All QFTs are Finite

Question

The usual Dyson expansion applied to quantum field theory for virtually all interesting interactions yields divergent loop integrals that must then be regularized and renormalized.

Suppose we make do without the Dyson expansion and interaction picture, and consider the following. I use the Schrodinger picture purely for convenience.

1. Construct a Hilbert/Fock space for our QFT $(\mathcal{H},⟨.|.⟩)$.

2. Represent our field operators $\hat{\varphi}(\vec{x})$ and $\hat{\pi}(\vec{y})$ such that $$[\hat{\varphi}(\vec{x}), \hat{\pi}(\vec{y})]=i\hbar\delta^{(3)}(\vec{x}-\vec{y}).$$

3. Construct the full Hamiltonian operator $\hat{H} = \hat{H}_0 + \hat{H}_I$ which we verify to be self-adjoint $\hat{H^\dagger}=\hat{H}$.

4. Define the time evolution operator $$\hat{U}(t_2,t_1)=\exp(i\hat{H}(t_2-t_1)/\hbar).$$

Since our Hamiltonian $\hat{H}$ is self-adjoint, we have that $$\hat{U^\dagger}(t_2,t_1)\hat{U}(t_2,t_1)=\hat{U}(t_1,t_2)\hat{U}(t_2,t_1)=\hat{1}$$ i.e. our time evolution operator $\hat{U}(t_2,t_1)$ is unitary. Thus for any two normalizable states $|\psi_1⟩,|\psi_2⟩\in\mathcal{H}$ where $|⟨\psi_1|\psi_1⟩|$, $|⟨\psi_2|\psi_2⟩| < \infty$, we have that the transition amplitude $|⟨\psi_2|\hat{U}(t_2,t_1)|\psi_1⟩|^2 < \infty$. Our transition amplitudes are finite just as they are in ordinary finite dimensional quantum mechanics.

By the naive construction above which parallels usual QM theory, obviously we must find that the exact transition amplitudes are finite even if the perturbative amplitudes deriving from the Dyson expansion are divergent. The conclusion is that the "exact non-perturbative" transition amplitudes are finite regardless of the dynamical content of the theory contained within the full Hamiltonian $\hat{H}$. Like this, $\varphi^6$ theory in 4D is finite, Quantum General Relativity is finite, the Fermi Theory of the weak interaction is also finite etc... Surely this cannot be true, right?

What goes wrong with the previous construction? There are of course additional subtleties relating to the choice of Hilbert space (by failure of Stone von Neumann), definition of inner product (for example there are no infinite dimensional translation invariant Lebesgue measures if we attempt a Schrodinger type inner product) further mathematical subtleties relating to proving self-adjointness, generalizing the spectral theorem, the definition of the time-evolution operator etc... but are any of these mathematical complexities sufficient to revoke the core of the argument above? If none of these issues are relevant, then where exactly does this go wrong?

Answer 1

The problems in QFT do not begin or end with the perturbative expansion. There are various difficulties relating to the procedure you're proposing. On a high level, there is a difficulty in defining what $H_I$ is. Haag's theorem is a result that implies that if you manage to successfully write any lorentz invariant QFT in terms of a Hamiltonian $H=H_0+H_I$ where $H_0$ is the free field Hamiltonian, then the new theory you wrote down is actually just a free field theory again.

The main issue is that unlike in finite dimensional quantum mechanics, the relation $[\phi(x), \pi(y)]=i\delta(x-y)$ does not uniquely specify the field operators $\phi, \pi$ up to unitary equivalence. There's an uncountable infinity of unitarily inequivalent pairs, and these generally correspond to different physics. In some sense, one can say that the different pairs correspond to having states with different macroscopic behavior. What Haag's theorem tells us is that an interacting field must live in a different representation of the fields than a free field. So you cannot pass between a free and interacting theory unitarily.

But all of the above things are abstract considerations. What this means in practice is that any naive interaction hamiltonian that you see in a QFT course or book is not actually going to be a sensible operator. Typically you will see an expression like $H_I = \int \phi^4(x) dx$. There are two glaring problems with this expression. First, a careful study of free fields will reveal that $\phi(x)$ is actually a rather singular operator-valued distribution. You can already see this on the level of the two-point function. $\langle{\phi(0,x)\phi(0,y)\rangle} = (-\Delta+m^2)^{-1/2}(x-y)$. This thing is singular as $x\to y$, so $||\phi(x)\Omega||^2 =\infty$. In other words, it does not even map the vacuum into a sensible vector in the Hilbert space. You can temporarily resolve this issue by regularizing the field. Take $\phi_g(x) = \int g(x-y)\phi(y)dy$. One can show that this now becomes a fine, self-adjoint operator when $g$ is nice. Secondly, even with the regularized fields, the integral $\int \phi_g(x)^4$ does not converge to a reasonable operator. You can regularize this by reudcing the integral to a finite volume, $H_{I, g, V} = \int_V \phi_g(x)^4 dx$. Now you're in good shape. You can write $H(g, V) = H_0 + H_i(g, V)$. You can show that this thing is self-adjoint with some (still nontrivial) work. However this defines a theory that is not lorentz-invariant and not translationally invariant.

