According to Quantum Field Theory, all 'particles' are oscillations in their own fields. And according to Vacuum Zero Point Energy and the Heisenberg Uncertainty Principle, these fields have fluctuations in time and can never have 'zero' energy. If we examine a sufficiently large amount of space, why don't we see particles occasionally being formed out of nowhere? Why don't we see random flashes of light in interstellar space because of the electromagnetic field getting enough energy to form a photon?
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3$\begingroup$ Possible duplicate of Are vacuum fluctuations really happening all the time? $\endgroup$– John RennieCommented Nov 11, 2018 at 15:22
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4 Answers
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QFT models each frequency that a particle can have by a QM harmonic oscillator. This has a ground state energy of $\frac{1}{2}\hbar \omega$ in which no particle is present. So the answer to your question is that even though the ground state energy is non-zero, there are no particles present.
It is perhaps interesting to take note of the cosmological constant problem ( https://en.wikipedia.org/wiki/Cosmological_constant_problem ). It states the integrated ground state energy of the EM field is $10^{123}$ $J/m^3$, which would imply a universe the size of a soccer ball. We all know it is at least the size of an entire stadium !
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According to Quantum Field Theory, all 'particles' are oscillations in their own fields.
This is not true: According to QFT all space time is covered by fields represented by the free particle solutions of their corresponding quantum mechanical equation: Dirac for fermions, Klein Gordon for bosons and quantized Maxwell for photons. There is a field throughout space-time for all particles in the table of the standard model of particle physics. Plain waves have oscillations but these are in a complex number space , not measurable in any way.
On these fields creation and annihilation operators propagate particles allowing for the possibility of writing Feynman diagrams to represent the integrals necessary in order to get measurable quantities: particles created and annihilated at the vertices of the diagram. These integrals predict measurable quantities for particle interactions and are validated by data up to now.
A free particle by itself cannot be represented by a plane wave, because free particles are localized in nature, see these single electron footprints. . To describe a single particle in the lab one needs a wave packet representation, i.e. a distribution of momenta so as to be able to get a mathematical model of a particle at a reasonable space-time location. Note that this means a given momentum with a spread commensurate with the Heisenberg uncertainty.
these fields have fluctuations in time and can never have 'zero' energy.
Energy has to be supplied in order to get measurable, observable effects from vacuum loops. All the listed "successes" inusing loop diagrams have energy supplied by incoming real particles.
Many physical effects attributed to zero-point energy have been experimentally verified, such as spontaneous emission, Casimir force, Lamb shift, magnetic moment of the electron and Delbrück scattering, these effects are usually called "radiative corrections".
you ask:
Why don't we see random flashes of light in interstellar space because of the electromagnetic field getting enough energy to form a photon?
Getting energy from where? The photon field itself has no momentum it is just a "coordinate system" for an incoming particle with momentum. For an operator to operate on the field and create particles, energy must somehow be supplied. (For a real incoming from space photon the chosen answer here is relevant)
The handwaving discussions of vacuum loops independent of real particles is about cosmological models, where energy will be supplied by dark energy or something like that, but these are research models, not standard ones.
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The fields (or coordinates and momenta) fluctuate; the energy does not. Vacuum is an eigenstate with a certain energy. Besides, by definition, it is the ground state - the lowest energy. What are going to observe there if there is no excitation by definition?
answered Nov 11, 2018 at 15:32
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$\begingroup$ You are going to observe that the universe is the size of a football😂 $\endgroup$– my2ctsCommented Nov 21, 2019 at 8:39
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The vacuum state is orthogonal to every vector containing particles so the probability of a transition is $0$: you cannot see particles created from the vacuum anywhere.
The operator number of particle admits the vacuum as 0 eigenvector, so measuring it you always obtain 0.
What you can see in the vacuum are for instance vacuum polarization phenomena which can be described (but in my view this is nothing but a powerful but unnecessary theoretical view) as interactions of real particles, e.g., photons with "virtual particles" of another field (Dirc field) which stays in the vacuum state. This phenomenon is actually strongly due to the type of interaction (I mean $\overline{\psi}\gamma_\mu\psi A_\mu$) between electrons and photons.
The energy of the vacuum state is undefined in QFT unless you use the first step of the renormalization procedure (normal ordering). This immediately leads to $0$ for the renormalized stress energy tensor of the free theory.
In curved spacetime, the picture becomes more complicated and, in principle, you can have some non-vanishing density which may contribute to the measured values of certain quantities, like the cosmological constant which can be viewed as a finite counterterm of the renormalization of the stress energy tensor in a Hadamard state.
answered Nov 11, 2018 at 15:54