I Relation between quantum fluctuations and vacuum energy?

Demystifier

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Is it meant then that the value of the field at a fixed point is a distribution (in the sense that repeated measurements won't give the same result, however the expectation value, the mean of the distribution, will be a fixed value, i.e. 0) rather than a fixed value?

Is the point, that due to the Heisenberg uncertainty principle, one cannot precisely determine the value of the field at a given point and as such its standard deviation is non-zero. So by saying that a quantum field is fluctuating is simply a handwavy statement of the fact that its value (in vacuo) at a given point in space-time is not prescisely determined (i.e. it is not well-defined) and this results in a non-zero vacuum energy?!
Yes and yes.
 
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No. The fields always fluctuate - their correlation functions are nonzero. But the vacuum energy does not - its standard deviation is exactly zero.
Wait, I'm confused now. When I said this earlier you said I was incorrect. My understanding before was that the standard deviation of a quantum field is non-zero even in vacuum due to the Heisenberg uncertainty principle (since ##\sqrt{\langle\hat{\phi}(x)\rangle^{2}}\sqrt{\langle\pi_{\phi}(x)\rangle^{2}}≥\hbar/2##), and so the vacuum energy of the field is a constant which is due to these fluctuations.

There is no freedom here: We have to normal order and do all renormalizations (not only removing the infinite vacuum energy, but other infinities as well), in order that the Hamiltonian agrees (up to a factor ccc) with the time component of the 4-momentum in a representation of the Poincare group.
Yes, I understand that, but I was under the impression that the reason we can legitimately do this is because the theory is not sensitive to absolute energies and this allows us to carry out such renormalization procedures?!
 
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Yes and yes
Ok, it's reassuring to know that I've at least understood one thing correctly.
 

Haelfix

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No. Vacuum energy is not a physical term at all, only one of popular science, needed there to be able to speculate about fancy properties of virtual particles.
Hmm?
Vacuum energy is a well defined concept in GR. Consider for concreteness the theory of a single scalar field with potential V(phi). The action is given by
[tex]S= \int d^{4}x\sqrt{-g}\left ( \frac{1}{2}g^{uv}\partial _{u}\Phi \partial _{v}\Phi -V\left ( \Phi \right )\right ) [/tex]
The corresponding energy momentum tensor is computed as:
[tex]T_{uv}= \frac{1}{2}\partial _{u}\Phi \partial _{v}\Phi + \frac{1}{2}\left (g^{\alpha \beta }\partial _{\alpha }\Phi \partial _{\beta }\Phi \right ) g_{uv} - V\left(\Phi\right) g_{uv}[/tex]
The lowest energy density configuration if it exists is obtained when both the kinetic and gradient terms vanishes, which gives us our definition for vacuum energy and implies for this particular theory that:
[tex]T_{uv}^{vac}\equiv -\rho _{vac} g_{uv} =-V\left(\Phi_{0}\right) g_{uv} [/tex]
Where [itex]\Phi _{0}[/itex] is the value that minimizes the potential. Note that this is not necessarily zero. More generally you can argue by lorentz invariance that the form for vacuum energy is unique and fixed exactly as above.
Note that since the equations are sourced directly by the energy momentum tensor, you cannot simply subtract away dangerous (potentially infinite) renormalizations unlike the case for non gravitational theories.
 
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A. Neumaier

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Vacuum energy is a well defined concept in GR.
In classical general relativity only, and there it surely does not fluctuate!

But we were talking about the quantum case. There your action and your energy-momentum tensor are ill-defined, and it is not known how to renormalize them in arbitrary order perturbation theory. There is not even a well-defined notion of a vacuum state - its place is taken by the big family of Hadamard states. Thus how can there be a well-defined quantum vacuum energy?
The lowest energy density configuration
is frame dependent, hence has even classically no proper physical meaning. It depends on space-time position, hence is not the vacuum energy, but the vacuum energy density. which is even nonrelativistically a different concept.

It is not even clear that ##T_{00}## is bounded below, so that a lowest energy density configuration would exist. The two derivative terms are Lorentzian, hence have no reason all all to be nonnegative, and minimal when they vanish.
 
