If quantum fields are operator valued distributions, why aren't they always smeared?

Question

I don't completely understand the distributional character of a quantum field because I never see them "smeared" in basic textbooks. As I understand it, if $\chi : \mathcal{F} \rightarrow \mathbb{K}$ such that

$$ \chi[f] = \int\limits_{-\infty}^{\infty} f(x) \chi(x) \text{d} x,$$

then $\chi$ is said to be a distribution smeared with $f \in \mathcal{F}$; where $\mathcal{F}$ is the space of scalar functions and $\mathbb{K}$ is a number field. Now, let's take a free Klein-Gordon quantum field as an example. This field is expressed as

$$\phi(\textbf{x}, t) = \frac{1}{(2 \pi)^3} \int \frac{1}{\sqrt{2 E_{\textbf{p}}}} \Big(a_{\textbf{p}} (t) e^{-i p_\mu x^\mu} + a^\dagger_{\textbf{p}} (t) e^{i p_\mu x^\mu} \Big) \text{d}^3 p.$$

What confuses me is that, unless I'm wrong, this field is an operator valued distribution. If that's the case, shouldn't $\phi$ be smeared with a scalar function in order for it to be well defined? Explicitly, shouldn't it be written something like this

$$\phi[f] = \int f(x) \phi (x) \text{d} x,$$

such that, for $|\psi \rangle$ that belongs in a symmetric Fock space

$$\phi[f] |\psi \rangle = \Bigg( \int f(x) \phi (x) \text{d} x \Bigg) |\psi \rangle ?$$

If that's the case, this confuses me further, since the expression inside the parenthesis is a number. How could $|\psi \rangle$ be evaluated then?

Answer 1

Yes, the quantum fields must be smeared in order to become well-behaved (symmetric, densely defined) operators (in the Hilbert space of the theory). In mathematically-minded textbooks it is the standard of their initial presentation.

Actually, referring to coherent states they (the bosonic ones) could be viewed as quadratic forms without smearing, but this is a very special case, even if it takes into account some classical interpretation of these fields.

The point is that smeared quantum fields are not sufficient to define all physically interesting observables. As a consequence unsmeared quantum fields have its own physical relevance, though this notion is affected by several mathematical issues.

For instance, the stress-energy tensor or (self-)interaction terms, e.g. $\phi(x)^4$, in Lagrangians are linear combinations of products of unsmeared field operators. Here, the smearing occurs after taking the product.

This notion is mathematically not well defined and it needs some care to be used. At the end of the day, ultraviolet renormalization has its roots in this fact.

Answer 2

The main point of quantum field theory is to study dynamics. The dynamics is given by interactions among the different fields. These interactions are point interactions. The way that these point interactions are modeled in the theory is with the aid of field operators that annihilate and create the different fields. To be able to represent point interactions, these field operators need to be defined at points and not be smeared out. Eventually, the interaction points are integrated over all space, which I guess takes over the role of the smearing process.

Answer 3

Yes, if we are mathematically careful about the distributional nature of a quantum field, then we should indeed consistently write quantum fields only as "smeared functions" $\phi(f)$.

But really your question isn't about quantum fields, it's just about how physicists treat distributions, namely by pretending they're functions. We do this everywhere, not just in QFT: You will see classical mass densities $\rho(x) = \delta(x)$ or quantum mechanical wave "functions" $\psi(x) = \delta(x)$. The Dirac delta, too, is a distribution and not a function. See also this answer of mine for more ruminations on whether it is "wrong" to treat the $\delta$-distribution this way, and how the "wrong" language is often just shorthand for something that is also rigorously true.

For comparison, just look at how far a typical QM treatment gets by pretending that the operators on the infinite-dimensional Hilbert spaces of QM are "just like matrices", ignoring the rich subtlety of functional-analytic problems that can arise for unbounded operators. Sure, there's some edge cases where you get nonsense, but the typical intro to QM never runs into these. If you accept the usual treatment of operators in QM, you can accept the lax treatment of quantum fields as functions rather than distributions, too!

In QFT, it turns out that the distributional nature is related to something else the physicist has to do, namely UV renormalization: We use "point" interactions of the form $\phi(x)^n$ to define the interaction terms of our theories, and such a product is in general ill-defined for distributions. The physicist ignores the distributional nature of the quantum fields, pretends $\phi(x)^n$ is a meaningful thing to work with and pays for it with "infinities" we then have to "renormalize away".

The mathematically careful can instead choose to work in causal perturbation theory due to Epstein and Glaser, where "infinities" never appear and the renormalization parameters that control the UV divergences in the usual physics approach instead parametrize the choices of the products of distributions order by order in perturbation theory.

Both pathways end up with the same recipe for computing real-world scattering amplitudes depending on the renormalization parameters.

Answer 4

I am studying these topics and I think that an answer from who is beginning to manage this field, could be helpful because often same questions arise for different people. You should always think of an operator as a functional. What I mean, is that the most important think in QFT are the Green functions, i.e., objects of the the form

$$G(x_1,\dots,x_n)=<\Omega| \Phi(x_1)\dots\Phi(x_n)|\Omega>$$

where $\Omega$ is some particle state of interest in a QFT (often taken to be the vacuum state of the theory, if any) and $x_1,\dots,x_n$ are spacetime points. Now, the Wightman theory (I suggest you to give al look to "Local Quantum Theory" of Rudolf Haag, one of the first chapters) things like

$$G(f_1,\cdots,f_n)=<\Omega|\Phi(f_1)\cdots\Phi(f_n)|\Omega>$$

become to be relevant, with $f_1,\dots,f_n$ some smearing functions is some functional space, and

$$\Phi(f_k)=\int_M f_k(x)\Phi(x)d^4x$$

where $M$ is the Minkowski spacetime. These objects are useful for two reasons:

1. Functions of field operators, generate the observable von Neumann algebra. Then, one focuses on von Neumann algebras generated by some special QFT (notice that $\Phi(f_k)$ is a function of $\Phi$).

2. If one wants to obtain information about the values which an observable can take, than there is the necessity to smear out the filed operator in such a way that the support of the smearing function relates to the sensibility of the measure. So, because you don't have exact values for an observables, than you smear them out.

To finish, $\Phi(f_k)$ is an operator, so it can act on kets. Take a closer look to the standard second quantization of Klein-Gordon filed to clarify further your doubts.