# Can energy be taken out of the QFT vacuum?

There have been recent questions about the vacuum. In my simplified knowledge the vacuum is like a ground state energy level, and also that there might even exist other lower energy levels than the vacuum we find ourselves in. The sea is a soup of created and annihilated pairs of virtual particles with virtual energies and momenta.

In a normal sea on earth, which also represents a ground state of water, energy can be taken out of random waves by the clever construction of valves that allow only one way motion of water. Is it conceivable that a gadget of similar function could be found for the vacuum sea, or is it forbidden by conservation laws?

My intuition tells me that it might be possible if GR is taken into account, but my physics knowledge does not stretch to support this.

Edit : An explanation of why I am asking this question:

Let me expand on the example of the sea. The energy from the waves comes from either tides, i.e. gravitational forces, or wind (temperature differentials). If these were missing the oceans would be like glass representing a unique ground state of the gravitational well of the earth.

In an analogy, a gravitational wave going through the vacuum would be supplying energy to the vacuum sea.

I have been thinking of this analogy ever since cold fusion surfaced and refuses to die out, the most recent one being discussed here too. Approached from nuclear physics orders of magnitude the claims seem preposterous. There are people though who believe they have results of extra energy over input energy, much more than chemical reactions could supply.

This set me thinking on vacuum energy and the analogy with getting energy from the sea. A crystal is a prime candidate for any exploration of such concepts and in all cold fusion "successful" results crystals have been used. Now if the effect depended on the vacuum and how much distorted it was by a passage of a gravitational wave at the time of the experiment, or the exact orientation of the crystal, or the type of impurities in the crystal ( F centers etc) one would expect to get haphazard results, and non repeatable by other experimenters.

Of course this would be the first experimental evidence of gravitational waves :).

Edit 20/7/12

Maybe I should clarify that an acceptable answer in the negative would be one based on conservation laws. I believe that data trumps theory, and next in line are conservation laws,because they are the distillation of an enormous amount of data. Some people seem to think that theoretical definitions can substitute for proof in physics, but physical theories change, solid data do not, and this is physics, not axiomatic mathematics.

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Wouldn't that be very similar to Maxwell's demon? – Willie Wong Jul 2 '11 at 14:44
If the vacuum is the ground state, what does it mean to have a lower state. Analogy would be a value smaller than the minimum. Also in the particle interpretation, what would it correspond to, negative number of particles? – MBN Jul 2 '11 at 16:32
– anna v Jul 2 '11 at 19:01
I looked at it but it doesn't help my confusion. What is the vacuum state? I thought it is the state killed by all the anihilation operators. What would a lower than that mean? – MBN Jul 3 '11 at 0:08

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@Marek The question is not about pumping energy from different vacua, although it might be related, if, for example, vacua had an "uncertainty principle" type of existence. The question relates to the marrying of QFT and gravity. Whether a gravitational wave would change the vacuum energy level as it is propagating in time so that energy might be captured of its passage, as in the sea wave analogy. – anna v Jul 4 '11 at 3:58
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## 7 Answers

Well, let us be honest here. This question was not supposed to be answered from the very beggining. First of all we don't know how to mix quantum mechanics and gravity. There is no good consistent theory for that. Another thing is that today "the Dirac sea" analogy is not considered to be a very good thing. It is a pre-QFT naiive picture. Finally we are supposed to talk about "waves" in this "sea"... While the "sea" itself is an obsolete analogy... And all that is in a context of non-existing theory... Come on...

Now. There is actually a formal way to answer the question, because the question is about the "QFT vaccuum". And the QFT vacuum have a precise definition. Which basically says that it is "something you cannot take energy from". Actually, we start from that definition to build QFT. So the answer is: "you cannot take energy out of the QFT vacuum by definition".

Maybe we are wrong to start from this definition. Maybe for quantum gravity we need another starting point. But then it wouldn't be a QFT -- it would be a new theory which will have QFT as a limiting case. And in the range of validity of QFT that "formal" answer will hold.

