Explore The Vacuum Fluctuation Myth in Quantum Theory
This Insight Article is a sequel of the Insight Articles ”The Physics of Virtual Particles” and “Misconceptions about Virtual Particles“ which make precise what a virtual particle is and what being real means, document some of the liberties taken in physics textbooks in the use of this concept, and mention the most prominent misuses. A further Insight Article, ”Vacuum Fluctuations in Experimental Practice”, shows the example of a recent article in the scientific literature how some authors claim the observation of vacuum fluctuations, justified only by superficial, invalid reasoning.
In short, the concept of virtual particles is well-defined and useful when restricted to its use in Feynman diagrams and associated technical discussions. But it is highly misleading when used to argue about vacuum fluctuations as if these were processes happening in space and time. The latter is a frequent misunderstanding, a myth that has not the slightest basis in particle physics. (The proper meaning of some terms related to the vacuum is explained at the end of ”The Physics of Virtual Particles”.)
The two articles mentioned do not, however, explain how it is possible that this misunderstanding is so widespread, and even serious experts resort to misleading imagery when explaining the subject to the general public. This is answered here at the example of Steve Carlip’s page on Hawkings radiation, where Steve Carlip, a well-known theoretical physicist working on quantum gravity, gave a lucid but completely mythical narrative about how vacuum fluctuations create Hawking radiation. This vacuum fluctuation myth comes from taking pieces of intuition and connecting them with a plausible narrative. The following is a reconstruction of the sort of thoughts that combine to justify the myth in the eyes of those who use this language. (For those interested, a non-mythical description of Hawking radiation is given by Sabine Hossenfelder here.)
The starting point is the sound knowledge that there are technical notions of vacuum fluctuations (= nonzero vacuum expectation values), virtual particles (=internal lines in a Feynman diagram), and that in bare quantum field theory with a cutoff, the vacuum is a complicated multiparticle state depending on the cutoff – though in a way that it diverges when the cutoff is removed, so that nothing physical remains. Then the question arises: Is is possible to convey a bit of this to ordinary people? It is highly unsatisfactory not to be able to talk about what one is doing in one’s research…
So one goes for analogies and images. Already calling internal lines ”virtual particles” is a step in this direction. Allow yourself a little more liberty and combine it with Feynman’s classical absorber theory of radiation; after all, Feynman also invented the diagrams bearing his name, possibly even inspired by this analogy. The lines defining the virtual particles look like world lines in a classical process, so why not interpret them (in one’s imagination) as the quantum remnants of the classical world lines of Feynman’s earlier (later abandoned) theory? This happy accident makes the story possible. It is not completely accurate but plausible (in the absence of the correction of the intuition by mathematical formulas) because both classical particles and virtual particles are represented pictorially by lines, and it is something that ordinary people can imagine. This is the beginning of the myth. An extra reassurance that you are on a good path is that the arrows that physicists draw on their diagrams (to indicate the sign of conserved quantum numbers) happen to match Feynman’s classical (later abandoned) idea that antiparticles are just particles moving backward in time.
To bring in more physics one has to be able to interpret complete Feynman diagrams. Tree diagrams are easy but bring in a new aspect. They talk about real and virtual particles. On an electron line containing two external vertices, the electron changes its status from being real (external) to being virtual (internal) and back (external) again. We learn from it a new fact – a virtual particle can become real, and conversely. The interpretation as world lines teaches us other things: A single Feynman diagram should in fact be considered just as a tiny snapshot of an extended web containing all particles in the universe; after all, world lines do not begin and end nowhere. Thus ”in reality” (meaning in the simplified virtual reality painted for the general public) all particles should be viewed as virtual until they are observed (where they obviously are real). This matches a version of the highly respected Copenhagen interpretation: Unobserved particles have a sort of ghost existence since properties emerge only when they are subjected to quantum measurement. You are pleased by this coincidence – it seems to say that there is a coherent story to be told. Also, since most of the lines in the Feynman diagram end, you have a layman’s picture for decaying particles: What you see in a bubble chamber is just a Feynman diagram made visible! This is the first serious manifestation of the myth. In spite of lacking any grounding in real physics (being grounded instead in the visual analogy), you feel entitled to make this identification – it serves your final goal to make some of the intricacies of microphysics accessible to the general public. No one there will ever ask how it can be that two virtual particles can bend as in a Feynman diagram with a loop – so that they find each other exactly at the right place and with exactly the right momentum to annihilate. Therefore such impossibilities – that would spoil the goal of giving a simplified picture of what happens – are silently swept under the carpet.
