# Relation between quantum fluctuations and vacuum energy?

• I
As far as I understand it, the non-zero vacuum energy attributed to a quantum field (at each point in space-time) is precisely due to the Heisenberg uncertainty principle (and the fact that the energy of the quantum field at each space-time point is quantised). Accordingly (in order to satisfy the Heisenberg uncertainty principle) the value of a given quantum field will fluctuate at each space-time point such that its value is not precisely determined. In this sense, the fluctuations of the quantum field (due to the uncertainty principle) can be interpreted as the source of its vacuum energy.
What particularly confuses me, if this is the case, is how a fluctuating value of the field results in a constant vacuum energy (or is the point that the average, i.e. expectation value of the energy due to these fluctuations a constant?)

My question is, is this what people mean when they speak of quantum fluctuations and vacuum energy? Are the two intrinsically related (a result of the Heisenberg uncertainty principle and quantisation), or have I completely misunderstood things?

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A. Neumaier
2019 Award
Quantum fields don't fluctuate (except figuratively in popular science accounts for lay people). They are just a bit uncertain in their value, which may deviate from the expectation value by up to a few standard deviations. See Chapter A8: Virtual particles and vacuum fluctuations of my theoretical physics FAQ.

Masroor Bukhari and vanhees71
Demystifier
Gold Member
When physicists say that quantum observables fluctuate, they don't mean that they change with time. Instead, as A. Neumaier said, it only means that their value is uncertain. That's why a fluctuation can be constant, and that's why fluctuations are closely related to uncertainty relations.

bhobba
When physicists say that quantum observables fluctuate, they don't mean that they change with time. Instead, as A. Neumaier said, it only means that their value is uncertain. That's why a fluctuation can be constant, and that's why fluctuations are closely related to uncertainty relations.
Is it meant then that the value of the field at a fixed point is a distribution (in the sense that repeated measurements won't give the same result, however the expectation value, the mean of the distribution, will be a fixed value, i.e. 0) rather than a fixed value?

Is the point, that due to the Heisenberg uncertainty principle, one cannot precisely determine the value of the field at a given point and as such its standard deviation is non-zero. So by saying that a quantum field is fluctuating is simply a handwavy statement of the fact that its value (in vacuo) at a given point in space-time is not prescisely determined (i.e. it is not well-defined) and this results in a non-zero vacuum energy?!

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A. Neumaier
2019 Award
Is it meant then that the expectation value of the field is a distribution rather than a fixed value?
The expectation value ##\langle\phi(x)\rangle## of a quantum field is typically an ordinary function of space-time position ##x##; the correlations are (nicely behaved) distributions only and become functions after smearing in space and time - just as for classical stochastic fields.

To see the meaning in statistical terms, look at the special case of 1+0-dimensional fields with discrete time (and no space) coordinates. The corresponding data are called time series. Consider the El Nino time series data (or climate data from the last 500 million years, or other history data lying in the past). They form a determined sequence but are typically analyzed in terms of a statistical model. The model predicts the trend (expectation values as a function of time) and the true values are close to the trend within at most a few standard deviations.

The only change in the quantum field case is that (due to the operator nature of the fields) one can no longer talk about true values; but talking about the uncertain values within (somewhat fuzzy) limits still makes perfect sense. The quantum uncertainty can be pictured in the same way as the uncertainty in the position or diameter of a cloud in the sky - these are intrinsically limited in accuracy by the extended and fuzzy nature of the cloud.

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Masroor Bukhari
The only change in the quantum field case is that (due to the operator nature of the fields) one can no longer talk about true values; but talking about the uncertain values within (somewhat fuzzy) limits still makes perfect sense. The quantum uncertainty can be pictured in the same way as the uncertainty in the position or diameter of a cloud in the sky - these are intrinsically limited in accuracy by the extended and fuzzy nature of the cloud.
Would what I put in my edit (apologies, I edited it before I received your reply) to my previous post be correct at all?

