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Quotation from the Wikipedia article about vacuum energy:

"The theory considers vacuum to implicitly have the same properties as a particle, such as spin or polarization in the case of light, energy, and so on."

In quantum field theory, why is vacuum considered to have the same properties as a particle?

Qmechanic
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    QFT has fields, not particles as they are seen in QM. With "vacuum" and "particles" only being labels for different states of the QFT fields. – M.S. Feb 26 '23 at 14:10
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    Defining a vacuum in QFT does not involve anything about particles. This sounds like a stream of consciousness from a wiki editor trying to make analogies. – Connor Behan Feb 26 '23 at 16:00
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    @ConnorBehan Isn't the vacuum in QFT typically defined as "the state containing no particles"? – HTNW Feb 26 '23 at 20:49
  • The quote is phrased poorly but there is nothing wrong about it. The vacuum state, as well as any excited state, is just a vector in your Hilbert space, the only difference being that the vacuum has lower energy than any other state. There is nothing else other than this, that makes the vacuum state different from particle states. The vacuum can have any of the properties we usually associate to particles, such as charge, momentum, etc. – AccidentalFourierTransform Feb 28 '23 at 18:51

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In nonrelativistic (particle-based) QM, many properties of particles cannot be explained. There is no reason that electrons are spin 1/2, or they must have antisymmetric wavefunctions while photons have symmetric ones, or even why particles of the same kind are indistinguishable. These are all put into the theory by hand. We add a "spin" coordinate to the wavefunction for every particle and impose symmetry constraints on multi-particle wavefunctions because these modifications give the right physics. But the changes have no mathematical backing (they are axioms, or more accurately "inputs to the model"), so we can't answer "why."

In QFT, a lot of these questions are answered. Each type of particle corresponds to a single field on spacetime. Every individual particle is a (quantum of) excitation on the corresponding field. (This explains why particles come in only a few "kinds", and are indistinguishable within their kind.) The apparent properties of particles derive from the properties of the field. Photons are spin 1 and bosons because their underlying field is a vector field, and electrons are spin 1/2 and fermions because their underlying field is a Dirac spinor field. It simplifies the problem that QM poses, of "why do the nonillions of fundamental particles I see every day come in only a few types?" down to just "why do these few fields have the types (scalar/spinor/vector/etc.) they have and the couplings they have?". (Of course, this residual question seems nigh-unanswerable.)

Once you move from particles to fields, the vacuum becomes more interesting. Every point of spacetime contains every field (this is just as true classically as it is in QFT). This is true also of "empty" spacetime, containing no particles. Such a vacuum will still have all the fields in them; they will just not be excited. So a statement like "electrons have spin 1/2" applies to all spacetime, including the vacuum, since the electron field is "attached" to all spacetime, including spacetime with no actual electrons.

The phrasing you find on Wikipedia is a bit misleading, IMO. The vacuum of course contains zero spin angular momentum, zero charge, etc. And the properties of the fields (their spins, their charges, etc.) hold in all spacetime, including non-empty spacetime.

HTNW
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  • If you consider very short time intervals, can the fields in vacuum not become non-zero? – Peter - Reinstate Monica Feb 27 '23 at 09:16
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    @Peter-ReinstateMonica Sure, but I think focusing on that is missing the point. The fields exist on spacetime regardless of their values, classically and in QFT, and they have properties (spin, couplings, ...) regardless of their values and again regardless of whether you're thinking classically or in QFT. The vacuum fluctuations in QFT should be thought of as confirmation of the continued existence of the fields in otherwise empty spacetime. – HTNW Feb 27 '23 at 15:10
  • I think you're overselling how arbitrary everything is. If states are vectors, then the space they live in has to be a representation of $SU(2)$, which restricts the values that spin can take. It doesn't tell us that electrons have to be spin 1/2, but once we measure that they do, we know that they have exactly spin 1/2 (not, say, a value very close to 1/2 or some such). – Charles Hudgins Feb 27 '23 at 22:36
  • @CharlesHudgins Good point. But you still need QFT to get from "this one electron has spin 1/2" to "all electrons are identical to this electron so they also have spin 1/2", and also to get from "electrons have spin 1/2" to "electrons obey the exclusion principle", which I would argue are the bigger "asks" you have to deal with when using "bare" QM. – HTNW Feb 27 '23 at 22:56
  • @HTNW I see what you mean. – Charles Hudgins Feb 27 '23 at 23:02
  • @HTNW Your last paragraph is wrong. The vacuum state will in general carry all sorts of charges. Only in trivially gapped theories is the vacuum a scalar under everything. – AccidentalFourierTransform Feb 28 '23 at 18:49
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In quantum mechanics, and physics in general, the math gives an accurate description of nature. People also explain with words what is going on and why the math applies. The concepts are often very different from everyday life and confusing. People simplify, and the explanations are often a compromise between accurate and understandable.

The wave-particle duality is a good example of this. A photon or electron is often described as a wave or a particle. They aren't really either one. They are something like both a classical wave and a classical particle, but also different. See How can a red light photon be different from a blue light photon?

QFT is based on the idea that space is filled with fields. Particles are excitations of the field.

In QFT, a vacuum isn't an empty region of space. For space to be empty, the fields would have to be $0$. In reality, they are continually fluctuating. This means the vacuum is full of virtual particles.

Virtual particles are another example of a compromise, but they too have particle-like and wave-like properties.

mmesser314
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The reason? As always in real physics, it's because it works, capturing real phenomena. Perhaps the clearest demonstration of this is the phenomenon of squeezed vacuum.

John Doty
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