The next thing is to interpret the bare multiparticle state. It is obviously a complex superposition of bare particles. Make the next move to identify bare particles with virtual particles; after all, both are unobservable but appear in some version of the formalism. Now you have the picture of the vacuum as teeming with particles. From the form of Feynman diagrams with one or more loops, you can read off that in order to make sense of the narrative these particles have to pop in and out of existence. This is the birth of the next item in the myth. That in a superposition nothing dynamical happens is a small nuisance that you happily sacrifice in order to be understandable to your intended audience. After all, you can now give an illusion of having conveyed something of the complexities of the naive perturbative approach without having to talk about perturbation theory. In addition, without asking for it, you have found an unexpected visual interpretation of the notion of a vacuum fluctuation: A teeming vacuum where particles constantly pop in and out of existence clearly fluctuates, and every single act of popping may rightfully be regarded as a fluctuation of the vacuum. Another piece of the myth has found its place. Never mind that there is not the slightest way of justifying this analogy on the level of mathematical formulas. What counts is how the picture appeals to the general public, and it is obvious that drastic simplifications are needed to achieve this goal.
Now one needs to worry about the basic principles of physics in all this. After all, one doesn’t want to talk about particles alone but conveys some general physics as well. Let us bring in conservation laws. Everyone knows that energy is conserved in Nature. But wait, doesn’t the creation of particles require some energy? Never mind, quantum mechanics comes to the rescue. People will have heard of the Heisenberg uncertainty relation, and if they haven’t this is an opportunity to make your audience acquainted with it. It states the intrinsic uncertainty of position and momentum in nonrelativistic mechanics. What does it tell about energy conservation? Nothing at all, but analogy comes to the rescue. In relativistic physics, time is the 4th coordinate of position, and energy the 4th coordinate of momentum. Thus we don’t make a big blunder if we consider a time-energy uncertainty relation. (Though in mainstream quantum physics, time is not an observable – it has a completely different status.) Uncertain energy can be liberally interpreted as a slightly inaccurate conservation law. After all, one can derive from quantum mechanics only that the expectation of the energy operator is conserved. Expectation brings to mind that whatever you measure inaccurately must be measured many times for getting an improved accuracy. Thus only the average energy needs to be conserved. Reinterpret the physical ensemble average (in the service of simplifying the physics to give your audience a coherent story) as an average in time.
Thus you found the solution: Energy can be borrowed for a short period of time if it is returned on average. The next item of the myth arrived. Now you are quite confident that you’ll be able to get a full and rich story (for laymen only, so all the small and big blunders made can be excused) and continue to turn it into something you’ll tell in public (or write in a book). You hope that the attentive audience will not ask where the energy is borrowed from, but unfortunately, you told the story first to a colleague with an unbiased mind and he insisted on that this should be clarified first. You need to look at some more pieces of information to get the next input. Fortunately, you soon find it: The zero-point energy of a harmonic oscillator had in the past always been ignored by saying that only energy differences are observable. Maybe it is the bank from which the virtual particles lining up for popping into existence can borrow their energy. And yes – it turns out that the bare quantum field has a huge amount of zero-point energy – an infinite amount if you take the physical limit. Clearly, this must be the source – and no ordinary person will be interested to question it. Thus the final piece of the myth arrived. You are happy – it will be a really good story conveying a lot of physics while still being understandable to ordinary people.
That there is no physical mechanism for how the borrowing works is a small nuisance that (for the layman) can be ignored – after all, they want a simple story that they can believe, not a technical discussion of all the problems involved – they know that quantum mechanics is full of unresolved problems. At this point, your story is already so convincing that you don’t mind that all observable quantities also become infinite in the limit considered and that when you instead do a proper renormalization (needed to get the high accuracy predictions quantum field theory is famous for) the whole capital of the vacuum energy bank shrinks to zero!
Now the particle philosophy for the laymen is essentially complete. Only a few – to laymen imperceptible – jumps of the imagination were needed in the service of understandability. Like in a cinema, where the pictures jump in discrete steps but provide a sufficient illusion for the audience to see a continuous story. To make sure that the audience, captured by the imaginative illusion, will not take it for physical reality, and to ensure that your status as a respected scientist is preserved, you begin with a caveat (like Steve Carlip did on his page on Hawkings radiation, in the inconspicuous first line after the heading ”An Incomplete Glossary” – long forgotten at the time the reader enters the mythical narrative linked to above): ”Be warned – the explanations here are, for the most part, drastic oversimplifications, and shouldn’t be taken too literally.” But in spite of this, you can instead be sure that most of your audience will ignore this sentence said in the first few seconds in favor of the nice mental pictures that you took a whole hour to explain and make intelligible.
When Hawking discovered what was later called Hawking radiation this picture for the general public was already well entrenched. So he only had to figure out how his discovery would fit in – and it fitted well. Instead of talking about gravitational energy (not visible, hence a sort of vacuum) creating a particle-antiparticle pair one partner of which escapes there is only a small step to saying what the educated general public expects. Since the particles are not (yet) observable by the faraway observer seeing only the radiation, they must be sold according to the philosophy developed above as virtual particles created (hence vacuum fluctuations in action). Years later, when one of the particles is finally observed by the faraway observer, it becomes real as a piece of the observable Hawking radiation.
