# “…Each point in space makes a contribution of E=ħω/2, resulting in a calculation of infinite zero-point energy in any finite volume..”?

Is there then infinite energy (in reality) in vacuum/virtual particles/zero point energy?(https://en.m.wikipedia.org/wiki/Zero-point_energy)

## 2 Answers

If we take the simplest case of a free scalar field and quantise it then we find the field has an infinite number of modes. These modes look like an infinite plane wave in position space, or if we Fourier transform into momentum space then they look like a harmonic oscillator at a point in the momentum space.

Incidentally I suspect the quote is getting a bit mixed up between position and momentum space, because the statement Each point in space makes a contribution of $E=\tfrac{1}{2}\hbar\omega$ only makes sense in momentum space.

Anyhow, each mode can contain no particles, one particle, two particles, and so on, and there are an infinite number of these modes. The energy of any particular mode increases with the number of particles in, which is as you would expect.

However, because these modes behave like simple harmonic oscillators that means even if a mode has no particles in it the energy of the mode is not zero. Instead it has a zero point energy $\tfrac{1}{2}\hbar\omega$, where $\omega$ is the frequency associated with the mode. And if we have an infinite number of modes, each with a non-zero energy, that means the total energy will be infinite. This is the source of the claim that the vacuum energy is infinite.

There are several ways round this. For example it's generally true in physics that we can only measure difference in energy so what we would actually observe is the energy of a quantum field relative to its zero point energy. That is, we subtract off the (infinite) zero point energy to make the vacuum energy zero. Then as we add particles to our field the energy increases from zero. This seems a bit cavalier, but what we're really saying is the zero point energy is an artefact of our mathematical model and not something real.

I think this is what the statement you quote is referring to, though we should note there is a second more pernicious source of infinite energy. The free field I've described so far is a simplified mathematical object that we use because it's easy to work with, but all real fields are interacting. In this case we need to consider the interactions when calculating the vacuum energy, and when we do this we also get an infinite energy, even after we've subtracted off the (infinite) zero point energy. This second infinity is fixed by renormalisation.

The energy of the vacuum is obviously not infinite because if it was it would cause an infinite spacetime curvature and the universe would not exist. I said in a previous paragraph that in physics we can only measure differences in energy, but the exception to this is general relativity. The spacetime curvature is related to the total energy density, and an infinite energy density would cause an infinite spacetime curvature. Even if the vacuum energy were very high but not infinite it would still create vastly more curvature than we observe.

Since the universe exists we have to conclude that the infinite zero point energy either doesn't gravitate for some unknown reason, or it's an artefact of our mathematical model and doesn't really exist. Likewise the infinite (or possibly just very large) interacting vacuum energy. I don't think we understand quantum field theory well enough to say for certain how this is resolved.

• "I suspect the quote is getting a bit mixed up between position and momentum space" I agree, though getting rid of UV divergences by introducing a cut-off is equivalent to a lattice discretization of space right ? After discretization (with a length parameter $a$), the amount of points in a finite volume is finite. In momentum space, this amounts to an upper bound $p < 1/a$, and the divergence is suppressed. The total energy should be $\propto V$ regardless. Would you agree with that ? – Lucas Gautheron Nov 23 '16 at 19:55
• So if i understood you well if"all real fields are interacting. In this case we need to consider the interactions when calculating the vacuum energy, and when we do this we also get an infinite energy, even after we've subtracted off the (infinite) zero point energy. This second infinity is fixed by renormalisation." then there's infinite energy in reality? – Noduagg Nov 23 '16 at 22:05
• @Noduagg: I've extended my answer to address your comment. – John Rennie Nov 24 '16 at 6:19

An experimentalist's answer:

The vacuum by definition in particle physics means "nothing", therefore working with energy and momentum operators on "nothing" one gets nothing, zero. The quote misunderstands the current standard model of particle physics, and this is the only tool there exists for measuring and predicting the behavior of particles which are the constituents of all matter.

Is there then infinite energy (in reality) in vacuum/virtual particles/zero point energy?

It is particles that carry momentum and energy in the quantum mechanical model of particle physics, and are the source of any fields, classical and quantum mechanical. In Quantum Field Theory, the mathematical statement is used that each particle in the standard model table each particle exists as a field over all spacetime, an electron field, a neutrino field,.... This is a mathematical construct, to illustrate the mathematics of QFT, which uses creation and annihilation operators on these field to represent the real particles and their interactions, so as to calculate cross-sections and lifetimes etc that can be validated experimentally.

Operating with the energy and momentum operators on the vacuum representation of the fields gives zero energy and zero momentum, there is nothing there, unless a real particle exists in the region.

Let us address the difference between real and virtual. These are the characterizations of the lines used in the schematics of Feynman diagrams,

Feynman diagrams make easy the calculations of QFT expressions so as to be compared with experiments. Real are the particles entering and leaving with external lines. They carry energy and momentum and are on mass shell , i.e. $E^2-p^2~ =~m^2$ ($c~=~1$ in this system). The mass of the particle. All internal lines are called virtual and are off mass shell, the mass is variable under the implied integration, they are place holders for the quantum number conservation of the named particle but no physical attributes can be identified with the virtual internal lines.

To have virtual particles one must have real particles entering or leaving an interaction region to supply the energy and momentum.

Vacuum loops, as in your question, do not exist because the integrals implied by the loop diagram would have zero both as upper and lower limit. The confusion arose because a single real particle traversing the vacuum does see vacuum loops , but the energy momentum balance within the integrals is supplied by the particle traversing the vacuum with creation and annihilation operators defining its path. The loops are real in the calculations as the Lamb shift calculations show, but the particles schematically depicted are not.

The confusion arises because people take the point of view of a traversing electron (for example), which does "see" virtual particle loops, for the electron the vacuum is populated, and they forget that the electron is supplying the energy momentum balance. These loops do add up to infinities that have to be corrected with renormalization tools in order to get predictions for experimental measurements.

So in conclusion, vacuum energy depends on the model used. I was taught nuclear physics using QFT tools ( back in 1961), and there, the vacuum energy was the real lowest energy of the nucleus under consideration. But applying QFT to elementary particles is a different story than a harmonic oscillator model. The elementary particle fields posited in QFT have zero energy and momentum by themselves, only a specific real particle propagating on these fields and exchanging energy and momentum within the Heisenberg's uncertainty interval can generate virtual loops.