I've started studying QFT this year and in trying to find a more rigorous approach to the subject I ended up find out lots of people saying that "there is no way known yet to make QFT rigorous when there are interactions".

As for the textbook approach, even without interactions it already seems not much rigorous, still, approaching it in the right way it seems to be possible to make it precise.

Now, the rigour issue with interactions in QFT isn't explained in the books I'm using, and I confess I still didn't get it.

I mean: some people say the problem are the Dyson's series that in QFT wouldn't converge, some people say the issue is that the Fock space representation cannot be built with interactions and hence particles don't even exist in this case. Some people even say that it is not even possible to describe the theory with Hilbert spaces. And there are quite a few more points people make on this matter.

My question here isn't "how to solve these issues" because it seems to me that up to this day no one knows this yet. My question is: what really is the problem in more concrete terms.

What are the problems that make QFT with interactions be non rigorous? How interactions causes these problems in contrast to free QFT?


3 Answers 3


There are many different problems with interactions; or, rather, many manifestations of the same problem.

For example, interactions are always non-linear in the equations of motion, e.g., $$ (\partial^2+m^2)\phi=\lambda\phi^3 $$ or similar equations for QED. As the operators are distributions, their products are ill-defined, and there is no rigorous method to make sense out of $\phi^3$. It is just a meaningless expression (up to physicists standards, one could at this point mumble something about normal ordering, but this doesn't really achieve much).

Only in the case $\lambda=0$ does the equation above have a meaningful interpretation: it becomes a well-defined differential equation for a distribution, which can be made very rigorous within the context of distribution theory. For general $\lambda$, the equation is just meaningless.

As free fields are well-defined and understood, one may attempt to fix the problem above by switching into the interaction picture, $$ \phi=U\Phi U^\dagger $$ where $\Phi$ is a free field, which is a well-defined object. Here Haag's theorem enters the picture and tells us that $U$ doesn't exist. Yet we physicist play to pretend that it does exist, and write $$ U=\mathrm {Te}^{iS_\mathrm{int}} $$ only to realise, later on, that $S_\mathrm{int}$ is plagued by divergences (for example, in the form of divergent counter-terms in the interaction Lagrangian). This is the price we must pay to have a finite $S$ matrix: as $U$ cannot possibly exist, we must encounter divergences in its very definition, or otherwise the theory would be utterly inconsistent. This is the point of view held by some people: QFT evades Haag's theorem through renormalisation, and only because the latter is an intrinsically ill-defined operation.

One may even try to give up on trying to formulate the theory from first principles, and just content ourselves by defining the theory by its Feynman rules. Barring aside the fact that the perturbative series is asymptotic, Feynman rules are meaningless too from a rigorous point of view. For one thing, they include propagators and products thereof; and these objects are distributions as well, so their product is ill-defined. This fact of course manifests itself once again through divergences: Feynman diagrams include all sorts of divergences, which cannot be accommodated within a mathematically rigorous theory. This approach is typically hopeless too.

The only way to really fix these problems is to work on a lattice. This is because when you go to a discrete space-time, distributions lose their singular nature, and you can use standard functions (i.e., the fields become operator-valued functions). For example, the Dirac delta in the r.h.s. of the canonical commutation relations becomes a Kronecker delta, so the l.h.s. loses its status of distribution. On more practical terms, when you work on a lattice everything is convergent, and so the theory makes sense, at least from a perturbative point of view. More fundamentally, when you work on a lattice, the degrees of freedom become finite, and so Haag's theorem doesn't apply anymore.

