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  What are quantum fields mathematically?

+ 10 like - 0 dislike

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?

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user Oliver Gregory
asked Jun 5 in Theoretical Physics by Oliver Gregory (50 points) [ no revision ]
Related: physics.stackexchange.com/q/155608/2451 and links therein.

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user Qmechanic

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user Count Iblis
maybe this will help ? arxiv.org/abs/1602.04182

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user anna v

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user md2perpe
Sadly, nobody knows.

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user tparker

4 Answers

+ 9 like - 0 dislike

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.

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user Slereah
answered Jun 5 by Slereah (470 points) [ no revision ]
The linear operators on the Hilbert space you speak of, are they the creation and annihilation operators from the second Quantization?

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user Oliver Gregory
@Slereah: yes except the target is not $\mathcal{B}(\mathscr{H})$. Even after smearing these field operators are unbounded and defined on a common dense domain.

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user Abdelmalek Abdesselam
Well, I didn't say it was a surjection :p

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user Slereah
surjection? I did not say your map is not surjective but rather it is not even a map.

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user Abdelmalek Abdesselam
+ 10 like - 1 dislike

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.

This post imported from StackExchange Physics at 2017-10-10 21:09 (UTC), posted by SE-user Conifold
answered Jun 5 by Conifold (120 points) [ no revision ]

Any QFT is supposed to be a theory (a bookkeeping) of evolutions of occupation numbers of particles. The field operators are supposed to contain time dependencies of the occupation number amplitudes. Unfortunately, in reality our equations contain also unnecessary effects (self-induction) and represent evident things as rarely happening ones (= soft mode excitations are considered perturbatively). I believe that in future QFT will be formulated in physically and thus mathematically correct way, as it is supposed to be. Currently it is not so, due to our errors.

@VladimirKalitvianski: QFT is quantum field theory, and hence a theory of field evolution. In the relativistic case, i is not connected to particle evolution. Indeed, by Haag's theorem, interacting relativistic quantum field theories have no Fock space realization (which would be necessary for multiparticle evolution), and particles make sense only asymptotically, as scattering in and out states. This is how they appear in collision experiments.

@ArnoldNeumaier: You and I, we have different backgrounds. I speak of particle occupation numbers and the corresponding time-evolving amplitudes. In my opinion, these amplitudes may be well defined at any time. Their time dependencies are determined with interactions, which also may be well defined at any time. It is like coupling two or more optic fibres, if you prefer some analogy. Currently we couple particles in a wrong way, thus "difficulties".

@VladimirKalitvianski: So as far as relativistic field theories are concerned, you are speaking of free quantum field theories, since only in these particle number is a well-defined concept. 

@ArnoldNeumaier: You are wrong. During interaction the particle occupation numbers are well defined; they just have time-dependent amplitudes.

@VladimirKalitvianski: What you claim only holds in the nonrelativistic setting, or with a cutoff. In relativistic QFT, defined by the requirement of Poincare invariance, Haag's theorem implies that number operators are defined only for free fields (as they require a Fock space for their definition). Under renormalization, number operators develop uncurable infinities.

@ArnoldNeumaier: Now you are right, except for your implied unambiguity of "relativistic QFT." As I said, your claim holds in the current relativistic QFT. More explanation of what I mean deserves another topic. See here.

+ 3 like - 0 dislike

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 Peirls-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.

answered Oct 12 by Urs Schreiber (5,715 points) [ revision history ]

With apologies, I find the approach to QFT in this answer and on nCatLab too formalistic. I don't see enough discussion of why the mathematical structure is the way it is (at least not that's accessible at the level of this PhysicsOverflow question, which I see as about why the mathematics is the way it is). It is, perhaps, not so much that QFT is unclear so much as that it is not simply stated and justified. Why is it that the basic questions asked above can be unclear after ten minutes reading to anyone who even thinks they might be able to understand QFT?

I don't see, for example (though this is my worry of quite a few years, not a worry posed by the question), why quantum fields should be linear maps (that is, distributions) from the test function space into the *-algebra of operators? Indeed, insisting on the use of bounded operators requires that the map from the test function space has to be nonlinear, either to unitaries, $(\lambda,f)\mapsto\exp{\mathrm{i}\lambda\hat\phi_{\!f}}$, or to resolvents, $(\lambda,f)\mapsto(\mathrm{i}\lambda+\hat\phi_{\!f})^{-1}$, so what constraints on nonlinearity can or cannot be justified and how?

