I'm confused as to how quantum fields are defined mathematically, and I've seen from questions on this site and Wikipedia articles that classical fields are just functions that output a field value for a given point in space input.

Is this the same for quantum fields? Are quantum fields too just functions? If so how do they account for laws of quantum mechanics?

I've also seen answers on here saying things about operator valued distributions, etc... Are these operators the creation and annihilation operators of the second quantization? Also if the field is a field of operators then how do we determine the value of the field at a point?

I have all these snippets of knowledge, and I'm not sure how they fit together to mathematically describe quantum fields.

Finally, I'm confused as to how it works with the rest of QFT, and I guess this is my main question; if a quantum field is just a field of creation and annihilation operators, or even some other operators, how do we define particles and their interactions? You always hear that "particles are just excitations in their quantum fields." But mathematically how does this work. And fit in with the other bits I've mentioned?


3 Answers 3


The replies which suggest that the answer to "What is a quantum field?" is unclear or even open are wrong.

The impression that this could be unclear is owed to the standard textbooks sticking to the heuristics that helped Tomonaga-Schwinger-Feynman-Dyson to guess the theory many decades back, but the mathematical nature of realistic quantum field theory was completely understood by the mid 70s and further developed since. A survey of the state of the art is at

First of all it is worthwhile to realize that there is a difference between a field configuration and an observable on the space of all field configurations.

A field itself, either in classical physics or in its quantization, is simply a function on spacetime, assigning to each spacetime point the "value" of that field at that point. Or rather, more generally it is a section of a bundle over spacetime, called the field bundle. For instance if the field bundle is a spin bundle then the field is a spinor, if it is the differential form bundle, then the field is a gauge potential as for electromagnetism, etc.

Now from the Lagrangian density one obtains two things: the equations of motion as well as a pre-symplectic form on the space of all those field histories which solve the equations of motion. This is called the covariant phase space of the theory.

An observable is a function on this covariant phase space. It sends any field history to a number, the "value of that observable on that field history". But since the covariant phase space is itself a space of functions (or rather sections), a function on it is a functional.

Among these are the "point evaluation functionals", i.e. the observables whose value on a field history is the value of that field at a given point. The business about distributions is simply that on these point evaluation functionals the Peierls-Poisson bracket is not defined (only its integral kernel is defined, which is what you see in the textbooks). So one restricts to those observables which are functionals on the space of field histories on which the Poisson bracket actually closes. These are smearings of the point evaluation functionals by compactly supported spacetime functions. So then a point evaluation functional becomes a map that once a smearing function has been specified yields an observable. This way already classical point evaluation field observables are distributions: "classical observable-valued distributions".

Now all that happens in quantization, is that the pointwise product algebra of functionals on the covariant phase space gets deformed to a non-commutative algebra. It is traditional to demand to represent this algebra inside an algebra of operators on a Hilbert space, but for the most part this is a red herring. What counts is the non-commutative algebra of quantum observables. For computing the predictions of the theory, its scattering amplitudes, it is not actually necessary to represent this by operator algebra.

Anyway, whether you like to represent the non-commutative algebra of quantum observables by operators or not, in any case the result now is that a point evaluation functional is something that reads in a smearing function and then produces the corresponding observable, exhibited now as an element of a non-commutative algebra. In this way quantum observables on fields are algebra-element valued (e.g. operator-algebra element valued) distributions.

And, yes, for free fields this does yield the familiar creation annihilation operators, for details on how this works see

There is detailed exposition of these questions at

Presently this is written up to the classical story. For the quantum theory check out the site again in two months from now.

