- #36
A. Neumaier
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No. The normal ordering does not eliminate the 2-particle creation term. (You must have been thinking of the nonrelativistic case!)Demystifier said:It also annihilates the vacuum.
No. The normal ordering does not eliminate the 2-particle creation term. (You must have been thinking of the nonrelativistic case!)Demystifier said:It also annihilates the vacuum.
A. Neumaier said:As mentioned by Witten on p.12, field creation and annihilation operators are not unitary operators, hence not physically realizable in time. So there is no interpretation problem.
The spectral theorem tells us that every self-adjoint operator can be constructed from projection operators (which are themselves self-adjoint) and every local operator can be constructed from local projection operators. So I would regard the application of local observables to be physically realizable, in principle.rubi said:The Reeh-Schlieder theorem says that you can reach all states by repeated application of local operators to the vacuum. However, it doesn't say anything about locality, because there is no physical process that is modeled by an application of a local operator to the state. All physical processes are modeled by unitary evolution or projection and both these operations respect locality. The Reeh-Schlieder theorem is just a mathematical fact about the cyclicity of the vacuum state with respect to local algebras.
But this is still not yet unitary!DrDu said:I thought Reh-Schlieder considers the algebra of the observables. Anyhow this should not make a difference as it should be possible to always find a hermitian operator which gives the same state when applied to the vacuum as an operator formed from only creation operators. E. g. use ##a+a^\dagger## instead of ##a^\dagger##.
True, but the unitary operators formed from local observables are also elements of the local algebra, so the argument should hold also for the unitaries. A simple example is again ##\exp(a+a^\dagger) = c \exp(a^\dagger)\exp(a)##, so that ##\exp(a+a^\dagger)## is up to a constant equivalent to ##\exp(a^\dagger)## when applied to the vacuum.A. Neumaier said:But this is still not yet unitary!
But it no longer generates the same state!DrDu said:True, but the unitary operators formed from local observables are also elements of the local algebra, so the argument should hold also for the unitaries. A simple example is again ##\exp(a+a^\dagger) = c \exp(a^\dagger)\exp(a)##, so that ##\exp(a+a^\dagger)## is up to a constant equivalent to ##\exp(a^\dagger)## when applied to the vacuum.
You are right. The irony is that I have shown it by myself a long time ago in https://arxiv.org/abs/hep-th/0202204 (e.g. Eqs. (131) and (204)) but in the meantime forgot it. Silly me!A. Neumaier said:No. The normal ordering does not eliminate the 2-particle creation term. (You must have been thinking of the nonrelativistic case!)
Demystifier said:That's very illuminating, so let us continue in the same spirit.
How about the charge current operator ##J^{\mu}(x)##? It also annihilates the vacuum. Would you say that ##J^{\mu}(x)## is also nonlocal?
Thank you again for the clarifications! By the same spirit, I meant discussion at the level which can be understood by a physicist who (like me) is not very fluent in axiomatic QFT.samalkhaiat said:You must be joking! I don’t see how this is in the same spirit as #34?
1) In #34 I was telling you about the separating property of the vacuum which is a direct consequence of the Reeh-Schlieder theorem, that is any (smeared) local operator, [itex]J \in \mathcal{A}(\mathcal{O})[/itex], annihilating the vacuum is itself vanishing, i.e., [itex]J|\Omega\rangle = 0, \ \Rightarrow \ J = 0[/itex]
2) Symmetry currents do not annihilate the vacuum state, their integrated charges do if the symmetry is not spontaneously broken: [tex]Q|\Omega \rangle = \int d^{3}x \ J^{0}(x) |\Omega \rangle = 0.[/tex] Given this together with above separating property of the vacuum S. Coleman proved his famous theorem: Symmetry of the vacuum is the symmetry of the world : [itex]Q|\Omega \rangle = 0, \ \Rightarrow \ \ \partial_{\mu}J^{\mu}|\Omega\rangle = 0[/itex]. Since, [itex]\partial_{\mu}J^{\mu}[/itex] is a local operator, then by the separating property, we obtain the conservation law [itex]\partial_{\mu}J^{\mu}=0[/itex].