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Hmm?
Vacuum energy is a well defined concept in GR. Consider for concreteness the theory of a single scalar field with potential V(phi). The action is given by
[tex]S= \int d^{4}x\sqrt{-g}\left ( \frac{1}{2}g^{uv}\partial _{u}\Phi \partial _{v}\Phi -V\left ( \Phi \right )\right ) [/tex]
The corresponding energy momentum tensor is computed as:
[tex]T_{uv}= \frac{1}{2}\partial _{u}\Phi \partial _{v}\Phi + \frac{1}{2}\left (g^{\alpha \beta }\partial _{\alpha }\Phi \partial _{\beta }\Phi \right ) g_{uv} - V\left(\Phi\right) g_{uv}[/tex]
The lowest energy density configuration if it exists is obtained when both the kinetic and gradient terms vanishes, which gives us our definition for vacuum energy and implies for this particular theory that:
[tex]T_{uv}^{vac}\equiv -\rho _{vac} g_{uv} =-V\left(\Phi_{0}\right) g_{uv} [/tex]
Where [itex]\Phi _{0}[/itex] is the value that minimizes the potential. Note that this is not necessarily zero. More generally you can argue by lorentz invariance that the form for vacuum energy is unique and fixed exactly as above.
Note that since the equations are sourced directly by the energy momentum tensor, you cannot simply subtract away dangerous (potentially infinite) renormalizations unlike the case for non gravitational theories.
This is my understanding of things - at least for semi-classical applications of GR, in which the gravitational sector is classical GR and the matter sector is quantum.
 

Haelfix

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That's right, rho(vac) is an energy density here, and it's form is invariant up to the usual ambiguities with pseudotensors. I agree the quantum case is more complicated, going beyond tree level you will get oscillator mode contributions and you will have to regularize the problem, which is much more involved than in flat space. Still it can be done, at least as an effective theory. Regarding whether T00 is bounded from below, that's not a problem provided you specialize to cases where the metric is static, and so has a form of time translation invariance (or to some sort of adiabatic regime in cosmological settings), then it just becomes a requirement on the form of the potential such that it satisfies the right energy conditions (like the weak energy condition)
 

A. Neumaier

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rho(vac) is an energy density
and the energy itself, i.e., the Hamiltonian (in a fixed frame), is the integral of the energy density over the Cauchy surface corresponding to a fixed time in the fixed frame. This energy is as ill-defined in the naive quantum version as it is in flat space, with an infinite (i.e., physically meaningless) zero-point energy. Thus renormalization is necessary, and in canonical quantum gravity this forces the vacuum energy (in a fixed frame) to be zero.

It is very difficult to say how frame transformations (local diffeopmorphisms) relate the renormalized Hamiltonians of different frames. (This seems to be related to a Tomonaga-Schwinger dynamics with respect to multifingered time.) Thus it is very difficult to say what the notion of vacuum could possibly mean in quantum gravity.
 

Haelfix

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Thus renormalization is necessary, and in canonical quantum gravity this forces the vacuum energy (in a fixed frame) to be zero.
Yes, but that is somewhat of a technicality. The underlying dynamics are all nicely hidden within the Hamiltonian constraint within that formalism. From that one construct what you would look for, which would be the appropriate vacuum expectation value, which is of course badly divergent.
 

A. Neumaier

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which is of course badly divergent.
But this means physically meaningless - which is the whole point of the present discussion. No physically meaningful definition has ever be propoed in the quantum case, and it is unlikely that there will be one. One will ultimately be able to renormalize observable stuff such as scattering amplitudes, but not unobservable stuff such as vacuum energy.
 
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Is it purely due to the running of the coupling with the energy scale then? (I know a very small amount about the renormalization group, but not much)
Read my introduction to renormalisation:
https://www.physicsforums.com/insights/renormalisation-made-easy/

Its a very very brief, but as far as it goes, correct introduction.

Then read:
http://arxiv.org/pdf/hep-th/0212049.pdf

What you wrote is basically correct, but its a lot more sophisticated than that - its do do with renormalisation group flow. What fooled physicists for a long time was many of the constants that appear in QFT equations like charge, mass and (as you point out) the coupling constant depends on the cutoff.

But it cant be pushed too far because of whats called the Landau pole:
https://en.wikipedia.org/wiki/Landau_pole

In fact it's an indication that QED is sick, and indeed we know long before the Landau pole is reached another theory takes over - the electroweak theory. My understanding is that also has a Landau pole and exactly how that is handled my knowledge comes up short - maybe some of the more knowledgeable posters here can comment.