A. Neumaier
2019 Award
Would what I put in my edit (apologies, I edited it before I received your reply) to my previous post be correct at all?
No. Vacuum energy is not a physical term at all, only one of popular science, needed there to be able to speculate about fancy properties of virtual particles.

Physical meaning has only the vacuum expectation value. It is everywhere in (flat) spacetime exactly zero for fields without spontaneously broken symmetry, and everywhere constant for fields with spontaneously broken symmetry (such as the Higgs field). The (uncertain) value of a field has meaning only in a universe filled with matter - but this is no longer a vacuum in the sense of quantum field theory, as now even the matter free regions contain nonzero fields. In such a universe one can consider measurement, and hence assign uncertain values at least to measurable fields like the electromagnetic field.

In quantum gravity, even the notion of vacuum itself becomes physically meaningless (as it then is coordinate system dependent).

No. Vacuum energy is not a physical term at all, only one of popular science, needed there to be able to speculate about fancy properties of virtual particles.
But in free-theory the Hamiltonian has a non-zero expectation value (before introducing normal-ordering). Is this simply an artefact of the maths? Also, what about loop corrections in perturbation theory?

A. Neumaier
2019 Award
But in free-theory the Hamiltonian has a non-zero expectation value (before introducing normal-ordering). Is this simply an artefact of the maths? Also, what about loop corrections in perturbation theory?
The vacuum state exists independent of perturbation theory. The latter is just one way to approximately compute its correlation functions.

The (physical = renormalized) Hamiltonian of a free relativistic quantum field theory is ##H=\int_{R^3} dp a^*(p)cp_0a(p)##, with zero vacuum expectation value. There is no mathematical artifact.

Note that an energy shift would violate Poincare invariance, which is one of the defining peroperties of the vacuum representation.

Note also that naive quantization, to which your remark perhaps refers) yields an additional infinite additive constant - showing that the naive procedure is physically meaningless.

Note also that naive quantization, to which your remark perhaps refers) yields an additional infinite additive constant - showing that the naive procedure is physically meaningless.
Yes, this is what I was referring to.

The vacuum state exists independent of perturbation theory. The latter is just one way to approximately compute its correlation functions.
I appreciate that, but aren't loops (at higher-order corrections in perturbation theory) interpreted as off-shell 'particles' (fluctuations in the quantum field) mediating the given interaction? Someone once told me that in the interaction picture their are no fluctuations, since the vacuum is equal to the free-theory vacuum, but in the Heisenberg picture, the interaction vacuum is non-trivial and hence particles can interact with the vacuum (and such interactions are interpreted as interactions with off-shell particles).

A. Neumaier
2019 Award
aren't loops (at higher-order corrections in perturbation theory) interpreted as off-shell 'particles' (fluctuations in the quantum field) mediating the given interaction? Someone once told me that in the interaction picture there are no fluctuations, since the vacuum is equal to the free-theory vacuum, but in the Heisenberg picture, the interaction vacuum is non-trivial and hence particles can interact with the vacuum (and such interactions are interpreted as interactions with off-shell particles).
With sloppy enough terminology, one can talk this way. But the talk does not hold water when one digs a bit deeper. Please read the Insight article on virtual particles linked to in my post #7 in this thread, and the other companion Insight article mentioned there. There is also a thorough discussion there; see especially post #58 and post #43 there!

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With sloppy enough terminology, one can talk this way. But the talk does not hold water when one digs a bit deeper. Please read the Insight article on virtual particles linked to in my post #7 in this thread, and the other companion Insight article mentioned there. There is also a thorough discussion there. Especially post #58 and post #43 there!
Thanks for the links. I've had a read of them and the point you make that vacuum fluctuations are equivalent to non-zero expectation values (arising due to some symmetry breaking process, right?).

It frustrates me that reading through a QFT textbook (Matthew Schwartz's) book, the author actually talks of vacuum energy as possible being meaningful in the context of gravitation, and many scientific papers on the subject of the cosmological constant problem discuss the vacuum energy of quantum fields and how they should contribute to the energy-momentum tensor and thus be a source of curvature. All of this makes learning the subject a very confusing process - I feel like it's misleading.