Demystifier said:It looks as if the Reeh-Schlieder theorem depends on special properties of the vacuum not shared by other states. But Haag, in his book "Local Quantum Physics" (2nd edition), at page 102 says:
(ii) Obviously in the theorem the vacuum may be replaced by any vector with
bounded energy.
Can you clarify that? Which aspects of Reeh-Schlieder theorem do and which do not depend on special properties of the vacuum?
samalkhaiat said:3) the set [itex]\{ \mathcal{A}|\Psi \rangle \}[/itex] is dense in [itex]\mathcal{H}[/itex].
Well, in one of the posts above you said that "symmetry of the vacuum is the symmetry of the world". Obviously, in this statement "vacuum" cannot be replaced by any state. And yet, you related this statement to the Reeh-Schlieder theorem, in which vacuum can be replaced by "any" state (satisfying some general conditions). I was hoping for some additional physical insights on that.samalkhaiat said:So, I don’t really know what exactly you meant by “aspects” of the theorem.
No, see #18 for the definition of [itex]\mathcal{A}[/itex] and [itex]\mathcal{A}(\mathcal{O})[/itex].DrDu said:Isn't this already the result one wants to proove?
Who said “any”? “Any” state is not Poincare’ invariant, “any” state is not an eignstate of [itex]P_{\mu}[/itex], “any” state is not that which makes [itex]\mathcal{H} = \overline{\mathcal{A}| \Omega \rangle}[/itex] and [itex]\{0\}[/itex] the only invariant representation sub-spaces of the field algebra.Demystifier said:Well, in one of the posts above you said that "symmetry of the vacuum is the symmetry of the world". Obviously, in this statement "vacuum" cannot be replaced by any state. And yet, you related this statement to the Reeh-Schlieder theorem, in which vacuum can be replaced by "any" state (satisfying some general conditions). I was hoping for some additional physical insights on that.
Well, Haag said "any state with bounded energy". Any eigenstate of [itex]P_{\mu}[/itex] with a finite eigenvalue [itex]p_{\mu}[/itex] has bounded energy, right? But such states are not Poincare invariant (unless [itex]p_{\mu}=0[/itex]). Do I miss something?samalkhaiat said:Who said “any”? “Any” state is not Poincare’ invariant, “any” state is not an eignstate of [itex]P_{\mu}[/itex], “any” state is not that which makes [itex]\mathcal{H} = \overline{\mathcal{A}| \Omega \rangle}[/itex] and [itex]\{0\}[/itex] the only invariant representation sub-spaces of the field algebra.
Yes and it is very trivial. I have already told you that shifting the energy by a constant does not change any thing. So, if you have the eignvalue equations [itex]\left( M_{\mu\nu}, P_{\mu}\right)|\Omega \rangle = (c_{\mu\nu},q_{\mu})| \Omega \rangle[/itex], you can always redefine the Poincare’ generators [itex]\left( M_{\mu\nu},P_{\mu}\right) \to \left( M_{\mu\nu}+ c_{\mu\nu},P_{\mu}+q_{\mu}\right)[/itex] to obtain [itex]\left( M_{\mu\nu},P_{\mu}\right)| \Omega \rangle = 0[/itex].Demystifier said:Do I miss something?
