Is there a local interpretation of Reeh-Schlieder theorem?

[A. Neumaier]

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!)

 

[DrDu]

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.

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 $a^\dagger$.

 

[DrDu]

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.

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.

 

[A. Neumaier]

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 $a^\dagger$.

But this is still not yet unitary!

 

[DrDu]

But this is still not yet unitary!

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]

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.

But it no longer generates the same state!

 

[Demystifier]

No. The normal ordering does not eliminate the 2-particle creation term. (You must have been thinking of the nonrelativistic case!)

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!

 

[samalkhaiat]

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?

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, $J \in \mathcal{A}(\mathcal{O})$, annihilating the vacuum is itself vanishing, i.e., $J|\Omega\rangle = 0, \ \Rightarrow \ J = 0$
2) Symmetry currents do not annihilate the vacuum state, their integrated charges do if the symmetry is not spontaneously broken: $$Q|\Omega \rangle = \int d^{3}x \ J^{0}(x) |\Omega \rangle = 0.$$ 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 : $Q|\Omega \rangle = 0, \ \Rightarrow \ \ \partial_{\mu}J^{\mu}|\Omega\rangle = 0$. Since, $\partial_{\mu}J^{\mu}$ is a local operator, then by the separating property, we obtain the conservation law $\partial_{\mu}J^{\mu}=0$.

 

[Demystifier]

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, $J \in \mathcal{A}(\mathcal{O})$, annihilating the vacuum is itself vanishing, i.e., $J|\Omega\rangle = 0, \ \Rightarrow \ J = 0$
2) Symmetry currents do not annihilate the vacuum state, their integrated charges do if the symmetry is not spontaneously broken: $$Q|\Omega \rangle = \int d^{3}x \ J^{0}(x) |\Omega \rangle = 0.$$ 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 : $Q|\Omega \rangle = 0, \ \Rightarrow \ \ \partial_{\mu}J^{\mu}|\Omega\rangle = 0$. Since, $\partial_{\mu}J^{\mu}$ is a local operator, then by the separating property, we obtain the conservation law $\partial_{\mu}J^{\mu}=0$.

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.

I have one additional question (and don't be mad at me if the question sounds stupid ).

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]

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?

Yes, shifting the energy by a constant has no effects on the spectrum condition, Poincare algebra, or on the domain of definition of the elements of the local observable algebra $\mathcal{A}(\mathcal{O})$.

In the proof of the Reeh-Schlieder theorem, any vector $|\Psi \rangle$ can play the role of the vacuum state as long as 1) it is unique up to a complex number, 2) belongs to the domain of definition (i.e., stable under the action) of $U(\Lambda , a)$, and 3) the set $\{ \mathcal{A}|\Psi \rangle \}$ is dense in $\mathcal{H}$. The theorem then tells you that the whole Hilbert space can be generated from such vector by applying a polynomial algebra $\mathcal{A}(\mathcal{O})$ defined on an arbitrarily small but open set $\mathcal{O}$ in Minkowski space time. Then, the theorem adds one more feature to the vacuum which is its separability. So, I don’t really know what exactly you meant by “aspects” of the theorem.

 

[DrDu]

Isn't this trivial? By the Reeh Schlieder Theorem, any state can be created from the vacuum by localized operators

3) the set $\{ \mathcal{A}|\Psi \rangle \}$ is dense in $\mathcal{H}$.

Isn't this already the result one wants to proove?

 

[Demystifier]

So, I don’t really know what exactly you meant by “aspects” of the theorem.

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]

Isn't this already the result one wants to proove?

No, see #18 for the definition of $\mathcal{A}$ and $\mathcal{A}(\mathcal{O})$.

 

[samalkhaiat]

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.

Who said “any”? “Any” state is not Poincare’ invariant, “any” state is not an eignstate of $P_{\mu}$, “any” state is not that which makes $\mathcal{H} = \overline{\mathcal{A}| \Omega \rangle}$ and $\{0\}$ the only invariant representation sub-spaces of the field algebra.

 

[Demystifier]

Who said “any”? “Any” state is not Poincare’ invariant, “any” state is not an eignstate of $P_{\mu}$, “any” state is not that which makes $\mathcal{H} = \overline{\mathcal{A}| \Omega \rangle}$ and $\{0\}$ the only invariant representation sub-spaces of the field algebra.