A second worry (again of mine, not of the questioner): the requirement that wavenumbers must lie in the forward lightcone is longstandingly justified by the correspondence principle, yet neither deterministic classical fields nor random fields impose such a constraint unless for mathematical reasons they decide to work with the analytic signal that can be obtained by applying the Hilbert transform to a real-valued signal. The wavenumber constraint lies at the heart of the mathematical structure of Axiomatic QFT, yet it is not fundamental to the construction of the Fock-Hilbert space of states, which might, pace C*-algebra fundamentalism, be thought a starting point for QFT (to save a long exposition, I have to cite my https://arxiv.org/abs/1709.06711).

@PeterMorgan the "approach" that I am talking about is the established one since the 1970s, culminating around Epstein-Glaser 73 and summarized in the seminal Erice proceedings Velo-Wightman 76. That it comes across as something exotic still in 2017 is a sign of the detachment of the QFT community from the foundations of their own subject, which are foundations that physicists have laid in the 70s, not mathematicians.  Mathematicians only started to take note of the beautiful subject of QFT-made-precise when Kreimer 97 appeared.

But you are right that the observables are not all linear functionals. (Did I say that?) Instead they are "multi-linear" as long as we are talking perturbation theory, i..e. sums of distributions on powers of the spacetime manifold.

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I find it helpful to think of QFT as a signal analysis formalism; we have to think about both how to model both creating and measuring physical states, which are physically realized as signals (if you think about CERN or indeed about any experiment, everything comes down to signals that are stored and subsequently analyzed, perhaps only a few or many millions of signals; much of the signal analysis is ad hoc, as when an electronic circuit decides when an event happens and records the time it happened instead of recording the signal in detail attosecond by attosecond, but the QFT part of the signal analysis is much more idealized and systematic). For an elementary starting point, think about the vacuum state, which maps an operator $\hat A$ to a complex number, which we can write as $\omega(\hat A)=\langle 0|\hat A|0\rangle$ (other states are possible as starting points, particularly thermal states, but not in this answer). All the operators $\hat A$ are generated by quantum field operators $\hat\phi_f$, by multiplying and adding operators together.

The $f$'s are called "test functions", but in signal analysis they would be called "window functions", which is useful insofar as they describe what "window" we're looking through when we record measurement results; we'll see below that we can also usefully call them "modulation functions", because they also describe modulations of the vacuum state. The "usual" quantum fields, which are indeed operator-valued distributions, can be obtained by taking a test function that we can loosely think of as zero except at a point $y$, say, where it's infinite, which we write as the Dirac delta function (in four dimensions), $\delta_y(x)=\delta^4(x-y)$, so we can write $\hat\phi(y)=\hat\phi_{\delta_y}$ (and we can do something similar for the fourier transform), but this object has to be used with considerable care because products such as $\hat\phi(y)\hat\phi(y)$ are not well-defined (there are various strategies to accommodate this, taking the quantum field to be generalized functions of various kinds instead of as distributions, or, if we do flagrantly use such products, we will have to adjust the results after the event by subtracting infinite numbers as necessary).

