Reeh-Schlieder theorem : what's the matter with the Taj Mahal?

[Husserliana97]

I'm currently struggling to understand the implications of the Reeh-Schlieder theorem - having little mastery of QFT formlism, let alone AQFT. Its connection with vacuum state entanglement, although I have an intuitive idea of it, is far from perfectly clear to me, at the moment. And then there's Reinhard Werner's statement that “By acting on the vacuum with suitable operations in a terrestrial laboratory, an experimenter can create the Taj Mahal on (or even behind) the Moon!” (quoted by Rainer Verch in this arXiv article [1]).

My question is twofold. Could you make this connection intelligible to me; how, in short, the theorem presupposes the entanglement of the vacuum state of a field? And explain it to me in a not too formal way. And secondly: how can we draw this implication that Werner seems to draw quite naturally? Do you have to be a supporter of Everett's theory to argue such as thing?

[1] https://arxiv.org/abs/gr-qc/0512053

[Moderator's note: PDF attachment removed and replaced with link to arxiv.]

 

[PeterDonis]

@Husserliana97 for future reference, please do not post papers as PDF attachments. Post links to where they can be found online. I have replaced the attachment in your OP with a link to the arxiv page for the paper.

 

[PeroK]

And then there's Reinhard Werner's statement that “By acting on the vacuum with suitable operations in a terrestrial laboratory, an experimenter can create the Taj Mahal on (or even behind) the Moon!” (quoted by Rainer Verch in this arXiv article [1]).

I'll believe that when I see it. It's not so easy to build the Taj Mahal on the dark side of the Moon!

 

[Husserliana97]

Well, the article makes it clear that the cost/effect ratio in attempting to create a given state (Taj Mahal on the Moon) by local operations (in a laboratory, say) is simply enormous. Such a creation therefore seems impossible to me FAPP - but not in principle, it seems...

 

[Demystifier]

See the paper by Witten: https://arxiv.org/abs/1803.04993
In short, the operator needed to act on the vacuum to create Taj Mahal is not a unitary operator, therefore the creation of Taj Mahal is physically impossible.

 

[Demystifier]

Could you make this connection intelligible to me; how, in short, the theorem presupposes the entanglement of the vacuum state of a field? And explain it to me in a not too formal way.

That's in fact very easy to explain. Consider two particles in nonrelativistic QM interacting via the harmonic oscillator potential proportional to $(x_1-x_2)^2$. Find the ground state of the system. The wave function of the ground state is a function $\psi(x_1-x_2)$, which is not a product of the form $\psi_1(x_1)\psi_2(x_2)$. Therefore the two particles are entangled in the ground state. Similarly, a free field can be viewed as an infinite collection of harmonic oscillators, so the fields at different points are entangled in the ground state.

 

[Husserliana97]

See the paper by Witten: https://arxiv.org/abs/1803.04993
In short, the operator needed to act on the vacuum to create Taj Mahal is not a unitary operator, therefore the creation of Taj Mahal is physically impossible.

I hear. But in an Everettian interpretation, wouldn't this operation be purely epistemic, rather than physical? A simple.acquisition of knowledge rather than a physical action on the state (exactly like measurement, associated with a reduction that is also non-unitary)?

 

[PeroK]

Well, the article makes it clear that the cost/effect ratio in attempting to create a given state (Taj Mahal on the Moon) by local operations (in a laboratory, say) is simply enormous. Such a creation therefore seems impossible to me FAPP - but not in principle, it seems...

There are two ways to look at such claims. The first is to take the current best theory and extrapolate it beyond any experimental corroboration. And claim that such extrapolations are valid. The second is to accept the current best theory as far as it goes. You can extrapolate to some extent - although what's a reasonable extrapolation is, of course, debatable.

In my opinion, the theoretical conclusion here seems quite vacuous. It's not something you can even begin to demonstrate.

Moreover, I wonder where the gravitational effects of the Earth and Moon enter the calculations?

 

[martinbn]

I'm currently struggling to understand the implications of the Reeh-Schlieder theorem - having little mastery of QFT formlism, let alone AQFT.

Why are you looking at this theorem if you have little understanding of QFT?

