The Vacuum Fluctuation Myth - Comments

[RockyMarciano]

No. Fluctuation in today's usage always means change, not just being uncertain!
This is only your private interpretation of the term. Never before I heard of someone talk about quantum vaccilations! And even that word means not just uncertainty but wafering uncertainty - a process in time!

The term uncertainty or indeterminacy is obviously related to change, change in measuring expectations, have you heard about statistical fluctuations referred to the uncertainty in measurements? I mean that's QM.
Frankly, it looks as though you stubbornly need to hold on to your straw man and the mantra "nothing fluctuates" (indeed a quite private interpretation), when everybody knows since Heisenberg's first modern quantum mechanics paper, the concept of conjugate observables fluctuating as described by Fourier transform coefficients and outlined by Born in its probabilistic rule to capture just that fluctuation in the measurement of noncommuting observables, so hardly my own private interpretation.

 

[A. Neumaier]

change in measuring expectations

This is nonsense. In a stationary setting (such as the ground state of a quantum system), expectations are constant, not fluctuating.

Measuring expectations means making a lot of individual measurements of different realizations of the same system and taking their mean. Each measurement deviates from the mean, and the minimal mean square deviation is quantified by the uncertainty relation.

The fluctuation is neither in the quantum system nor in the expectation but in the series of measurements. It is due solely to the measurement process. It comes from the fact that each time a different particle is measured.

But when you measure a field there is only one field so nothing that could fluctuate. Unless the field itself fluctuates - i.e., changes its values rapidly in time like in turbulence. But a free field in the vacuum state is far from turbulent.

 

[RockyMarciano]

This is nonsense. In a stationary setting (such as the ground state of a quantum system), expectations are constant, not fluctuating.

I wasn't using the word expectation technically, as referred to expectation values there. I was referring to what you explain below(so I'm happy that you got my meaning right after all) and this applies irrespective of the states being stationary or not.
By the way, even if using your biased concept of the word fluctuation referring only to oscillation, are you saying that stationary waves are not oscillating? I guess they are not waves then either.

Measuring expectations means making a lot of individual measurements of different realizations of the same system and taking their mean. Each measurement deviates from the mean, and the minimal mean square deviation is quantified by the uncertainty relation.

I see you have finally understood what quantum fluctuations mean, This minimal mean square deviation refers to something changing(in this case the conjugate noncommuting variables), or there would be no minimal square deviation. So see, it was not so difficult to see how it is this uncertainty that using the statistical language of QM is referred by some people as quantum fluctuation

The fluctuation is neither in the quantum system nor in the expectation but in the series of measurements. It is due solely to the measurement process. It comes from the fact that each time a different particle is measured.

Sure, that is where the fluctuation called quantum fluctuation in QM is(I'm leaving out the quantum field extension of the term for reasons I explained in a previous post) and yes, that is the process it is due to.

 

[mfb]

The Heisenberg uncertainty relation is not about quantum fluctuations but about the intrinsic uncertainty in measuring noncommuting observables.

I would argue that it is not about the measurement. The state itself has this uncertainty. If it would be an issue of the measurement, you could try to find measurements that avoid this uncertainty (like measuring entangled particles and so on - I'm sure you know all those ideas). But we know that no measurement, no matter how clever, can avoid it.

 

[stevendaryl]

I am not sure that there is any actual content to the question: "Does anything really fluctuate in the ground state?" To answer it requires going beyond the wave function to an interpretation.

 

[A. Neumaier]

This minimal mean square deviation refers to something changing(in this case the conjugate noncommuting variables), or there would be no minimal square deviation.

No. The noncommuting observables are fixed, hence do not change ate all; in the textbook case of Heisenberg's uncertainty relation, they are always $p$ and $q$. . Only the measurement results change, under repeated measurement of similarly prepared systems. As long as nothing is measured, nothing fluctuates!

are you saying that stationary waves are not oscillating?

They are oscillating, of course, but not fluctuating. In physics, the latter means unpredictable changes. The oscillations of a stationary wave are very predictable.

 

[A. Neumaier]

I would argue that it is not about the measurement. The state itself has this uncertainty.

