I've been studying QFT for almost a year now but am still fairly unclear on the basic ontology of the theory. Here's what I'd consider the "basic ontology" of non-relativistic quantum mechanics:

  1. A particle is represented by a wavefunction $\psi(\vec{x},t)$, whose square modulus is the probability density of finding the particle at a certain point in space and moment in time.
  2. The Schrodinger Equation tells us how $\psi(\vec{x},t)$ evolves in time.
  3. Each observable has a corresponding operator whose eigenvalues are the possible outcomes of a measurement. To find the probability of measuring this outcome for a particle in the state $\psi(\vec{x})$ at time $t$, write $\psi(\vec{x})$ as a sum of weighted eigenfunctions of the operator, and take the the square modulus of the coefficient on the eigenfunction associated with the eigenvalue you want.

You could tweak this to cover more general physical systems than a particle, or avoid demanding that we speak only in terms of the position representation, but this gets the basic idea across.

I've never seen anything remotely like this for QFT. Is, say, a free electron in QFT, represented by an operator field $\phi(x)$, or a Fock space state $\left|\phi\right>$, or a combination of the two? Does everything above for NRQM remain valid with some minor tweaks, or is it completely obsolete in QFT? If the time-evolution equations of QFT (Klein-Gordon, Dirac, etc.) determine a field's evolution, then what determines a state's evolution?

  • $\begingroup$ With some exceptions (Lamb shift and such), in QFT, you are mostly focussed on scattering of asymptotic states, essentially in momentum space, or lifetimes. So your observables are quite different and excessive attention to localized states is largely pointless. $\endgroup$ Apr 12, 2019 at 21:20
  • 4
    $\begingroup$ > excessive attention to localized states is largely pointless. Where in the question is WillG asking about localized states? Why would asking about them be pointless? $\endgroup$ Apr 12, 2019 at 21:36
  • 5
    $\begingroup$ I'd guess that the states we "focus on" is largely irrelevant to the basic ontology of a theory. $\endgroup$
    – WillG
    Apr 12, 2019 at 21:41
  • 1
    $\begingroup$ From a condensed matter perspective, you can start with a Hamiltonian and construct the path integral. In this lens the quantized Hamiltonian is more fundamental, so everything just carries over from QM $\endgroup$
    – Aaron
    Apr 13, 2019 at 0:14
  • 1
    $\begingroup$ @d_b That would be an interesting complement to my answer. I cited a review by Monnier of the Atiyah–Segal functorial approach (arxiv.org/abs/1903.02828) but didn't explain what it is. I just left it as click-bait. $\endgroup$ Apr 13, 2019 at 3:36

1 Answer 1


I'm interpreting the question like this:

What are the general principles of QFT, conceptually?

Not sure if this is the kind of answer the OP is looking for, but I'll give it a try and see how it's received. This is a non-perturbative perspective.

QFT refines quantum theory

QFT is a refinement of the general principles of quantum theory. The general principles of quantum theory say that observables (measurable things) are represented by operators acting on a Hilbert space; but they don't say much about what kinds of observables a model should include. Specifying what the basic observables (measurable things) are, and which operators represent them, is the task of specifying a model. QFT does this in a relatively systematic way, as explained below.

Once the observables have been specified, the rules are the same as usual. Whenever an observable is measured, we can project the state onto one of the observable's eigenspaces, with relative frequencies determined by Born's rule. That is, after a measurement, we replace $|\psi\rangle\to P_n|\psi\rangle$, where $P_n$ is the projection operator onto the observable's $n$-th eigenspace, with relative frequencies $\langle\psi| P_n|\psi\rangle$, just like we learned to do in introductory QM.

Observables in QFT are tied to spacetime, not to particles

For the sake of answering the present question, I'll contrast QFT with another class of models that I'll call "quantum mechanics." Sometimes "quantum mechanics" is used as a synonym for the general principles of quantum theory, but that's not how I'm using the words here.

  • In the class of models that I'll call "quantum mechanics," observables are tied to particles.

  • In QFT, observables are tied to regions of spacetime.

Conceptually, this is probably the most important thing to understand about QFT: it doesn't have observables tied to particles. In QFT, particles are phenomena that can occur, and deciding precisely which phenomena should be called "particles" can be a messy business (except in trivial models).

In QFT, observables are associated with regions of spacetime. For the sake of being concise, I'll pretend that we can associate observables with points of spacetime, ignoring the many mathematical problems this causes. In QFT, the association between observables and regions (or points) of spacetime is the data that defines a specific model. This association is usually required to satisfy some basic conditions, like these:

  • The time-slice principle: All observables can be expressed in terms of those associated with a neighborhood of any single time. (I'm referring to the Heisenberg picture here, so observables are parameterized by time, and states are not. The Schrödinger picture will be mentioned below.)