Thus if you want to summarize to lay people the Hawking effect in a single phrase, what is more, natural than to say that ”vacuum fluctuations cause the Hawking radiation” without repeating the warning that this ”shouldn’t be taken too literally”?
Full Professor (Chair for Computational Mathematics) at the University of Vienna, Austria
Arnold, You are talking about the mathematical "vacuum" right and not the real world vacuum where the existence of particles and radiating fields complicate things and make it actually seething.
[QUOTE="vanhees71, post: 5643596, member: 260864"]If the two observables are not compatible generally you can't do that.”Yes, so there is no state with the required properties and therefore nothing that could be prepared. I would not see this as a limitation of what can be prepared, since it is clear that only states can be prepared. The Heisenberg uncertainty relation just gives a necessary condition for the existence of a state with given uncertainties.[QUOTE="vanhees71, post: 5643596, member: 260864"]what do you define as "fluctuation", and I think the usual meaning of the word is”I had cited in post #32 a number of dictionaries that define what the usual meaning is. They all agree that it means a kind of wavering uncertainty, not that something is just uncertain or undefined. For example, the age of the universe does not fluctuate though it is uncertain to us. Only our estimates of it fluctuate in the course of time. In quantum mechanics, it is always the measurement results that fluctuate (in a series of experiments), not the quantities themselves. The latter have no values when unobserved (in the Copenhagen interpretation), one makes no statement at all about them (in the minimal interpretation; you should know that!). In a few interpretations one can make assertions about some unobserved values – e.g., in Bohmian mechanics about positions (and hence velocities and momenta – all exactly and in simple solvable instances not fluctuating!) or in my thermal interpretation (where most values are intrinsically approximate, again not fluctuating!).Thus the fluctuation is only present in the measured ensemble, where they have the same statistical nature as in classical stochastic ensembles – that each realization differs a bit from each other one.
[QUOTE="A. Neumaier, post: 5643588, member: 293806"]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.”I don't understand this. Of course, if the two observables are compatible you can (at least in principle) prepare the system in a common eigenstate, and then both observables have a determined value. If the two observables are not compatible generally you can't do that. That's the content of the uncertainty relation.The next question then is, what do you define as "fluctuation", and I think the usual meaning of the word is that any quantity that has an undefined value due to the prepared state fluctuates, and the fluctuation is characterized by the standard deviation of the corresponding probability distribution.
[QUOTE="vanhees71, post: 5643536, member: 260864"]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. [QUOTE="mfb, post: 5643548, member: 405866"]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.
[QUOTE="A. Neumaier, post: 5643530, member: 293806"]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.
[QUOTE="A. Neumaier, post: 5643530, member: 293806"]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.
[QUOTE="mfb, post: 5643527, member: 405866"]I never said fluctuation, I said uncertainty.”Okay, I edited my post to remove your name.
[QUOTE="vanhees71, post: 5643510, member: 260864"]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.[QUOTE="mfb, post: 5643507, member: 405866"]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.) 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.
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!
I never said fluctuation, I said uncertainty.
I agree with mfb that 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.
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!
[QUOTE="A. Neumaier, post: 5643399, member: 293806"]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.
[QUOTE="vanhees71, post: 5643431, member: 260864"]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.
[QUOTE="A. Neumaier, post: 5643408, member: 293806"]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}##$$[hat{E}_i(t,vec{x}),hat{B}_j(t,vec{y})]=-mathrm{i} epsilon_{ijk} partial_k delta^{(3)}(vec{x}-vec{y}).$$
[QUOTE="A. Neumaier, post: 5643399, member: 293806"]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).
[QUOTE="vanhees71, post: 5643003, member: 260864"]the electromagnetic field fluctuates”What do you mean by that phrase, apart from that its measured values are inherently uncertain? [QUOTE="ftr, post: 5643034, member: 465682"]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. [QUOTE="vanhees71, post: 5643352, member: 260864"]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.
[QUOTE="mfb, post: 5642971, member: 405866"]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.
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.
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
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.htmlIt'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.
[QUOTE="A. Neumaier, post: 5642758, member: 293806"]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?
[QUOTE="RockyMarciano, post: 5642783, member: 585697"]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 to use the English language correctly. Saying [QUOTE="RockyMarciano, post: 5642774, member: 585697"]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. And these inferences come out true if the theory is correct. If we would have to check all inferences from a theory in order to trust it 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. [QUOTE="RockyMarciano, post: 5642783, member: 585697"][…] 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.
[QUOTE="A. Neumaier, post: 5642777, member: 293806"]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?
[QUOTE="A. Neumaier, post: 5642765, member: 293806"]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.
[QUOTE="RockyMarciano, post: 5642774, member: 585697"]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!
[QUOTE="A. Neumaier, post: 5642767, member: 293806"]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 wich 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.
[QUOTE="RockyMarciano, post: 5642761, member: 585697"]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!
[QUOTE="RockyMarciano, post: 5642761, member: 585697"]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.
