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Is the vacuum empty and boring?

It depends on the observables you want to learn about

One of the tricks by which the proponents of various reactionless drives brainwash the gullible laymen is the suggestion that they can create a self-accelerating spaceship because instead of "directly" violating the momentum conservation law, they are "pushing against the vacuum".

So some defenders of this pseudoscience were writing such things in the thread. For a while, I was patient but my patience has been depleted for a long time. I just can't comprehend how someone may fail to see that "violating the momentum conservation law" and "pushing against the vacuum" are exactly the same thing expressed in different words – and they are (or it is) forbidden in Nature.

After all, the vacuum is "nothing", and if "nothing" can absorb your money, then the money is lost, isn't it? "Absorbed by nothing" and "disappear" are synonymous terms, too. At least, these insights looked crystal clear to me when I was 8 or so. A discussion about them has never looked like a scientific one to me. It was always a discussion about which child was left behind. If someone can get deluded and "approve" a violation of the momentum conservation law just because it was renamed to "pushing against the vacuum", he or she clearly didn't understand that law at all.

The law holds and whether you decide to express is in different words cannot make a difference.

You could object that my reasoning above depended on the assumption that "the vacuum is nothing" or that "the vacuum is empty". But doesn't quantum mechanics teach us that it is not empty, after all? You can surely find lots of these comments everywhere – e.g. in the Empty Space Is Not Empty YouTube video – and those statements, when properly understood, are true.

So doesn't it allow the vacuum to "absorb" the momentum without creating any new particles or objects? No, it doesn't.

Newton's, Maxwell's, Einstein's vacuum

A short history about the scientific history of the vacuum follows. In Newton's mechanics, the vacuum was empty (and a completely rigid, non-dynamical, and flat Euclidean arena) by construction. There was nothing to carry any information. The information was carried by the coordinates and velocities of the particles that could have been added on top of the empty space. Each particle added 6 coordinates to the phase space. But the vacuum meant no particles, and it meant no coordinates and nothing to talk about.

Later, people realized that the empty space had fields, e.g. electromagnetic fields, in it. That's why electrically charged objects and magnets could attract or repel each other even when the vacuum was in between them. They would quickly learn that the "actual" vacuum had the extra condition that the vectors like \(\vec E = \vec B = 0\). The fields are set to zero in the vacuum. Otherwise the space isn't empty.

(Later, Einstein added the metric tensor \(g_{\mu\nu}\) among the fields in the vacuum, and the value of this tensor has to be nonzero. The vacuum admitted gravitational waves, it could have been curved, even when there was no "matter" in the given region. One must be careful whether the term "the vacuum" allows the nonzero curvature or not. I don't want to discuss general relativity here.)

But before they talked about the vacuum in this simple way, a big portion of the top physicists went through the 19th fad of the luminiferous aether. The observation they started with was that the sound was oscillating air (or another carrier) and because light is composed of waves as well, these waves must be an oscillating medium (not necessarily air), too. The medium had to penetrate solids and the vacuum, too. It was named after the least tangible and most spiritual classic element of ancient philosophy (or ancient pseudoscience), the aether. The adjective "luminiferous" meant that it "carried the light".

James Clerk Maxwell was among the big shots who have found this picture intriguing if not unavoidable. This theory – making light and sound really similar – predicted the aether wind. The speed of light should depend on the speed of the observer – on his relative speed with respect to the aether. Just like you feel the wind if you open the window in your fast car (the air is moving relatively to you), you should feel the aether wind (the aether is moving relatively to you). The aether wind was found not to exist by the Morley-Michelson experiment. Einstein could have used this experimental result to discover relativity – except that Einstein was always indicating that he didn't need it and he wasn't actively aware of the Morley–Michelson experiment at all. Unless Einstein was fraudulently hiding the knowledge he has used, in order to look more ingenious than he was, he was an example of the power of pure thought.

