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

• Featured

#### Demystifier

2018 Award
Non-philosophically inclined experts in relativistic QFT often insist that QFT is a local theory. They are not impressed much by arguments that quantum theory is non-local because such arguments typically rest on philosophical notions such as ontology, reality, hidden variables, or the measurement problem.

But how about the Reeh-Schlieder theorem? The Reeh-Schlieder theorem (recently explained in a relatively simple way by Witten in https://arxiv.org/abs/1803.04993 ) is an old theorem, well-known in axiomatic QFT, but relatively unknown in a wider QFT community. In somewhat over-simplified terms (see Sec. 2.5 of the paper above), the theorem states that acting with a suitable (not necessarily unitary) local operator (e.g. having a support only on Earth) on the vacuum, one can create an arbitrary state (e.g. a state describing a building on Mars). As all other theorems that potentially can be interpreted as signs of quantum non-locality, this theorem is also a consequence of quantum entanglement. However, unlike other theorems about quantum non-locality, this theorem seems not to contain any philosophical concepts or assumptions. The theorem is pure mathematical physics, without any philosophical excess baggage.

What I would like to see is how do non-philosophically inclined experts in QFT, who insist that relativistic QFT is local, interpret Reeh-Schlieder theorem in local terms? How can it be compatible with locality?

Last edited:
Related Quantum Physics News on Phys.org

#### Peter Morgan

Gold Member
One trouble with the Reeh-Schlieder theorem is that the Wightman axioms, say, that are required to prove it are not satisfied by interacting Lagrangian fields. Hegerfeldt proves something similar with much more restricted assumptions, almost only the spectrum condition (that the energy-momentum must be in or on the forward light-cone), which an interacting quantum field ought to satisfy, but it seems often to be argued that this is not an example of nonlocality, and perhaps the proof is not compelling because it can be dismissed on a quick reading as being about nonrelativistic QM.
More constructively, the 2-point Wightman function (the non-time-ordered 2-point Vacuum Expectation Value) has to have the Källén-Lehmann form even for an interacting field (the scalar boson case, to keep at least some simplicity), so it's definitely non-zero at space-like separation (and so is the time-ordered VEV). Microcausality shows up in this context as the imaginary part of the 2-point VEV being zero at space-like separation, but the real part is always non-zero except on the light-cone, where it's undefined (there's a delta function on the light-cone, as well as inverse square scaling at small invariant distance from the light-cone).
Even for Gibbs' states of the classical Klein-Gordon field there are nonlocal correlations, however, even though the advanced and retarded propagators of the classical KG dynamics are definitely zero at space-like separation, so we can't just point to nonlocal correlations and say, "look, nonlocality!" For the classical KG equation, it's not the advanced or retarded propagator that shows up when we compute the Gibbs' state, but (at space-like separation) we obtain the inverse fourier transform of $(\vec k\cdot\vec k+m^2)^{-1}$, corresponding to different boundary conditions. The Gibbs' state, it should also be noted, is either a consequence of an infinite-time convergence to equilibrium because of non-KG dynamical perturbations (one supposes), or else it's just a superdetermined-by-minimum-free-energy initial condition.
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", J. Math. Mech. 14 (1965) 629. 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.

#### A. Neumaier

how do non-philosophically inclined experts in QFT, who insist that relativistic QFT is local, interpret Reeh-Schlieder theorem in local terms? How can it be compatible with locality?
The theorem is essentially a consequence of analyticity, where knowledge of a function in some neighborhood implies knowing it everywhere.

The theorem is compatible with locality since it is proved from the Wightman axioms, which are based on locality.

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.

#### Demystifier

2018 Award
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.
This indeed is a good argument that there is no physical non-locality. But since the creation and annihilation operators are well defined mathematically, one could still argue that there is something non-local about QFT in its mathematical formulation.

#### A. Neumaier

This indeed is a good argument that there is no physical non-locality. But since the creation and annihilation operators are well defined mathematically, one could still argue that there is something non-local about QFT in its mathematical formulation.
Well, mathematically, all nonlocal theories ever discussed were created locally from some desk. But this has no philosophical consequences.