[QUOTE="stevendaryl, post: 5642687, member: 372855"]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.
[QUOTE="A. Neumaier, post: 5642749, member: 293806"]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.
[QUOTE="mfb, post: 5642685, member: 405866"]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. [QUOTE="mfb, post: 5642685, member: 405866"]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?
[QUOTE="RockyMarciano, post: 5642653, member: 585697"]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![QUOTE="RockyMarciano, post: 5642653, member: 585697"]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.
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.
[QUOTE="A. Neumaier, post: 5642233, member: 293806"]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.
Arnold, maybe it doesn't matter and you probably know, if you post anywhere on Wikipedia, and you are not a registered user, or signed in, your IP address is visible to anybody and everybody ?This little critter, called SineBot, will automatically sign a message with ones registered user name, or in your case… your IP address.I blacked yours out, but see…
[ATTACH=full]110351[/ATTACH]
[QUOTE="A. Neumaier, post: 5642623, member: 293806"]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.
[QUOTE="RockyMarciano, post: 5642618, member: 585697"]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.
[QUOTE="A. Neumaier, post: 5642564, member: 293806"]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.
The passage''In terms of virtual particles, the propagator at spacelike separation can be thought of as a means of calculating the amplitude for creating a virtual particle-antiparticle pair that eventually disappear into the vacuum, or for detecting a virtual pair emerging from the vacuum. In Feynman's language, such creation and annihilation processes are equivalent to a virtual particle wandering backward and forward through time, which can take it outside of the light cone.''from the same wikipedia article is also misleading. Creation and annihilation operators only exist for time-like, on-shell momenta; hence the associated creation and annihilation processes all refer to real particles.
[QUOTE="ftr, post: 5642575, member: 465682"]What do you think this statement is saying”As stated, it is meaningless since ##Phi(x)## is not an operator, hence not an observable. As remarked in your quote, one has to use a smeared version (averaging over a small open region in space-time) to produce an operator. Even with this amendment, the statement is misleading. The ''vacuum value'' is not a commonly used expression. The nearest expression with a formal meaning is the vacuum expectation value, but this is completely determined and hence certain. What is probably meant is that if one could measure the local value of a smeared field in the vacuum state (don't ask how this ever can be done, as the vacuum contains no particles, hence no observer), the result would have a significant uncertainty. This is formally true if one assumes (as is commonly done) that the Born interpretation holds in this (counterfactual) case.But in the statement quoted, the mistake already pointed out is made, that uncertainty and fluctuation are equated. This turns an unconspicuous statement that some measurement result has a significant uncertainty (we know this holds for most measurements) into the remarkable and wrong statement that the value fluctuates with the implied, equally wrong consequence that the vacuum is ''active''. This is typical for the exaggerations made when turning banal news into exciting stories for everyone.
Arnold What do you think this statement is saying then in wiki"So what does the spacelike part of the propagator represent? In QFT the vacuum is an active participant, and particle numbers and field values are related by an uncertainty principle; field values are uncertain even for particle number zero. There is a nonzero probability amplitude to find a significant fluctuation in the vacuum value of the field Φ(x) if one measures it locally (or, to be more precise, if one measures an operator obtained by averaging the field over a small region)."https://en.wikipedia.org/wiki/Propagator
[QUOTE="RockyMarciano, post: 5642325, member: 585697"]Fluctuation is a word that is synonim both of oscillation and of indeterminacy or uncertainty.”No. Fluctuation in today's usage always means change, not just being uncertain! All the major dictionaries agree on that:http://www.dictionary.com/browse/fluctuation1. continual change from one point or condition to another.2. wavelike motion; undulation.3. Genetics. a body variation due to environmental factors and not inherited.http://dictionary.cambridge.org/dictionary/english/fluctuatefluctuate: to change, especially continuously and between one level or thing and anotherhttps://en.oxforddictionaries.com/definition/fluctuationAn irregular rising and falling in number or amount; a variationhttps://www.merriam-webster.com/dictionary/fluctuate1. to shift back and forth uncertainly 2. to ebb and flow in waveshttps://www.vocabulary.com/dictionary/fluctuationThe noun fluctuation refers to the deviations along the path from one point to another. We see frequent fluctuationsin the stock market, as prices go up or down, and also in the weather, which is always changing.http://www.macmillandictionary.com/dictionary/british/fluctuationfrequent changes in the amount, value, or level of somethingEven wikipeedia describes it as a change, though in a completely unscientific manner (not surprisingly, since it also promotes lots of other nonsense about virtual particles):''In quantum physics, a quantum fluctuation (or quantum vacuum fluctuation or vacuum fluctuation) is the temporary change in the amount of energy in a point in space […] 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.''The quality of the Wikipedia statement can be assessed from the second sentence quoted, which is absurd. The vacuum state is always the eigenstate of the number operator with exactly zero particles. There is no uncertainty in the number of particles, since it is an eigenstate.[QUOTE="RockyMarciano, post: 5642325, member: 585697"]what fluctuates(vacillates i.e. it is intrinsically uncertain) is precisely the noncommuting observables.”This is only you 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!