  • $\begingroup$ Is there any reason (other than love for the continuum) why physicists do not switch to a theory on a lattice? $\endgroup$
    – user126422
    Commented May 3, 2017 at 0:25
  • $\begingroup$ For distributions in general, powers are not defined, but there are subspaces of distributions where powers are defined. Can't one work on such a subspace? $\endgroup$
    – md2perpe
    Commented May 3, 2017 at 6:02
  • 3
    $\begingroup$ @WillyBillyWilliams modern searches for Lorentz Violation still haven't resulted in any discoveries afaik. Lorentz violation is the main problem with lattices. By the way, if one adopts the Loop Quantum Gravity formalism, field theory on a lattice actually can be Lorentz-invariant. This is because lattices aren't predefined in LQG, but exist all at once as degrees of freedom of quantum spacetime. But this is a separate research programme as it unavoidably involves gravity. $\endgroup$ Commented May 3, 2017 at 6:26
  • $\begingroup$ Back in the day, I only saw lattice computations of QCD bound states and the like (hadron masses etc.). Has there ever been an analytic computation of some scattering cross section (say QED or electroweak) in a lattice-regulated QFT? Would that correspond to a non-perturbative ("all orders") result? $\endgroup$
    – Toffomat
    Commented May 3, 2017 at 11:24
  • 3
    $\begingroup$ @WillyBillyWilliams very good question! (and very good answer by S.P.). I'd like to add that one of the main reasons not to use lattice field theory all the time is because it is very complicated from a practical point of view. For example, you cannot use Fourier integrals, but you have to use Fourier series instead; propagators have a more complicated expression; you lose translation invariance; there are issues with fermion doubling; etc. $\endgroup$ Commented May 3, 2017 at 19:25

Defining products of distributions can be done rigorously using as input the operator product expansion for pointwise correlation with suitable remainder bounds. This is the object of my article "A Second-Quantized Kolmogorov-Chentsov Theorem". A quick account can be found in the Oberwolfach report for this recent workshop.

Let me take a trivial example, a Gaussian field with scaling dimension $\Delta<\frac{d}{4}$. The Euclidean propagator is $$ C(x,y)=\langle \phi(x)\phi(y)\rangle\sim \frac{1}{|x-y|^{2\Delta}} $$ at short distances. There is a mathematically well-defined probability measure $\mu_C$ on the space $S'(\mathbb{R}^d)$ of temperate distributions such that $$ \int \phi(f)\phi(g) d\mu_C(\phi)=\int C(x,y) f(x)g(y)\ dx\ dy $$ for any two test functions $f$ and $g$. Pick your favorite mollifier $\rho$, i.e., a smooth compactly supported function with $\int \rho(x)\ dx=1$. Define the rescaled version $\rho_{\epsilon}(x)=\epsilon^{-d}\rho(\epsilon^{-1}x)$ and consider the regularized field $$ \phi_{\epsilon}=\rho_{\epsilon}\ast \phi $$ obtained by convolution with the rescaled mollifier. Define the (recentered) tentative pointwise square $$ :\phi_{\epsilon}^2:(x)=\phi_{\epsilon}(x)^2- \mathbb{E} \phi_{\epsilon}(x)^2 $$ where the expectation is with respect to the measure $\mu_C$. Then it is not terribly hard to prove that $:\phi_{\epsilon}^2:$ converges when $\epsilon\rightarrow 0$ in every Lebesgue space $L^p(S'(\mathbb{R}^d),{\rm Borel}, \mu_C)$, $p\ge 1$, and almost surely to a random distribution which can be interpreted as the renormalized (Wick) square of the original field $\phi$. The result of the paper above is a generalization of this construction to non-Gaussian, i.e., interacting fields.


Two books by Scharf (‘Finite Quantum Electrodynamics, 2nd ed. 1995‘ and ‘Quantum Gauge Theories – A True Ghost Story, 2001‘) on causal perturbation theory give you as much mathematical precision on perturbative renormalized quantum field theory as you might wish. See my introduction to causal perturbation theory for an overview of what is going on there.

Causal perturbation theory is just standard renormalized perturbation theory made mathematically precise. But as any perturbation theory it at first only yields a formal power series in the (renormalized) coupling constant. For QED and similar theories, this series has convergence radius zero, hence does not yet define a function. They cannot be evaluated at a finite value of the coupling, except approximately.

The unsolved problems are in finding a nonperturbative version of the theory that produces unique functions of the renormalized coupling constants that upon expansion reproduce the power series.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.