A. Neumaier
2019 Award
Thanks for the links. I've had a read of them and the point you make that vacuum fluctuations are equivalent to non-zero expectation values (arising due to some symmetry breaking process, right?).
No. Fluctuations are expressed not in the field expectations but in the field correlations. They are present no matter whether or not symmetries are broken. They are even present in a single harmonic oscillator if you call the ground state the vacuum state (as often done in quantum optics)!

One subtracts the expectation value if it is nonzero to get the broken symmetry theory for the redefined fields. The resulting Hamiltonian (in momentum space, in normally ordered form) is then renormalized to define the vacuum of the interacting theory. In this now physical theory, vacuum expectations of the physical (=renormalized) fields are again zero! The vacuum fluctuations refer to the nonzero correlation functions, which (after a Fourier transform) are of Kallen-Lehmann type and whose poles define the particle masses.

It frustrates me that reading through a QFT textbook (Matthew Schwartz's) book, the author actually talks of vacuum energy as possible being meaningful in the context of gravitation, and many scientific papers on the subject of the cosmological constant problem discuss the vacuum energy of quantum fields and how they should contribute to the energy-momentum tensor and thus be a source of curvature. All of this makes learning the subject a very confusing process - I feel like it's misleading.
It is very frustrating and misleading, and it takes time for everyone to separate the chaff from the wheat, both being contained in the typical textbooks to various extent. Note that saying ''possibly'' in a text means that what follows is speculation. Stuff for the imagination pointing towards unresolved problems where people replace missing knowledge by (more or less incorrect) intuitive pictures of all sorts.

But this is nearly inevitable in research where the foundations are unknown - it is fishing in the dark, hoping to catch a fish through clever imagination where they might be found. Research into the unknown is always a frustrating adventure where one has to follow many wrong leads - until one happens to find one that opens the sight to a solution....

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vanhees71, bhobba and Mentz114
No. Fluctuations are expressed not in the field expectations but in the field correlations. They are present no matter whether or not symmetries are broken. They are even present in a single harmonic oscillator if you call the ground state the vacuum state (as often done in quantum optics)!

One subtracts the expectation value if it is nonzero to get the broken symmetry theory for the redefined fields. The resulting Hamiltonian (in momentum space, in normally ordered form) is then renormalized to define the vacuum of the interacting theory. In this now physical theory, vacuum expectations of the physical (=renormalized) fields are again zero! The vacuum fluctuations refer to the nonzero correlation functions, which (after a Fourier transform) are of Kallen-Lehmann type and whose poles define the particle masses.
Are you referring to propagators of fields (correlations between their values at two different spacetime points)?

How should I interpret phenomena such as vacuum polarisation in QED without fluctuations of the interaction vacuum? (Apologies for my ignorance)

A. Neumaier
2019 Award
How should I interpret phenomena such as vacuum polarisation in QED without fluctuations of the interaction vacuum?
The following description taken from wikipedia is nonsense: ''the vacuum between interacting particles is not simply empty space. Rather, it contains short-lived "virtual" particle–antiparticle pairs (leptons or quarks and gluons) which are created out of the vacuum in amounts of energy constrained in time by the energy-time version of the Heisenberg uncertainty principle. After the constrained time, which is smaller (larger) the larger (smaller) the energy of the fluctuation, they then annihilate each other.''

The temporal story told is completely fictitious and has not the slightest backing in the formalism of quantum field theory. It is an imaginative (and imaginary) animation of the Feynman diagrams for vacuum fluctuations along the lines stated in post #58 of the other thread mentioned above. There is no uncertainty principle for virtual particles, and no formally meaningful way to assign a concept of time to them.