[QUOTE="RockyMarciano, post: 5642325, member: 585697"]This is purely semantic but both the insight and thread are about semantics so why not get it right?. Fluctuation is a word that is synonim both of oscillation and of indeterminacy or uncertainty. All it means in the quantum context is the Heisenberg indeterminacy of the ground state, and what fluctuates(vacillates i.e. it is intrinsically uncertain) is precisely the noncommuting observables. Of course many people by extension thinks about something moving or oscillating, that I guess it is what you understand if you disregard the meaning of fluctuation as vacillation/indeterminacy. Since Heisenberg indeterminacy lies at the heart of the quantum departure from classical physics, quantum fluctuations by extension are also referred by many as this departure from classicality.On the other hand if one is strict with the math not even the fields or the waves actually oscillate, since the math always describes a rigid picture, a shortcoming of analysis. But this should show just how ridiculous can blind strictness get.”But it seems vacuum fluctuation is mentioned more in QED and you are talking about "quantum fluctuations" https://en.wikipedia.org/wiki/Quantum_fluctuationhttps://en.wikipedia.org/wiki/QED_vacuumhttps://en.wikipedia.org/wiki/Vacuum_polarizationso some "fluctuation" is related to vacuum in the vicinity of interactions others to an otherwise empty interstellar vacuum. It sound like many concepts being mixed up.
[QUOTE="A. Neumaier, post: 5642233, member: 293806"]The Heisenberg uncertainty relation is not about quantum fluctuations but about the intrinsic uncertainty in measuring noncommuting observables. Nothing fluctuates there.”This is purely semantic but both the insight and thread are about semantics so why not get it right?. Fluctuation is a word that is synonim both of oscillation and of indeterminacy or uncertainty. All it means in the quantum context is the Heisenberg indeterminacy of the ground state, and what fluctuates(vacillates i.e. it is intrinsically uncertain) is precisely the noncommuting observables. Of course many people by extension thinks about something moving or oscillating, that I guess it is what you understand if you disregard the meaning of fluctuation as vacillation/indeterminacy. Since Heisenberg indeterminacy lies at the heart of the quantum departure from classical physics, quantum fluctuations by extension are also referred by many as this departure from classicality.On the other hand if one is strict with the math not even the fields or the waves actually oscillate, since the math always describes a rigid picture, a shortcoming of analysis. But this should show just how ridiculous can blind strictness get.
[QUOTE="ftr, post: 5642272, member: 465682"]So Arnold, Virtual particles don't exist so how do two charged particles interact at distance?”Oh please, did you even touch any QFT book?
So Arnold, Virtual particles don't exist so how do two charged particles interact at distance? So what does hold an electron in an orbit even if not classical?
[QUOTE="Haelfix, post: 5642224, member: 4167"]I have heard the word used more when discussing things like barrier penetration in nrqm. So an author will write something like "classically you will never measure a particle here, but b/c of 'quantum fluctuations' or 'quantum jitters' you will see a tunneling phenomenon on the other side and the nonzero possibility for the detection of a particle".”Though this is somewhat unrelated to the present topic, let me mention that quantum tunneling is a misnomer. It is motion over the barrier and not through the barrier. For the ''tunneling'' probability tends to zero as the barrier gets higher, and is zero when the barrier is infinitely high. No matter how long a tunnel through the barrier would have to be! Thus it is like the motion of a classical particle with a random kinetic energy – it has a small probability of being kicked over the barrier and ending up outside the well it was in originally.Again nothing that fluctuates!
[QUOTE="RockyMarciano, post: 5642213, member: 585697"]To me quantum fluctuations are defined by the fact that the ground state in qm must also obey the Heisenberg principle”The Heisenberg uncertainty relation is not about quantum fluctuations but about the intrinsic uncertainty in measuring noncommuting observables. Nothing fluctuates there.
[QUOTE="A. Neumaier, post: 5642088, member: 293806"]. The quantum mechanical ground state is dynamically completely inert under the quadratic Hamiltonian that defines the oscillator. Nothing fluctuates. “Yep, they are clearly stationary states in the nrqm case. So, I would say I have heard the word used more when discussing things like barrier penetration in nrqm. So an author will write something like "classically you will never measure a particle here, but b/c of 'quantum fluctuations' or 'quantum jitters' you will see a tunneling phenonemon on the other side and the nonzero possibility for the detection of a particle". So there the word would presumably mean some sort of deviation from classical expectations.In the context of inflation, the same sort of pedagogical word choice is frequently used informally in the context of a potential term for a scalar field, where you have either tunneling between false and true vacuums, or alternatively where you have oscillatory behavior at the bottom of a well analogous to the phenomenon which leads to the longitudinal mode in the Higgs phenomenon.