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Thanks for regurgitating my answer.............. – Chris Gerig Jul 19 '12 at 22:12
See my comment to @ChrisGerig . There are theories mooted which do have several vacua, QFT would be working on one of those, at the moment the one we find ourselves in. I believe my simplistic analogy of the tides should give an intuition. – anna v Jul 20 '12 at 3:09
"it would be a new theory which will have QFT as a limiting case." This is what I am exploring with this question. "And in the range of validity of QFT that "formal" answer will hold." Do you have a strong argument that you know how that limiting case is approached? For example, superconductivity , quantum mechanics on kilometer scales, shows that it is not dimension that defines absolutely the limit between QM and classical.There is the added ingredient of complexity of bulk matter and its different responses. – anna v Jul 21 '12 at 4:04
@Kostya , if you expand a bit your last paragraph I would accept your answer. I am in disagreement with the answer chosen by the bounty setter. It is not the answer of a physicist, but of a mathematician. A theory can be beautiful and fully consistent in its definitions. The results may agree with data to the available limit of measurements, but that does not mean that the definitions define nature, as progress in physics has shown us again and again. It is at the limits of the validity of a theory that it may be tested against nature with surprising results, for the theory. – anna v Jul 25 '12 at 3:10
@annav I've tried to expand my last paragraph. But I can't. I just really have nothing else to say on the point: whatever I try, I either already said that or don't know what I'm talking about. I should note that I'm a bit formal here as well -- I'm answering the question. I'm not participating in any discussion or "exploration" or whatever. Let me finally stress that I'm against this distinction you draw between "mathematicians" and "physicists". Intuition is great. But if rationality tells that intuition is wrong -- have courage to leave it. – Kostya Jul 25 '12 at 14:29
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No you cannot take energy out of the vacuum, BY DEFINITION.

Using this analogy with a "sea" is nonsense, because the statement "vacuum is a sea of virtual particles" is also an ambiguous statement not to be taken literally.

The definition of vacuum is ground-state energy, and so if you could take energy away from it, then the energy of the vacuum would be lowered, contradicting the fact that it's already in the lowest possible state.

This is also the reason why spontaneous absorption does not occur with electrons/atoms (whereas spontaneous emission can occur).

I want to point out why this question is not closed yet:
Anna's train of thought is "The answer does not satisfy me because it is dependent on definitions which change as theories change."
BUT, the question itself is dependent on definitions... you can't ask about apples and then say 'well maybe they can be oranges'. If you can pose a question about 'energy' and 'vacuum' then they have to be defined.
I now vote to close!

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The ground state energy is the energy of the quantum mechanical system whose potentials one knows and has solved. It is different for different initial set ups. I am talking of something like the Theory of Everything, which in some speculative theories does have several ground states. – anna v Jul 20 '12 at 3:05
p.s. BTW, nature does not follow definitions, it is observers who make them, and new definitions can supersede or embed old ones. – anna v Jul 20 '12 at 3:20

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Well, usually the physics of tomorrow hides in the metaphysics of today, though I do not think of this as metaphysics. Rather at the frontier of physics. Your answer does not satisfy me, because it says: these are the definitions, do not bother me with facts. We would have never gotten quantum mechanics if questioning were stop by current definitions. – anna v Jul 20 '12 at 5:57

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The answer does not satisfy me because it is dependent on definitions which change as theories change. Are you saying that gravitational waves do not carry energy by "definition" and it is futile to try and discover them? – anna v Jul 20 '12 at 6:30

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For you to even begin talking to anyone, you have to specify what a vacuum is and what energy is, otherwise your question doesn't even make sense, and no logical statements can be made! Therefore, once you bring in the words "energy" and "vacuum", then a priori they have a definition which your question must adhere to. And this question has an answer... the end. – Chris Gerig Jul 20 '12 at 6:38
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Peter Milonni has devoted a lot of work to this area, i will recommend this book: The Quantum Vacuum, it basically gives a overview on lot of physical systems where vacuum field effects are dominant, and how certain boundary conditions can effect them to produce unexpected effects

the expansion of fields around the vacuum in harmonic modes with quantized particle population of the modes is only valid when the physics can be fairly approximated in terms of the vacuum eigenmodes (which are the regular field wave functions). Of course, when there are special boundary conditions, this doesn't change too much since we still have eigenmodes (which correspond to the boundary geometry) but still can talk about particle populations

when the system (meaning, the boundary conditions) are dynamic (moving mirrors, or superconductor walls moving over the superconductivity phase) i don't think there is an authoritative answer about the validity of these approximations. So in short, there could be interesting things to say about the vacuum dynamics when a more friendly framework exists to make computations in highly dynamical regimes

I made a somewhat relevant question a while ago about casimir walls that melted and froze back with an oscillatory transversal magnetic field.

i'm sorry, i'm not addressing your larger question, as to what relevant things change when taking GR in consideration. Hopefully someone more knowledgeable on the matter will.