#### Demystifier

2018 Award
The theorem is compatible with locality since it is proved from the Wightman axioms, which are based on locality.
I am not convinced that all axioms are based on locality. In particular, the axiom that there exists an invariant state (called vacuum) looks quite non-local to me. Intuitively, this axiom says that there exists a state that everywhere looks the same.

#### Demystifier

2018 Award
The theorem is essentially a consequence of analyticity, where knowledge of a function in some neighborhood implies knowing it everywhere.
Isn't then analyticity itself a kind of mathematical non-locality?

#### A. Neumaier

the axiom that there exists an invariant state (called vacuum) looks quite non-local to me. Intuitively, this axiom says that there exists a state that everywhere looks the same.
All states of any field theory are nonlocal, since they specify the field everywhere. But this is an irrelevant meaning of nonlocality. The latter means either nonlocal dynamics (not valid in relativistic QFT by definition) or nonlocal correlations (valid by direct construction). Locality of the dynamics and nonlocality of the correlations are fully compatible.
Isn't then analyticity itself a kind of mathematical non-locality?
Yes. But arbitrarily small deviation in the given neighborhood may have arbitrarily large consequences at any given other point, so this has no physical relevance.

#### A. Neumaier

am not convinced that all axioms are based on locality.
Fort compatibility it is sufficient that the dynamical locality axiom is among the Wightman axioms, and that there are examples satisfying these axioms.

#### Demystifier

2018 Award
nonlocal correlations (valid by direct construction)
What do you mean by "direct construction"?

#### Demystifier

2018 Award
Yes. But arbitrarily small deviation in the given neighborhood may have arbitrarily large consequences at any given other point, so this has no physical relevance.
This sheds some light on the origin of Reeh-Schlieder theorem, but why does it not have physical relevance?

By the way, this reminds me of the butterfly effect in chaos theory. Would you say that butterfly effect also does not have physical relevance?

#### Demystifier

2018 Award
All states of any field theory are nonlocal, since they specify the field everywhere. But this is an irrelevant meaning of nonlocality. The latter means either nonlocal dynamics (not valid in relativistic QFT by definition) or nonlocal correlations (valid by direct construction). Locality of the dynamics and nonlocality of the correlations are fully compatible.
Whether or not nonlocality of correlations requires some kind of nonlocal dynamics (perhaps at the level of hidden variables) is a part of usual philosophical discussions about the meaning of Bell theorem, which I want to avoid in this thread.

#### A. Neumaier

What do you mean by "direct construction"?
e.g., two entangled photons a la Bell.
why does it not have physical relevance?
because in observable physics, consequences must depend smoothly on causes.
Whether or not nonlocality of correlations requires some kind of nonlocal dynamics (perhaps at the level of hidden variables) is a part of usual philosophical discussions
It involves no philosophy, only shut-up-and-calculate, since free QED (which is dynamically local) suffices to do the argument in the traditional, purely mathematical way.

#### Demystifier

2018 Award
because in observable physics, consequences must depend smoothly on causes.
Why?

#### A. Neumaier

Because one can prepare and measure only to finite precision.

#### Peter Morgan

Gold Member
because in observable physics, consequences must depend smoothly on causes
The Wightman axioms enforce analyticity — in other words significantly more than smoothness, which finite precision could never confirm or deny. How significant is it that thermal and other non-vacuum states require, so to speak, less analyticity?
Also, smoothness wouldn't be expected of a classical ideal model, particularly in statistical mechanics, where discontinuity is commonplace. We might expect that smoothness would reappear if one modeled more accurately, but in practice we work with the discontinuous model because it is accurate enough.