The Heisenberg equations of motion (see below) are expressions of this principle.

  • In relativistic QFT, we impose Einstein causality (aka microcausality): If two points are separated by a spacelike interval, then the associated observables commute with each other.

The Einstein causality principle prevents faster-than-light communication. In non-relativistic QFT (or in lattice constructions of "relativistic" QFT), we can relax this to: If two observables are associated with different points at the same time, then they commute with each other. By the way, non-relativistic QFT overlaps with what I called "quantum mechanics" above. More about this below.

To make contact with experiment, we need to know what particles a given QFT predicts, and how they behave. This can be worked out explicitly in trivial models, where "trivial" means "the particles don't interact with each other," but it is very difficult to work out explicitly in non-trivial models. More about this below.

Observables are built from field operators

Observables in QFT are typically constructed in terms of fields, which of course is where the name quantum field theory comes from. Fields, like observables, are tied to spacetime. For example, a Dirac spinor field is an operator $\psi_n(\mathbf{x},t)$ parameterized by a point in space $\mathbf{x}$ and a time $t$ and a spinor index $n$, which would take values $n\in\{1,2,3,4\}$ in four-dimensional spacetime. (That's a coincidence, by the way; in $N$-dimensional spacetime, the number of components of a Dirac spinor grows exponentially with increasing $N$.)

Field operators don't necessarily need to satisfy the same basic conditions that observables do. In particular, we can have fermion fields, which don't commute with each other at spacelike separations, even though observables constructed from those fields should still commute with each other at spacelike separation. That's why observables must involve a product of an even number of fermions fields, never an odd number.

In most models, observables are constructed with the help of gauge fields, with the understanding that observables are invariant under gauge transformations even though the fields from which they are constructed are not. There's a lot more to be said about that subject, far too much to say here.

The vacuum state, and states with particles

Here's another basic condition that is usually imposed, at least when the spacetime is flat:

  • The spectrum condition: The Hamiltonian $H$, the operator that generates time-translations of all observables, must satisfy $\langle\psi|H|\psi\rangle\geq 0$ for all state-vectors in the Hilbert space. In other words, the energy must be non-negative (or at least bounded from below, in which case we can add an inconsequential constant to $H$ to make it non-negative). A lowest-energy state is called a vacuum state, at least if it also satisfies something called the cluster property that I won't describe here.

It's not yet clear how the spectrum condition should be generalized for generic curved spacetimes. There is a promising idea called the "microlocal spectrum condition," but this is still an active area of research today. This subject is important because knowing what state should be used as the/a vacuum state is a prerequisite for defining what a "particle" is. Particles are things that can be counted, and the vacuum state should have none of them. (That rule is broken in curved spacetime. I won't go into that here, but I described a pragmatic approach in another answer.)

Here's the idea: If $|0\rangle$ is the vacuum state, then an observable $D$ that is constructed from field operators localized in a given region $R$ and that satisfies $D|0\rangle=0$ could be used as a model of a particle-counting device localized in $R$ — except that such a thing is mathematically impossible in relativistic QFT, because of the famous Reeh-Schlieder theorem. The best we can do is construct a local observable that approximately annihilates the vacuum state. That's part of why defining what "particle" should mean in QFT is a bit messy.

When analyzing trivial models, we can get around this by considering non-local particle-counting operators. The recipe is to express a given field operator as a sum of positive- and negative-frequency terms, called creation and annihilation operators. (These operators are necessarily non-local in space.) From these, we can construct $n$-particle states and particle-counting operators, as described in many textbooks. In non-trivial models, this becomes much more difficult. This might be the main reason why QFT is so hard to learn.

In strictly non-relativistic QFT, those complications go away, and we can explicitly construct $n$-particle states even in non-trivial models. Since particle-number is conserved in non-relativistic QFT, we can even consider a sub-model consisting of only states with a given number of particles. For states with no more than one particle of each species, the result is what I'm calling "quantum mechanics," in which observables can be tied to individual particles.$^\dagger$

$^\dagger$ If we consider states having more than one particle of the same species, even if their total number is fixed, then observables must still be tied to spacetime. The traditional way of saying this is to say that the particles are "indistinguishable."

The equations of motion

If the time-evolution equations of QFT (Klein-Gordon, Dirac, etc.) determine a field's evolution, then what determines a state's evolution?