It's very hard to talk about the details of the aether. FitzGerald and others have even constructed a working prototype out of wheels and gears (does anyone have a picture of that device?) that behaved according to Maxwell's equations at long distances. But how could it penetrate the solids? Clearly, our modern quantum mechanical theories of solids don't allow any material to get through. The theory couldn't have worked well and it's sort of meaningless to ask any details about how the theory worked – because it didn't. ;-)

Albert Einstein has therefore made the vacuum truly empty once again. There was no "material" that could pick a preferred reference frame in the vacuum. The electromagnetic fields live "directly" in the vacuum and the equations they obey – Maxwell's equations – are exactly invariant under the Lorentz transformations. Equally importantly, the "state" of the vacuum is unique and perfectly well-defined, including by the conditions \(\vec E = \vec B = 0\). The vacuum carries no information.

But the scientific view on the question "is the vacuum really empty" changed once again when quantum mechanics, especially quantum field theory, came to the scene. The vacuum is the arena of some frantic activity. The animated GIF above visualizes (and one opens a can of worms if he "visualizes" quantum mechanics, as I will emphasize again) the energy density in the vacuum according to QCD (quantum chromodynamics), the theory of fields that transmit the color force between the quarks.

QCD is frequently picked to visualize the vacuum because of two related reasons: the equations describing the QCD fields are nonlinear, and therefore showing interestingly behaving blobs; the behavior of QCD is often calculated on big computers which is why such animations are naturally available.

The dimensions at the picture above are of order one femtometer and the time scale in which the blobs appear and disappear is comparable to the time that light needs to fly over one femtometer, perhaps \(10^{-23}\) seconds. You could have seen the animation in popular talks of Wilczek, your humble correspondent, or in the YouTube video mentioned above.

So the vacuum looks like a boiling soup, you could say, and the boiling soup may surely absorb the momentum from a Q-thruster (quantum vacuum plasma thruster) which allows it to accelerate without a reaction. Or does it?

Again, of course, this is forbidden. The video above shows some local observable that I will call \(\rho(x,y,z,t)\), an energy density. To properly answer questions about any quantum mechanical system, including the QCD vacuum, you need to use quantum mechanics and find the appropriate operators that represent the observable quantities. Indeed, \(\hat\rho(x,y,z,t)\) is one of those operators.

Now, the QCD vacuum \(\ket 0\) is not an eigenstate of \(\hat\rho(x,y,z,t)\) for some choice of the point i.e. of \((x,y,z,t)\). What does it mean when it is not an eigenstate? It means that you cannot assume the density \(\rho(x,y,z,t)\) to be a uniquely well-defined \(c\)-number. Instead, you need to decompose the vacuum state into a superposition of eigenstates of \(\hat\rho(x,y,z,t)\). The amplitudes may be squared and interpreted as a probability distribution – this is how it always works in quantum mechanics. Under some rules that you would have to learn in detail, the visualization above shows some "typical" values in the distribution that you could get if you were measuring \(\rho(x,y,z,t)\) at reasonably spaced points and at a reasonable frequency of measurements.

You may ask whether the vacuum "carries" the complicated information about the particular animation shown by the animated GIF (or a somewhat different but qualitatively similar process). If it were the case, the vacuum would suddenly carry the information – which the classical vacuum of Newton and Einstein didn't. But does it? The answer is that such information – the precise shape of the blobs – only "exists" if you measure it! If you don't measure it, there is no privileged size or shape or dance of the blobs. This is an important point and it holds for all systems described by quantum mechanics.

Because \(\ket 0\) isn't an eigenstate of \(\hat\rho\), the behavior of \(\hat\rho\) (if you decide to talk about it or measure it!) is complicated. Well, it is perhaps more correct to say that the quantity is not predictable with certainty. But that doesn't mean that nothing about the vacuum is predictable with certainty!