In principle the (pure or mixed) state can be prepared and measured to arbitrary accuracy (e.g., by quantum tomography), hence has no uncertainty in itself. It just encodes the uncertainty revealed by potential measurements.

IIf it would be an issue of the measurement, you could try to find measurements that avoid this uncertainty (like measuring entangled particles and so on - I'm sure you know all those ideas). But we know that no measurement, no matter how clever, can avoid it.

How do we know it? Only because the Heisenberg relations says so. This means that no matter how hard we try, we cannot avoid uncertainty in individual measurements.

We can say the position of a particle is uncertain but this doesn't make it fluctuating. If we measure once we don't see any fluctuation, just a definite value deviating from the mean. measuring twice is usually impossible (except for nondemolition measurements of conserved variables - which then don't fluctuate by definition).

For stationary beams (i.e., identically prepared particles), fluctuations are always fluctuations of measurement results on different realizations of the system. But if no measurements are taken there are no fluctuations. In some very respectable interpretations of quantum mechanics it is even meaningless to assign definite values to a system when it is not measured! How can a nonexistent value fluctuate?

 

[RockyMarciano]

No. The noncommuting observables are fixed, hence do not change ate all; in the textbook case of Heisenberg's uncertainty relation, they are always $p$ and $q$. . Only the measurement results change, under repeated measurement of similarly prepared systems.

For most physicists I know the Heisenberg's uncertainty is intrinsic and independent of the measurement process

As long as nothing is measured, nothing fluctuates!

If nothing is measured you don't have a physical theory.

They are oscillating, of course, but not fluctuating. In physics, the latter means unpredictable changes. The oscillations of a stationary wave are very predictable.

You have managed to shift the meaning you attributed to the word fluctuation to make it indistinguishable with mine here(unpredictability in the sense of uncertainty in position and momentum, if you apply the word to the quantum uncertainty relations you'll know what serious people mean by "quantum fluctuations" in nrqm). But I'm sure you think you are right and I'm wrong so feel free to go on with your "nothing flutuates" mantra.

 

[A. Neumaier]

I am not sure that there is any actual content to the question: "Does anything really fluctuate in the ground state?" To answer it requires going beyond the wave function to an interpretation.

Possibly yes. If this is true then asserting that quantum fluctuations exist has no content either. Thus quantum fluctuations do not exist in any meaningful sense. Except perhaps in interpretations such as Bohmian mechanics that postulate the existence of variables that are in principle unobservable - since everything observable must be formulated in orthodox quantum mechanics in order not to make deviating predictions.

 

[A. Neumaier]

For most physicists I know the Heisenberg's uncertainty is intrinsic and independent of the measurement process

But then the uncertainty relation is a purely theoretic statement about mathematical expectation values - not one about fluctuations where something changes in an unpredictable way. The only way to relate it to change is by relating it to changing measurement results.

 

[A. Neumaier]

If nothing is measured you don't have a physical theory.

We have a good physical theory about what happened inside stars and on the Earth eons ago.

Long before there were physicists to measure anything and long before people who dared to assert that if nothing is measured you don't have a physical theory!

 

[RockyMarciano]

We have a good physical theory about what happened inside stars and on the Earth eons ago.

Long before there were physicists to measure anything and long before people who dared to assert that if nothing is measured you don't have a physical theory!

You are confusing measurements with observers. My statement just indicates that without measurement(as a concept which is nothing but the possibility of interaction) you don't have a physical theory, only a mathematical construct instead with no connection with nature, that seems to be what you have in mind. Without interactions there are no physics.
By the way, only measurements, either direct or indirect can tell us if our theory of the star's core is good or not. One doesn't need to go inside the star, that is a quite naive view of what measurements are.

 

[A. Neumaier]

measurement(as a concept which is nothing but the possibility of interaction)

This is again a very nonstandard way of using physical terms. Nobody but you calls the mere possibility of interaction a measurement. Please do some serious reading about measurement before continuing discussion!

 

[RockyMarciano]

But then the uncertainty relation is a purely theoretic statement about mathematical expectation values - not one about fluctuations where something changes in an unpredictable way. The only way to relate it to change is by relating it to changing measurement results.

