Insights The Vacuum Fluctuation Myth - Comments

[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.

 

[vanhees71]

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.

 

[A. Neumaier]

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.

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.

I had citedhttps://www.physicsforums.com/posts/5642564/bookmark 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, let alone 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.

 

[ftr]

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.

 

[A. Neumaier]

Even wikipedia describes it as a change, though in a completely unscientific manner (not surprisingly, since it also promotes lots of other nonsense about virtual particles):

I just noticed that the German version of wikipedia is far better than the English version on vacuum fluctuations, virtual particles, and the like!

 

[vanhees71]

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.

Sure, that's how observables are defined, at least for me as a proponent of the minimal interpretation.

 

[vanhees71]

I just noticed that the German version of wikipedia is far better than the English version on vacuum fluctuations, virtual particles, and the like!

Indeed, that's rare!

 

[Nugatory]

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.

Some of the frustration with this topic comes, I think, from that balance being different with lay audiences who will never do the calculations and with serious students. For the former, the heuristic means that the vacuum is full of particle-antiparticle pairs appearing and annihilating themselves, real objects that just happen to have a very short lifetime. It's fair to dismiss that as a "myth" and (as a volunteer mythbuster at PF) I'm comfortable assigning it a fairly high negative value.

 

[A. Neumaier]

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.

I am talking about what the quantum field theoretic textbooks call the vacuum. Mostly (not to complicate things) the vacuum according to the standard model, i.e., in a flat space-time, with a nonaccelerated observer. Both the free vacuum (in a Fock space) and the interacting vacuum (in a renormalized theory such as the standard model).

The real world vacuum must also account for gravitation, and for the lack of consensus about quantum gravity it is difficult to say much definite about that. But some things seem to be firmly established in (semiclassical) quantum gravity (in curved space-time, but without dynamical quantization of the gravitational field), and are consistent with what I am saying:

In quantum gravity, the notion of vacuum (and hence of particles) is an observer-dependent notion. In a generally covariant description it is impossible to formulate the particle concept; only fields make sense. Particles appear only when modeled in the rest frame of a particle detector. Thus it seems that it is the particle detector (commonly called the observer) that turns fields into particles (by creating spots on a screen, peaks of a current, clicks in a counter, tracks in a fluid or a wire chamber). Characteristic for this is the Unruh effect: What appears to an observer A at rest (in its frame) as a vacuum [the observer excepted - which is acceptable in a cosmological setting] appears to a uniformly accelerated observer B as a thermal bath of particles. The basic reason is that in a system that appears as a vacuum to the observer at rest, the accelerated observer B is surrounded in its own rest frame not by a vacuum but by a strong gravitational field (created by the inertial forces) that excites the detector. Thus general covariance implies the observer dependence of the notion of vacuum. (Something similar happens in the Hawking effect for black holes.)

If one tries to interpret the Unruh effect in terms of a seething vacuum it is paradoxical that the first observer sees and observed nothing of this seething, while the accelerated observer observes it. It is far more natural to explain everything in terms of the inertial forces, where it is clear that not the vacuum seen by A but the uniform acceleration (which requires energy input) creates the conditions leading to the detector response.

In more technical terms: It is well-known that in curved space-time there is no generally covariant vacuum state, and that its place is taken by the class of Hadamard states, which transform into each other under arbitrary diffeomorphisms (coordinate transformations). These Hadamard states are seen by each observer (defined by a world line) at a particular time (selecting a point $x$ in space-time) as an external (classical) gravitational field in the Minkowski space tangent to the space-time manifold at $x$. The observer interprets everything in terms of a traditional quantum field theory on this tangent space, where the typical scattering calculations for finding cross sections are performed.

In most Hadamard states, the resulting gravitational field is nonzero, hence the system is not in a vacuum state, no matter which observer interprets it. In some special Hadamard states there are a minority of very special observers (on a set of measure zero) who would see a true vacuum (like observer A in the above, standard description of the Unruh effect). These observers are related to each other by a Lorentz transformation, so that they agree on what happens within the effects known from special relativity. All other observers - the overwhelming majority - don't see these special Hadamard states as anything special but are (in their rest frame) immersed in a nonzero gravitational field.

 

[PeterDonis]

it is not about physics, but the use of English words.

Wouldn't the best way to deal with that be to taboo those English words and restate everything in terms of math? This thread seems to me to have way too many posts arguing about terminology instead of physics; as far as I can tell everyone agrees on the physics.

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

I would state this a bit differently: I would say that because English is vague, unlike math, there is no one "correct" use of English words to describe the physics. That's why, when we really have to be precise, we use math.