In particular, the vacuum is – by definition – an eigenstate of the Hamiltonian (the total energy of everything)\[

\hat H \ket 0 = \lambda \ket 0

\] with the lowest allowed eigenvalue \(\lambda\) which we have to identify as \(\lambda=0\) if the theory preserves the Lorentz symmetry (zero is the most natural choice for other reasons and in other contexts, too; we don't do cosmology here so we neglect the tiny cosmological constant). So if you measure the total energy of the vacuum, the result is 100% guaranteed to be zero! The decomposition of \(\ket 0\) into the eigenstates of \(\hat H\) is trivial; only one term contributes. The resulting probabilistic distribution is therefore peaked at zero, the guaranteed value of the total energy.

Also, we have \([\hat H,\hat{\vec p}]=0\). The energy commutes with the three components of the momentum. This allows us to simultaneously diagonalize them. Relativistic causality implies that \(E\geq \abs{\vec p}\) and because \(E=0\) for the vacuum \(\ket 0\), it follows that \(|\vec p| = 0\), too. The vacuum is already a state in the "simultaneous diagonalization" of the energy and momentum and it's therefore the eigenstate of the momentum, too:\[

\hat{\vec p} \ket 0 = 0

\] The vacuum carries no total momentum and you may be sure about it – much like you may be sure about the energy. So all the "absorption of momentum by the vacuum" is simply impossible. The momentum of the vacuum is a fixed and given constant (namely zero). Its being fixed means that it cannot change. The fact that it cannot change means that the vacuum cannot "absorb" it from objects. All the reactionless and "pushing against the vacuum" devices are impossible.

Vacuum isn't plasma

One intermezzo. The Q-thruster, a brother of the EM drive, is also said to be related to the "plasma" in the vacuum because "the vacuum is a plasma of a sort". This claim reveals yet another, mostly independent stupidity of the proponents of this pseudoscience because the vacuum isn't a plasma in any physical sense at all.

In physics, plasma is just an ionized gas. Like a gas, it contains a nearly freely flying particles, but unlike gas, these particles carry the electric charge (of both signs). The plasma behaves in various ways etc. However, the vacuum doesn't carry any physical particles. Whenever you measure the number of physical particles in any state, by the operator \(\hat N_i = \hat a^\dagger_i \hat a_i\), you get zero with 100% certainty once again.

Also, a plasma, due to its high energies of the particles etc., admits a pretty good classical description (the uncertainty principle ceases to be constraining if the momenta are high and the distances between the particles are long). You may explain the applicability of classical physics by the high temperature of the plasma. If something is characteristic about the "phenomena in the quantum vacuum", it is that this vacuum is "as quantum i.e. non-classical as you can get". After all, it's the state that you get at the coolest possible temperature, the absolute zero.

So this presentation of the vacuum as a "plasma" is completely wrong from every possible perspective. The plasma is rather close to an ideal gas. The vacuum is nothing of the sort. The plasma is hot, the vacuum is the coldest thing. And so on. Amusingly enough, I think that they say that the vacuum is a "plasma" because they actually confuse "plasma" as understood in physics with "plasma" as observed by biologists, e.g. blood plasma, cytoplasm, and similar "plasmas". The vacuum looks like a boiling soup and the boiling soup is like a blood plasma. But blood plasma isn't a physicist's plasma!

It's also completely wrong and stupid – for different reasons – to say that the QCD vacuum, to be specific, is analogous to the blood plasma. But I can't correct all wrong statements that these people have said because their number is basically infinite, as expected from their stupidity which is unbounded from above. At some point, I would be "correcting" every possible sequence of words one may construct because that's almost exactly what they are doing.

The total energy cannot be viewed as a "composite function" in the classical sense

I want to return to the QCD vacuum and its having "certainly vanishing total energy and momentum" yet "uncertain and fluctuating energy density (and other local fields)". If you think in terms of classical physics, the combination of these two claims could look like a contradiction to you.