It's the other way around, and this argument can be found in standard textbooks like Peskin&Schröder: Because time evolution with using ##\sqrt{\hat{\vec{p}}^2+m^2}## as an Hamiltonian in a putative 1st-quantization formulation of relativistic QT (which I call relativistic QM) leads to non-locality and breaks causality even for free fields, one concludes that one has to include the negative-frequency modes into the came, and then the observation of a stable world, i.e., the boundedness of the Hamiltonian of particles from below, forces us to use the 2nd-quantization formulation, i.e., QFT, which I'd call the only physically sensible relativistic QT we know of. It also allows for microcausality and validity of the linked-cluster theorem for the S-matrix, which clearly shows that interactions in QFT are indeed described as local interactions. There are no "spooky actions at a distance" as Einstein thought in the modern formulation of QFT, aka the "Standard Model".Peter Morgan said:What I think is curious about the quantized KG field, and might turn at least some heads that might not be turned by any of the above discussion, is that the real part of its 2-point Wightman function is the inverse Fourier transform of ##\left(\sqrt{\vec k\cdot\vec k+m^2}\right)^{-1}##, where the operator ##\sqrt{-\vec\partial\cdot\vec\partial+m^2}## is said to be "anti-local" by mathematicans (see my Physics Letters A 338 (2005) 8–12, arXiv:quant-ph/0411156, and, much more definitively, I.E. Segal, R.W. Goodman, "Anti-locality of certain Lorentz-invariant operators", http://www.jstor.org/stable/24901461. This last might possibly be the reference for some physicists to be confronted with, because its conclusions are quite similar to Hegerfeldt's conclusions.)
At the end of the day, I think all the nonlocality can be attributed to boundary and initial conditions, which, not being dynamical, for some people makes it not nonlocality. After dark, however, this somewhat pushes us towards the introduction of some form of superdeterminism, so perhaps it's just that you take your choice of poison.
String theorists would disagree.vanhees71 said:QFT, which I'd call the only physically sensible relativistic QT we know of
The presence/absence of spooky action at a distance has nothing to do with the fact that QFT interactions are local.vanhees71 said:It also allows for microcausality and validity of the linked-cluster theorem for the S-matrix, which clearly shows that interactions in QFT are indeed described as local interactions. There are no "spooky actions at a distance" as Einstein thought in the modern formulation of QFT, aka the "Standard Model".
Anything published in a physics journal is - physics.vanhees71 said:Since when is string theory physics? SCNR.
Any potential displays spooky acts at a distance!Demystifier said:nonlocal Coulomb-like potentials.
Which ontic/hidden-variable interpretation of which local relativistic QFT do you have in mind?Demystifier said:If one accepts an ontic/hidden-variable interpretation of QT, then the spooky action at a distance is present even in local QFT.
I agree with all of this, which is an entirely consistent way of discussing QFT, but it's a perspective that I consider to be laden with conventions. What might be called the "Einstein conventions" are also entirely consistent, and we can rigorously transform from one to the other (arguably this is what is done in my arXiv:1709.06711 for the free EM, Dirac, and complex KG quantum and random fields, which is being not discussed here on PF; I'll propose that the math of the exact transformations there implicitly defines what the Einstein conventions might be), but within the Einstein conventions there is a precise kind of Lorentz invariant nonlocality and other properties are transformed (including that the positivity of the quantum Hamiltonian operator becomes the positivity of the Hamiltonian function). The conventions you are pressing for, almost insisting upon, which might be crudely stated as the Correspondence Principle and all its consequences, have been supremely successful for the last 90 years, but I suggest that a significant part of the progress in our understanding of and in our ability to engineer using quantum physics over the last 30 years, say, has been through considering alternative conventions, in some of which the effective nonlocality of a state can be considered something of a resource.vanhees71 said:It's the other way around, and this argument can be found in standard textbooks like Peskin&Schröder: Because time evolution with using ##\sqrt{\hat{\vec{p}}^2+m^2}## as an Hamiltonian in a putative 1st-quantization formulation of relativistic QT (which I call relativistic QM) leads to non-locality and breaks causality even for free fields, one concludes that one has to include the negative-frequency modes into the came, and then the observation of a stable world, i.e., the boundedness of the Hamiltonian of particles from below, forces us to use the 2nd-quantization formulation, i.e., QFT, which I'd call the only physically sensible relativistic QT we know of. It also allows for microcausality and validity of the linked-cluster theorem for the S-matrix, which clearly shows that interactions in QFT are indeed described as local interactions. There are no "spooky actions at a distance" as Einstein thought in the modern formulation of QFT, aka the "Standard Model".
There's of course the usual caveat that there's no mathematically rigorous proof for the existence of interacting QFTs in (1+3)-space-time dimensions, and purists might say that we cannot be sure that QFT really works as expected and suggested by renormalized perturbative QFT.