Are you referring to propagators of fields (correlations between their values at two different spacetime points)?
Field propagators used in perturbative calculations are Fourier transforms of the correlation functions of free fields, not of the physical fields. Thus they are interpreted in terms of fluctuations of the free fileds, whereas the interactions are treated perturbatively.

The correlation functions of the interacting electromagnetic fields are expressed nonperturbatively in terms of the photon self-energy, which is a real property of the real fields. The existence of vacuum polarization is equivalent to the fact that the photon self-energy is nonzero. This is the only physically valid interpretation of vacuum polarization!

The perturbative computation of the self-energy (and hence of the vacuum polarization) involves Feynman diagrams containing virtual processes of increasing order, representing high-dimensional integrals giving contributions to the self-energy. But these virtual processes are not processes happening in time, but book-keeping devices for getting the calculations correct. Physically, the self-energy is a nonlocal operator on the system Hilbert space that can be decomposed in many ways into pieces without an independent meaning. Only the total operator makes physical sense. The operator defined by the 1-loop Feynman diagram is just the simplest approximation of this operator.

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bhobba
bhobba
Mentor
How should I interpret phenomena such as vacuum polarisation in QED without fluctuations of the interaction vacuum? (Apologies for my ignorance)
Virtual particles are simply pictorial representations of terms in a Dyson series that appears in Feynman diagrams.

The picture of charge screening due to vacuum polarization is just a pictorial heuristic to explain why the bare charge blows up as the energy scale increases - its not what is actually going on.

The other thing to note is virtual particles do not appear in lattice gauge theory, strongly suggesting they are simply artifacts of the perturbative methods used, which is hardly surprising considering they are simply representations of terms in a pertubative series.

Thanks
Bill

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secur
vanhees71
Gold Member
2019 Award
Quantum fields don't fluctuate (except figuratively in popular science accounts for lay people). They are just a bit uncertain in their value, which may deviate from the expectation value by up to a few standard deviations. See Chapter A8: Virtual particles and vacuum fluctuations of my theoretical physics FAQ.
...and particularly the energy of the vacuum state, i.e., the ground state of the quantum-field theoretical model does not fluctuate, but it is exactly determined since the vacuum state is an eigenstate of the Hamiltonian. Usually one normalizes the vacuum energy to 0 for convenience. The absolute value of energy is not observable (except in General Relativity, but there it's the greatest riddle of all theoretical physics today, having to do with the smallness of the cosmological constant, which cannot be explained by some "natural" (symmetry) principle yet; the Standard Model predicts a value which is ##10^{120}## orders of magnitude too large).

bhobba and QuantumQuest
vanhees71
Gold Member
2019 Award
Are you referring to propagators of fields (correlations between their values at two different spacetime points)?

How should I interpret phenomena such as vacuum polarisation in QED without fluctuations of the interaction vacuum? (Apologies for my ignorance)
Very simple. There are no such fluctuations in the vacuum. QED vacuum polarization can only be detected by putting at least one charge somewhere and measure the electromagnetic field, but that's no more vacuum but there's at least this charge present. What's called "vacuum polarization" is the self-energy of the photon field and manifests itself, e.g., by deviations of the field of a resting point charge from the classical solution, i.e., the Coulomb field. As far as I know, these deviations have never been directly detected, but they are measurable in terms, e.g., of the Lamb shift of the hydrogen spectral lines, and this is one of the best agreements between theory in experiment in entire science.

bhobba
A. Neumaier
2019 Award
Usually one normalizes the vacuum energy to 0 for convenience. The absolute value of energy is not observable
This is not true in the present context.

In nonrelativistic mechanics, only energy differences are measurable. But in relativistic mechanics, the energy operator must satisfy the Poincare relations, hence cannot be shifted by an arbitrary constant. Thus absolute energies are meaningful. For a translation invariant state (i.e., the vacuum state) this fixes the vacuum energy at exactly zero, without any freedom or fluctuations.