Mixing nrqm and qft may lead to confusions. To me quantum fluctuations are defined by the fact that the ground state in qm must also obey the Heisenberg principle, that is what in graphic language fluctuates in quantum fluctuations. When going to the quantum field picture there are so many things that change(for one thing position is no longer a operator while what used to be states are operators,etc…) that there is no longer a good mathematical translation of this, and the vacuum state of the field theory doesn't qualify when formally defined. There, no more mystique.
[QUOTE="HyperStrings, post: 5641896, member: 611083"], what is causing the ''field theoretic effects'' that oscillate the atom”Well, the interaction with the crystal, or if you wish, the field defined by it cause these effects. Switch the interaction or the mean field off and the effect is gone. This proves that these are the responsible agents. Not mystical quantum fluctuations.
[QUOTE="Haelfix, post: 5641926, member: 4167"]if the origin of the word in textbooks is precisely when discussing simple harmonic oscillators, particles in a box, and other simple nonrelativistic quantum mechanics. There it would presumably mean a fluctuation relative to a classical zero.”I have seen the word used only in the context of (relativistic or nonrelativistic) quantum field theory. It doesn't make sense for a harmonic oscillator or a particle in a box. The quantum mechanical ground state is dynamically completely inert under the quadratic Hamiltonian that defines the oscillator. Nothing fluctuates. There is an uncertainty about the values of observables not commuting with the energy, but this is because it is impossible to measure them more accurately, not because these would fluctuate in time. The traditional interpretations refrain from saying what happens in between measurememt; none of them claims that these observables have all the time exact but fluctuating values.
[QUOTE="A. Neumaier, post: 5641886, member: 293806"]A nonzero vacuum expectation value doesn't mean in any sense that the vacuum is fluctuating. Otherwise the ground state of a single harmonic oscillator would also be fluctuating….”We completely agree then, although i'm now wondering if the origin of the word in textbooks is precisely when discussing simple harmonic oscillators, particles in a box, and other simple nonrelativistic quantum mechanics. There it would presumably me a fluctuation relative to a classical zero.
[QUOTE="A. Neumaier, post: 5641782, member: 293806"] Analogies don't create truth” There is a good physics lecture on 'The Dangers of Analogies' and I agree with you, we should be very discerning of analogy. Though, in the lecture/paper he gives rules to how to properly use analogies safely when discussing/teaching physics and admits that sometimes there is no other way to explain something. Just as all Gaileleo had to prove his theory was an analogy. Its a double edged sword, as we must also be aware, we are heading into a time of where experiments will need to be able to represent complex planckian scale effects, so we may very well have to bite the bullet and start to understand analogous experiments. As it is very possible for them to be useful. With that being said, I will try to be more discreet and specific as I can respect your position.[QUOTE="A. Neumaier, post: 5641782, member: 293806"]Quantum fluctuations are everywhere,”So, in your opinion, what is causing the ''field theoretic effects'' that oscillate the atom, 'up the slope of the bowl' in a dead, zero point BEC well?The paper observed the BEC well oscillations, mathematically predicted the subtle nuance of effects of acoustic oscillations that match the 'vibration' of the field to exactly match the 'field theoretic effect' of the oscillations or quantum fluctuations of the atoms in the BEC well. Then showed, precisely, those same exact quantum fluctuations, with their mathematically predicted phonon model. I have tried to put together any various combination of small changes in the atomic structure that would allow the equivalent of such oscillations but I can't find any correlation that would create these specific, exactly replicated, oscillations. So I am in agreement with the paper.[QUOTE="Haelfix, post: 5641816, member: 4167"](and then there are all the complicated renormalization scheme caveats associated with what we mean by this”Which brings me to another point that, we are using normalization because of planck scale discrepencies, and 'science' is okay with that, but if you try to isolate those discrepencies with a mathematical application of a 'mistake fixing, re-normalization', science is not okay with that? The very process of normalization is in essence, 'blurring the clarity of the image'. Then a proper re-normalization can result with 'sharpening of the image'.As I see it, the vibration of the atoms field creates virtual phonon wave oscillations, the atoms wiggle up the bowls slope and the quantum fluctuations are the wiggle of those vibrations.
[QUOTE="Haelfix, post: 5641816, member: 4167"]whether its just a matter of terminology. I'm fine with saying the quantum vacuum is a subtle creature, and it's a little hard to define what a 'fluctuation' actually means”That's the problem. It is nowhere meaningfully defined, but used a lot in informal talk. One can meaningfully talk about fluctuations of an observable ##X##, meaning that the variance of ##X## is nonzero, but there is no formal definition of vacuum fluctuation, hence it is meaningless. How can you meassure something that doesn't even have a proper definition? One can measure spectral shifts, or forces, but claiming that in this way one has measured vacuum fluctuations needs more than saying that some nonzero vacuum expectation is used somewhere in the calculations. For the latter is the case in any perturbative computation of anything in quantum field theory, hence doesn't say anything nontrivial.The informal meaning most often used, e.g., in [4], that these fluctuations consist of virtual particles popping in and out of existence, is plainly wrong. [QUOTE="Haelfix, post: 5641816, member: 4167"]one can show that instantons can contribute to the QCD vacuum.”This isn't about vacuum fluctuations, but about the proper definition of what the vacuum state means in QCD. [QUOTE="Haelfix, post: 5641816, member: 4167"]we can actually demonstrate that the vacuum is nonzero in certain cases.”A nonzero vacuum expectation value doesn't mean in any sense that the vacuum is fluctuating. Otherwise the ground state of a single harmonic oscillator would also be fluctuating….
