynman rules for the vertex is, according to my calculations, given by $\int dz dw \Omega(z,w)$, where one specifies a vertex with two points $z,w$ and joins each one to two lines. This causes loop factors of the form $\delta(\mathbf{x},\mathbf{x})$ which don't really make sense in the general case. In case $X$ is discrete it should be the Kronecker delta so it's just 1. In case $X=\mathbb{R}^n$ it should be the usual Dirac delta, so that's problematic to regularize.

3. The attempt at a solution

The only cause for divergences is in those factors, and they are in front of every diagram with a loop. This means that if I place a formal cutoff $\Lambda$, pretending to work in $d$-dimensional momentum space and write $\int d^d p \sim \Lambda^d$, subtracting the "divergences" with counterterms cancels every loop diagram, leaving me with tree diagrams that are zero because of the delta functions in the propagators. The only non-zero amplitudes in the theory are diagrams with loops, and my interpretation is that "physical" particles are $\bar{\psi}\psi$ pairs.

Knowing that this form of the propagator can be traced back to the absence of spatial derivatives, I tried to add a formal laplacian term in the action, with a parameter $\epsilon$ in front. Pretending again to be in $d$ dimensions, the basic loop integral evaluates to $\sim (m^2/\epsilon)^{d/2}$ which is again an overall divergent factor. Straight dimensional regularization of $\int d^d p$ gives zero.

I tried to modify the kinetic Lagrangian in a way that could extract a sensible finite part from $\delta(\mathbf{x},\mathbf{x})$, without success. I tried to treat delta functions and propagators as distributions, as measures, as operators, but in those formalisms it's difficult to define loops. I read something about casual perturbation theory and the R- and W-operations, and applying that scheme to the "squared delta" in this case just gives me an arbitrary constant. I'm supposed to fix it with some renormalization condition, but I don't know what I should do.

Last edited: Jul 27, 2014
2. Aug 20, 2014

### Greg Bernhardt

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

3. Aug 21, 2014

### rubbergnome

Thanks for responding. I actually used a modified Zeta Regularization technique. First of all I imposed that space must be a $d$-dimensional subspace of $\mathbb{R}^n$. Then I defined the $\delta$ distribution via the usual "Fourier transform of 1", but integrating in $d$ dimensions, even though the integrand if a function on $\mathbb{R}^n$ (with dimensional regularization). I checked that the resulting distribution acts as a Dirac delta: $$\int d^d x \delta(x,y)f(y)=f(x)$$ for sufficiently nice functions $f$. Then, to regularize the divergent integral for $\delta(x,x)$ I used a zeta regularization recursive formula for polynomial integrals, studied by Garcia, Elizalde and others. It's the second formula in this paragraph

http://en.wikipedia.org/wiki/Renormalization#Zeta_function_regularization

and it gives finite results for $d>0$. In the zero dimensional case I impose the distribution to be the Kronecker delta, so that everything is consistent. Isn't it? I don't really know. The thing is that the only regularization method that gives finite, non zero results in this case seems to be this one.