 

[Husserliana97]

I don't see how this question is relevant. It was a way of specifying that an answer referring to too advanced concepts, such as those of the AQFT, would be indigestible to me. Nevertheless, one can be interested in the (in this case fascinating) implications of a theorem, and try, at least initially, to build up an approximate intuition that is not misleading (one will then have to go further, but it's a first step in my opinion). Demystifier has tried his hand at this, and I thank him for it. I also find Witten's article (at least paragraph 5, which is fortunately the one I'm most interested in) very enlightening.
But this kind of personal question, on the other hand, doesn't help at all.

 

[PeroK]

Nevertheless, one can be interested in the (in this case fascinating) implications of a theorem, and try, at least initially, to build up an approximate intuition that is not misleading (one will then have to go further, but it's a first step in my opinion).

This is not an implication in the sense of something that will happen. Also, the more extravagent the extrapolation, the more sensitive the conclusion is to small variations in the theory.

Suppose tomorrow I started to play chess where I simply choose my move from all legal options - with all being equally likely. It's possible, of course, that I could become undefeated world champion over the next five years using this technique. There's no reason I can't always choose the best move in every position. What are the implications of that for the game of chess?

And, indeed, in what way has our understanding of the game of chess developed from this implication?

 

[Husserliana97]

I meant “implication” in the logical sense: the theorem implies the entanglement of the field in the vacuum state.
But as far as the Taj Mahal is concerned, I have to admit that your example speaks for itself...!

 

[martinbn]

I don't see how this question is relevant. It was a way of specifying that an answer referring to too advanced concepts, such as those of the AQFT, would be indigestible to me. Nevertheless, one can be interested in the (in this case fascinating) implications of a theorem, and try, at least initially, to build up an approximate intuition that is not misleading (one will then have to go further, but it's a first step in my opinion). Demystifier has tried his hand at this, and I thank him for it. I also find Witten's article (at least paragraph 5, which is fortunately the one I'm most interested in) very enlightening.
But this kind of personal question, on the other hand, doesn't help at all.

I was only asking for your motivation. If you want to study number theory you don't start with the proof of Fermat's theorem.

 

[Husserliana97]

I was only asking for your motivation. If you want to study number theory you don't start with the proof of Fermat's theorem.

Well then, my apologies for this over-interpretation of your previous comment.
I'm interested in quantum entanglement in all its forms, without restricting myself to Bell pairs - whose formalism, for that matter, I believe is within my grasp. I read in certain articles (and in Susskind's lectures) that the vacuum state of a field is presumed to be entangled; and in the course of my research I came across this theorem, which seems to be supported by vacuum entanglement, without constituting a sufficient proof of course. But the statement about the Taj Mahal surprised me. Hence my question.

 

[PeterDonis]

in an Everettian interpretation, wouldn't this operation be purely epistemic

No. In an Everettian interpretation, all operations have to be unitary, since the basic premise of that interpretation is that everything is unitary evolution, all the time; any non-unitary operation is simply out of bounds. It's only in "collapse is a real thing" interpretations where the possibility of non-unitary operations even has to be entertained at all.

 

[Husserliana97]

No. In an Everettian interpretation, all operations have to be unitary, since the basic premise of that interpretation is that everything is unitary evolution, all the time; any non-unitary operation is simply out of bounds. It's only in "collapse is a real thing" interpretations where the possibility of non-unitary operations even has to be entertained at all.

Operators corresponding to physical quantities are not unitary (they are hermitian). Furthermore, the reduction (generally) associated with measurement violates the unitarity of the evolution operator.
So the Q.M. has 2 parts, deterministic unitary evolution when there is no measurement, the movement is continuous and preserves the norm;
the second part tells us that the non-deterministic character must also be taken into account during measurements by projections (non-unitary operators on the eigenvectors of the observable being measured).
So everything is unitary in QM as long as we don't measure anything (or only things that are perfectly known, such as when we repeat the same measurement).
Do we agree with this premise?

It seems to me that what is physical (or "ontic"), in Everett's view, is what happens "beyond" the measurement. The fact remains that when we carry out a measurement, we obtain information; admittedly always partial, because of decoherence, which produces an appearance of non-unitary evolution, or an appearance of reduction (the underlying unitarity is preserved, but completely escapes us, the macroscopic observers). That was basically what I wanted to say.

