# I Relation between quantum fluctuations and vacuum energy?

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1. May 20, 2016

### Frank Castle

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?

Last edited: May 20, 2016
2. May 20, 2016

### A. Neumaier

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.

3. May 20, 2016

### Demystifier

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.

4. May 20, 2016

### Frank Castle

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?!

Last edited: May 20, 2016
5. May 20, 2016

### A. Neumaier

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.

Last edited: May 20, 2016
6. May 20, 2016

### Frank Castle

Would what I put in my edit (apologies, I edited it before I received your reply) to my previous post be correct at all?

7. May 20, 2016

### A. Neumaier

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).

8. May 20, 2016

### Frank Castle

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?

9. May 20, 2016

### A. Neumaier

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.

10. May 20, 2016

### Frank Castle

Yes, this is what I was referring to.

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).

11. May 20, 2016

### A. Neumaier

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!

Last edited: May 20, 2016
12. May 20, 2016

### Frank Castle

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.

13. May 20, 2016

### A. Neumaier

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 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....

Last edited: May 20, 2016
14. May 20, 2016

### Frank Castle

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)

15. May 20, 2016

### A. Neumaier

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.

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.

Last edited: May 21, 2016
16. May 20, 2016

### Staff: Mentor

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

Last edited: May 20, 2016
17. May 21, 2016

### vanhees71

...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).

18. May 21, 2016

### vanhees71

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.

19. May 21, 2016

### A. Neumaier

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.

Last edited: May 21, 2016
20. May 21, 2016

### vanhees71

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).