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vanhees71
Gold Member
2019 Award
How that? In special relativity, I can add a constant to the Hamiltonian without changing anything on the dynamics. Why do you say that doesn't hold in the QFT case either? If I don't normalize the vacuum energy to 0, all that happens is that (in the Schrödinger picture) the vacuum state ket (and all other state kets) get an additional common phase factor ##\exp(-\mathrm{i} E_0 t)##, which is just a common phase factor and thus not observable. It doesn't even lead to a change in time dependence of the states at all, because not the state kets represent the states but the corresponding rays (or the projection operators as statistical operators, where the common phase cancels either).

bhobba and Demystifier
A. Neumaier
2019 Award
In special relativity, I can add a constant to the Hamiltonian without changing anything on the dynamics.
No. The relativistic energy is by definition the zero component of the 4-momentum vector, multiplied by ##c##. A free relativistic particle satisfies ##p^2=m^2##, and ##H=cp_0## by definition. One cannot change ##p_0## by a constant shift without altering everything.

Of course, one could redefine ##H## to be ##cp_0+E_0## with an arbitrary nonzero constant ##E_0##, without changing the dynamics. But this modified Hamiltonian is no longer the relativistic energy!

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vanhees71
Gold Member
2019 Award
Well, the absolute value of the single-particle energy above (!!!) the vacuum energy is fixed to ##E=E_{\text{vac}}+c \sqrt{m^2 c^2+\vec{p}^2}##, but it's still only defined up to the constant vacuum-energy value. Of course, if you want to have the single-particle energy together with it's momentum forming a four-vector you must set ##E_{\text{vac}}=0##. That's a nice argument for setting the vacuum energy to 0. It's however, still arbitrary. Note that manifest Lorentz covariance is broken anyway by using the canonical formalism.

The picture of charge screening due to vacuum polarization is just a pictorial heuristic to explain why the bare charge blows up as the energy scale increases - its not what is actually going on.
Is it purely due to the running of the coupling with the energy scale then? (I know a very small amount about the renormalization group, but not much)

In nonrelativistic mechanics, only energy differences are measurable. But in relativistic mechanics, the energy operator must satisfy the Poincare relations, hence cannot be shifted by an arbitrary constant. Thus absolute energies are meaningful. For a translation invariant state (i.e., the vacuum state) this fixes the vacuum energy at exactly zero, without any freedom or fluctuations.
I have to admit, I don't quite understand why Poincaré invariance requires that the vacuum energy is zero?! I mean, the vacuum state of a theory can have non trivial structure a priori, right? Also, in special relativity, I can shift the energy by a constant amount and it doesn't affect the dynamics of the theory, which suggests that absolute values of energy are unobservable, hence why we can normal order fields and remove the vacuum energy. However, in general relativity, if I shift the energy by a constant amount then this does affect the dynamics of the theory (since gravity is sensitive to absolute energies) hence why the presence of vacuum energy should be reviewed and considered much more carefully in this case.

Very simple. There are no such fluctuations in the vacuum.
Would it be at all correct to say that a given quantum field is not fluctuating, but due to the Heisenberg uncertainty principle, a given quantum field and its conjugate momentum do not have well-defined values. When people talk of vacuum fluctuations are they are referring to this fact, but trying to make the information more palatable for their audience, since it is easier to conceptualise something fluctuating rather than it not having well-defined values (since classically, where the analogues are position and momentum, these always have well-defined values)?!

A. Neumaier
2019 Award
Would it be at all correct to say that a given quantum field is not fluctuating
No. The fields always fluctuate - their correlation functions are nonzero. But the vacuum energy does not - its standard deviation is exactly zero.

You can add as much talk as you like to make things palatable for you or for anyone else, but this doesn't change the physics, which is in the formal relations.
why we can normal order fields and remove the vacuum energy.
There is no freedom here: We have to normal order and do all renormalizations (not only removing the infinite vacuum energy, but other infinities as well), in order that the Hamiltonian agrees (up to a factor ##c##) with the time component of the 4-momentum in a representation of the Poincare group.

A. Neumaier