After answering a question on quantum fluctuations Could quantum fluctuations spawn real matter?, I got into conversation with E. D. Kramer (to whom thanks) and in the end it may be that we had a genuine disagreement, or else we were each making correct but different assertions but using words like ‘particle’ differently. I felt that in the end it would be better to put my understanding down as a question, and thus ask others to critique it.

This concerns the questions, does the quantum vacuum fluctuate, and can particles appear in otherwise empty space as a result of such fluctuation?

This post is not just yet another query about a perennial question. It is an attempt to provide a clear answer.

My answer to both questions is ‘no’. I don’t intend by this to ignore or denigrate the work of experts in quantum field theory (QFT) who sometimes use such language in an effort to convey physical intuition. I want to present here a semi-technical summary of the situation in QFT, and a statement of what the physical implications are. My question is: is the following right?

The treatment of a collection of interacting quantum fields in flat spacetime is mathematically difficult and puzzling, with the result that one is not always certain of the physical implications. I assert that it is true to say that energy is conserved in quantum field theory (QFT) and this is the bottom line for me. It follows that if there is a large empty spacetime with just the collection of interacting quantum fields in their communal ground state, then this state of affairs will persist. As time goes on, the state will not change into one in which there are excitations or superpositions of excitations from this ground state. This statement is uncontroversial (I think!?) and easy to prove. All you need to do is to note that the ground state $|\Omega\rangle$ of the complete set of interacting fields, whatever that may be, is by definition an eigenstate of the full Hamiltonian, and therefore it is completely static apart from an unobservable global phase.

The last phrase in the previous paragraph shows that the quantum vacuum does not fluctuate. The phrase ‘superpositions of excitations’ used in the previous paragraph is what the everyday term ‘particles’ refers to. Therefore particles cannot appear in otherwise empty space with no cause other than the full interacting fields initially in their joint ground state.

Now let’s take this a little further.

We don’t have just $|\Omega \rangle$ in the actual physical cosmos. We have a universe with stuff, i.e. excitations, in it. So what happens then?

To make this more concrete, let’s consider what happens when we introduce something like a single atom in its internal ground state. In QFT a single atom is itself a highly complex excitation of several fields which no one knows how to write down in a clear way. One can then ask whether or not such an atom will remain in its ground state, or will it perhaps undergo Rabi oscillations as it interacts with the surrounding vacuum? The answer depends on how you think the atom gets expressed by the formalism. Are we talking about a ‘bare’ atom or a ‘dressed’ atom? The ‘bare’ atom is an atom imagined as if it could somehow avoid the coupling to the surrounding fields. The ‘dressed’ atom is one in which the coupling has been taken into account.

If you imagine that the interactions can be switched on at some particular time, then the bare atom will undergo Rabi oscillations from that time onward owing to its interaction with the surrounding field which is otherwise (i.e. apart from this bare atom) in its ground state. The dressed atom, on the other hand, is a system which incorporates both the bare atom and, to some extent, the surrounding field, and it has as its eigenstates superpositions of those of the joint system: (bare atom + surrounding field). Such a dressed atom does not Rabi flop in the vacuum. I think. I admit my knowledge of QFT is not sufficiently confident to be completely sure about this. But I think that if such a dressed atom were to get excited by interaction with the surrounding field, when the latter has no further excitations other than those required to make the dressed atom, then something impossible would result. The impossible result is that the dressed atom could then subsequently emit a photon and I think that such a sequence of events does not conserve energy overall. But Schrodinger’s equation says energy is conserved for an isolated system (the isolated system here being the entire set of fields including the parts needed to make the atom, all in a flat spacetime).

For a second example, just consider a single particle such as an electron travelling along initially in a momentum eigenstate. I think that if there are no other excitations around, other than those required to make a real (dressed) electron, then the electron will conserve momentum. In other words its wavevector is constant, or in everyday language it will travel in a straight line. Is that right? If it is right then clearly the vacuum is not causing anything to happen to such an electron. (But you can find myriad statements on the web and in popular books along the lines that such an electron is working its way through a soup of fluctuations).

Finally, let us note that a possible room for misunderstanding here is that the word ‘particle’ has more than one meaning. In everyday language the word ‘particle’ is a shorthand for a physically realistic thing, a thing that can move from one place to another and cause a detector to click. In QFT however, the word 'particle' is often used to mean ‘an excitation of the free field’. From a physical point of view, the free field is a fiction and so are its excitations. It is an important mathematical tool, somewhat like a component in a Fourier analysis, but of course no field really has its interactions with other fields turned off. I think that some of the written statements asserting that the quantum vacuum has particles milling about in it is an attempt to give a physical interpretation to the fact that $| \Omega \rangle$ (the ground state of the full theory) is not equal to $|{\bf 0}\rangle$ (the ground state of the free theory). One can note that, for any given free-field state with $n \ne 0$ particles, one has $\langle n | \Omega \rangle \ne 0$. Then one may assert ‘if a measurement of the particle number were to be performed, there is a non-zero probability that the outcome will be $n$’ and then one may say ‘so at any moment $n$ particles can pop out of the vacuum’. The hidden pitfall here is that ordinary particle detectors do not detect excitations of the free field (i.e. the mathematical abstractions). They detect some sort of property of the full interacting fields. An apparatus that could really discern one excitation of the free field completely from another would be very weird, quite unlike an ordinary particle detector, and I guess, but do not claim to know, that it would itself provide whatever energy requirements are needed to get a non-zero reading from the vacuum.