Well, Haag said "any state with bounded energy". Any eigenstate of $P_{\mu}$ with a finite eigenvalue $p_{\mu}$ has bounded energy, right? But such states are not Poincare invariant (unless $p_{\mu}=0$). Do I miss something?

 

[samalkhaiat]

Do I miss something?

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 $\left( M_{\mu\nu}, P_{\mu}\right)|\Omega \rangle = (c_{\mu\nu},q_{\mu})| \Omega \rangle$, you can always redefine the Poincare’ generators $\left( M_{\mu\nu},P_{\mu}\right) \to \left( M_{\mu\nu}+ c_{\mu\nu},P_{\mu}+q_{\mu}\right)$ to obtain $\left( M_{\mu\nu},P_{\mu}\right)| \Omega \rangle = 0$.

 

[vanhees71]

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.

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.

 

[Demystifier]

QFT, which I'd call the only physically sensible relativistic QT we know of

String theorists would disagree.

 

[vanhees71]

Since when is string theory physics? SCNR.

 

[Demystifier]

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".

The presence/absence of spooky action at a distance has nothing to do with the fact that QFT interactions are local.
- If one accepts a suitable orthodox/minimal interpretation of QT, then the spooky action at a distance is absent even in non-relativistic QM with nonlocal Coulomb-like potentials.
- If one accepts an ontic/hidden-variable interpretation of QT, then the spooky action at a distance is present even in local QFT.

 

[Demystifier]

Since when is string theory physics? SCNR.

Anything published in a physics journal is - physics.

 

[A. Neumaier]

nonlocal Coulomb-like potentials.

Any potential displays spooky acts at a distance!

If one accepts an ontic/hidden-variable interpretation of QT, then the spooky action at a distance is present even in local QFT.

Which ontic/hidden-variable interpretation of which local relativistic QFT do you have in mind?

 

[Peter Morgan]

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.

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.

 

[Demystifier]

A ny potential acts at spooky distance!

What is ny potential?

Which ontic/hidden-variable interpretation of which local relativistic QFT do you have in mind?

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.

 

[martinbn]

He means any, not a ny.

 

[Demystifier]

A ny potential acts at spooky distance!

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]

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 you could at least link to key posts in these threads.

 

[A. Neumaier]

He did not use this expression to describe Newtonian mechanics.

But Newton himself was already dissatisfied with action at a distance, and general relativity has eliminated this spooky feature.

Quantum mechanics reintroduced immediate effects at a distance; they are inherent to Born's probability interpretation of $|\psi(x)|^2$. Einstein therefore objected to the probabilistic interpretation of QM.

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.

 

[martinbn]

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.

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]

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.

Are you saying that relativistic QFT cannot explain the experiments that show violation of Bell inequalities?

 

[Demystifier]

Actually it is Newtonian gravity that has action at a distance (whether one finds it spooky or not).

Of course.

Quantum mechanics has no action at a distance (but I agree with Einstein that it is spooky).

Quantum mechanics has correlations at a distance. Would you agree that those correlations are spooky?

 

[Peter Morgan]

Quantum mechanics has correlations at a distance. Would you agree that those correlations are 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]

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.

Those correlations can be explained by local deterministic beables. The Bell-type correlations cannot. That's why the latter are much spookier.

 

[Peter Morgan]

Those correlations can be explained by local deterministic beables. The Bell-type correlations cannot. That's why the latter are much spookier.

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:

From this perspective, the violation of Bell-type inequalities is a signal that at least some of the measurements are incompatible, so that noncommuting operators must be used in models of those measurements, which in turn can be taken to be a signal that somewhere in the experimental procedure a transformation from one basis to another was implicitly or explicitly introduced. That's classically as straightforward as it is for quantum theory. 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.
Experimental violations of Bell inequalities are most often associated with eigenspaces of noncommuting operators that act as generators of representations of the Euclidean rotation group (usually in conditions where nonrelativistic approximations are entirely adequate). Such, generated by the Poisson bracket, are classically as natural as they are for quantum theory.

 

[Demystifier]

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.

What Bell has shown is that quantum nonlocality is not like this classical non-spooky "nonlocality".