Is the Vacuum Really Empty? Exploring Vacuum Fluctuations and Quantum Mechanics

[Haorong Wu]

TL;DR Summary

Is the statement that the vacuum is not empty wrong?

Hello, there. From another post, I read the insights Learn the Physics of Virtual Particles in Quantum Mechanics and Explore The Vacuum Fluctuation Myth in Quantum Theory.

Also, I sometimes see a statement that the vacuum is not empty from some media. Most often, it says the vacuum can create or annihilate particles due to vacuum fluctuation. It seems that the statement is elusive. The vacuum, which is often referred to as some space avoiding any matter, could not actually create or annihilate any real particles. Hence, could I say the vacuum is actually empty?

Also, could I say that the vacuum fluctuation is related to the vacuum state, not the empty space or the free space, and it states that the electric field operator has a nonzero variation in the vacuum state, and the free space itself plays no important role here?

 

[vanhees71]

Indeed, by definition the vacuum, i.e., the ground state of the Standard Model, is indeed really empty, and it's empty by construction.

What's however an observational fact are quantum fluctuations, but these are fluctuations of observables of "something" and not the truly empty vacuum!

 

[Haorong Wu]

Indeed, by definition the vacuum, i.e., the ground state of the Standard Model, is indeed really empty, and it's empty by construction.

What's however an observational fact are quantum fluctuations, but these are fluctuations of observables of "something" and not the truly empty vacuum!

So when we say fluctuation, should we explicitly say what is fluctuating? And the particle number operators do not fluctuate? I mean $\left < 0 \right | n \left | 0 \right>=\left < 0 \right | n^2 \left | 0 \right>=0$.

 

[vanhees71]

Indeed, you have to say what's fluctuating and how to measure it. Whenever you measure something, of course, you need to use some measurement device. Then of course it's no vacuum anymore, because the measurement device consists of the particles. If the particle number is an observable than it may well fluctuate too, depending on the situation.

 

[Haorong Wu]

Indeed, you have to say what's fluctuating and how to measure it. Whenever you measure something, of course, you need to use some measurement device. Then of course it's no vacuum anymore, because the measurement device consists of the particles. If the particle number is an observable than it may well fluctuate too, depending on the situation.

Regarding the particle number, $n=a^\dagger a$, since $\left < 0\right | a^\dagger a\left | 0 \right >=\left < 0\right |( a^\dagger a)^2\left | 0 \right >=0$, I think it could not fluctuate in vacuum state.

 

[Haorong Wu]

I must say it somehow makes me feel sad. The fantasy from my childhood is broken today.

 

[A. Neumaier]

Regarding the particle number, $n=a^\dagger a$, since $\left < 0\right | a^\dagger a\left | 0 \right >=\left < 0\right |( a^\dagger a)^2\left | 0 \right >=0$, I think it could not fluctuate in vacuum state.

Yes, that's why particles cannot be created in a true vacuum.

I must say it somehow makes me feel sad. The fantasy from my childhood is broken today.

That's the price of growing up.

 

[Haorong Wu]

Hi, @A. Neumaier. Your insights are informative. I still have confusion about vacuum and vacuum state. The vacuum state is the state with the lowest energy and contains no physical particles, but according to vacuum fluctuation, there exists a varying electric field, so the space is not actually empty. Therefore, the vacuum state has nothing to do with the actual vacuum, is that right?

Also, I do not understand why Hawking radiation is just a mathematical tool. Is that because we observe black holes are emitting thermal radiation and in building a suitable model, Hawking radiation is proposed but this is by no means what really happens near black holes? In other words, is it true that we know the radiation phenomena but we do not know what really happens there?

Thanks!

 

[DennisN]

Is that because we observe black holes are emitting thermal radiation and in building a suitable model, Hawking radiation is proposed but this is by no means what really happens near black holes?

As far as I know, Hawking radiation has not yet been observed. Hawking radiation is a prediction of theoretical physics.

 

[weirdoguy]

Also, I do not understand why Hawking radiation is just a mathematical tool.

Virtual particles are, not Hawking radiation.

 

[Haorong Wu]

Thanks, @DennisN, and @weirdoguy. I am still studying the black hole part, and have not reached the QFT in curved spacetime part. I have read another page, here. So the Hawking radiation is actually created at a distance from the event horizon, and it is depended on how different observers define particles (whatever that means. I do not follow that part.)

So how should I introduce Hawking radiation to other non-experts in a physically-meaningful language?

 

[phinds]

So how should I introduce Hawking radiation to other non-experts in a physically-meaningful language?

The heuristic description of Hawking Radiation as "virtual particles" was invented by Hawking himself because, he said, there just didn't seem to be any other way to explain in English that which really can only be explained in math.

SO ... use the hueristic and explain that it's only to give sort of an idea of what's going on and if they want more they have to learn the math.

 

[Haorong Wu]

The heuristic description of Hawking Radiation as "virtual particles" was invented by Hawking himself because, he said, there just didn't seem to be any other way to explain in English that which really can only be explained in math.

SO ... use the hueristic and explain that it's only to give sort of an idea of what's going on and if they want more they have to learn the math.