So that's basically it. That's my understanding and I am asking if it is right. To really pin this down, I finish by repeating the physically sensible thought-experiment which I alluded to above. If at time $t \rightarrow -\infty$ you put a hydrogen atom in its ground state in otherwise empty space, with the quantum fields in whatever state is compatible with the presence of a hydrogen atom and nothing else, then later on, as time $t \rightarrow \infty$, can there be a hydrogen atom plus a photon, with the fields otherwise in whatever state is compatible with the presence of a hydrogen atom and a photon? I mean, throughout, a real, dressed hydrogen atom, not the mathematical fiction of a bare one. (The concept behind the thought-experiment is that the hydrogen atom gets excited by interacting with the surrounding vacuum and then emits a photon).

This thought-experiment is one way to try to put physical meaning to assertions that the quantum vacuum itself can cause particles to be formed or detected in otherwise empty space, or that the quantum vacuum 'contains' such particles. I note that in the calculation of Unruh radiation, there is no excitation expected for an inertially moving detector (e.g an atom). Could I be missing that perhaps that is just a first-order approximation? I note also that in the phase transitions envisaged in Big Bang cosmology, each one involves fields far from their ground state.

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    $\begingroup$ I also got into an argument with the same user and I have the same position as you -- while some introductory QFT texts carelessly identify particles with free theory eigenstates, that's not how particles are defined in any other context. If you have an electron sitting in space, where "electron" is defined the way everybody defines it, it'll stay an electron forever. $\endgroup$
    – knzhou
    Commented Nov 30, 2018 at 19:48
  • $\begingroup$ This is just to say thankyou for responding; I am genuinely both interested and unsure if this is an agreed thing. I feel more and more sure that it is right, but then how to account for the great morass of thoroughly misleading statements and pictures on blogs etc. by scientifically literate commentators? $\endgroup$ Commented Dec 1, 2018 at 22:20
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    $\begingroup$ Almost nobody interested in science is literate in quantum mechanics, and almost nobody interested in quantum mechanics is literate in quantum field theory. So some wrong statements are passed down by oral tradition. The idea that particles pop in and out of empty space because of "vacuum fluctuations" probably originated from BKS theory and "old fashioned perturbation theory", (obsolete) predecessors to modern QM and QFT where energy wasn't conserved. Once a claim is out there it just gets repeated. $\endgroup$
    – knzhou
    Commented Dec 1, 2018 at 22:26

1 Answer 1


This new question is about flat spacetime, and we seem to be in agreement that nothing happens in the flat-spacetime vacuum state; photons do not spontaneously appear, nor does anything else.

QFT in a curved spacetime background is a different story, and since this new question was spawned by a previous discussion about QFT in curved spacetime, I'll offer a few comments about how that situation is different.

Energy is normally defined via the Hamiltonian. In a spacetime background that has a timelike Killing vector field (aka time-translation symmetry), such as in flat spacetime, we can define the Hamiltonian to be the operator that generates translations along that special timelike vector field. Once we have the Hamiltonian, it defines what we usually mean by "energy" in QFT.

But in a curved spacetime background that does not have any timelike Killing vector field, what operator should we single out and declare to be the "Hamiltonian" (or "energy operator")? What criterion would we use? Without a clear answer to that question, we have no clear criterion to define which state should represent the "vacuum state." According to pages 6-8 in "Axiomatic quantum field theory in curved spacetime" (https://arxiv.org/abs/0803.2003),

...one of the clear lessons of the study of free quantum fields in curved spacetime is that, in a general curved spacetime, there does not exist a unique, `preferred' vacuum or other state.

In contrast to the flat spacetime case, QFT in a generic curved spacetime has no conserved energy, so considerations similar to the ones used to understand ordinary spontaneous symmetry breaking (What does spontaneous symmetry breaking have to do with decoherence?) may become relevant.

Now, back to the flat spacetime case:

in the calculation of Unruh radiation, there is no excitation expected for an inertially moving detector (e.g an atom).

The vacuum state in flat spacetime is equivalent to a state with "particles" when described in Rindler coordinates, but this by itself does not directly imply that anybody is actually going to experience any "particles". These are just two equivalent descriptions of the same state, as indicated in the excerpt.

However, an accelerating particle detector in flat spacetime is a different story. A particle detector (or other object) can't be accelerating unless something is making it accelerate. When this is taken into account, calculations (which I have not done myself) have concluded that an accelerating detector does detect particles in the "vacuum state" (if we can still use that term for a state that contains an instrument strapped to a rocket engine), even if it's only accelerating for a brief time:

  • Anglin (1993), "Influence functionals and the accelerating detector", Physical Review D, 47: 4525-37

According to another analysis, an inertial bystander sees the same process as the emission of a particle:

  • Unruh and Wald (1984), "What happens when an accelerating observer detects a Rindler particle", Physical Review D, 29: 1047-1056

None of this contradicts the original assertion that nothing "happens" in the flat spacetime vacuum state, because that state does not contain any particle detectors (or any other matter), accelerating or otherwise.

  • $\begingroup$ Interesting but I would like to understand what made you think the question was about curved space - I cant see that anywhere in the question? I thought he meant a single atom in the whole universe and nothing else. $\endgroup$
    – isometry
    Commented Nov 30, 2018 at 19:36
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    $\begingroup$ Just to be clear: I completely agree that things go differently in curved spacetime and your reply is helpful and interesting to me concerning that case. It is not an answer to the question I asked though, in that my question concerns flat spacetime and inertially moving detectors. (The discussion was in comments to my answer to previous questions). $\endgroup$ Commented Nov 30, 2018 at 20:09
  • $\begingroup$ @DanYand I suggest you might add a line at the start saying something like "broad agreement for flat spacetime, and now I will say interesting things to show why curved spacetime is a whole new story" $\endgroup$ Commented Nov 30, 2018 at 20:12

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