[QUOTE="A. Neumaier, post: 5641772, member: 293806"]It is a very useful (and historically sanctioned) tool to capture the imagination of an audience without presenting any formula, although it does not resemble at all what happens. The latter is discovered only if one wants to see what the talk means – and one discovers that it means nothing. ''vacuum fluctuations'' are just a buzzword for ''field theoretic effects'', nothing more.”I'm a little uneasy by some of these discussions, b/c its a little difficult to know what the claim is, and whether its just a matter of terminology. I'm fine with saying the quantum vacuum is a subtle creature, and it's a little hard to define what a 'fluctuation' actually means considering that we are by assumption talking about stationary states (and then there are all the complicated renormalization scheme caveats associated with what we mean by this). However, i'm a little uneasy by the implication that the 'real' thing when properly understood is trivial or empty. That has definitely not been demonstrated.The only way you can measure the quantum vacuum, is by doing an experiment, and it's always an inferred counterfactual property (this is what we would measure if we thought the vacuum was trivial or empty). So when people talk about polarizing the vacuum like for the Lamb shift, its basically about feynman diagrams that sort of look like bonafide vacuum diagrams except that there are external legs present, and there is always a strong background field (as well as the presence of a hydrogen atom as a spectator). Of course real vacuum diagrams have no external legs, and there is no background field present, so it is technically true that the presence of a measured nonzero effect for the Lamb shift doesn't necessarily tell us about the nature of the real quantum vacuum.Then there is the temptation to think that the nonzero quantum vacuum is then only some sort of perturbative artifact. But this is also wrong. For instance, one can show that instantons can contribute to the QCD vacuum. Also, one of the virtues of supersymmetric theories, is that they are sometimes exactly solvable, and we can actually demonstrate that the vacuum is nonzero in certain cases.I guess I don't understand what the claim is specifically.
[QUOTE="HyperStrings, post: 5641758, member: 611083"]I don't see how your retort disproves acoustic quantum fluctuations.”Quantum fluctuations are everywhere, but calling an unexcited crystal a quantum vacuum, as the authors of your source [4] do, is quite a misnomer. Analogies don't create truth. ''The authors measured an energy shift in the presence of the quantum vacuum, finding a value in good agreement with theoretical expectations. In addition to providing the first quantitative measurement of the phononic Lamb shift, the result confirms the validity of the theoretical framework that describes the effect. ''The author measured an energy shift in the presence of a crystal (producing of course an interaction that changes all energy levels), finding a value in good agreement with theoretical expectations. But this sounds too unexciting to be worth reporting.
[QUOTE="stevendaryl, post: 5641755, member: 372855"]Of course you (and A. Neumaier) are right. However, it is interesting that respectable physicists (most recently, I saw a video lecture by Alan Guth where this happens) very often present their informal reasoning in terms of vacuum fluctuations. It seems like it's a useful heuristic for reasoning about what's possible, even though the mathematical details, when you actually try to calculate things, don't actually resemble the "fluctuation" reasoning much at all.”It is a very useful (and historically sanctioned) tool to capture the imagination of an audience without presenting any formula, although it does not resemble at all what happens. The latter is discovered only if one wants to see what the talk means – and one discovers that it means nothing. ''vacuum fluctuations'' are just a buzzword for ''field theoretic effects'', nothing more.
They were virtual phonons, and I do not ascribe to the theory that phonons are particles, or even 'virtual particles'. though it is upsetting you didn't check the other papers. These fluctuations were not to be meant as a cohesive description upon one anothers papers. They do certainly outline quantum fluctuations. I don't see how your retort disproves acoustic quantum fluctuations. Considering you didn't take the time to read them nor cite any paper yourself, I find your assessment, illogical.