 

[PeterDonis]

Operators corresponding to physical quantities are not unitary (they are hermitian).

An operator can be both. Indeed, in the Everett interpretation, they have to be both, since everything is unitary time evolution.

the second part tells us that the non-deterministic character must also be taken into account during measurements by projections

There are no such things in the Everett interpretation. "Measurement" in that interpretation is just unitary evolution that entangles a "measured system" and a "measuring device".

when we carry out a measurement, we obtain information; admittedly always partial, because of decoherence, which produces an appearance of non-unitary evolution, or an appearance of reduction (the underlying unitarity is preserved, but completely escapes us, the macroscopic observers). That was basically what I wanted to say.

This is fine, but it contradicts the two statements from you that I quoted above.

 

[Demystifier]

I was only asking for your motivation. If you want to study number theory you don't start with the proof of Fermat's theorem.

But if you want to find a motivation for studying number theory, the Fermat's theorem is exactly what you often start with.

 

[Husserliana97]

Indeed --

An operator can be both. Indeed, in the Everett interpretation, they have to be both, since everything is unitary time evolution.


There are no such things in the Everett interpretation. "Measurement" in that interpretation is just unitary evolution that entangles a "measured system" and a "measuring device".


This is fine, but it contradicts the two statements from you that I quoted above.

Indeed, I should have said that the non-deterministic character was also an appearance, in the Everett framework. My initial paragraph seems to me to correspond to the instrumental, or conventional, point of view.

But I confess that I find it hard to see how observables, by which I mean Hermitian operators, could also be unitary. It seemed to me that the two qualifications were opposed.

 

[Demystifier]

The strange thing about the RH theorem is not that you can create something as complex as Taj Mahal. From elementary QM we know that we can create any state of the harmonic oscillator by acting with creation operators on the vacuum. What is strange is that we can create something there on another place by acting only locally here on this place.

 

[Demystifier]

But I confess that I find it hard to see how observables, by which I mean Hermitian operators, could also be unitary. It seemed to me that the two qualifications were opposed.

They are opposed. Only a trivial unit operator is both hermitian and unitary. But both unitary and hermitian operators play a role in measurement. The process of measurement (in MWI) is unitary, but the observable measured by this process is hermitian. The difference is in how you measure vs what you measure. Also note that any unitary operator can be written in the form $U=e^{iA}$, where $A$ is some hermitian operator.

 

[Husserliana97]

Indeed --

They are opposed. Only a trivial unit operator is both hermitian and unitary. But both unitary and hermitian operators play a role in measurement. The process of measurement (in MWI) is unitary, but the observable measured by this process is hermitian. The difference is in how you measure vs what you measure.

That's a much better formulation than mine (in my terminology, ontic = that which is measured), and I humbly concur. But I was under the impression that PeterDonis was arguing something else.

 

[Husserliana97]

The strange thing about the RH theorem is not that you can create something as complex as Taj Mahal. From elementary QM we know that we can create any state of the harmonic oscillator by acting with creation operators on the vacuum. What is strange is that we can create something there on another place by acting only locally here on this place.

Certainly -- this creation being an attestation of the intricate nature (and therefore not local in this sense) of the regions of the field in the vacuum state, do we agree?

Yet in that case, it seems strange to me that some authors take this Taj Mahal's example

 

[PeterDonis]

Only a trivial unit operator is both hermitian and unitary.

Huh? Unitary means the conjugate transpose is the inverse. Hermitian means it's its own conjugate transpose. So if it's Hermitian and also its own inverse, it's unitary.

An obvious non-trivial example is the Pauli spin matrices:

https://en.wikipedia.org/wiki/Pauli_matrices

 

[PeterDonis]

I was under the impression that PeterDonis was arguing something else.

Your impression is correct. See my post #24 just now.

 

[Demystifier]

Huh? Unitary means the conjugate transpose is the inverse. Hermitian means it's its own conjugate transpose. So if it's Hermitian and also its own inverse, it's unitary.

An obvious non-trivial example is the Pauli spin matrices:

https://en.wikipedia.org/wiki/Pauli_matrices

I stand corrected.