# Vacuum in interacting QFT and physical particles

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## Main Question or Discussion Point

After some reading, I'm quite confused about the vacuum state in interacting QFT. I've read the @A. Neumaier post on "The Vacuum Fluctuation Myth," where he notes that "bare quantum field theory with a cutoff, the vacuum is a complicated multiparticle state depending on the cutoff – though in a way that it diverges when the cutoff is removed, so that nothing physical remains." This is for an interacting theory. There have also been short, relevant previous discussions here and here, also regarding interacting theories.

In non-interacting theory, the vacuum state an eigenstate of the particle number operator (eigenvalue zero). Ok, all good. Now, for an interacting theory, you introduce a cutoff allowing you to work in Fock space, and end up with a complicated superposition of bare-particle Fock states... so, now if I were to apply the particle number operator repeatedly in this case, I would get random numbers of bare particles ... but we don't observe this (if we did, it would imply the vacuum contained particles).

Here are my questions:

1. As @A. Neumaier alludes to above - if we remove the cutoff and apply renormalization, does the complex multiparticle state disappear? It seems yes, but in another post @Avodyne noted that this my not possible be in 3 + 1D?

2. Does the physical vacuum of interacting QFT then always contain zero physical particles? That is, is the physical vacuum state an eigenstate of the particle number operator, with eigenvalue zero? (using the interacting Hamiltonian of course)

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king vitamin
Gold Member
Now, for an interacting theory, you introduce a cutoff allowing you to work in Fock space, and end up with a complicated superposition of bare-particle Fock states... so, now if I were to apply the particle number operator repeatedly in this case, I would get random numbers of bare particles ... but we don't observe this (if we did, it would imply the vacuum contained particles).
It wouldn't imply that the vacuum contained particles, it would imply that the vacuum had "bare particles" whatever that means. If the vacuum is Lorentz invariant, it would still carry zero momentum and be a singlet under all other conserved (not spontaneously broken) symmetries. Particle number is not a symmetry in an interacting QFT (in fact some QFTs do not even admit particle-like excitations).

Here you are taking an operator which is a symmetry of one Hamiltonian and applying it to a different Hamiltonian without that symmetry - you shouldn't be surprised when that operator does not have a useful meaning anymore.

Does the physical vacuum of interacting QFT then always contain zero physical particles? That is, is the physical vacuum state an eigenstate of the particle number operator, with eigenvalue zero? (using the interacting Hamiltonian of course)
I sort of answered this above - interacting QFTs do not have a conserved particle number (or even particles necessarily), so there is no operator such as the one you are referring to.

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Thanks @king vitamin.

For my own clarity - the physical vacuum, in which all of our experiments take place, is considered to be interacting, correct? I'm not sure about the implications of not having a conserved particle number? (what is means precisely)

For reference:

I didn't add an additional bit of important text from @A. Neumaier's discussion: "With cutoff one can work in a Fock space created in terms of bare particles, and with a well-defined normally ordered Hamiltonian that describes the interacting dynamics. The physical vacuum is then the ground state of this Hamiltonian. It contains zero physical particles, but expressed in terms of the particles of the bare theory, it is a very complicated multi-bare-particle state."

king vitamin
Gold Member
Yes, if by "physical vacuum" you mean the Standard Model, it is certainly interacting. (And almost all other physical systems described by QFT are also interacting.)

This discussion depends a little on your preferred definition of "particles." In a free theory, you can define a particle number operator and it's really clear what they are. In an interacting QFT, I usually think of particles defined in a less precise way in terms of "long-lived" excitations whose energy is "approximately" additive. Here I've put vague words in scare quotes to emphasize that the definition isn't extremely precise, but if a theory is perturbatively accessible from a free QFT then usually some notion of particles can be used to define things like S matrix elements between different particles. (But even here, there are non-zero matrix elements between states with a different numbers of particles, so we don't have a conservation of particle number.)

Some interacting QFTs have low-lying excitations with a definite energy and infinite lifetime, like the state of a single electron or a single proton in the Standard Model for example. But these states are protected from decay by conservation laws like electric charge and baryon number.

(Finally, I know that Weinberg's QFT textbook defines a particle to be an irreducible representation of the Poincaré group. Under this definition, the vacuum itself is also "particle," so I don't think this sort of definition is what you're looking for.)

A. Neumaier
2019 Award
That is, is the physical vacuum state an eigenstate of the particle number operator, with eigenvalue zero? (
The physical vacuum state contains zero particles in the noninteracting case. In the interacting case, the physical vacuum state belongs to a Hilbert space in which no particle number operator exists, hence one cannot talk about the number of particles. But the asymptotic vacuum state is an eigenstate of the S-matrix, which is an operator on the asymptotic Fock space. In this state, all quantum numbers zero. In particular, this means that zero particles go in, and zero particles go out.

• vanhees71 and dextercioby
@A. Neumaier could you define the asymptotic vacuum state? (although we cannot talk of particle number, we don't find the interacting vacuum to be full of physical, dressed particles all by itself... and the bare particle state disappears during renormalization, but in what way?)

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Thanks @king vitamin. How does one describe what is in the interacting vacuum? (of the Standard Model)

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A. Neumaier
2019 Award
could you define the asymptotic vacuum state?
It is the standard vacuum state in a free Fock space, with one field per asymptotic particle (including bound states). It is related to the interacting theory by Haag-Ruelle theory. The latter cannot be explained in a few lines, but you are welcome to study it

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king vitamin