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

  • Thread starter Demystifier
  • Start date
  • Featured

Peter Morgan

Gold Member
244
39
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 completeness of the algebra of operators generated by a Wightman operator-valued distribution ##\hat\phi(x)## means that we can construct the vacuum projection operator ##|0\rangle\langle 0|## (the spectrum is ##\{0,1\}##, so this is in ##\mathcal{B(H)}##, although I've never seen an explicit construction). This is a projection but it's definitely not a local operator.
We also know, however, that no operator in a local algebra ##\mathcal{A(O)}## annihilates any given state with bounded energy (Haag LQP, Ch. II, Theorem 5.3.2), which the vacuum projection operator does, so although Reeh-Schlieder proves that all states in the Hilbert space ##\mathcal{H}## can be approximated by the action of a local algebra ##\mathcal{A(O)}## on the vacuum state, not all operators in ##\mathcal{B(H)}## can be approximated by operators in ##\mathcal{A(O)}##.
I think your claim, @rubi, that "physical processes are modeled by unitary evolution or projection" would be better stated as "physical processes are modeled by unitary evolution or local projections".
A separate difficulty, however, is that unitary evolution is not local insofar as the generators of time-like translations do not commute with ##\mathcal{A(O)}##; one uses in quantum field theory not advanced or retarded propagators, but instead the time-ordered Wightman function of the free scalar field (where we can state something concrete) is the Feynman propagator, a Greens function corresponding to different boundary conditions, so that it does not respect locality quite so much (which comes down to analyticity again). Of course there are cases when boundary conditions are such that classical physics also uses noncausal propagators, so this is not anything new to QFT.
 

rubi

Science Advisor
847
348
The completeness of the algebra of operators generated by a Wightman operator-valued distribution ##\hat\phi(x)## means that we can construct the vacuum projection operator ##|0\rangle\langle 0|## (the spectrum is ##\{0,1\}##, so this is in ##\mathcal{B(H)}##, although I've never seen an explicit construction). This is a projection but it's definitely not a local operator.
We also know, however, that no operator in a local algebra ##\mathcal{A(O)}## annihilates any given state with bounded energy (Haag LQP, Ch. II, Theorem 5.3.2), which the vacuum projection operator does, so although Reeh-Schlieder proves that all states in the Hilbert space ##\mathcal{H}## can be approximated by the action of a local algebra ##\mathcal{A(O)}## on the vacuum state, not all operators in ##\mathcal{B(H)}## can be approximated by operators in ##\mathcal{A(O)}##.
How is this relevant to my comment? I never claimed that you can approximate ##\mathcal B(\mathcal H)## by local operators.

I think your claim, @rubi, that "physical processes are modeled by unitary evolution or projection" would be better stated as "physical processes are modeled by unitary evolution or local projections".
That's just a special case of my "claim", which is not really a claim but rather an axiom of quantum mechanics.

A separate difficulty, however, is that unitary evolution is not local insofar as the generators of time-like translations do not commute with ##\mathcal{A(O)}##
That would not make any sense, because that would mean that ##\mathcal A(\mathcal O)## were conserved quantities, i.e. the fields were not dynamical. The relation between the field algebra and the Poincare group (and hence time evolution) is made precise by one of the Wightman axioms, which just says that the fields transform appropriately under Poincare transformations (see samalkhaiats post). Nothing is non-local about this transformation behaviour. It's the same as in classical field theory. If the fields did commute with ##H##, then this transformation behaviour would be spoiled.
 

Peter Morgan

Gold Member
244
39
That's just a special case of my "claim", which is not really a claim but rather an axiom of quantum mechanics.
"physical processes are modeled by unitary evolution or local projections" is a restriction of "physical processes are modeled by unitary evolution or projection". Not all projections "respect locality", only local projections "respect locality". I take it that the vacuum projection operator is a counterexample to "[all] projections respect locality".
Granted that it's not a "claim"; I think I would call it an "empirical principle", though "axiom" will do well enough.
The relation between the field algebra and the Poincaré group (and hence time evolution) is made precise by one of the Wightman axioms, which just says that the fields transform appropriately under Poincaré transformations (see samalkhaiats post). Nothing is non-local about this transformation behaviour.
The other aspect of the relation between the field algebra and the Poincaré group is the spectrum condition, which has consequences for analyticity and for locality.
 

rubi

Science Advisor
847
348
"physical processes are modeled by unitary evolution or local projections" is a restriction of "physical processes are modeled by unitary evolution or projection".
That doesn't make it less true. Physical processes are modeled by unitary evolution or projections. There was no need to be more precise, because my argument was just that that the repeated application of local operators doesn't model any physical process and hence the Reeh-Schlieder theorem is consistent with locality.