The formulation described above uses the Heisenberg picture, in which fields (and observables) are parameterized by time but states are not. With some assumptions about the structure of the model, we can switch to the Schrödinger picture, in which states are parameterized by time but observables are not. In the Schrödinger picture, the equation that describes how states evolve in time is just the usual Schrödinger equation $$ i\frac{d}{dt}\big|\psi(t)\big\rangle = H\big|\psi(t)\big\rangle $$ where $H$ is the Hamiltonian, which is an operator expressed in terms of the same field operators from which all other observables are constructed. As usual, it is the observable associated with the system's total energy. This is the same Hamiltonian that we use in the Heisenberg picture to describe the time-dependence of a field operator $\phi$: $$ i\frac{\partial}{\partial t}\phi(\mathbf{x},t) =\big[\phi(\mathbf{x},t),\,H\big]. $$ I wrote the time-derivative as a partial derivative here, because field operators are also parameterized by the spatial coordinates. The relationship between the Schrödinger and Heisenberg pictures is the same as it is in "quantum mechanics." The nice thing about using the Heisenberg picture in QFT is that it treats the space and time coordinates in a more balanced way: field operators (and observables) are parameterized by both of them. That makes general principles like Einstein causality much easier to express.

The Heisenberg equations of motion and commutation relations for the field operators are typically constructed using the canonical quantization recipe, starting with a "classical" Lagrangian. (I put "classical" in quotes because it may involve anticommuting fermion fields.) However, we can also have non-Lagrangian QFTs — something that would seem very mysterious if we thought of canonical quantization as the definition of QFT.

Alternative perspectives

In the formulation described above, observables are the main actors. There are other formulations, like the path-integral formulation, which can be more convenient for calculating things like correlation functions. Correlation functions implicitly contain everything there is to know about the model, and they are especially convenient for studying scattering processes — after using some subtle trickery (like the LSZ reduction formula) to relate them to the theory's particles.

The path-integral formulation suggests another way of thinking about QFT, one that opens the door to new kinds of insights. For people who already know the basics of category theory, a relatively concise introduction to the idea can be found in "A Modern Point of View on Anomalies," https://arxiv.org/abs/1903.02828.

Even though it's been around for a long time, the best way of thinking about QFT might be something we haven't conceived yet. This sentiment was expressed by the mathematical physicist Yuji Tachikawa in a presentation that starts with these slides (after a nostalgic introduction):

Title slide: What is quantum field theory?

Next slide: I don't know.

Next slide: THE END. Thank you for listening.

  • $\begingroup$ Wow, thank you for this long and informative response. I think the most important statement in this for me was that particles in QFT are not things but phenomena. Is the following summary correct? "QFT includes both states (vectors of a Hilbert space) and operator fields. The one that changes with time depends on which picture is used (Schrodinger vs Heisenberg), but either way the relationship between state and field determines probabilities of observables. The measurement of a particle at some place is one such observable." $\endgroup$
    – WillG
    Apr 13, 2019 at 20:22
  • $\begingroup$ Also is it then fair to say that the popular explanation, "a particle is an excitation of a field," is not accurate in general? In the Schrodinger picture, for example, the field is constant, so couldn't be "excited." Shouldn't we say something more like, "a particle is a spike in the expectation value of the number operator near a certain spacetime point"? $\endgroup$
    – WillG
    Apr 13, 2019 at 20:30
  • $\begingroup$ @WillG The summary in your first comment is indeed correct, and it probably would have made a much better answer than mine! The popular explanation quoted in your second comment has some truth to it, but it's oversimplified. The correspondence between field-species and particle-species is not always one-to-one. Your "spike in the expectation value" is good, although it begs the question of which operator qualifies as the number operator. We know the answer in trivial models (I mean trivial in a technical sense, not a pejorative sense), but usually not in non-trivial models. $\endgroup$ Apr 13, 2019 at 21:24
  • $\begingroup$ Interesting. So I guess choosing a particle number operator is like choosing which operators in NRQM correspond to which physical observables? Like, why $i\partial / \partial t$ represents momentum: because QM gives the freedom to choose which operators we want for each physical observable when building a specific model, and some models just seem to work? $\endgroup$
    – WillG
    Apr 13, 2019 at 22:10
  • $\begingroup$ @WillG After a model is constructed from field operators, we know how all of its observables are localized in spacetime. (That's more or less what constructing a model means in QFT.) Once we've done that, we can ask which of its observables has the right properties to act like particle-counters. In principle, that interpretation something we can deduce from the model, not something that we need to choose or assume. $\endgroup$ Apr 13, 2019 at 23:18

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.