Vacuum in interacting QFT and physical particles

In summary: I think. But I'm not entirely sure. In any case, I don't think this is really relevant here.)In summary, an interacting QFT with a cutoff does not necessarily have a physical vacuum state which contains zero particles.
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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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  • #2
asimov42 said:
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.

asimov42 said:
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."
 
  • #4
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.)
 
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asimov42 said:
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.
 
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  • #6
@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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asimov42 said:
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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  • #9
asimov42 said:
Thanks @king vitamin. How does one describe what is in the interacting vacuum? (of the Standard Model)

That depends on what exactly you mean by something being "in" a quantum state. You can certainly do things like compute entanglement entropy (for various choices of entangling cuts), compute expectation values/correlation functions with respect to the state, write down the wave functional for the state in the Schrödinger representation, etc. Is anything like this what you're looking for with this question?
 

1. What is the concept of vacuum in interacting QFT?

The concept of vacuum in interacting quantum field theory (QFT) refers to the lowest energy state of the system, which is not truly empty but rather contains quantum fluctuations. In interacting QFT, the vacuum is not stable and can be excited by interactions with other particles.

2. How do physical particles emerge from the vacuum in interacting QFT?

In interacting QFT, physical particles emerge from the vacuum through a process known as particle creation. This occurs when the vacuum is perturbed by interactions, causing the creation of particle-antiparticle pairs. These particles can then interact with each other, giving rise to the physical particles we observe.

3. What role does the Higgs field play in the vacuum of interacting QFT?

In interacting QFT, the Higgs field is responsible for giving mass to particles. The Higgs field permeates throughout the vacuum and interacts with other particles, giving them mass through the Higgs mechanism. Without the Higgs field, particles would be massless and the universe would look very different.

4. Can the vacuum of interacting QFT ever be completely empty?

No, the vacuum in interacting QFT is never truly empty. Even in the absence of particles, the vacuum contains quantum fluctuations that constantly arise and disappear. These fluctuations contribute to the energy of the vacuum and have observable effects on physical systems.

5. How does the vacuum of interacting QFT relate to the vacuum of non-interacting QFT?

The vacuum in interacting QFT is significantly different from the vacuum in non-interacting QFT. In non-interacting QFT, the vacuum is stable and empty, while in interacting QFT, the vacuum is unstable and contains quantum fluctuations. Additionally, the vacuum in interacting QFT can give rise to particles, while the vacuum in non-interacting QFT cannot.

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