Agreed. I said most physicist, not necessarily my opinion.

 

[RockyMarciano]

This is again a very nonstandard way of using physical terms. Nobody but you calls the mere possibility of interaction a measurement. Please do some serious reading about measurement before continuing discussion!

If you really think measurement has nothing to do with interacting is true that is not worth to continue the discussion. I didn't define it as "the possibility of interaction", I said that measurement implied the possibility of interaction, how do you measure anything without interacting with it?

 

[A. Neumaier]

measurement [...] I didn't define it as "the possibility of interaction", I said that measurement implied the possibility of interaction, how do you measure anything without interacting with it?

Well, then learn tocorrectly use the English language. Saying

measurement(as a concept which is nothing but the possibility of interaction)

will be understood by everyone as implying ''measurement is nothing but the possibility of interaction''!

Measurement is needed to check theories. But when they are checked they are applied in many, many unchecked and uncheckable instances, to make inferences. These inferences come out true if the theory is correct. If, in order to trust a theory, we would have to check all inferences from it by new measurements, the theory would be more than worthless.

Thus a good theory works correctly no matter whether or not something is measured, and is applied no matter whether or not something is measured. In particular quantum mechanics. But the only times something fluctuates (outside of turbulent or stochastic processes) is when one actually makes multiple measurements on similarly prepared systems. Thus fluctuation is related to actual measurement and not to the mere possibility of interaction.

[...] is not worth to continue the discussion.

True, and I'll stop the discussion with you here since it is boring to others to have to endlessly repeat myself.

 

[mfb]

In principle the (pure or mixed) state can be prepared and measured to arbitrary accuracy (e.g., by quantum tomography), hence has no uncertainty in itself. It just encodes the uncertainty revealed by potential measurements.

How do you prepare a state that has well-defined position and momentum at the same time? As in: if you choose to measure one of them at random, you can be sure what you will measure. If you agree that this is impossible, how is that not an uncertainty of the state itself?

 

[vanhees71]

Well, now some are eliminating too many fluctuations. I agree with the statement that it's not the vacuum that fluctuates. To the contrary it's the very state that is stable under time evolution. There's nothing and it stays nothing, and this nothing is Poincare invariant. You cannot "perturb the nothing" without introducing something, and that's the key to resolve the quibbles the formal-QFT lovers (and I'm counting myself to them). Just omit "vacuum" and just say fluctuations, and indeed everything is fine, i.e., indeed the electromagnetic field fluctuates as well as the charges in a well-defined sense, i.e., the vacuum expectation value of e.g., the electromagnetic field vanishes but not its square.

The Lambshift is not due to fluctuations of the vacuum but the fluctuations of the charges (in the case of the hydrogen atom of the protons and electrons) and the electromagnetic field, and quantitatively these fluctuations are defined within perturbation theory, which can be very elegantly and precisely expressed in terms of Feynman diagrams which have a certain intuitive appeal in the sense of mechanisms like "exchange of fields" (propagator lines) and "quantum fluctuations" (loops of propagator lines).

However, what's observable are the asymptotic free states. In the case of the Lambshift that's the initially somehow excited hydrogen atom which spontaneously emits a photon (em. wave) whose energy can be measured and doing this accurately enough (as did Lamb and Retherford in their very famous measurement, but not measuring spectral lines in the visible light but using a microwave resonator) you find the Lamb shift. As you see, you never ever can measure the vacuum state itself since of course you have to introduce something to measure it. In this case it's a hydrogen atom and a microwave resonator to excite the corresponding states. See, e.g.,

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/lamb.html

It's far from being vacuum! It's a beam of hydrogen atoms, microwave fields and a lot of other equipment to figure out the quantum corrections to the tree-level (Dirac quantum mechanics approximation), parts which in the theoretical analysis in terms of loop corrections of perturbation theory then are colloquially called "vacuum polarization", "vertex correction", etc. but one shouldn't take this too literally, what's done is to use an analysis in terms of perturbation theory, use a lot of tricky calculational tools to renormalize the first infinite integrals in a proper and physically sensible way etc.