Thanks
Bill
 
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A. Neumaier

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whats called the Landau pole [...] In fact it's an indication that QED is sick,
The Landau pole is a perturbative phenomenon, and it is unknown whether it has any nonperturbative (i.e., physical) impact. It is speculation only (though widely thought to be credible) that this indicates a sickness of QED or electroweak theory. This is called the triviality problem - it is a completely open problem, even for ##\phi^4## theory! See here for my view of this problem.
 
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Read my introduction to renormalisation:
https://www.physicsforums.com/insights/renormalisation-made-easy/

Its a very very brief, but as far as it goes, correct introduction.

Then read:
http://arxiv.org/pdf/hep-th/0212049.pdf

What you wrote is basically correct, but its a lot more sophisticated than that - its do do with renormalisation group flow. What fooled physicists for a long time was many of the constants that appear in QFT equations like charge, mass and (as you point out) the coupling constant depends on the cutoff.

But it cant be pushed too far because of whats called the Landau pole:
https://en.wikipedia.org/wiki/Landau_pole

In fact it's an indication that QED is sick, and indeed we know long before the Landau pole is reached another theory takes over - the electroweak theory. My understanding is that also has a Ladau pole and exactly how that is handled my knowledge comes up short - maybe some of the more knowledgeable posters here can comment.

Thanks
Bill
Thanks for the links and info! So is the Landau pole of a theory the limit (as the energy scale increases) at which the coupling becomes infinite and thus the theory breaks down? If a theory doesn't have a Landau pole then as the energy scale increases the coupling becomes weaker and weaker, and in this limit the theory is said to be asymptotically free, and is thus stable at all energy scales (as I understand it QCD is such an example)?!

In terms of fluctuations of quantum fields, is any of what I wrote in post #27 correct? I have to admit I still have some confusion over it as there have been I few slightly conflicting posts (possibly some details lost in translation).
 
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Thanks for the links and info! So is the Landau pole of a theory the limit (as the energy scale increases) at which the coupling becomes infinite and thus the theory breaks down?
Well the usual perturbation methods break down - what that means, from my understanding, is still unclear.

In terms of fluctuations of quantum fields, is any of what I wrote in post #27 correct? I have to admit I still have some confusion over it as there have been I few slightly conflicting posts (possibly some details lost in translation).
Lets go back to basics. QM doesn't say whats going on when not observed so what does that imply about quantum fluctuations when not observed?

In QFT things are so far removed from usual experience all sorts of things are grasped at to have some of intuition, but its not really true.

The energy of a quantum field actually turns out to be infinite which is silly. Even sillier is, since the energy zero point is arbitrary, subtracting infinity from it and setting that to zero. The way its handled lies in something called normal ordering which you can look into.

It's important to understand that one can do QED without these infinities:
https://www.amazon.com/dp/0486492737/?tag=pfamazon01-20

It is pretty clear these days its an artifact of the mathematical methods used.

Thanks
Bill
 
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Lets go back to basics. QM doesn't say whats going on when not observed so what does that imply about quantum fluctuations when not observed?
So is the notion of a quantum fluctuation (in the ground state) deeply connected to the Heisenberg uncertainty principle, in that the values of the field, ##\hat{\phi}(x)## and its conjugate momentum, ##\hat{\pi}(x)## are not well-defined, and so we analogise this to our classical way of thinking in describing this as the quantum field fluctuating?!
I get that a quantum field fluctuates in the sense that obeys an equation of motion (some sort of wave equation) and hence quantised fluctuations propagating through the field correspond to quanta of that field - measurements of such quantisation being quantified by correlation functions of the field at different points. But what about in vacuum state of the quantum field? Is it simply that the field is in a fixed state, but that the uncertainties of its value (due to the uncertainty principle), we visualise this as the field itself fluctuating (so-called vacuum fluctuations of the field)?!
 
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So is the notion of a quantum fluctuation (in the ground state) deeply connected to the Heisenberg uncertainty principle, in that the values of the field, ##\hat{\phi}(x)## and its conjugate momentum, ##\hat{\pi}(x)## are not well-defined, and so we analogise this to our classical way of thinking in describing this as the quantum field fluctuating?!
A quantum field is an operator at a certain point in space time. Do position operators fluctuate, do momentum operators fluctuate? Are operators not well defined? Think about it. What fluctuates with quantum operators? Hint - its not called an observable for nothing.