The other aspect of the relation between the field algebra and the Poincaré group is the spectrum condition, which has consequences for analyticity and for locality.
The spectrum condition isn't concerned with the field algebra, but only with the the spectra of the Poincare generators (more specifically the translations), independent of the field algebra. It has no negative consequences for locality either.
 

Demystifier

Science Advisor
Insights Author
2018 Award
10,222
3,088
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.
This indeed is an argument that QFT physics is local, but it doesn't convince me that QFT mathematics is local.
 

rubi

Science Advisor
847
348
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

Science Advisor
Insights Author
2018 Award
10,222
3,088
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

Science Advisor
Insights Author
1,616
806
This indeed is an argument that QFT physics is local, but it doesn't convince me that QFT mathematics is local.
Is [itex]\mathcal{L}_{int} = J^{\mu}(x)A_{\mu}(x)[/itex] “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 [itex]\hat{O}|\Omega \rangle = 0[/itex], where [itex]\hat{O}[/itex] is a self-adjoint operator on [itex]\mathcal{H}[/itex], then the theorem tells you that either [itex]\hat{O} = 0[/itex] or else [itex]\hat{O} \not\in \mathcal{A}(\mathcal{O})[/itex], i.e., the observable [itex]\hat{O}[/itex] is not a local observable. For example: since the number operator in relativistic QFT can annihilate the vacuum state [itex]|\Omega \rangle[/itex], the theorem tells you that there is no local number operator in relativistic QFT.
 

Demystifier

Science Advisor
Insights Author
2018 Award
10,222
3,088
Is [itex]\mathcal{L}_{int} = J^{\mu}(x)A_{\mu}(x)[/itex] “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 [itex]\hat{O}|\Omega \rangle = 0[/itex], where [itex]\hat{O}[/itex] is a self-adjoint operator on [itex]\mathcal{H}[/itex], then the theorem tells you that either [itex]\hat{O} = 0[/itex] or else [itex]\hat{O} \not\in \mathcal{A}(\mathcal{O})[/itex], i.e., the observable [itex]\hat{O}[/itex] is not a local observable. For example: since the number operator in relativistic QFT can annihilate the vacuum state [itex]|\Omega \rangle[/itex], 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 ##:\; :##?
 
Last edited:

DrDu

Science Advisor
6,006
745
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

Science Advisor
6,006
745
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

Science Advisor
Insights Author
6,842
2,781
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

Science Advisor
6,006
745
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

Science Advisor
Insights Author
6,842
2,781
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

Science Advisor
Insights Author
2018 Award
10,222
3,088
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! o0)
 

samalkhaiat

Science Advisor
Insights Author
1,616
806
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, [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].
 
Last edited:

Demystifier

Science Advisor
Insights Author
2018 Award
10,222
3,088
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].
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 :smile: ).

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?
 
Last edited:

samalkhaiat

Science Advisor
Insights Author
1,616
806
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 [itex]\mathcal{A}(\mathcal{O})[/itex].

In the proof of the Reeh-Schlieder theorem, any vector [itex]|\Psi \rangle[/itex] 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 [itex]U(\Lambda , a)[/itex], and 3) the set [itex]\{ \mathcal{A}|\Psi \rangle \}[/itex] is dense in [itex]\mathcal{H}[/itex]. The theorem then tells you that the whole Hilbert space can be generated from such vector by applying a polynomial algebra [itex]\mathcal{A}(\mathcal{O})[/itex] defined on an arbitrarily small but open set [itex]\mathcal{O}[/itex] 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

Science Advisor
6,006
745
Isn't this trivial? By the Reeh Schlieder Theorem, any state can be created from the vacuum by localized operators
3) the set [itex]\{ \mathcal{A}|\Psi \rangle \}[/itex] is dense in [itex]\mathcal{H}[/itex].
Isn't this already the result one wants to proove?
 

Demystifier

Science Advisor
Insights Author
2018 Award
10,222
3,088
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

Science Advisor
Insights Author
1,616
806
Isn't this already the result one wants to proove?
No, see #18 for the definition of [itex]\mathcal{A}[/itex] and [itex]\mathcal{A}(\mathcal{O})[/itex].
 

samalkhaiat

Science Advisor
Insights Author
1,616
806
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 [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

Science Advisor
Insights Author
2018 Award
10,222
3,088
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.
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?
 

Want to reply to this thread?

"Is there a local interpretation of Reeh-Schlieder theorem?" You must log in or register to reply here.

Related Threads for: Is there a local interpretation of Reeh-Schlieder theorem?

Replies
3
Views
2K
Replies
3
Views
1K
Replies
26
Views
3K
Replies
56
Views
4K
Replies
24
Views
5K
Replies
32
Views
7K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top