 

[A. Neumaier]

I would say that because English is vague, unlike math, there is no one "correct" use of English words to describe the physics. That's why, when we really have to be precise, we use math.

With some proper care, one can use the English language in an astonishingly precise way, and doing this is usually of much help. Mathematicians (like me) like to be very precise, not only in their formulas (where it is a must) but also in the informal language and imagery that goes with it.

This is why mathematicians never generate the same amount of public interest (precision is an antidote against http://sensationalism ) as physicists even when they try to be popular. It is also the ultimate reason why mathematics is far more precise than theoretical physics. However, there are parts of theoretical physics (such as classical Lagrangian and Hamiltonian mechanics or quantum optics proper) where the English language is used in a far less misleading way as it is done in the popular quantum myths.

 

[PeterDonis]

With some proper care, one can use the English language in an astonishingly precise way

I agree that this can be done, and inside a particular professional community, it is reasonable to expect it to be done. But PF is not such a community; there are people here from various professional communities, but there are also people here who are not math or science professionals at all. So at the very least, you are going to have people who are used to different usages of ordinary language to refer to precise concepts, and in many cases you will have people who don't know any of the precise professional technical terms in any field. In such a case I would argue that it is often better to just admit up front that ordinary language is inadequate and to make sure to be clear about what precise concepts you are referring to, expressed in mathematical terms.

 

[A. Neumaier]

in many cases you will have people who don't know any of the precise professional technical terms in any field. In such a case I would argue that it is often better to just admit up front that ordinary language is inadequate and to make sure to be clear about what precise concepts you are referring to, expressed in mathematical terms.

Wouldn't those you address in the first of the quoted sentence be lost when you do the second? One needs some mediation between the two, to make the shift from being used only to ordinary language to getting used to the math easier. Just because PF caters for different groups of people one also needs different ways of trying to say the same.

 

[ftr]

Thanks for the detailed answer, I was just wondering now that you mentioned quantum gravity, in string theory the landscape problem is interpreted as different universes. So can different universes have different vacua. Moreover, in LQG space itself is seen as fluctuating which I presume is the quantum analog of GR, or is that a myth also.

Edit: I guess you are not against the vacuum having intrinsically a constant scalar(or vector) field of sort.

 

[mfb]

An uncertainty in the position means that the position is not known exactly.

But that's what we were talking about all the time? "The state has an intrinsic position uncertainty", or more general "The state has an intrinsic position/momentum uncertainty". The position/momentum uncertainty is a property of the state.

 

[PeterDonis]

Wouldn't those you address in the first of the quoted sentence be lost when you do the second?

It's true that people who don't understand the professional jargon might not understand the math either, but that just means they are going to have to do more work themselves, to acquire the necessary background. Telling them to be sure to use a certain English word a certain way won't help, because they don't have access to the technical concept that it refers to. The only reason professionals can use English words to name certain technical concepts is that they already understand the technical concepts using math, so they can all agree on what a particular English word or expression means.

What I am saying is that if your goal is to make lay people, who don't understand the math, correctly understand physics when expressed in ordinary language instead of math, I'm not sure that goal is achievable. But if your goal is to make lay people, who don't understand the math, understand that they don't understand the physics, and shouldn't try to reason based on ordinary language descriptions that might not correctly express the physics, I think that's a more modest goal that might be achievable.

 

[RockyMarciano]

I get the impression that it is not only a semantic problem behind this, it is a conceptual divide. It hinges critically on whether one understands the meaning of the outcomes of EPR experiments and accepts what they imply or not.
Briefly, the outcomes of those experiments require anyone who understands them to abandon local realism. Traditionally this requirement used to be separated on a choice between giving up locality or giving up classic realism, but let's say everybody here accepts QFT and relativity(wich everybody should) so that leaves as only choice giving up classic realism.

In the context of this thread giving up classic realism is equivalent to disregard states as entities separated from their measurments. If one does this the alleged distiction between uncertainty in the state versus uncertainty in the measurement, and the rejection of the word fluctuation to refer to Heisenberg's uncertainty are not possible.
Fortunately most people in this thread seem to understand and assume the Bell theorem as per the emprical results of EPR experiments. I can understand that those who don't will have a hard time accepting or understanding what serious physicists mean when they refer to quantum fluctuations(basically refer to obeying the Heisenberg uncertainty in different contexts), because for them the statistical fluctuation from noncommuting relations refers only to measurements separated from states, they give wavefunctions an ontological existence that is classically separated from measurements.
I can see how a pure mathematician could disregerd experimental evidence though, I would not expect it from physicists.

 

[zonde]

I would state this a bit differently: I would say that because English is vague, unlike math, there is no one "correct" use of English words to describe the physics. That's why, when we really have to be precise, we use math.