The vacuum state provides a probability distribution for any self-adjoint operator $\hat A$ (that is, any operator for which $\hat A^\dagger=\hat A$, which I'll assume you know of from QM). For free bosonic quantum fields, for which we have a good mathematical definition (the other elementary example being free fermion quantum fields), the characteristic function for $\hat\phi_f$,
$$\langle 0|\mathrm{e}^{\mathrm{i}\lambda\hat\phi_f}|0\rangle=\mathrm{e}^{-\frac{1}{2}\lambda^2(f^*,f)},$$
allows us to work out the probability distribution for $\hat\phi_f$ by fourier inverse transform, provided $\hat\phi_f^\dagger=\hat\phi_{f^*}=\hat\phi_f$,
$$\langle 0|\delta(\hat\phi_f-v)|0\rangle=\frac{1}{\sqrt{2\pi(f,f)}}\mathrm{e}^{-\frac{v^2}{2(f,f)}},$$
indeed with the commutator $[\hat\phi_f,\hat\phi_g]=\hat\phi_f\hat\phi_g-\hat\phi_g\hat\phi_f=(f^*,g)-(g^*,f)$ it allows us to work out a probability distribution for any other self-adjoint operator. The object $(f,f)$ is the crucial geometric object for a free QFT; for the Klein-Gordon field it is
$$(f,g)=\hbar\int\tilde f^*(k)\tilde g(k)2\pi\delta(k{\cdot}k-m^2)\theta(k_0)\frac{\mathrm{d}^4k}{(2\pi)^4},$$
while for the EM field it is
$$(f,g)=-\hbar\int k^\mu\tilde f_{\mu\alpha}^*(k)k^\nu\tilde g_\nu^{\ \alpha}(k)2\pi\delta(k{\cdot}k)\theta(k_0)\frac{\mathrm{d}^4k}{(2\pi)^4},$$
both of these being positive semi-definite sesquilinear forms, which in math we would also call pre-inner products, both also being translation and manifestly Lorentz invariant, and both also being zero except when the wave-number $k^\mu$ is in the forward light-cone. A final property is that the commutator $[\hat\phi_f,\hat\phi_g]=(f^*,g)-(g^*,f)$ is (somewhat non-obviously, but as it has to be to ensure the kind of causality required in QFT) zero whenever the supports of $f$ and $g$ are space-like separated (the support of $f$ is the region of space-time where $f(x)\not=0$). Notice that $(\delta_x,\delta_x)$ is undefined ---loosely we can say it's infinite, so that the variance of the probability distribution for $\hat\phi(x)$ is infinite---, which is the most graphic reason for thinking we can't measure $\hat\phi(x)$.

The vacuum state allows us to construct a Hilbert space of what I find it helpful to call "modulated states", with the simplest example being
$$\omega_g(\hat A)=\frac{\langle 0|\hat\phi_g^\dagger\hat A\hat\phi_g|0\rangle}{\langle 0|\hat\phi_g^\dagger\hat\phi_g|0\rangle}=\frac{\langle 0|\hat\phi_g^\dagger\hat A\hat\phi_g|0\rangle}{(g,g)},$$
for which we obtain a different probability distribution for $\hat\phi_f$,
$$\omega_g(\delta(\hat\phi_f{-}v)){=}\frac{\langle 0|\hat\phi_g^\dagger\delta(\hat\phi_f{-}v)\hat\phi_g|0\rangle}{(g,g)}\qquad\qquad\qquad\qquad$$
$$=\frac{1}{\sqrt{2\pi(f,f)}}\left[1{-}\frac{\textstyle|(f,g)|^2}{\textstyle(f,f)(g,g)}\left(1{-}\frac{\textstyle v^2}{\textstyle(f,f)}\right)\right]\mathrm{e}^{-\frac{v^2}{2(f,f)}}.$$
Notice the very significant fact that what we have modulated is not the field, but the probability distributions. QFT is an essentially stochastic theory. Notice the $v^2$ term in the above expression (not in the exponential, in the part that changed): if we work out what the probability distribution is for more elaborately modulated states, we'll find $v^4$ terms, $v^3$ terms, any power you like. In the way I've presented QFT (not the only way it can be done, of course), the highest power is one measure of how excited the state is. Ways of talking about QFT are important, IMO; I think it's good to talk about modulating a "state" to obtain a different "state" (not about modulating "the field"), which determines how the expected measurement results (for many different $\hat A$'s) will be modulated relative to the original state.

QFT proper takes the test function space to be infinite-dimensional, most often what's called the Schwartz space of functions that are smooth both in real space and in wave-number space, but all the above construction works quite well if we just take the test function space to contain just a few test functions, say $\{f_1, f_2, f_3\}$ (making sure that $(f_1,f_1)$, $(f_2,f_2)$, and $(f_3,f_3)$ are all finite), because we have preserved manifest Lorentz covariance.

There are lots of details that I've not covered, but for me this is the most helpful way I know to understand QFT (and also helpful when thinking about alternative ways to introduce interactions, on which I've not touched at all, although interacting theories are still about measuring and modulating states). Good luck, but, finally, a disclaimer: the above math can all be checked, it's OK, but the formalism and language I've used is very different than you will see and hear if you watch videos of lectures on QFT from the Perimeter Institute, say (and many others are available on YouTube), so use the above with discretion.

answered Oct 12 by Peter Morgan (1,120 points) [ revision history ]
edited Oct 13 by Peter Morgan

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