[QUOTE="bhobba, post: 5641716, member: 366323"]Its a total myth that quantum fluctuations, virtual particles yada, yada, yada exist. Its part of the pictorial language that has grown up with Feynman diagrams – but are really just terms in a Dyson series:https://en.wikipedia.org/wiki/Dyson_series“Of course you (and A. Neumaier) are right. However, it is interesting that respectable physicists (most recently, I saw a video lecture by Alan Guth where this happens) very often present their informal reasoning in terms of vacuum fluctuations. It seems like it's a useful heuristic for reasoning about what's possible, even though the mathematical details, when you actually try to calculate things, don't actually resemble the "fluctuation" reasoning much at all.For an example from Guth, http://web.mit.edu/physics/people/faculty/guth_alan.html“One of the intriguing consequences of inflation is that quantum fluctuations in the early universe can be stretched to astronomical proportions, providing the seeds for the large scale structure of the universe. The predicted spectrum of these fluctuations was calculated by Guth and others in 1982. These fluctuations can be seen today as ripples in the cosmic background radiation, but the amplitude of these faint ripples is only about one part in 100,000. Nonetheless, these ripples were detected by the COBE satellite in 1992, and they have now been measured to much higher precision by the WMAP satellite and other experiments. The properties of the radiation are found to be in excellent agreement with the predictions of the simplest models of inflation.”
[QUOTE="HyperStrings, post: 5641635, member: 611083"]Quantum fluctuations are also observed through the lamb shift [4].”The paper [4] states: ''It arises because zero-point fluctuations of the electromagnetic field in vacuum perturb the position of the hydrogen atom’s single bound electron.'' The truth – shown by all sources that actually do the calculations – is that the electromagnetic field interacting with the hydrogen atom’s single bound electron introduces radiation corrections into the corresponding Dirac equation. According to standard quantum mechanical perturbation theory, these corrections result in the Lamb shift. Nothing with vacuum fluctuations or ''virtual photons popping in and out of existence'' as your source claims.I didn't check the other papers but it is unlikely that they provide harder evidence for what you want to imply.
[QUOTE="HyperStrings, post: 5641635, member: 611083"] Quantum fluctuations are also observed through the lamb shift [4].”Gee – I always thought it was simply including further terms and re-normalization.Its a total myth that quantum fluctuations, virtual particles yada, yada, yada exist. Its part of the pictorial language that has grown up with Feynman diagrams – but are really just terms in a Dyson series:https://en.wikipedia.org/wiki/Dyson_seriesThanksBill
It was a very strong description. I was curious about a few pieces in your explanation. In, the paper, Direct Observation of Quantum Phonon Fluctuations in a One-Dimensional Bose Gas[1] it is shown that quantum fluctuations result from 'acoustic vibrations'. Lending strong correlation with the analog of the fluctuation in very still and ice cold states. It is then repeated in 2016 [2] as "we prove that quantum fluctuations stabilize the ultracold gas far beyond the instability threshold imposed by mean-field interactions.". With analog vibrations having a seemingly strong subadvity for quantum fluctuations and the experiment involving analog hawking radiation of black holes using "dumb holes"[3], we could say the quantum fluctuation is relative to the energy of the black hole. Quantum fluctuations are also observed through the lamb shift [4].[1] http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.108.225306[2] http://journals.aps.org/prx/abstract/10.1103/PhysRevX.6.041039[3] http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.85.4643[4]http://physics.aps.org/articles/v9/139
[MEDIA=youtube]X6HobTJ2jnk[/MEDIA]Minute 48 and 49.
[QUOTE="Collin237, post: 5623544, member: 577701"]Ideas about black holes, however, are argued on merely theoretical merit.”False. As an example, the "chirp" of GW150914 depends on the horizon radius of the merged black hole. If you don't have a horizon, or have it in another place, the chirp no longer matches the GR predictions.Now I will make a prediction – when confronted with the evidence, your reaction will not be "I was wrong". It will be some argument that somehow this doesn't count.
Observed orbital motion around massive objects that appear as black holes of some sort kind of suggest black holes as predicted by GR.
[QUOTE="Drakkith, post: 5623454, member: 272035"]The same is true of stars, nebulas, exoplanets, galaxies, etc.”Those are taken seriously as objects. If a physicist has a new idea about them, it's explored, refined, etc., and goes through an ordinary Kuhnian trial. Ideas about black holes, however, are argued on merely theoretical merit.[QUOTE="Nugatory, post: 5623426, member: 382138"]Only if you're willing to put Hawking's derivation (the real thing,”What else would you call it? Already in just the abstract, the remarks about "entropy" are a huge red flag.
I usually steer away from such insights because to me the specialists writing them assume too much prior knowledge in both the content and notation department.This insight I got a lot out of, thank you.
[QUOTE="Collin237, post: 5623403, member: 577701"]All we know about black holes is that there are things that match the description as deduced from billions of miles away.”The same is true of stars, nebulas, exoplanets, galaxies, etc.
[QUOTE="Collin237, post: 5623403, member: 577701"]But Hawking radiation itself is a myth. A”Only if you're willing to put Hawking's derivation (the real thing, not the heuristic upon which Professor Neumaier is heaping scorn) in the category of "myth".
But Hawking radiation itself is a myth. All we know about black holes is that there are things that match the description as deduced from billions of miles away.
Nice article!
Nice Insight Arnold!
The article is a slightly polished version of my much-liked posting at https://www.physicsforums.com/posts/5453034