 

[ftr]

so is this statement in wiki also wrong

"In the modern view, energy is always conserved, but because the particle number operator does not commute with a field's Hamiltonian or energy operator, the field's lowest-energy or ground state, often called the vacuum state, is not, as one might expect from that name, a state with no particles, but rather a quantum superposition of particle number eigenstates with 0, 1, 2...etc. particles."

https://en.wikipedia.org/wiki/Quantum_fluctuation

 

[vanhees71]

No, by definition the vacuum state is the state of lowest energy. You have to be a bit careful, however which vacuum you refer to. Of course the free-particle vacuum is not the same as the fully interacting one! It's a highly non-trivial issue.

 

[A. Neumaier]

How do you prepare a state that has well-defined position and momentum at the same time? As in: if you choose to measure one of them at random, you can be sure what you will measure. If you agree that this is impossible, how is that not an uncertainty of the state itself?

There are no such states, so your first question is a meaningless request. The conundrum you pose arises from mixing classical thinking (where in the deterministic case a state means specified values of $p$ and $q$) and quantum thinking, where a state means something quite different.

In general, both in classical and quantum mechanics, a state is a positive linear functional on the observable algebra. In quantum mechanics, the latter is the algebra of linear operators on a Schwartz space (as in the case of $p$ and $q$), and states are therefore in 1-1 correspondence with density operators, positive linear integral operators of trace one. This density operator can be prepared and measured to arbitrary accuracy for sources producing sufficiently small systems such as photons or electrons. One cannot require more about preparing or measuring a state. Thus there is no uncertainty in the state itself.

However, there is an uncertainty in prediction the value of $p$ and $q$ from any exactly known state, given by the uncertainty relation. No matter how accurately the state is known, the values of $p$ and $q$ in a joint measurement cannot be predicted better than within this uncertainty.

 

[A. Neumaier]

the electromagnetic field fluctuates

What do you mean by that phrase, apart from that its measured values are inherently uncertain?

so is this statement in wiki also wrong

"In the modern view, energy is always conserved, but because the particle number operator does not commute with a field's Hamiltonian or energy operator, the field's lowest-energy or ground state, often called the vacuum state, is not, as one might expect from that name, a state with no particles, but rather a quantum superposition of particle number eigenstates with 0, 1, 2...etc. particles."

https://en.wikipedia.org/wiki/Quantum_fluctuation

Yes, as I had mentioned already in post #32.

No, by definition the vacuum state is the state of lowest energy. You have to be a bit careful, however which vacuum you refer to. Of course the free-particle vacuum is not the same as the fully interacting one! It's a highly non-trivial issue.

Not even the Hilbert spaces are the same, so one cannot express the objects in the interacting theory in terms of those of a free theory, except asymptotically.

But the vacuum in an interacting theory still contains no particles in any meaningful sense.

The natural $N$-particle states (if one wants to define them at all) are - both in the free and in the interacting case - the states obtained by acting upon the vacuum with integrals over products of $N$ renormalized field operators. In the free case one can use the CCR or CAR to make a clean decomposition of these integrals into integral over normally ordered products of creation and annihilation operators and only the pure creation terms contribute. In the interacting case, this decomposition is no longer useful as the positive and negative frequency parts of the renormalized fields have no longer nice commutation properties.

 

[vanhees71]

There are no such states, so your first question is a meaningless request. The conundrum you pose arises from mixing classical thinking (where in the deterministic case a state means specified values of $p$ and $q$) and quantum thinking, where a state means something quite different.

In general, both in classical and quantum mechanics, a state is a positive linear functional on the observable algebra. In quantum mechanics, the latter is the algebra of linear operators on a Schwartz space (as in the case of $p$ and $q$), and states are therefore in 1-1 correspondence with density operators, positive linear integral operators of trace one. This density operator can be prepared and measured to arbitrary accuracy for sources producing sufficiently small systems such as photons or electrons. One cannot require more about preparing or measuring a state. Thus there is no uncertainty in the state itself.

However, there is an uncertainty in prediction the value of $p$ and $q$ from any exactly known state, given by the uncertainty relation. No matter how accurately the state is known, the values of $p$ and $q$ in a joint measurement cannot be predicted better than within this uncertainty.