#### Peter Morgan

Gold Member
A separate aspect of nonlocality in QM/QFT is the commonplace use of the vacuum projection operator by practicing physicists. Whenever a physicist discusses a transition probability, meaning $\bigl|\langle\phi|\psi\rangle\bigr|^2=\langle\psi|\phi\rangle\langle\phi|\psi\rangle$, the measurement $|\phi\rangle\langle\phi|$ in the state $\hat A\mapsto\langle\psi|\hat A|\psi\rangle$, they are implicitly using a modulated form of the vacuum projection operator $|0\rangle\langle 0|$, thereby erasing all the careful construction of microcausal locality. $|\phi\rangle\langle\phi|$ does not commute with any local observable.
This is such a commonplace that I doubt it can be eradicated in general use, but it seems too free and easy with nonlocality that the implicit use of the vacuum projection operator is a step in the most common computations that show that QM violates Bell-CHSH inequalities [of course one has constructions that violate Bell-CHSH inequalities that are entirely local, such as Landau's Phys Lett A 120, 54(1987), "On the violation of Bell's inequalities in quantum theory", so this is merely a criticism of using the vacuum projection operator when it does not have to be used.]

#### samalkhaiat

Non-philosophically inclined experts in relativistic QFT often insist that QFT is a local theory. They are not impressed much by arguments that quantum theory is non-local because such arguments typically rest on philosophical notions such as ontology, reality, hidden variables, or the measurement problem.

But how about the Reeh-Schlieder theorem? The Reeh-Schlieder theorem (recently explained in a relatively simple way by Witten in https://arxiv.org/abs/1803.04993 ) is an old theorem, well-known in axiomatic QFT, but relatively unknown in a wider QFT community. In somewhat over-simplified terms (see Sec. 2.5 of the paper above), the theorem states that acting with a suitable (not necessarily unitary) local operator (e.g. having a support only on Earth) on the vacuum, one can create an arbitrary state (e.g. a state describing a building on Mars). As all other theorems that potentially can be interpreted as signs of quantum non-locality, this theorem is also a consequence of quantum entanglement. However, unlike other theorems about quantum non-locality, this theorem seems not to contain any philosophical concepts or assumptions. The theorem is pure mathematical physics, without any philosophical excess baggage.

What I would like to see is how do non-philosophically inclined experts in QFT, who insist that relativistic QFT is local, interpret Reeh-Schlieder theorem in local terms? How can it be compatible with locality?
I am having trouble with your notion of locality or the lack of it in QFT. Locality in QFT refers to all (localizable) quantities that can be measured in practice, i.e., the self-adjoint elements of the polynomial algebra $\mathcal{A}(\mathcal{O})$ generated by operators of the form $$\int d^{4}x_{1} \cdots d^{4}x_{n} \ \varphi_{k_{1}} (x_{1}) \cdots \varphi_{k_{n}}(x_{n}) \ f(x_{1} , \cdots , x_{n}) , \ \ n = 0 , 1 , 2 , \cdots$$ with $\mathcal{O}$ being a (bounded) open set in $\mathbb{R}^{(1,3)}$ and $f \in \mathcal{D} (\mathcal{O}^{n})$. When $\mathcal{O} \subset \mathbb{R}^{(1,3)}$ is a finite space-time region, the self-adjoint elements of $\mathcal{A}(\mathcal{O})$ are called local observables. When $f \in \mathcal{S} (\mathbb{R}^{4n})$, the resulting polynomial algebra is called the field algebra $\mathcal{A}$.
In theories with positive-definite Hilbert space $\mathcal{H}$, one can show that the following statements are equivalent to one another:
i) Uniqueness and cyclicity of the vacuum: There exists a unique vector $\Omega$ invariant under any translation $P_{\mu}\Omega = 0$, which is cyclic with respect to $\mathcal{A}$, i.e., $\mathcal{A}\Omega$ is dense in $\mathcal{H}$ with respect to an admissible topology $\tau$: $\overline{\mathcal{A}\Omega}^{\tau} = \mathcal{H}$.
ii) Irreducibility of the field algebra $\mathcal{A}$.

The R-S theorem states that for any open set $\mathcal{O}$ in space-time, the equality $$\overline{\mathcal{A}(\mathcal{O})\Omega}^{\tau} = \overline{\mathcal{A}\Omega}^{\tau} = \mathcal{H} ,$$ holds for any admissible topology $\tau$. In English, the theorem asserts that if the field algebra is irreducible (or cyclic with respect to the vacuum $\Omega$) then the set of states obtained by applying a polynomial algebra of fields (taken from arbitrarily small but open set in space-time) to $\Omega$ also spans the whole Hilbert space $\mathcal{H}$. In other words, the theory says that the physical states in $\mathcal{H}$ cannot be localized although the observables can. So, the non-local features are carried by the vacuum.