If you have something like a soup and its individual regions are "uncertain", then their union must probably be "uncertain" as well. Imagine that you consider the total energy of a Klein-Gordon field\[

H = \int d^3 x \left[ \frac{m^2\phi^2}{2} + \frac{(\nabla \phi)^2}{2} \right]

\] In classical physics, you are encouraged to think that \(\phi(x,y,z,t)\) (and its derivatives) are the "independent variables" and \(H\) is a result of a calculation, a function of these independent variables that is no longer independent. So if the values of \(\phi(x,y,z,t)\) are "uncertain", then the value of the function \(H\) must be uncertain as well.

Well, this isn't really true even in classical field theory. The uncertainty about \(\phi\) may be written in terms of probabilistic distributions and the probabilistic distribution may be made to vanish whenever \(H\neq 0\) – so that despite the uncertainty in \(\phi(x,y,z,t)\), we may be certain that \(H=0\). This "making \(H\) vanish" corresponds to incorporating some (strong yet delicate) correlations between the values of individual independent variables. In classical physics, it could be unnatural to arrange the distribution to obey this condition – incidentally, it would be more similar to the "microcanonical" distribution (the total energy is given) while most people consider the "canonical" distribution more natural.

But in quantum mechanics, it's extremely natural to consider this would-be "microcanonical" distribution for \(\phi(x,y,z,t)\) for which \(H=0\) is certain. Why? Because physically, the function \(H\) is actually more important than the value of the "would-be independent" observables \(\phi(x,y,z,t)\). It's the total energy that Nature wants to minimize. So when the energy dissipates and disappears from a region and it cools down, the region unavoidably approaches the eigenstate of \(H\) with the lowest possible eigenvalue, namely zero. The values of the local fields may be unpredictable, uncertain, and oscillating, but the value of \(H\) becomes simple.

I said that this certainty that the total \(H=0\) is somewhat similar to the "microcanonical" ensemble in (even classical) statistical physics. If you add the energy to one region, you must subtract the same amount from others so that the total energy is still zero. But I must also emphasize that quantum field theory makes these compensations much more accurately. The fluctuations of \(\hat\rho(x,y,z,t)\) at two different points are not only "inversely correlated" but they are "perfectly entangled" so that the state \(\ket 0\) remains the precise energy eigenstate. It's not just the expectation value of the energy that is zero. The energy is zero with certainty. The probabilistic distribution is a delta-function located at zero.

Entanglement is omnipresent in quantum mechanics. For example, the simplest textbook example of entanglement is Bell's state, the singlet state of two spins\[

\ket{0;0} = \frac{ \ket{\uparrow}\ket{\downarrow}-\ket{\downarrow}\ket{\uparrow} }{\sqrt{2}}

\] If you measure any component of the total angular momentum \(\vec J\), you get zero. People who still think classically find this strange and surprising and unnatural and they emit lots of these adjectives that create a strange atmosphere. But Nature finds it perfectly unsurprising, generic, standard, consistent, common sense, and essential for the Universe to function. Such entangled states are everywhere.

The vacuum state of the empty space is analogous. It entangles all the operators such as the field operators at different points in such a way that the total \(H\) and the total \(\vec p\) are equal to zero.

People thinking classically and looking at quantum mechanics usually want to imagine that the wave function is at the root of the Universe. They want to believe that the wave function is something like a classical field. Well, it isn't. At some point, when they grow up (and most of them never do), they may accept that the observables – the Hermitian operators – are the players that really make the physics.

But even when they do, they tend to assume that the "elementary" operators that we would use as the coordinates of the phase spaces in classical physics – such as \(\hat x,\hat p\) in mechanics or elementary fields \(\hat \phi(x,y,z,t)\) in quantum field theory – are the natural primary objects, and all the other operators are just some "derived" or "secondary" functions or functionals of the "primary" observables. They think that all these functions of \(\hat x\) and \(\hat p\) are kind of "redundant" – everything about them may be reduced to \(\hat x\) and \(\hat p\).