For a very good treatment of scattering theory and the issue of wave packets and the uncertainty issue, see Messiah, Quantum Mechanics (it's non-relativistic, but the basic definitions are valid also in the relativistic case).

 

[vanhees71]

What do you mean by that phrase, apart from that its measured values are inherently uncertain?

The question is about my statement that the electromagnetic field fluctuates. That's very clear since there's an uncertainty relation for the em. field, which follows immediately from the canonical equal-time commutator relations of $\vec{A}$
$$[\hat{E}_i(t,\vec{x}),\hat{B}_j(t,\vec{y})]=-\mathrm{i} \epsilon_{ijk} \partial_k \delta^{(3)}(\vec{x}-\vec{y}).$$

 

[A. Neumaier]

The question is about my statement that the electromagnetic field fluctuates. That's very clear since there's an uncertainty relation for the em. field, which follows immediately from the canonical equal-time commutator relations of $\vec{A}$
$$[\hat{E}_i(t,\vec{x}),\hat{B}_j(t,\vec{y})]=-\mathrm{i} \epsilon_{ijk} \partial_k \delta^{(3)}(\vec{x}-\vec{y}).$$

But uncertainty is not the same as fluctuation. The latter is about an unpredictable process in time or space; the former is about the impossibility of an exact joint measurement.

 

[mfb]

One cannot require more about preparing or measuring a state.

You could require more, and in classical mechanics you can have more: no uncertainty about the possible measurement results for position of momentum. Quantum mechanics tells us that is impossible. That is exactly the uncertainty in the state I mentioned. This is getting a discussion about semantics, but I see the misconception "the uncertainty principle is just our inability to measure better" often, and I think it arises from this difference.

 

[vanhees71]

The uncertainty relations are about standard deviations of quantities, which is what's usually understood if you talk about fluctuations. The uncertainty relation is not about the impossibility of an exact joint measurement but an exact joint preparation. We have discussed this at length in this forum for years now!

 

[stevendaryl]

Whether a nonzero standard deviation implies fluctuation is a matter of semantics. But as I said, the talk about "fluctuations" is just a heuristic way to talk and think about various quantum phenomena in a qualitative way. In computing actual numbers, the heuristic is not good enough, and the detailed calculations don't explicitly involve fluctuations, at all. I think everyone agrees on those facts. The disagreement is about whether the heuristic itself has any value. There are two sides of this question: (1) On the plus side, does the heuristic help in suggesting new phenomena that can then be investigated more rigorously? (2) On the minus side, does the heuristic lead us astray, in the sense of suggesting that things ought to be possible, when they really aren't? The fact that the detailed calculations don't involve fluctuations at all to me isn't an example of the heuristic being misleading, as long as everyone is clear that it is only a heuristic.

 

[mfb]

I never said fluctuation, I said uncertainty.

 

[vanhees71]

Well, calculating loop diagrams, i.e., correlation functions of field operators implicitly imply the calculation of fluctuations. It's very obvious in the many-body case, where the Kubo formula of transport coefficients shows that you exactly do this!

 

[A. Neumaier]

The uncertainty relation is not about the impossibility of an exact joint measurement but an exact joint preparation.

What is an exact joint preparation?

One can prepare states but not measurement results; the uncertainty relations only refer to the latter.

You could require more, and in classical mechanics you can have more: no uncertainty about the possible measurement results for position of momentum. Quantum mechanics tells us that is impossible. That is exactly the uncertainty in the state I mentioned.

But it is not an uncertainty in the state but an uncertainty in the possible measurement results! You say this yourself. An uncertainty in the state would mean an uncertainty about which state it is. There is no such uncertainty in principle.

The classical analogue of a quantum state is a classical probability distribution. If one has a Gaussian distribution with given mean and variance then the distribution is completely certain although the realizations described by it have uncertainty. But this does not allow one to talk about uncertain probability distributions in this case - this means something completely different, namely uncertainty about the parameters of the distribution. Of course, measuring anything will always leave this sort of uncertainty about the true parameters, but this uncertainty can be made arbitrarily small by obtaining sufficiently many realizations. This is what I mean by saying that there is no such uncertainty in principle.