Now it turns out if you chug through the math (do Fourier transforms and stuff) its exactly the same as one of the equivalent ways of doing QM:
http://www.colorado.edu/physics/phys5260/phys5260_sp16/lectureNotes/NineFormulations.pdf

See the second quastisation formulation. That's how particles enter into it but now the number of particles is not fixed.

I think you will benefit from studying the following book where a lot of these issues are examined in detail:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

Thanks
Bill
 
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A. Neumaier

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Is it simply that the field is in a fixed state, but that the uncertainties of its value (due to the uncertainty principle), we visualise this as the field itself fluctuating (so-called vacuum fluctuations of the field)?
Yes. If the field is in the vacuum state, this is the complete, objective description of the field, and nothing at all goes on. Otherwise it is in a state with a (fixed or indefinite) particle content, only then something can happen. Again, the state describes everything that can objectively be said about the system.

Whatever is visualized in quantum field theory is added heuristics to aid human limitations in working with the formal content. Don't put too much into the visualization! It is just the way physicists familiarize themselves (and different physicists do it differently) with the formal machinery. To do anything in quantum field theory one needs to work with the latter, and the intuitive picture may be a help (in the earliest stages) or an obstacle (in an intermediate stage) for using this machinery. Ultimately the intuitive picture no longer matters - then one can say one has understood the formalism.
 
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Whatever is visualized in quantum field theory is added heuristics to aid human limitations in working with the formal content. Don't put too much into the visualization! It is just the way physicists familiarize themselves (and different physicists do it differently) with the formal machinery. To do anything in quantum field theory one needs to work with the latter, and the intuitive picture may be a help (in the earliest stages) or an obstacle (in an intermediate stage) for using this machinery. Ultimately the intuitive picture no longer matters - then one can say one has understood the formalism.
:biggrin::biggrin::biggrin::biggrin::biggrin::biggrin::biggrin::biggrin::biggrin::biggrin:

Thanks
Bill
 
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A quantum field is an operator at a certain point in space time. Do position operators fluctuate, do momentum operators fluctuate? Are operators not well defined? Think about it. What fluctuates with quantum operators? Hint - its not called an observable for nothing.

Now it turns out if you chug through the math (do Fourier transforms and stuff) its exactly the same as one of the equivalent ways of doing QM:
http://www.colorado.edu/physics/phys5260/phys5260_sp16/lectureNotes/NineFormulations.pdf

See the second quastisation formulation. That's how particles enter into it but now the number of particles is not fixed.

I think you will benefit from studying the following book where a lot of these issues are examined in detail:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

Thanks
Bill
I get that the field operators themselves do not fluctuate, but what about their vacuum expectation values, won't they in general be of the form ##\langle\hat{\phi}(x)\rangle =f(x)##?!

I've been reading through Quantum Field Theory for the gifted amateur and I haven't been able to find anything on quantum fluctuations yet (I'm up to page 140 so maybe I haven't gotten far enough in to the text yet).
I get that in the context of QFT (where special relativity applies) that the vacuum energy contribution is physically meaningless since only energy differences are physically observable, hence we can legitimately apply a normal ordering procedure to obtain a renormalizable physical theory. At least I think that's right?!

My original confusion over the subject arose from reading a few texts on trying to solve the cosmological constant problem. Many of the author's note that vacuum energy is not necessarily ignorable in the realm of general relativity. If we assume that the gravity sector is classical, i.e. described by GR and the mate sector is quantum, i.e. described by QFT, then if the vacuum energy is physical it should be a source of curvature. The problem is, they attribute the source of this vacuum energy as being from the "quantum fluctuations in the matter fields" , which I'm assuming they mean fluctuations in their vacuum expectation values (since the source is from the vev of their energy-momentum tensor). This is primarily what has caused confusion for me.
 
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I get that the field operators themselves do not fluctuate, but what about their vacuum expectation values, won't they in general be of the form ##\langle\hat{\phi}(x)\rangle =f(x)##?!
Expectations of what? What needs to be done in QM before you get an outcome? I basically told you the answer in a previous post. I can easily tell you, and you will likely kick yourself if I did, but its a point so basic, and so often ignored in these discussions you need to see if for yourself. Its one of those things Dr Neumaier mentioned as part of its heuristics - but it doesn't make the statement the vacuum is a broiling sea with constantly changing energy from the Heisenberg uncertainty principle as per the link below correct:
https://en.wikipedia.org/wiki/Quantum_fluctuation

It a common, no very common, heuristic but you should immediately see its problem - it goes right to the foundations of QM.

