One of the main causes which leads to ill-defined loop integrals in Quantum Field Theory is that the variables of a Field Theory, $\varphi(x)$ for instance, are Quantum Fields which are governed by creation and annihilation operators which fulfill commutation relations which contain distributions

$$[ a(\mathbf{k}), a^\dagger(\mathbf{k}')] = \delta(\mathbf{k} - \mathbf{k}').$$

Therefore the Quantum Fields $\varphi(x)$ have to be considered as distributions and products of distributions, as they appear in the Lagrangian are not properly defined.

On the other hand when the path integral approach is used it is said (see Peskin& Schroeder, or Srednicki, or A.Zee's QFT in a nutshell) that the expression

$$Z(J) = \int D\varphi \exp{i \int d^4 x\left[\frac{1}{2}((\partial \varphi)^2 -m^2\varphi^2) +J\varphi\right]}$$

consists of ordinary fields which don't have operator character (they are not quantized), therefore do not contain distributions. However, upon evaluation the path integral of the free scalar theory as written above we already get (see A.Zee for instance):

$$Z(J) = {\cal{C}} \exp{ \left[-\frac{i}{2}\int \int d^4x d^4 y J(x) D_F(x-y) J(y)\right]}$$

where $D_F(x-y) $ is the scalar Feynman propagator which is well known to have distribution character.

How is it possible that from an apparent non-distributive approach (ansatz) one ends up with distributions? Even worse, for an interactive theory we get the whole series of Feynman-diagrams including those with loops which are infinite if they are not regularized.

So what is the underlying cause that also in the path integral approach we get results which only make sense if they are regularized? Is it the path integral measure $D\varphi$ which is mathematically not properly defined which leads to the same problems as those experienced in QFT based on second quantization?

  • $\begingroup$ The physical interpretation is fairly trivial I would say: trying to express the scattering from infinity to infinity in an interacting theory is doomed from the beginning. No matter how small the interaction, an interacting physical vacuum is perfectly opaque in that limit. It does not have asymptotic states. In practice it doesn't matter, of course. The spatial scale for the interactions that are relevant at the LHC, for instance, is much smaller than the size of the beam pipe while the entire universe is still transparent for the photons that are being generated. $\endgroup$ Commented May 29, 2023 at 14:35
  • $\begingroup$ @FlatterMann I consider this as not so trivial what you've expressed. Could you develop a bit further? What do you mean with the interacting vacuum being perfectly opaque and on the other hand an entire universe that is transparent for photons? Is it due to the "effect" that occurs upon computing the Rutherford formula which requires a non-zero mass photon? $\endgroup$ Commented May 29, 2023 at 17:00
  • $\begingroup$ It simply means that the actual physical calculation for an LHC experiment (and every other currently conceivable experiment as far as I know) can have an IR and UV cutoff and still deliver useful predictions about what we expect to measure. The mathematical complications with these formulas are not relevant to their usefulness. If the perturbation theory does not converge, then it still doesn't mean that the theory doesn't work. It simply means we have to work non-perturbatively, but I don't know enough about those techniques to say much about the physical equivalents. $\endgroup$ Commented May 29, 2023 at 17:17

1 Answer 1


$$ \intop_{-\infty}^{\infty} d\omega\, e^{i \omega x} = 2\pi \delta(x) $$

How is it possible that starting with a non-distributional anzatz I arrived at a distribution (a Dirac delta function)? There isn’t even a path integral in here, just a normal 1-d integral.


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