Math is no replacement for English. These are two totally different things that have different functions.
One of the functions of ordinary language is to name things. Math has no such function.

Besides it's physicists themselves that have messed up English in physics. The usage of word "state" as statistical distribution is totally confusing not only for lay people but for physicists themselves. The word "state" has very important but different meaning as current physical configuration for some potentially changing situation. Historically it was state vector that was understood with the word "state" and there the correspondence is rather intuitive and clear.

 

[PeterDonis]

One of the functions of ordinary language is to name things. Math has no such function.

Really? What are mathematical symbols? They are names for things.

it's physicists themselves that have messed up English in physics.

This assumes that there is some one "right" way to describe physics in English (or any other ordinary language). There isn't. Ordinary language is based on ordinary experience, but physics is based on experiences that are not ordinary--if they were, we wouldn't need elaborate physical theories. The best we can do is to agree on some consistent terminology, at least in a particular field. But the terminology only helps if you understand the concepts it is referring to. And once you understand them, you understand that no ordinary language description is really the "right" one, because the concepts are not the ones that our ordinary language was built to express.

Historically it was state vector that was understood with the word "state"

Historically "state" has had a bunch of different meanings, depending on the theory. Picking one particular meaning from one particular formulation of one particular theory and saying that is the "right" one does not strike me as a fruitful way to proceed.

 

[vanhees71]

But that's what we were talking about all the time? "The state has an intrinsic position uncertainty", or more general "The state has an intrinsic position/momentum uncertainty". The position/momentum uncertainty is a property of the state.

Well, although I'm not a native English speaker I'd formulate it more precisely as: Any quantum state implies uncertainties of position and momentum. This in turn implies fluctuations in the sense of an ensemble interpretation of probabilities. How else would you define fluctuations?

You have the same notion also in classical statistical mechanics: A phase-space distribution function implies an uncertainty in energy or momentum and thus implies ("thermal") flucutations of these quantities.

 

[zonde]

Really? What are mathematical symbols? They are names for things.

There are symbols in physics theories that correspond to physically measurable things. Mathematical statements however do not depend on the correspondence we attach to mathematical objects. In that sense symbols that are used as placeholders for mathematical objects are not names for anything.

This assumes that there is some one "right" way to describe physics in English (or any other ordinary language). There isn't. Ordinary language is based on ordinary experience, but physics is based on experiences that are not ordinary--if they were, we wouldn't need elaborate physical theories. The best we can do is to agree on some consistent terminology, at least in a particular field. But the terminology only helps if you understand the concepts it is referring to. And once you understand them, you understand that no ordinary language description is really the "right" one, because the concepts are not the ones that our ordinary language was built to express.

No, my statement assumes that there is "wrong" way to describe physics in English. And don't forget that there is experimental side to physics. This side of physics needs ordinary language along with mathematical language.

Historically "state" has had a bunch of different meanings, depending on the theory. Picking one particular meaning from one particular formulation of one particular theory and saying that is the "right" one does not strike me as a fruitful way to proceed.

I am speaking about Quantum theory. And any physics theory has to establish correspondence with physical reality. So obviously physical reality needs description that is independent from particular physics theory. Statement that state vector (or density matrix) describes the state is such a correspondence rule IMO as "state" is primarily concept of physical reality and only secondarily concept of theory as much as theory corresponds to physical reality.

 

[vanhees71]

Math is no replacement for English. These are two totally different things that have different functions.
One of the functions of ordinary language is to name things. Math has no such function.

Besides it's physicists themselves that have messed up English in physics. The usage of word "state" as statistical distribution is totally confusing not only for lay people but for physicists themselves. The word "state" has very important but different meaning as current physical configuration for some potentially changing situation. Historically it was state vector that was understood with the word "state" and there the correspondence is rather intuitive and clear.

It's the other way around: Plane everyday languages (it's not restricted to English of course) are no replacement for math ;-).

The usage of the word "state" in the context of QT is not confusing but the essence of its content. A state is defined operationally as an equivalence class of prepartation procedures and the knowledge about the state implies the knowledge of probababilities (and only probabilities!) for outcomes of measurements, given the preparation of the measured system in this particular (pure or mixed) state.

I don't care about history when it comes to the scientific content of physics. The state never was understood as the state vector but as an equivalence class of state vectors, called rays. There are some textbooks that are imprecise with this, and that leads to a lot of confusion. The most general definition of a quantum state in the formalism is of course the Statistical Operator which includes both pure states (i.e., the Stat. Op. is a projector) and mixed states (describing the situation that one has only incomplete knowledge about the quantum state as is usually the case for macroscopic systems).