Its even used, incorrectly, to explain spontaneous emission:
http://www.calphysics.org/articles/Fain1982.pdf

It is wrong - the real answer is its because an electron is not an isolated system and is entangled with the vacuum so is not in a stationary state. But vacuum fluctuations is very commonly used as a heuristic such as in the above. It's still wrong and I am hoping you can see why it wrong. Then you can see why using it as a heuristic usually helps, but its crucial to understand its a heuristic - that it most definitely is NOT what is going on.

At least I think that's right?!
Basically - yes. That book explains normal ordering in detail.

My original confusion over the subject arose from reading a few texts on trying to solve the cosmological constant problem. Many of the author's note that vacuum energy is not necessarily ignorable in the realm of general relativity. If we assume that the gravity sector is classical, i.e. described by GR and the mate sector is quantum, i.e. described by QFT, then if the vacuum energy is physical it should be a source of curvature. The problem is, they attribute the source of this vacuum energy as being from the "quantum fluctuations in the matter fields" , which I'm assuming they mean fluctuations in their vacuum expectation values (since the source is from the vev of their energy-momentum tensor). This is primarily what has caused confusion for me.
That is not the source of energy in QFT. The ground state has infinite energy which is obviously an absurdity. A more careful look has it as zero.

Thanks
Bill
 
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vanhees71

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Although A. Neumair states otherwise, I still think that indeed there is no way to fix the absolute value of energy in special relativity. You have to define the Hamiltonian, and already for free fields you get a divergent result, but the trouble is due to the problem of multiplying field operators at the same space-time point, because the canonical commutators shows that they are distribution valued operators. For a field component ##\hat{\phi}## and its canonical conjugate momentum ##\hat{\Pi}## you have the canonical equal-time commutator
$$[\hat{\phi}(t,\vec{x}),\hat{\Pi}(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}).$$
Using the mode decomposition of free fields in terms of annihilation and creation operators show, however, that you can introduce the norma-ordering procedure to define a finite Hamiltonian up to an additive c-number constant, but this additive constant is irrelevant for the dynamics and thus can be neglected. Then you define the absolute value of the energy of the vacuum state, which is the ground state of the free-field system, to 0.
 

A. Neumaier

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I still think that indeed there is no way to fix the absolute value of energy in special relativity. You have to define the Hamiltonian, and already for free fields you get a divergent result
The infinities and the resulting ambiguity appear only in a naive Lagrangian approach to field theory, where already the Lagrangian is ill-defined, i.e., mathematically meaningless.

In axiomatic quantum field theory, which is consistent with all our knowledge about quantum fields (though interacting models in 4D haven't yet been constructed rigorously), a field theory is defined by a family of vacuum N-point functions satisfying certain axioms, among them translation invariance. Given these N-point functions there is a canonical way to construct a Hilbert space with a representation of the Poincare group, satisfying the traditional commutation relations. It is these commutation relations that do not allow one to shift the Hamiltonian, defined as the generator of the time translations.

The vacuum is defined as a translation invariant state (not only up to phase), hence the Hamiltonian has the eigenvalue zero in this state. For a free field, the Hamiltonian turns out uniquely to be ##H=\int Dp a^*(p)pa(p)##, where ##Dp## is the invariant measure integrating over momenta on the appropriate mass shell. This is the unique manner of proceeding in a manifestly covariant way; there is no freedom except for choosing mass and spin. Nowhere is an infinity or a vacuum energy encountered, no field operators are multiplied at the same position. The canonical commutation relations follow from the representation theory of the Poincare group and the connection between spin and statistics. The details are in Weinberg's QFT book, surely a standard work. Causal perturbation theory extends this perturbatively to the interacting case. Again, nowhere is an infinity or a vacuum energy encountered, no field operators are multiplied at the same position.

If you introduce a nontrivial zero-point energy, Lorentz covariance implies in another frame nontrivial zero-point momenta. But there is no covariant way to do this consistently., quite apart from the question what such a zero-point momentum should mean. The vacuum doesn't move! This shows that in a relativistic context the zero-point energy is a spurious concept.
 