What is ny potential?A. Neumaier said:A ny potential acts at spooky distance!
E.g. Bohmian interpretation of relativistic QFT. But that's not the topic of this thread, I already said a lot about those things in other threads.A. Neumaier said:Which ontic/hidden-variable interpretation of which local relativistic QFT do you have in mind?
It is important where the word "spooky" is put. The expression is "spooky action at a distance", not the "action at spooky distance". The expression is used by Einstein to describe features related to the problem of quantum measurement, wave-function collapse, (in)completeness of QM and such. He did not use this expression to describe Newtonian mechanics.A. Neumaier said:A ny potential acts at spooky distance!
But you could at least link to key posts in these threads.Demystifier said:What is ny potential? E.g. Bohmian interpretation of relativistic QFT. But that's not the topic of this thread, I already said a lot about those things in other threads.
But Newton himself was already dissatisfied with action at a distance, and general relativity has eliminated this spooky feature.Demystifier said:He did not use this expression to describe Newtonian mechanics.
Actually it is Newtonian gravity that has action at a distance (whether one finds it spooky or not). Quantum mechanics has no action at a distance (but I agree with Einstein that it is spooky).Demystifier said:It is important where the word "spooky" is put. The expression is "spooky action at a distance", not the "action at spooky distance". The expression is used by Einstein to describe features related to the problem of quantum measurement, wave-function collapse, (in)completeness of QM and such. He did not use this expression to describe Newtonian mechanics.
Are you saying that relativistic QFT cannot explain the experiments that show violation of Bell inequalities?A. Neumaier said:But relativistic QFT has eliminated again the action at a distance. It uses Born's rule only for asymptotic scattering results in the rest frame of the scattering event, not for the interpretation of arbitrary observables. Indeed, observables defined by smeared relativistic fields do not have a meaningful operational Born interpretation.
Of course.martinbn said:Actually it is Newtonian gravity that has action at a distance (whether one finds it spooky or not).
Quantum mechanics has correlations at a distance. Would you agree that those correlations are spooky?martinbn said:Quantum mechanics has no action at a distance (but I agree with Einstein that it is spooky).
There are correlations at a distance in classical equilibrium Gibbs states, even for local dynamics such as the KG equation, either because equilibrium states are the consequence of infinitely long-time relaxation processes or else just because of nonlocal boundary conditions, translation invariance, minimum free energy, or some equivalent constraint. The former suggests there is no spookiness, the latter could be said to be spooky, but which one uses depends on tractability during computation as much as on philosophical niceties.Demystifier said:Quantum mechanics has correlations at a distance. Would you agree that those correlations are spooky?
Those correlations can be explained by local deterministic beables. The Bell-type correlations cannot. That's why the latter are much spookier.Peter Morgan said:There are correlations at a distance in classical equilibrium Gibbs states, even for local dynamics such as the KG equation, either because equilibrium states are the consequence of infinitely long-time relaxation processes or else just because of nonlocal boundary conditions, translation invariance, minimum free energy, or some equivalent constraint. The former suggests there is no spookiness, the latter could be said to be spooky, but which one uses depends on tractability during computation as much as on philosophical niceties.
Violations of Bell-type inequalities can be seen as more-or-less natural by noting that they are a consequence of incompatible measurements, noncommuting operators, or transformations between different basis elements (take your choice between those three). I take the nonlocality not to be at the mathematical center of the story. I suggest the following reference:Demystifier said:Those correlations can be explained by local deterministic beables. The Bell-type correlations cannot. That's why the latter are much spookier.
What Bell has shown is that quantum nonlocality is not like this classical non-spooky "nonlocality".Peter Morgan said:Of course, if the change of basis was something like a Fourier transform, which is about as maximally nonlocal as a transformation can be, then yes there's nonlocality, but it's a relatively demystified (I often want to use the word) spookiness. In classical signal analysis, the Wigner function turns up as soon as one considers time-frequency distributions.