10
$\begingroup$

I'm a mathematician trying to learn quantum field theory. This question has two parts: first, I want to double check that I'm thinking about the surrounding issues correctly, after that I'll ask my main question.

It's taken me a while to understand how I should think about particle states in an interacting quantum field theory and how they relate to the field operators. Here's what I think is true; please tell me if I'm wrong!

  1. When we pass from a free to an interacting theory, no one has a description of the state space that's as nice as the Fock space picture. Instead, we simply posit that the combined spectrum of the momentum operators $P^\mu$ contains some isolated hyperboloids $p^2=m^2$. (This is the famous "mass gap".) We decide to call the corresponding eigenstates "one-particle states with mass $m$". This name is somewhat sensible since, in particular, these states will be eigenstates of the Hamiltonian $H=P^0$, so they are stable and have the sorts of momenta we would expect from a particle of mass $m$. These states have very little to do with any one-particle states for any free theory.

  2. For a field operator $\phi$, look at the Källén-Lehmann representation of the corresponding two-point function. If $|p\rangle$ is a one-particle state for which $\langle p|\phi(0)|\Omega\rangle\ne 0$, it turns out there will be corresponding pole in the two-point function and vice versa. (When this condition is met, $\phi$ is an "interpolating field" for $|p\rangle$.) We can use this to construct a one-particle state out of an appropriately smeared version of $\phi$ applied to the vacuum; this cruicially requires the isolatedness of the eigenvalue mentioned above.

  3. If the interpolating field is one of the "basic" fields appearing in the Lagrangian, the state we build in this way is an "elementary" particle, and if it's some more complicated combination of products of basic fields it's a "composite" particle. For a given one-particle state, there are in general lots of different fields you could use for this purpose, but we always assume there is at least one.

  4. We can build $n$-particle states by doing a similar smearing trick with well-separated smearing functions and letting $t\to\pm\infty$; in particular, unlike the one-particle states, these multiple-particle states are only really meaningful in this asymptotic limit. But once we have these, we can define the S-matrix in terms of their overlaps, and we can write down the LSZ formula and a bunch of Feynman diagrams and we're off to the races.

My main question (aside from "is everything I just said correct?") is why should we expect these assumptions to hold? In particular, the parts that are the least obvious to me are (a) why should the spectrum of the $P^\mu$'s contain these isolated hyperboloids, and (b) why should I assume that the corresponding states all have interpolating fields as in (2) above?

I'm aware that there are some problems extracting a completely rigorous version of this whole story and that massless particles, gauge symmetries, and so on introduce complications; that's not what I'm asking about. I'm asking for intuition: right now these assumptions --- the mass gap and the existence of interpolating fields --- feel arbitrary to me, and I'm wondering if anyone has a way of thinking about the logical structure that makes it feel more natural.

Improve this question
$\endgroup$
10
  • $\begingroup$ You might want to have a look at Section 3 of www-m5.ma.tum.de/foswiki/pub/M5/Allgemeines/WojciechDybalski/… $\endgroup$
    – Abdelmalek Abdesselam
    Feb 23, 2019 at 23:49
  • $\begingroup$ Thanks for the reference! On a first reading, though, it seems like it doesn't quite address the same question. I think I'm convinced that a sensible scattering theory can be derived from (among others) the mass gap assumption; I'm asking why I should expect that assumption to hold in the first place. $\endgroup$
    – Nicolas Ford
    Feb 24, 2019 at 0:21
  • $\begingroup$ We don’t expect the properties to hold in general, just in some models. As you pointed out, it is an open question whether Yang-Mills is one of them. But if you read through the millennium prize problem description, you’ll notice that mass gap is just part 2 of the problem, part 1 being to define a theory to begin with... We are at a point where we don’t even know for sure if QFT exists, and there are no physical reasons why it must (QFT is just an approximation to quantum gravity). All we can really do is run lattice simulations, and there’s numerical evidence that your properties do hold. $\endgroup$
    – Prof. Legolasov
    Feb 24, 2019 at 2:53
  • 1
    $\begingroup$ @InertialObserver: I took a class in grad school that was taught out of Srednicki, which left me pretty confused about questions of this form. I've sort of cobbled together the picture I currently have from a bunch of places, most recently Duncan's "Conceptual Framework of Quantum Field Theory". This is part of what makes me worried that I'm understanding something wrong! $\endgroup$
    – Nicolas Ford
    Feb 24, 2019 at 7:14
  • 1
    $\begingroup$ Good question, one reason why there shd not be a continuum of one electron states witj different mass is that these are not observed in experiment. Usually if a theory is accepted to describe a domain of reality some statements (which shd be theorems) are infered from the experimental ovservations. $\endgroup$
    – lalala
    Mar 14, 2019 at 10:24

