Divergences in the sense of QFT

1. Nov 13, 2008

mhill

if we consider the propagators and other Fourier integrals in the sense of 'distribution' then are all the divergences that appear in QFT (quantum field theory) due to the divergent quantities

$$\delta ^{k} (0)$$

that is my idea, all the divergences appear because in the commutation relations

$$[\Psi (x) , \Psi (y) ] = \delta (x-y)$$

appear the dirac delta function an its derivatives, or in the mathematical sense all the divergencies are proportional to the 'value'

$$\delta ^{k} (0)$$ , here 'k' means the k-th derivative of the delta function

2. Nov 13, 2008

schieghoven

Derivatives of the delta function are defined according to rules similar to integrating by parts, for example
$$\int f(x) \delta'(x-x_0) dx = - \int f'(x) \delta(x-x_0) dx = -f'(x_0)$$​
and higher derivatives accordingly. There's no inconsistency and I don't think that these are the cause of any of the (UV- or IR-) divergences of QED. One problem with distributions is that in general there is no consistent way to multiply them, which is kind of what we try to do (I think) in some of the loop diagrams -- maybe this has something to do with it.

Dave

3. Nov 13, 2008

DarMM

Indeed, multiplying distributions is essentially the cause of all divergences in the path integral approach. See the text by Glimm and Jaffe, "Quantum Physics, A Functional Integral point of view."