Of course, many aspects of particle physics can be explained using the assumption that particle states are localized excitations of the vacuum, i.e., they arise by applying local field operator to the vacuum as per Wightman. Similarly, in the algebraic setting of Haag, Roberts and others, it is assumed that a particle state cannot be distinguished from the vacuum by measurements done in the space-like complement of sufficiently large but bounded regions of space-time. Particles carrying a non-gaugable-symmetry charge, such as iso-spin, strangeness, baryon number or lepton number, fit well into this scheme. But, we know this method does not work in gauge theories: For example in QED, Gauss’ law allows the charged states to be distinguished from the vacuum in the causal complement of any bounded region, because it is possible to evaluate the charge by measuring the flux through a very large sphere. Thus, in gauge field theories charged states are not localizable.

#### A. Neumaier

smoothness wouldn't be expected of a classical ideal model, particularly in statistical mechanics, where discontinuity is commonplace. We might expect that smoothness would reappear if one modeled more accurately, but in practice we work with the discontinuous model because it is accurate enough.
In statistical mechanics, everything is analytic until one takes the thermodynamic limit (a conventional approximation), in which case discontinuities arise in some relationships, due to phase transitions. But the sensitivity in analytic continuation (and hence in the Reeh-Schlieder nonlocality) is very different - it is a generic ill-posedness, not just on a set of measure zero.

#### Peter Morgan

Gold Member
the theorem asserts that if the field algebra is irreducible (or cyclic with respect to the vacuum $\Omega$) then the set of states obtained by applying a polynomial algebra of fields (taken from arbitrarily small but open set in space-time) to $\Omega$ also spans the whole Hilbert space $\mathcal{H}$. In other words, the theory says that the physical states in $\mathcal{H}$ cannot be localized although the observables can. So, the non-local features are carried by the vacuum.
"the physical states in $\mathcal{H}$ cannot be localized although the observables can" is a nice accessible summary of the Reeh-Schlieder theorem, but I think the difficulty that Demystifier has focused on is a consequence of particle physicists not believing that interacting quantum fields in the Lagrangian formalism fall under the Wightman or Haag-Kastler axioms.
It can't be the case that irreducibility (or cyclicity) is sufficient on its own, because one at least needs there to be a concept of space-time, which irreducibility does not speak to, but can any theory to which a physicist's interacting quantum field can be an asymptotic expansion —with all the mathematical troubles that different types of regularization and renormalization imply— be proved to satisfy the requirements for a derivation of the Reeh-Schlieder theorem? Because of Hegerfeldt, and also because of Arnold Neumaier's point above,
The theorem is essentially a consequence of analyticity,
it seems that a representation of the translation group and the spectrum condition alone might be sufficient axioms to prove that "the physical states in $\mathcal{H}$ cannot be localized although the observables can", independently of irreducibility, additivity, microcausality, representation of the Lorentz group, and primitive causality. [The difficulty with Hegerfeldt is that none of his many papers are definitively enough stated or proved. Have a look at his paper arXiv:quant-ph/9809030, for example, which I believe is his most recent. Because of the way Hegerfeldt does things, one hears it said that Hegerfeldt only applies to nonrelativistic QM, however it more seems that positive frequency is sufficient, the (stronger) spectrum condition is not required. The connection of positive frequency with analyticity through the Hilbert transform is relatively more elementary than working with the spectrum condition.] It seems that indeed physicists might agree that any theory they would be willing to accept would have to satisfy positive energy/the spectrum condition, because that is associated with stability.

#### A. Neumaier

The connection of positive frequency with analyticity through the Hilbert transform
it is crucial for causality, bot in the nonrelativistic and the relativistic case. in the relativistic case, the spectral condition of Wightman is essential for Poincare invariance.
there to be a concept of space-time, which irreducibility does not speak to
But the Poincare group present produces flat space-time through the Fourier transform.