OK. I will introduce Hawking effect to others and give them the caveat that it is just a fantastic story, and do not take it seriously.

 

[A. Neumaier]

according to vacuum fluctuation, there exists a varying electric field

No. Vacuum fluctuations do not say that anything fluctuates in time. They are just a vivid (and misleading) way of talking about the n-point functions figuring in perturbative quantum field theory. There is nothing real about them; they are just a computational crutch.

Therefore, the vacuum state has nothing to do with the actual vacuum, is that right?

No. The vacuum state is the unique state without particles.

Hawking radiation is actually created at a distance from the event horizon, and it is depended on how different observers define particles (whatever that means. I do not follow that part.)

It is the attempt to put mathematical formulas into language, giving the appearance of understandability.

The choice of a curved coordinate system for spacetime determines what the concept of particles means on the formal level. But as regards observations, we are bound to observe in the coordinate system distinguished by the solar system.

 

[Haorong Wu]

No. Vacuum fluctuations do not say that anything fluctuates in time. They are just a vivid (and misleading) way of talking about the n-point functions figuring in perturbative quantum field theory. There is nothing real about them; they are just a computational crutch.

Sorry. I do not understand this. In Quantum Optics by Marlan O.Scully, it reads that for a single-mode linearly polarized field operator $$E(\mathbf r,t)=\mathcal E a e^{-ivt+i\mathbf k \cdot \mathbf r}+h.c.,$$ its expectation value vanishes, i.e., $$\left < n \right | E \left | n \right >=0.$$ However, the expectation for $E^2$ is given by $$\left < n \right | E^2 \left | n \right > = 2 |\mathcal E|^2 (n+\frac 1 2).$$
Then from probability theory, the variation of $E$ is not equal to zero. Doesn't that mean the electric field varies in spacetime?

No. The vacuum state is the unique state without particles.

So, the vacuum state does not even contain any electric field? Otherwise, electric fields may be regarded as matter. I think I will understand this sentence better after I understand the previous one.

Thanks, @A. Neumaier .

 

[A. Neumaier]

Then from probability theory, the variation of $E$ is not equal to zero. Doesn't that mean the electric field varies in spacetime?

No. This variation is not a variation in time. Time doesn't figure at all in the formulas.

So, the vacuum state does not even contain any electric field? Otherwise, electric fields may be regarded as matter. I think I will understand this sentence better after I understand the previous one.

It is the state where the measurable electromagnetic field strength is identically zero (if we could measure it). Nothing to oscillate, no photons, no matter.

 

[Haorong Wu]

It is the state where the measurable electromagnetic field strength is identically zero (if we could measure it).

But if its strength is identically zero, why its second-moment $\left < E^2\right>$ is not zero?

Is it because of the uncertainty principle? After all, $E^2$ could be related to the energy, and there is some uncertainty between the measure outcome of energy and the measure time gap.

But the expression for $\left < E^2\right>$ has no $\Delta t$ part, so maybe not.

 

[A. Neumaier]

But if its strength is identically zero, why its second-moment $\left < E^2\right>$ is not zero?

Is it because of the uncertainty principle? After all, $E^2$ could be related to the energy, and there is some uncertainty between the measure outcome of energy and the measure time gap.

But the expression for $\left < E^2\right>$ has no $\Delta t$ part, so maybe not.

The formal expectation is not a statistical expectation. One cannot make many copies of the vacuum and there observe $E$ to take an average over many $E^2$. One cannot even make a single copy. Thus trying to give it a statistical interpretation is meaningless.

 

[Haorong Wu]

The formal expectation is not a statistical expectation. One cannot make many copies of the vacuum and there observe E to take an average over many E^2. One cannot even make a single copy. Thus trying to give it a statistical interpretation is meaningless.

I am not familiar with the concept of formal expectation. Could you suggest some paper I could read regarding it? Thanks!

 

[A. Neumaier]

I am not familiar with the concept of formal expectation. Could you suggest some paper I could read regarding it? Thanks!

Formal expectation is just the mathematical expression $\langle A\rangle =$tr$\rho A$, without a probabilistic interpretation, hence formal. It satisfies all the mathematical properties used to manipulate it, but is otherwise abstract. Just as a wave function may be formally regarded as a vector, an element of a vector space, though 'vector' originally meant a 3-dimensional vector pointing somewhere.

Thus there is nothing to read about. Isn't it clear from the nature of the vacuum that an interpretation of your expression as an expectation in the statistical sense is meaningless?

 

[Haorong Wu]

Formal expectation is just the mathematical expression $\langle A\rangle =$tr$\rho A$, without a probabilistic interpretation, hence formal. It satisfies all the mathematical properties used to manipulate it, but is otherwise abstract. Just as a formal vector is an element of a vector space, e.g. a wave function, though 'vector' originally meant a 3-dimensional vector pointing somewhere.

Thus there is nothing to read about. Isn't it clear from the nature of the vacuum that an interpretation of your expression as an expectation in the statistical sense is meaningless?