1 Answer 1

Reset to default
5
$\begingroup$

An interacting quantum field theory on Minkowski space should have a representation of the Poincare group as unitary operators, and the irreducible representations of the Poincare group are exactly the stable one-particle states. This point of view is due to Wigner. There can be both massive and massless irreducible representations, the mass gap isn't crucial for the idea. There is also no fundamental distinction between elementary and composite one-particle states.

A general state will break up into irreducible components under the action of the Poincare group. This is manifestation of the idea that you can describe an interacting state by a collection of asymptotic one particle states. It depended on there being a representation of the Poincare group, which in turn depends on the Minkowski space setting.

If we have this Hilbert space structure with products of irreducible representations, we can define raising and lowering operators and then we can in turn collect these into fields with appropriate transformation properties. This is the point of view of Weinberg's QFT volume 1, which you might be interested in.

But in practice we simply assume that an operator (which might be a combination of 'elementary' fields) has a non-vanishing matrix element between the vacuum and any state which is consistent with its transformation properties. Although they necessarily exist, it is rather hard to choose an operator by accident which violates this assumption (it's like guessing the exact wave functions of an anharmonic oscillator).

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thanks for the answer! I think I like this perspective. A couple questions for clarification: (a) Is the main difference between this picture and the one in my question just that the "isolatedness" of the hyperboloids I mentioned in point 1 is less relevant than I thought? (b) There isn't always a state which has a non-vanishing overlap with the field applied to the vacuum, right? I think this is what happens with quarks. (c) It seems like this perspective critically depends on flat spacetime. Is there no coherent notion of one-particle states in curved spacetime? $\endgroup$
    – Nicolas Ford
    Mar 15, 2019 at 19:54
  • $\begingroup$ @NicolasFord, The mass gap is not relevant for the idea of a one-particle state, but it is relevant for the idea that a particle corresponds to a pole in the two-point function. Yes there are lots of problems with the ideas of particles and the vacuum in curved spacetime. There are also problems dealing with flat but compact spacetime. $\endgroup$
    – octonion
    Mar 15, 2019 at 20:29
  • $\begingroup$ @NicolasFord, No there isn't always a particle corresponding to a operator. What I was getting at is if we do have a particle in the spectrum, we can use any combinations of fields that has compatible symmetry properties and generically it will generate that particle (and other excited or multiparticle states) when acting on the vacuum. So we can learn about that particle by studying correlation functions of that operator. We don't have to pick a specially chosen interacting field that only creates that particle. $\endgroup$
    – octonion
    Mar 15, 2019 at 20:32
  • $\begingroup$ @octonion, wouldn't it be more correct to say that single-particle states correspond to representations of the Wigner's Little Group of the Poincare group? Saying that a single-particle state is a representation of the Poincare group is like saying that we can assign it both position and momentum values (which could be done in a classical theory). BTW, I would also add a tensor product with all other charges... $\endgroup$
    – mavzolej
    Aug 11, 2021 at 23:12
  • $\begingroup$ @mavzolej, The single particle states have an energy-momentum index and this lets you define a representation of the full Poincare group (translations and boosts and all). You don't need to be able to assign a position, you can still define a translation as $e^{ix\hat{p}}$. Wigner's Little Group is about those Poincare group transformations which leave the energy-momentum index fixed (for a particular reference state) and so it gets at the additional internal degrees of freedom which aren't covered by the energy-momentum part. $\endgroup$
    – octonion
    Aug 13, 2021 at 5:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.