The task from here is to try and take limits that remove these cutoffs. If you just directly remove them, then you're back to square one because everything will clearly blow up again. Thus, instead you try and make the other parameters in the theory, change as a function of $g,V$ in such a way that certain physical quantities such as the mass remain finite as a function of $g,V$. This is a better candidate for your theory, but you still cannot take the limit directly because Haag's theorem is still around. Using methods that are rather technical, the structure of the approximate ground states $\Omega_{g, V}$ can be used to construct a new Hilbert space (Read: new field operator representation) that will be the final home of the new quantum field theory. This total process is a very very rough summary of the process of non-perturbative Hamiltonian renormalization. Once it is carried out, what you obtain is a totally finite quantum theory like any other, with Hamiltonian, self adjoint observables, a Schrodinger equation, and no nonsense.

The program of study in which this procedure is carried out is known as constructive quantum field theory. In practice, it is very, very hard. It has been successfully done in dimensions less than 4. Thus, there is a variety of field theories that are finite in the sense that you mention, especially in $1+1D$. The problem is that the singularities get much worse as a function of dimension. $d=4$ is in some sense the borderline case of maximal difficulty before we start getting impossibility theorems. We have no successful examples of $d=4$ interacting quantum field theories. The best candidates for this procedure are the non-abelian Yang-Mills theories thanks to asymptotic freedom, but actually carrying this out is an enormous technical feat.

Answer 2

Constructing rigorously a quantum field theory is a very complicated endeavour, so there are many points at which the naive approach fails.

At point $(1)$, the difficulty is not finding the Hilbert space and its scalar product, because every infinite dimensional separable Hilbert spaces are unitarily equivalent. What is difficult is finding a Hilbert space $\mathcal H$ with a unitary action of the Poincaré group $\mathcal P\to U(\mathcal H)$ and relativistic quantum field operators and an invariant (and cyclic) vacuum state $|\Omega\rangle$. For free fields, this can be done using a Fock space construction. Attempting to construct rigorously a Hilbert space of wave-functionals as square-integrable (in some suitable sense) function(al)s over the space of configuration of fields seems very difficult. I don't know if this has been done even in simple cases.

At point $(2)$, $\hat\varphi(\vec x)$ and $\hat \pi(\vec x)$ are not well-defined operators, even for a free field theory : the delta function in the canonical commutation relations implies that they should be (operator-valued) distributions. For free fields and a Fock space, this can be done in the Heisenberg picture. Given a smooth function $f(\vec x,t)$ rapidly decreasing at infinity, one can define an operator $\hat \phi_f = \int f(\vec x,t)\hat\phi(\vec x,t)\text d\vec x\text dt $.

At point $(3)$, writing $H = H_0 + H_I$ looks like some kind of interaction picture, which does not exist for QFTs due to Haag's theorem. One of the concrete problems one would face when trying to do this is that only operators of the form $\hat \phi_f$ are well defined, so there is not obvious and easy way to define an operator like $H_I = \int \hat \varphi(\vec x)^3 \text d\vec x$.

Answer 3

This is a short answer that doesn't discuss the full issues, but may shed light on the fact this does not work.

Consider a free Klein-Gordon theory in flat spacetime. We know very well how to construct the Hilbert space and how to construct a vacuum. Everything is as simple as you may want. Still, computing the expected value of the Hamiltonian leads to an infinite value. You need to perform normal-ordering in order to get a sensible result.

From one point of view, this happens because the ground state of the harmonic oscillator has a non-vanishing energy, and the Klein-Gordon field is made of infinitely many of those, so you get an infinite value. This value clearly does not gravitate, so it should be discarded.

From a different point of view, one must recall that quantum fields are actually operator-valued distributions, and hence the product of two fields at any single point is often ill-defined. One then needs to get rid of the singular behavior (for example by means of normal ordering) to get to a sensible result.

Interestingly, the singular behavior occurs only at the UV limit. It seems either renormalization is essential, or quantum field theory breaks down at some sufficiently high energy scale (and then maybe spacetime is discrete, or string theory is the correct description, or something else happens).

Answer 4

There are several problems in what you describe, so much that one could write a book out of it.

The most fundamental one is the assumption that the normalizable states exist and that they belong to you Fock space. That is generally false for every interacting theory. This assumption is true almost exclusively for free theories, and those are indeed finite.

Interacting theories do not possess those states (they dont even possess an Hilbert space to be fair) so the assumption of having normalizable interacting states in the Fock space is simply untrue.