Graduate Is there a local interpretation of Reeh-Schlieder theorem?

[dextercioby]

This indeed is an argument that QFT physics is local, but it doesn't convince me that QFT mathematics is local.

In what way would "mathematics" be local?

 

[rubi]

This indeed is an argument that QFT physics is local, but it doesn't convince me that QFT mathematics is local.

I too don't know what you mean by "mathematics is local", but you can also perform similar non-local mathematical operations in classical relativistic field theory. For instance, take a field configuration with compact support, compute the Fourier transform, multiply that by some function with compact support and then compute the inverse Fourier transform. You will get a field configuration with support everywhere. I wouldn't call that "non-local mathematics" and I don't think it is problematic.

 

[Demystifier]

In what way would "mathematics" be local?

In this context, math would be local if all operators (not only unitary ones) with support in A would (by acting on vacuum) create states that empirically differ from vacuum only in A.

 

[samalkhaiat]

This indeed is an argument that QFT physics is local, but it doesn't convince me that QFT mathematics is local.

Is \mathcal{L}_{int} = J^{\mu}(x)A_{\mu}(x) “local or global mathematics”, whatever that means?
Look, the best way to look at the Reeh-Schlieder theorem is to understand it as a NO-GO theorem in relativistic QFT: If you encounter equation of the form \hat{O}|\Omega \rangle = 0, where \hat{O} is a self-adjoint operator on \mathcal{H}, then the theorem tells you that either \hat{O} = 0 or else \hat{O} \not\in \mathcal{A}(\mathcal{O}), i.e., the observable \hat{O} is not a local observable. For example: since the number operator in relativistic QFT can annihilate the vacuum state |\Omega \rangle, the theorem tells you that there is no local number operator in relativistic QFT.

 

[Demystifier]

Is \mathcal{L}_{int} = J^{\mu}(x)A_{\mu}(x) “local or global mathematics”, whatever that means?
Look, the best way to look at the Reeh-Schlieder theorem is to understand it as a NO-GO theorem in relativistic QFT: If you encounter equation of the form \hat{O}|\Omega \rangle = 0, where \hat{O} is a self-adjoint operator on \mathcal{H}, then the theorem tells you that either \hat{O} = 0 or else \hat{O} \not\in \mathcal{A}(\mathcal{O}), i.e., the observable \hat{O} is not a local observable. For example: since the number operator in relativistic QFT can annihilate the vacuum state |\Omega \rangle, the theorem tells you that there is no local number operator in relativistic QFT.

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? It is built from the field operator $\psi(x)$ as $J^{\mu}=\,:\!\bar{\psi}\gamma^{\mu}\psi\!:$ . Does it mean that $\psi(x)$ is also nonlocal? Or perhaps the nonlocality is in the normal ordering $:\; :$?

 

[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.