 

[zonde]

It's the other way around: Plane everyday languages (it's not restricted to English of course) are no replacement for math ;-).

Certainly. However math depends on ordinary language while ordinary language does not depend on math. ;)

I don't care about history when it comes to the scientific content of physics. The state never was understood as the state vector but as an equivalence class of state vectors, called rays.

Well, it seems you are right. Historically state was associated with energy states of electrons in atoms. At least it seems that way after glancing at Schrodinger's paper (1926).

 

[stevendaryl]

Well, although I'm not a native English speaker I'd formulate it more precisely as: Any quantum state implies uncertainties of position and momentum. This in turn implies fluctuations in the sense of an ensemble interpretation of probabilities. How else would you define fluctuations?

You have the same notion also in classical statistical mechanics: A phase-space distribution function implies an uncertainty in energy or momentum and thus implies ("thermal") flucutations of these quantities.

It's a little more subtle than that. If you have an ensemble of a million human beings, there will be a nonzero standard deviation for the height, but that doesn't imply that anybody's height is fluctuating. On the other hand, if the quantity \frac{d (height)}{dt} has a nonzero standard deviation, as well, then that would support the claim that heights are fluctuating.

 

[stevendaryl]

It's a little more subtle than that. If you have an ensemble of a million human beings, there will be a nonzero standard deviation for the height, but that doesn't imply that anybody's height is fluctuating. On the other hand, if the quantity \frac{d (height)}{dt} has a nonzero standard deviation, as well, then that would support the claim that heights are fluctuating.

In the quantum case, though, it seems interpretation-dependent. According to some interpretations, no physical variable has a value until it is measured, so the fact that \frac{dQ}{dt} has a nonzero standard deviation doesn't imply that Q is fluctuating, only that if you ever happen to measure \frac{dQ}{dt}, you will likely get something nonzero.

 

[atyy]

The other way to see the post is that it justifies the myth - one just has to accept the path integral picture, and an interpretation of the path integral picture. So the myth is not a myth, provided we accept that it describes the path integral picture and not the canonical picture. In other words, it is not a myth, provided we add the words "shouldn’t be taken too literally". Already, in Copenhagen, the wave function is not taken literally. So quantum mechanics is intrinsically mythical. There is nothing wrong with adding the path integral as metamyth.

 

[OCR]

This side of physics needs ordinary language along with mathematical language.

We used to think that if we knew one, we knew two, because one and one are two. We are finding that we must learn a great deal more about 'and'.

An interesting thread, please... carry on.

 

[stevendaryl]

The other way to see the post is that it justifies the myth - one just has to accept the path integral picture, and an interpretation of the path integral picture. So the myth is not a myth, provided we accept that it describes the path integral picture and not the canonical picture. In other words, it is not a myth, provided we add the words "shouldn’t be taken too literally". Already, in Copenhagen, the wave function is not taken literally. So quantum mechanics is intrinsically mythical. There is nothing wrong with adding the path integral as metamyth.

I guess there are quantum fundamentalists, who take it literally, and quantum Unitarians, who take it all metaphorically.

 

[A. Neumaier]

You have the same notion also in classical statistical mechanics: A phase-space distribution function implies an uncertainty in energy or momentum and thus implies ("thermal") fluctuations of these quantities.

Except that therrmal fluctuations in classical statistical mechanics are usually regarded (by invoking the ergodic hypothesis) as happening in time, Thus they are regarded as true fluctuations. While in quantum mechanics such a view is not really well-defined.

 

[A. Neumaier]

There are symbols in physics theories that correspond to physically measurable things. Mathematical statements however do not depend on the correspondence we attach to mathematical objects. In that sense symbols that are used as placeholders for mathematical objects are not names for anything.

Most definitions in mathematics define language naming things. The concept of group, of multiplication, of a field, a vector space, a vector, a set ..., the symbol + * / etc. All are creating descriptive language.

 

[PeterDonis]

Mathematical statements however do not depend on the correspondence we attach to mathematical objects.

Um, what? A mathematical symbol refers to a mathematical object. That's why we use it.

my statement assumes that there is "wrong" way to describe physics in English.

My response still applies with this interpretation.

don't forget that there is experimental side to physics. This side of physics needs ordinary language along with mathematical language.

Experimental apparatus can be described mathematically; in fact it has to be in order to compare experimental results with theoretical predictions. One does need a correspondence between mathematical symbols and actual objects in the laboratory (e.g., this 4-vector corresponds with this measuring device sitting in the lab).

any physics theory has to establish correspondence with physical reality.

No, any physics theory has to establish correspondence with the experimental evidence we use to test it. What, if any, correspondence it has with "physical reality" is a question of philosophy or metaphysics, not physics.