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Expectations of what? What needs to be done in QM before you get an outcome? I basically told you the answer in a previous post. I can easily tell you, and you will likely kick yourself if I did, but its a point so basic, and so often ignored in these discussions you need to see if for yourself. Its one of those things Dr Neumaier mentioned as part of its heuristics - but it doesn't make the statement the vacuum is a broiling sea with constantly changing energy from the Heisenberg uncertainty principle as per the link below correct:
https://en.wikipedia.org/wiki/Quantum_fluctuation
Is the point that you have to interact with the system in order to make a measurement of it therefore entangling the state of the system you are measuring with whatever measurement "probe" you are using?

t a common, no very common, heuristic but you should immediately see its problem - it goes right to the foundations of QM.

Its even used, incorrectly, to explain spontaneous emission:
http://www.calphysics.org/articles/Fain1982.pdf

It is wrong - the real answer is its because an electron is not an isolated system and is entangled with the vacuum so is not in a stationary state. But vacuum fluctuations is very commonly used as a heuristic such as in the above. It's still wrong and I am hoping you can see why it wrong. Then you can see why using it as a heuristic usually helps, but its crucial to understand its a heuristic - that it most definitely is NOT what is going on.
I wish lecturers would be more upfront about this - at least in my experience, not nearly enough emphasis is made that this notion of quantum fluctuations is purely a heuristic aid to understand calculations.

Is it simply that higher-order corrections in perturbation theory, to correlation functions such as ##\langle 0\lvert\hat{\phi}(x)\hat{\phi}(y)\rvert 0\rangle##, are referred to as "quantum fluctuations" since they are represented by loops in the corresponding Feynman diagrams and so are heuristically thought of as virtual particles being created and annihilated in vacuum?
 

A. Neumaier

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Is it simply that higher-order corrections in perturbation theory, to correlation functions such as ##\langle 0\lvert\hat{\phi}(x)\hat{\phi}(y)\rvert 0\rangle##, are referred to as "quantum fluctuations" since they are represented by loops in the corresponding Feynman diagrams and so are heuristically thought of as virtual particles being created and annihilated in vacuum?
Conventional Feynman diagrams represent time-ordered expectation values, not Wightman correlation functions. (The latter need the CTP formalism.) The term fluctuations derives from the informal generalization of the fact that fluctuating time series also have correlation functions. Identifying these fluctuations with virtual particles popping in and out of existence for a short time is additional but unrelated imagery for these fluctuations, as described here.

If you really want to understand quantum physics, concentrate on the formulas, and view the talk about it only as a very loose and fallible guide.
 
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Is the point that you have to interact with the system in order to make a measurement of it therefore entangling the state of the system you are measuring with whatever measurement "probe" you are using?
Bingo.

QM is a theory about observations. Whats going on when not observed it says nothing about. The uncertainty relations only applies to observations. Applying it to the vacuum ground state means if you observe it then the results you get follows from the uncertainty relations - but only if you observe it. If you don't it says a big fat nothing. The vacuum state isn't in a constant fluctuations etc etc.

What going on and why its a valuable heuristic to look at it that way is from the Von Neumann regress where collapse can be basically placed anywhere so if you think of it at the vacuum then generally no harm is done and it a good way of getting an intuitive grip. But that is NOT what is happening.

I wish lecturers would be more upfront about this - at least in my experience, not nearly enough emphasis is made that this notion of quantum fluctuations is purely a heuristic aid to understand calculations.
Don't worry about it. It happens in physics all the time. You start out with stuff of dubious validity that gets corrected later, either explicitly or you are supposed to cotton onto it yourself. But what you have done is developed intuition which actually is more important. The only issue is not realizing whats going on and even then it generally causes issues only when discussing foundational issues rather than actually solving problems. Guess what most discuss here :smile::smile::smile::smile::smile:.

The other group it causes issues with is those that actually think - so give yourself a pat on the back. I first came across this when studying how transistors work. They work via holes which was explained to be the absence of electrons. It was a big fact crock of the proverbial. It annoyed the bejesus out of me - it was so obviously wrong. Then as a footnote in a textbook I read the alert reader will recognize it as nonsense but rest assured a QM analysis shows quasi particles exist that act like that.

Thanks
Bill
 
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