However, this is still a completely wrong way of thinking about observables. In quantum mechanics, no observables are more "primary" than others. Every observable is in principle equally fundamental. Every Hermitian operator on the Hilbert space corresponds to an observable. Every such observable may be measured by a particular procedure. And the procedure cannot be reduced to the measurement of "elementary" observables such as \(\hat x\) and \(\hat p\) – or \(\hat \phi(x,y,z,t)\) and \(\partial_0\hat\phi(x,y,z,t)\) – because those can't be measured at the same moment! That's exactly what the uncertainty principle says. So generic observables can never be reduced to the measurement of a "set of elementary observables such as \(\hat x\) and \(\hat p\)". The complementarity is unavoidable and implies that you basically need a completely new apparatus to measure every new observable.

In many cases, physical questions depend on the values of observables – such as \[

\hat{\vec J}=\hat{\vec J}_1 + \hat{\vec J}_2

\] in the case of Bell's state, or \(\int\hat \rho(x,y,z,t)\) in the case of a quantum field – which used to be considered "composite" or "secondary" or "not elementary" in classical physics. In classical physics, their non-elementary status led you to expect that their values had to be uncertain or oscillating if the values of the "elementary" observables were uncertain or oscillating.

In most of the important situations of quantum mechanics, it's simply not the case. The observed ket vectors may often be eigenstates of these previously "composite" observables which means that you may be sure about their values – even though the values of the formerly "elementary" quantities are deeply uncertain or oscillating! This possibility not only trivially follows from quantum mechanics. It is really a typical situation that is important almost everywhere in quantum physics.

In particular, I mentioned the energy and the total angular momentum as examples of these formerly "composite" operators whose values are often perfectly known in quantum mechanics. It is no coincidence that it is the conserved quantities whose values are often exactly well-known.

When we discuss the momentum conservation (or any other conservation law), this insight guarantees that the momentum can't be "lost" to the vacuum despite the fact that you may find lots of observables whose values are very uncertain and oscillating (or "would be", if you decided to measure them) because the vacuum isn't their eigenstate.

You should better start to think quantum mechanically. Observables, and not wave functions, are the root players in physics. And all observables – all Hermitian operators – must be considered equally elementary or equally primary or equally "independent variables". If you have three observables such as \(\hat z=\hat z(\hat x,\hat y)\) that may be reverted to \(\hat x=\hat x(\hat y,\hat z)\) or \(\hat y=\hat y(\hat z,\hat x)\), these three prescriptions are always equally good and it is possible that the state vector you deal with is an eigenstate of \(\hat x\) but not \(\hat y,\hat z\). But it is also possible that it is an eigenstate of \(\hat y\) and not \(\hat x,\hat z\), and so on. Regardless of your previous relationships to these three observables, none of the options is more natural than others. All states (and their superpositions) in the Hilbert space are equally allowed – which also says that the eigenvectors of all conceivable Hermitian operators are as good as eigenvectors of all others.

With some experience in quantum physics, you will find all the statements above completely self-evident, and you will also be able to see that some people are trying to "pump" classical physics into their reasoning all the time whenever it seems that they have a problem with the facts above.

Quantum mechanics works perfectly, it works for all systems in the Universe exactly, and it leads to many phenomena that weren't possible in any classical theory. However, it also 100% confirms certain laws that have already existed in classical physics. And the momentum and energy conservation laws belong to the latter category. For quantum mechanics to give us answers, we must first know what are the questions – and it is the subject, the observer, who must know what questions he actually wants to be answered (and what he will perceive).

Questions about values of local fields may produce uncertain and fluctuating answers in quantum field theory (these predictions only become physical if you actually measure these observables: you may never assume that particular preferred values "objectively exist" without any measurement). But if you are an observer who cares about the total energy or the total momentum (or the total angular momentum etc.) of the vacuum, the situation is very different. Quantum field theory unambiguously says (and there is no contradiction with the previous sentence) that the vacuum is empty and boring, after all.

And that's the memo.

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