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    $\begingroup$ I think you're correct, but I'd want to add a comment emphasizing that the answer to OP's question "What are quantum fields mathematically?" is "There are no quantum fields". They aren't defined, any more than one needs to define the points of a noncommutative manifold. The things usually called 'quantum fields' in the elementary textbooks are in fact quantum observables. $\endgroup$
    – user1504
    Commented Oct 12, 2017 at 14:36
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    $\begingroup$ Yes! That's the right perspective. On the other hand, in good situations the non-commutative space defined as the formal dual of the non-commutative algebra of quantum observables does in fact concretely exist, not as a manifold, but as a Lie groupoid: [ctd below...] $\endgroup$ Commented Oct 12, 2017 at 17:04
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    $\begingroup$ @FelixMarin , in "QED: The Strange Theory of Light and Matter" Feynman recognizes the incompleteness of the theory at his time, when in Chap. 4. "Loose Ends" he writes: $\endgroup$ Commented Oct 16, 2017 at 8:55
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    $\begingroup$ "...is technically called 'renormalization.' But no matter how clever the word, it is what I would call a dippy process! Having to resort to such hocus-pocus has prevented us from proving that the theory [...] is self-consistent. It's surprising that the theory still hasn't been proved self-consistent one way or the other by now; [...] What is certain is that we do not have a good mathematical way to describe the theory of quantum electrodynamics: such a bunch of words..." $\endgroup$ Commented Oct 16, 2017 at 8:56
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    $\begingroup$ All this hocus-pocus goes away once one realizes the precise concept of quantum fields. This is not so much about mathematics, as just about clear concepts of physics. $\endgroup$ Commented Oct 16, 2017 at 8:56

The definition of a quantum field depends slightly on the formalism that you adopt, but globally, quantum fields are defined as operator-valued distributions. That is, if you have a quantum field $\Phi$, it is defined as

$$\Phi : \mathscr D(\mathcal M) \to\mathcal B(\mathscr H)$$

It maps smooth functions with compact support on the spacetime manifold to linear operators on the Hilbert space where your quantum theory is defined. By some abuse of notation, we sometimes write it as $\Phi(x)$, although this is only well defined if the distribution is also itself a smooth function.

This has some difficulties associated to it (since distributions can't be multiplied together easily, and QFT involves a lot of products of fields), meaning one has to use such methods as wavefront sets and renormalizations to make sense of everything.

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    $\begingroup$ I wish there was some lecture notes or books that lay out these concepts clearly for the QFT that we learn, for example, from Peskin-Schroeder. $\endgroup$
    – rainman
    Commented Sep 5, 2021 at 20:39
  • $\begingroup$ Could you specify what $\mathscr D$ is exactly? Smooth functions on spacetime to what? to $\mathbb C$? $\endgroup$
    – user56834
    Commented May 24 at 4:07

There is no mathematically sound formulation of realistic QFT yet so at this point we have no real answer to your question. The QFT that physicists use to make predictions is in the so-called Lagrangian formulation, which is a heuristic framework for obtaining perturbative expansions using Feynman diagrams. There is also algebraic or axiomatic QFT, mathematically well-defined but so far confined to free theories and toy models. The idea is that QFT must satisfy a list of axioms, the Wightman axioms being most commonly used, and the challenge is to construct realistic theories that satisfy them. Mathematically constructing a Yang-Mills theory with a mass gap is one of the Millenium problems.

In algebraic QFT fields are identified with operator-valued distributions, and the Fock space picture is a dual representation of them. This duality is similar to the Schrödinger vs Heisenberg pictures in quantum mechanics. The idea is that the Hilbert space of quantum fields, as distributions associated to localized regions of spacetime, is unitarily equivalent to the Fock space, where creation and annihilation operators are defined, and which is much more commonly used in practice. That is the Fock space of second quantization, so those operators are not the same as the field operators, which are quantized versions of classical fields (intuitively, the Fock space operators are "global" whereas the field operators are "localized"):

"Fortunately, the operators on a QFT Hilbert space include a set of field operators. If a particular wave equation is satisfied by a classical field $\phi(x)$, it will also be satisfied in operator equation form by a set of operators $\widehat{\phi}(x)$ on the state space of the quantized version of the field theory. Speaking somewhat imprecisely, $\widehat{\phi}(x)$ acts like a field of operators, assigning to each point x an operator with expectation value $(\psi,\widehat{\phi}(x)\psi)$. As the state evolves dynamically, these expectation values will evolve like the values of a classical field. The set of field operators is sometimes called the operator-valued quantum field. One caveat that will be important later: Strictly speaking, we cannot construct a nontrivial field of operators $\widehat{\phi}(x)$ defined at points. But it is possible to define a “smeared” quantum field by convolution with test functions.