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-1: This is nonsense. You can't take energy out of the vacuum, and this is conservation laws. – Ron Maimon Sep 4 '11 at 23:22

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@Ron, there is no such thing as an energy conservation law in GR. – lurscher Sep 5 '11 at 13:53
you aren't doing GR. – Ron Maimon Sep 5 '11 at 13:56
GR effects should manifest locally as out-of-equilibrium boundary conditions, which is why my answer tried to make those points. – lurscher Sep 5 '11 at 14:17
GR effects do not manifest locally--- the lack of conservation in GR can be attributed to new material entering a region on cosmological scales, or leaving it. The conservation of energy in QFT vacuum is a fact, and it is impossible to pump energy out unless the vacuum is unstable and you then wreck the whole universe by tunneling to a lower energy state. – Ron Maimon Sep 5 '11 at 14:25
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The ground state (=vacuum) is an eigenstate of the system Hamiltonian. The wave function is unique, no energy uncertainty exist. In this sense there is no fluctuation of energy to harvest. "Fluctuations" exist for non commuting with the Hamiltonian variables but it cannot be used to get some energy for the reason given above.

In GR the energy is not conserved and this is not connected to the vacuum.

Edit: in heat machines one tries to increase the temperature difference $T_{hot}-T_{cold}$ to increase the machine efficiency. Vacuum is the coldest body ever known. So the vacuum "cleaner" efficiency cannot exceed zero.

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Your last sentence is the crux of the matter. Why wouldn't general relativity create fluctuations in the vacuum, in a TOE? – anna v Jul 2 '11 at 18:40
Because GR is not a unique theory of gravity. There are theories compatible with a flat space-time with the energy/momentum conservation laws. – Vladimir Kalitvianski Jul 2 '11 at 18:59
It certainly is not a heat machine. @Vladimir , the system of comments does not accept your name in the beginning of the comment!! – anna v Jul 3 '11 at 19:18

In theory, QFT vacuum state is isotropic and invariant for all observers, as the base of all ladder operator algebras. It does not have any features other than the fluctuations. Hence, you cannot extract any more energy from it.

In practice, extracting energy out of the vacuum is not only possible, but it has been already achieved in laboratory using optical parametric amplifiers (NOPA); they are used for squeezing input light, but when there is nothing in the input, the output is a field that has lower average standard deviation of the energy than the normal vacuum in the range of frequencies where the amplifier is active. If we assign to the normal vacuum zero energy, then this "squeezed vacuum" field must have negative energy.

Of course, let's make perfectly clear that this mechanism cannot be used to extract useful energy, since the energy spent pumping the amplifiers greatly exceed anything that you could extract.

“In theory, theory and practice are the same. In practice, they are not.” ― Albert Einstein

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The concept of squeezing does NOT truly beat the "standard quantum limit", because although it squeezes one variable, the other canonical-variable increases. – Chris Gerig Jul 19 '12 at 23:02
@ChrisGerig, "it squeezes one variable, the other canonical-variable increases" that is correct. But do not forget you can choose what canonical variables you use. In most experimental setups, the variables chosen are amplitude versus phase. In this case, amplitude-squeezed vacuum has smaller amplitude fluctuations, which implies smaller average energy – diffeomorphism Jul 20 '12 at 16:15
@ChrisGerig, there is however, a valid critisism, if we accept that some form of quantum inequalities must hold, then the negative energy regions need to be overcompensed by regions with extra positive energy, so even the time-averaged energy density is nonnegative or the space-averaged energy density is nonnegative, but not both – diffeomorphism Jul 20 '12 at 16:17

There has been a proposal to use Casimir force and meta materials to build a vacuum energy extractor. Casimir force applied on parallel plates made of "normal" material pushes the plates outwardly and inwardly for metamaterials. I do not know if the idea has yet been tested.

http://physicsworld.com/cws/article/news/2007/may/02/casimir-force-could-drive-tiny-ratchets

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The vacuum state has a property named passivity, which means that any local operation onto the vacuum state does not extract but inject energy to the system. This implies that energy cannot be taken out of the vacuum only by local operations. However, if we adopt local operations and classical communication (LOCC), a part of zero-point energy of the vacuum state can be extracted. The scheme is called quantum energy teleportation. More information is available in wikipedia and a review article by Hotta, who first proposed the concept: http://www.tuhep.phys.tohoku.ac.jp/~hotta/extended-version-qet-review.pdf .

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