#### Peter Morgan

Gold Member
But the Poincare group present produces flat space-time through the Fourier transform.
Absolutely, but if the translation group is sufficient and the proof is independent of whether the Lorentz symmetry is present, then best not to include the Lorentz group. Irreducibility is a purely analytic property of the algebra of observables that doesn't imply any geometrical properties, not even the translation group.
One source of wiggle room for physicists is that one cannot be absolutely certain of Poincaré covariance for interacting Lagrangian quantum fields because neither regularization nor renormalization is absolutely certain to preserve Lorentz covariance; the translation group is more certain (although even that might be denied by those who say that we can't have a conversation with anything less than full quantum gravity).

Last edited:

#### A. Neumaier

neither regularization nor renormalization is absolutely certain to preserve Lorentz covariance
With causal renormalization theory one can be absolutely certain. One cannot dispense with it in elementary particle physics.

#### samalkhaiat

it seems that a representation of the translation group and the spectrum condition alone might be sufficient axioms to prove that "the physical states in $\mathcal{H}$ cannot be localized although the observables can", independently of irreducibility, additivity, microcausality, representation of the Lorentz group, and primitive causality.
For any $|\Phi \rangle \ , |\Psi \rangle \in \mathcal{A}| \Omega \rangle$, the spectrum postulate can be written (in a form which is suitable for Hilbert space with indefinite metric) as $$\int d^{4}a \ e^{-i p \cdot a} \ \langle \Phi | U(a) | \Psi \rangle = 0, \ \ \mbox{if} \ \ \ p \not\in \overline{\mathcal{V}}_{+} = \{ q \in \mathbb{R}^{4}; q_{0} \geq 0, \ q^{2} \geq 0 \},$$ where $U(a)$ is the $U(1 , a)$ element of the group of unitary operators $U(\Lambda , a)$ on $\mathcal{H}$, i.e., the groups homomorphism $U: \ ISO(1,3) \to U( \mathcal{H})$. Using translational invariance of the vacuum, we may define a “function” $W_{k_{1} \cdots k_{n}}(y_{1}, \cdots y_{n-1})$ by $$W_{k_{1} \cdots k_{n}}(x_{1}-x_{2}, \cdots , x_{n-1}-x_{n}) = \langle \Omega | \varphi_{k_{1}}(x_{1}) \cdots \varphi_{k_{n}}(x_{n})| \Omega \rangle ,$$ and rewrite the spectrum condition in the form $$\int d^{4}y_{1} \cdots d^{4}y_{n-1} \ e^{i \sum_{j = 1}^{n-1}q_{j}y_{j}} \ W_{k_{1} \cdots k_{n}} (y_{1} \cdots y_{n-1}) = 0, \ \ \mbox{if} \ \ q_{j} \not\in \overline{\mathcal{V}}_{+}. \ \ (1)$$ Next, we would like to continue $W_{k_{1}\cdots k_{n}}(y_{1}, \cdots , y_{n-1})$ analytically to the complex analytic function $W_{k_{1}\cdots k_{n}}(z_{1}, \cdots , z_{n-1})$ in the permuted extended tube. For this we need to combine the spectrum postulate (1) with the postulate of Poincare’ covariance: $$U^{\dagger}(\Lambda , a) \varphi_{a}(x) U(\Lambda , a) = D_{a}{}^{b}(\Lambda) \varphi_{b} \left( (\Lambda , a)^{-1}x \right) ,$$ and the postulate of microcausality (the local commutation relations): $$[\varphi_{i} (x) , \varphi_{j} (y)] = 0, \ \ \ \mbox{when} \ \ (x-y)^{2} < 0 .$$ The above together with the vacuum postulate allows us to prove the Reeh-Schlieder theorem. If $\mathcal{H}$ has a positive-definite inner product, then the vacuum postulate can be replaced by the irreducibility of the field algebra $\mathcal{A}$. The theorem can also be proved for Hilbert space with indefinite metric. In this case, however, the vacuum postulate is no longer equivalent to the irreducibility of $\mathcal{A}$.

#### rubi

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.

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

### 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