[...] We need an interpretation of field-theoretic states to determine which physically contingent facts they represent. In single-particle QM, a state is a superposition of states with determinate values for the theory’s observables (e.g. position and momentum)... in field theories we’re interested in systems that take on values for some field $\phi(x)$ and its conjugate momentum $\pi(x)$. So, when quantizing a field theory, we should just do to the field what we did to the mechanical system to generate QM. Impose commutation relations on $\phi(x)$ and $\pi(x)$, and move our states to the Hilbert space of wavefunctionals ($\Psi(\phi)$) that describe superpositions of different classical field configurations.

The equivalence to Fock space picture can be proved for free QFT, but axiomatic QFT has difficulties incorporating interactions or defining position operators. Because of this some argue that neither quantum field nor Fock space/particle interpretations can survive in a mathematically mature QFT, see e.g. Baker's Against Field Interpretations of Quantum Field Theory, from which the above quote is taken.

Wallace has a nice review In defence of naivete: The conceptual status of Lagrangian QFT that analyzes the mathematical structure of QFT as it is practiced, and argues to the contrary that it can be seen as a valid approximation of what algebraic QFT may one day yield. If that is the case then operator-valued distributions and Fock space states, interpreted as particle states, will be effective realizations of what quantum fields "are" at low energy levels:

"We have argued that such QFTs can be made into perfectly well-defined quantum theories provided we take the high-energy cutoff absolutely seriously; that the multiple ways of doing this are not in conflict provided that we understand them as approximations to the structure of some deeper, as yet unknown theory; that the existence of inequivalent representations is not a problem; that a concept of localisation can be defined for such theories which is adequate to analyse at least some of the practical problems with which we are confronted; and that the inexactness inherent in that concept is neither unique to relativistic quantum mechanics, nor in any way problematic.

  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Conifold
    Commented Jul 26, 2017 at 19:59
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    $\begingroup$ You write: "There is no mathematically sound formulation of realistic QFT". This is just not true, not for the perturbative QFT which is the one that most people mean when they say "QFT". See here: physicsforums.com/insights/paqft-idea-references $\endgroup$ Commented Oct 12, 2017 at 7:46
  • $\begingroup$ Perturbative theory is not mathematically sound unless there is a non-perturbative formulation that it is an approximation of and a proof of convergence in some sense. There are no such formulations or proofs for the QFT-s of the Standard Model, and it is not even clear if such formulations exist. It is possible that QFT-s will have to be embedded into theories of a different type (string, etc.) to have a non-perturbative formulation. This does not preclude, of course, heuristic perturbative theories from applying some rigorous mathematical methods (graphical, combinatorial, algebraic, etc.). $\endgroup$
    – Conifold
    Commented Oct 18, 2017 at 19:28
  • $\begingroup$ there is a perfectly good definition of non-perturbative completion of pQFT (ncatlab.org/nlab/show/C%2A+algebraic+deformation+quantization) and the concept of what a quantum field theory actually is is never open. What you have in mind is that it is hard to construct complete examples of non-perturbative quantum field theory. But this is very different from allegedly not knowing what quantum fields are, mathematically. The difference is the same as that between not having a proof of the Riemann hypothesis and not knowing what a prime number is. $\endgroup$ Commented Oct 18, 2017 at 20:12
  • $\begingroup$ [ctd.] We know what quantum fields are, mathematically, just as well as we know what prime numbers are. This is not in contradiction to the fact that we neither have a proof of the Riemann hypothesis nor a construction of non-toy non-perturbative quantum field theory. $\endgroup$ Commented Oct 18, 2017 at 20:14

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