Exactly the same holds for the quantum state. One can determine it (i.e., the parameters characterizing it) with arbitrarily high precision by considering sufficiently many realizations. Then one knows everything one likes about the quantum state (just as one knows given a classical probability distribution everything about the state of a classical stochastic system). Thus the quantum state is as certain as anything can ever be! But one still has uncertainty about the actual values of the realizations.

Note: If one would reserve the word state to pure states one could say that in a mixed state there is uncertainty about which pure state is meant. (Something like this is assumed in discussions about proper and improper mixtures.) But it is impossible to specify this uncertainty in any statistical way since a mixed state of full rank can be decomposed into pure states containing with a nontrivial coefficient an arbitrary given pure state. Thus I think that this sort of uncertainty in a state is misguided. I cannot perceive of any other potential meaning of uncertainty in a state.

 

[stevendaryl]

I never said fluctuation, I said uncertainty.

Okay, I edited my post to remove your name.

 

[vanhees71]

What is an exact joint preparation?

One can prepare states but not measurement results; the uncertainty relations only refer to the latter.

Such a preparation is that of a state, where both observables have determined values. If the operators corresponding to these observables don't commute you cannot do so, and that's the real meaning of the Heisenberg-Robertson uncertainty relation,
$$\Delta A \Delta B \geq \frac{1}{2} |\langle[\hat{A},\hat{B}]|.$$

But it is not an uncertainty in the state but an uncertainty in the possible measurement results! You say this yourself. An uncertainty in the state would mean an uncertainty about which state it is. There is no such uncertainty in principle.

Sure, the uncertainty is in the observables, not the state, which is determined by the preparation procedure, but you can measure either A or B as precisely as you are technically able to. It's not restricted by any uncertainty relation. In fact you have to measure the observables with a significantly higher precision than the standard deviations due to the prepared state to verify this uncertainty relation.

The classical analogue of a quantum state is a classical probability distribution. If one has a Gaussian distribution with given mean and variance then the distribution is completely certain although the realizations described by it have uncertainty. But this does not allow one to talk about uncertain probability distributions in this case - this means something completely different, namely uncertainty about the parameters of the distribution. Of course, measuring anything will always leave this sort of uncertainty about the true parameters, but this uncertainty can be made arbitrarily small by obtaining sufficiently many realizations. This is what I mean by saying that there is no such uncertainty in principle.

Exactly the same holds for the quantum state. One can determine it (i.e., the parameters characterizing it) with arbitrarily high precision by considering sufficiently many realizations. Then one knows everything one likes about the quantum state (just as one knows given a classical probability distribution everything about the state of a classical stochastic system). Thus the quantum state is as certain as anything can ever be! But one still has uncertainty about the actual values of the realizations.

Note; If one would reserve the word state to pure states one could say that in a mixed state there is uncertainty about which pure state is meant. (Something like this is assumed in discussions about proper and improper mixtures.) Though it is impossible to specify this uncertainty in any statistical way since a mixed state of full rank can be decomposed into pure states containing with a nontrivial coefficient an arbitrary given pure state. Thus I think that this sort of uncertainty in a state is misguided. I cannot perceive of any other potential meaning of uncertainty in a state.

I fully agree with this of course.

 

[mfb]

An uncertainty in the state would mean an uncertainty about which state it is.

That's the part where we disagree - and it is not about physics, but the use of English words.

 

[A. Neumaier]

Such a preparation is that of a state, where both observables have determined values.

OK, so you talk about the preparation of a nonexistent object, not of a state with particular properties. Of course, nonexistent things cannot be prepared.

That's the part where we disagree - and it is not about physics, but the use of English words.

Yes, in both cases it is a matter of the correct use of English words.

An uncertainty in the position means that the position is not known exactly. An uncertainty in a measurement result means that the measurement result is know available to more than a certain precision. By the same token, an uncertainty in the state means that the state is not known exactly?

Why should it mean something completely different, namely (as you take it to mean) the uncertainty of something deduced from a computation involved in that state? It is not the English language that would make it mean that.