OK. Leave math aside. Physically, $\left < E \right >=0$ while $\left < E^2 \right >\ne 0$ for vacuum state is just stating that there is no electric field in vacuum, while there still exists some (vacuum) energy. No fields, no variations, no fluctuations, only a little energy. That is all. Is that right? (It now seems right to me.)

 

[A. Neumaier]

OK. Leave math aside. Physically, $\left < E \right >=0$ while $\left < E^2 \right >\ne 0$ for vacuum state is just stating that there is no electric field in vacuum, while there still exists some (vacuum) energy. No fields, no variations, no fluctuations, only a little energy. That is all. Is that right? (It now seems right to me.)

No. $\langle E(x)\rangle$ is the electrical field at $x$ and can be (approximately) measured, though in general a smeared result is obtained in an actual measurement. In vacuum it is exactly zero. The field strength is also measurable and vanishes in a vacuum. In optics, some of the formal correlations (quadratic expressions at different arguments) are measurable, e.g., as polarization, but the vacuum has helicity 0, hence no polarization.

Your other expression has no measurable meaning, hence no physical meaning. It is just a mathematical expression that can be formed. The name fluctuations refers to it being nonzero, but this word is just a label given to it (because of the formal analogy to the statistical variance) to be able to talk about it.

 

[vanhees71]

But the energy-momentum tensor of the em. field has to be normal ordered to make sense. Thus $\langle u_{\text{em}} \rangle=\frac{1}{2} \langle :\vec{E}^2+\vec{B}^2: \rangle = 0$.

The vacuum is really (!) empty. It's the only state where in fact nothing (!) fluctuates.

Quantum fluctuations however indeed are observable and give rise to, e.g., the anomalous magnetic moment of the electron (deviation of the gyrofactor from 2 due to radiative corrections) or the Lamb shift of hydrogen lines (also due to radiative corrections).

 

[A. Neumaier]

Quantum fluctuations however indeed are observable and give rise to, e.g., the anomalous magnetic moment of the electron (deviation of the gyrofactor from 2 due to radiative corrections) or the Lamb shift of hydrogen lines (also due to radiative corrections).

These are not observed quantum fluctuations but observed properties of the effective QED interaction computed in low order perturbation theory with the help of vacuum expectation values. The fluctuations refer to the latter and are an artifact of perturbation theory.

When observing the half-life of a radioactive isotope nobody says that we observed vacuum fluctuations, although the latter (i.e., certain vacuum expectation values) would be involved in a perturbative computation started with QFT. That many give a different ontological status to the mathematical tools used in QFT than to those used in QM is just because of the muddled history of the subject...

 

[vanhees71]

I'd say the radioactive decay is also a quantum (not vacuum!) fluctuation. I'd also say that the word "vacuum fluctuation" is a contradictio in adjecto.

 

[A. Neumaier]

I'd say the radioactive decay is also a quantum (not vacuum!) fluctuation. I'd also say that the word "vacuum fluctuation" is a contradictio in adjecto.

What is a quantum fluctuation in precise terms?

 

[vanhees71]

If a state is not an eigenstate of the Hamiltonian its energy has an indetermined value and thus the energy "fluctuates". Take the example of an unstable nucleus, where the unstability is due to $\beta$ decay. Usually you consider the nucleus initially in an energy eigenstate (e.g., the ground state) concerning the strong interaction and take the weak interaction as the "perturbation". So initially the nucleus is not an eigenstate of the exact Hamiltonian, which includes the strong interaction, and that's why $\beta$ decay occurs with some time-dependent transition probability. In this sense the $\beta$ decay is described as a "quantum fluctuation".

 

[A. Neumaier]

its energy has an indetermined value and thus the energy "fluctuates".

Ah, so you take fluctuation just to mean not being determined.

The standard meaning involves actual motion:
https://www.dictionary.com/browse/fluctuation
Therefore I avoid the word fluctuation completely unless time is involved.

 

[vanhees71]

Fluctuations usualy means statistical fluctuations of any kind. In quantum theory if the system under consideration is not in a state, where an observable doesn't take a determined value, then the value of this observable fluctuates.

In a narrower sense with "quantum fluctations" you refer to fluctuations of the energy, i.e., whenever the system is not in an eigenstate of the Hamiltonian, quantum fluctuations occur, i.e., fluctuations of the energy occur.

That is, why I think that "vacuum fluctuations" strictly speaking make no sense, because if the system (particularly a system described by quantum fields) is prepared in the vacuum state (the ground state of the Hamiltonian of the QFT), there cannot be energy fluctuations, because the energy takes a determined value and as an energy eigenstate of a system that doesn't depend explicitly on time, this is a stationary state and thus the energy does not fluctuate.

 

[A. Neumaier]

In a narrower sense with "quantum fluctations" you refer to fluctuations of the energy, i.e., whenever the system is not in an eigenstate of the Hamiltonian, quantum fluctuations occur, i.e., fluctuations of the energy occur.

But then all quantum mechanical predictions apart from spectra are due to quantum fluctuations. Not a useful concept, in my view.

 

[vanhees71]

I usually don't use the phrase "quantum fluctuations", but it's regularly used in the literature. So one has to know what's usually meant by the authors using it.

 

[gary350]

I would assume outer space is not empty.