Vacuum energy is based on Heisenberg's energy–time uncertainty principle which says that $\Delta E$ can exist for $\Delta t$. This $\Delta E$ can produce electron positron pair for $\Delta t$ by which time they annihilate. My question / confusion is that upon annihilation a pair of photons are produced, and this photon energy continues to exist beyond $\Delta t$, perhaps forever. Is this correct understanding, as this seems to imply that $\Delta E$ can exist forever?

  • $\begingroup$ This is an improper application of the uncertainty principle. For the correct meanings of $\Delta t$ and $\Delta E$ in the "energy-time uncertainty principle", see this answer by joshphysics $\endgroup$ – ACuriousMind Aug 11 '19 at 19:38

Answer using QFT in flat spacetime

In quantum field theory in flat spacetime, the vacuum is a stationary state. It's energy is well-defined: $\Delta E=0$ in a stationary state. This is consistent with $\Delta E\Delta t\gtrsim\hbar$, because $\Delta t$ is undefined for a stationary state: in a stationary state, nothing "happens," ever. The energy of a stationary state is permanent.

To get a state in which something happens, we need to consider a superposition of different stationary states having different energies. If the spread in the energies involved in the superposition is $\Delta t$, then the time-scale over which the state can change in time is $\Delta t\sim \hbar/\Delta E$. A stationary state has $\Delta E=0$, so $\Delta t$ is "infinite": the state never changes. Nothing happens.

What about curved spacetime?

This answer is based on QFT in flat spacetime, which ignores gravity. In this obviously-incomplete model, the overall energy of the vacuum state has no physical significance: it is unobservable. We can shift it arbitrarily without affecting any testable predictions, just by adjusting the constant term in the lagrangian.

The situation changes when gravity is taken into account, because then the vacuum energy does have observable effects. Physics doesn't yet have a satisfying resolution of the vacuum-energy issue in that context. We don't even know exactly what "vacuum" means in that context.

Perspective about virtual particles

Feynman diagrams and the "virtual particle" language come from perturbation theory. Perturbation theory is a useful computational method, because it relates vacuum expectation values of the full theory (which is mathematically difficult) to various combinations of "vacuum" expectation values of an easier theory, but the vacuum state in the easier theory is a different state than the vacuum state in the full theory. Roughly speaking, the bubble diagrams illustrated in the answer by G. Smith represent terms in the perturbative expansion of the normalization factor that relates these two different states.

We have some idea of how to include gravity in perturbation theory, but this doesn't solve the "vacuum energy" problem. For that, we need a better understanding of quantum gravity than we currently have.

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  • $\begingroup$ Why is there a 40-orders-of magnitude discrepancy between the vacuum energy required by QM and that which cosmological observations have established? Common sense decrees that one or the other must be wrong, but no one is willing to say which because this is a taboo subject.. Taboo subjects are not good for science. $\endgroup$ – Michael Walsby Aug 10 '19 at 12:16
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    $\begingroup$ @MichaelWalsby It isn't taboo, just not directly relevant to answering the question that was asked here. Notice that I didn't say what the value of $E$ is in the vacuum state; I only said that it's perfectly well-defined. At least in flat spacetime, the vacuum state is, by definition, the state of lowest energy. That's its definition. The actual value of that lowest energy is irrelevant in flat spacetime; it only matters when gravity is considered, but addressing that properly requires a quantum theory of gravity. It's not taboo at all: we all agree that ordinary QFT is not the final word. $\endgroup$ – Chiral Anomaly Aug 10 '19 at 13:47
  • $\begingroup$ I wasn't criticising your answer or you personally, but just remarking on the fact that either QM must be wrong on this particular matter or the cosmological observations must be wrong, and in my experience no one is willing to say which. I suspect that whoever tells us which will find himself unpopular with one side or the other, and few people want to be unpopular, $\endgroup$ – Michael Walsby Aug 10 '19 at 14:07
  • $\begingroup$ In fact, it is this 40-orders-of magnitude discrepancy that prompted my question. I have related question, is flat space-time even a valid concept? I can see that over short delta-t timescale for pair production that flat space-time assumption could apply, but over longer time scales the rate of expansion of the universe has changed, and does this not invalidate the flat space-time concept? $\endgroup$ – FritzS Aug 10 '19 at 19:12
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    $\begingroup$ @FritzS You are right that spacetime is not flat, but flat spacetime is an excellent approximation in laboratory situations where quantum effects are typically observed. QFT in flat spacetime works great for particle physics, and classical (non-quantum) general relativity works great for large-scale gravitational physics, and we just don't have easy access to situations where both of those effects are simultaneously important. They must coexist somehow in the real world, of course, but we don't know exactly how yet, so we don't have a definitive answer to the vacuum-energy issue. $\endgroup$ – Chiral Anomaly Aug 11 '19 at 15:00

No, your understanding is incorrect. The “vacuum bubble” Feynman diagrams contributing to the vacuum energy density don’t have any external lines. All particles are virtual, not real.

The following image shows two of an infinite number of such diagrams. In the simplest such diagram (the loop at the top), an electron and positron simply emerge from the vacuum and return to the vacuum. There are no real photons produced, and the process is not really annihilation.

In more complicated vacuum diagrams, such as the second loop, all photons are virtual, not real.

enter image description here

The image is from this discussion of vacuum energy. (The wavy line is actually a virtual graviton, not a virtual photon, but a similar diagram with a virtual photon would exist.)

If the vacuum were constantly producing real photons from virtual electrons and positrons, we would presumably have observed them!

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  • $\begingroup$ Would it contradict any known principle (other than Occam's razor) to assume that, since off-shell electrons/positrons are not limited to a minimum energy of 0.511 KeV, presumably their thermal distribution could exist but much cooler and become almost undistinguishable from, and "clouded" by the CMB? $\endgroup$ – lurscher Aug 10 '19 at 2:42
  • $\begingroup$ @lurscher Well, I think it contradicts quantum field theory. Virtual particles don’t have thermodynamics as far as I know (although you can certainly formulate QFT at non-zero temperature). $\endgroup$ – G. Smith Aug 10 '19 at 3:11
  • $\begingroup$ since you said "we would presumably have observed them!" I just wanted to interject that depending on what the rate of "vacuum leakage" of unbalanced legs happening at cosmological scales, the thing to be observed could be there, but masked by the CMB $\endgroup$ – lurscher Aug 10 '19 at 13:37
  • $\begingroup$ The related question has been answered, "No, the 0.511MeV photons have not been observed". My intuition told me that these pair production "quantum fluctuations" could not be adding energy to the universe, but I have much to study to comprehend this answer. I have a follow-up question on pair production as the basis of Hawking radiation. At what point does the virtual particle that escapes the event horizon become a real particle; and does the other virtual particle that falls into the black hole become a real particle? Is there such a thing as transitioning from virtual to real? $\endgroup$ – FritzS Aug 10 '19 at 23:36
  • $\begingroup$ @FritzS The virtual particle story of Hawking radiation is a simplification. Please see physics.stackexchange.com/a/252236/123208 $\endgroup$ – PM 2Ring Aug 11 '19 at 15:38

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