Do we need a reference frame in Quantum Hilbert space?

[vanhees71]

... and the arena of perturbative relativistic quantum field theory: Fock spaces are separable Hilbert spaces. This is independent of regularization.

But this is not the sense of separability used in the earlier discussion of this thread.

Of course, with separability here I mean the mathematical notion, i.e., that there exists a complete countable orthonormal system of vectors. I was not sure about the Fock space of free particles in the infinite-volume limit since the natural basis is the occupation-number basis with respect to a single-particle basis, for which you usually use momentum-spin eigenstates, which are generalized vectors with the momenta being continuous variables, i.e., the general occupation number vector is
$$|\{N(\vec{p},\sigma) \}_{\vec{p} \in \mathbb{R}^3, \sigma \in \{-s,-s+1,\ldots,s-1,s\}},$$
which is uncountable.

In a finite box with periodic boundary conditions, the momenta are discrete and thus the occupation-number basis countable.

Of course, one could think to use other true single-particle bases (like harmonic-oscillator states, although I'm not sure, whether such a simple thing unambigously exists in the relativistic case).

 

[Robert Shaw]

Of course, with separability here I mean the mathematical notion, i.e., that there exists a complete countable orthonormal system of vectors. I was not sure about the Fock space of free particles in the infinite-volume limit since the natural basis is the occupation-number basis with respect to a single-particle basis, for which you usually use momentum-spin eigenstates, which are generalized vectors with the momenta being continuous variables, i.e., the general occupation number vector is
$$|\{N(\vec{p},\sigma) \}_{\vec{p} \in \mathbb{R}^3, \sigma \in \{-s,-s+1,\ldots,s-1,s\}},$$
which is uncountable.

In a finite box with periodic boundary conditions, the momenta are discrete and thus the occupation-number basis countable.

Of course, one could think to use other true single-particle bases (like harmonic-oscillator states, although I'm not sure, whether such a simple thing unambigously exists in the relativistic case).

That's a good point. Separability is a mathematical property.

The physical status of separability is not clear.

My example of Earthlings and Martians above is mathematically correct but raises a question about the physics.

Do you have any suggestions about the Earthlings and Martians point above?

 

[vanhees71]

What Einstein meant by "inseparability" was entanglement. The inseparability of QT was his real objection against QT, as according to himself has not become clear in the (in)famous EPR paper, which "hides the point behind erudition". In 1948 he wrote a much clearer paper (which is however in German), where he made this point clear:

http://dx.doi.org/10.1111/j.1746-8361.1948.tb00704.x

 

[Robert Shaw]

I'm wondering if the confusion may be due to the two different notions of product.
Shaw claims to be using Cartesian product ×, and rubi is using tensor product ⊗.
Now if H = ℝ² then H×H = H⊗H = ℝ⁴, however [1,0]×[0,1] = [1,0,0,1], whereas [1,0]⊗[0,1] = [0,1,0,0], and [1,0,0,1] isn't separable in H⊗H.

In H×H all members are separable, and in H⊗H only the tensor product of vectors are separable. Thus my guess is Shaw meant to say tensor rather than Cartesian.

It would nice if Shaw gave concrete examples (in H⊗H) for the Martian and the Earthian, so I could see clearly what he is talking about.

Alice and Bob are partial observers of states in a toy quantum Universe and widely discussed in the literature of Quantum Mechanics.

They see some states as separable and others as entangled. For Alice the partial measurement suboperator is diag(1,1,-1,-1) and Bob is diag(1,-1,1,-1).

Martians Ylc and Zog see the world differently. For Ylc adiag(1,1,1,1) and Zog adiag(1,-1,-1,1) ...antidiagonal matrices. Their eigenvectors are the maximally entangled states for Alice and Bob.

A state is prepared which is observed by Alice and Bob to be separable.

Ylc and Zog observe the same state and find it to be entangled.

 

[rubi]

A state is prepared which is observed by Alice and Bob to be separable.

Ylc and Zog observe the same state and find it to be entangled.

It doesn't matter who observes the particle. Any observer will conclude that Alice's particle is entangled with Bob's particle. As I said, unitarily equivalent descriptions will not change any physical fact.

 

[Robert Shaw]

It doesn't matter who observes the particle. Any observer will conclude that Alice's particle is entangled with Bob's particle. As I said, unitarily equivalent descriptions will not change any physical fact.

To support the statement that that "it doesn't matter who observes" please take my example of measurement operators for Earthpersons Alice, Bob, and Martians Ylc and Zog and calculate what the Martians observe when they take partial measurements of a state that the Earthpersons observe as separable.

 

[rubi]

To support the statement that that "it doesn't matter who observes" please take my example of measurement operators for Earthpersons Alice, Bob, and Martians Ylc and Zog and calculate what the Martians observe when they take partial measurements of a state that the Earthpersons observe as separable.

Well, I have explained it already in my post #27. Vectors transform as $\psi^\prime = U\psi$ and the observables transform as $O^\prime = U O U^{-1}$. You can now calculate:
$\left<\phi^\prime,O^\prime \psi^\prime\right> = \left<U\phi,U O U^{-1} U \psi\right> = \left<\phi,U^\dagger U O U^{-1} U \psi\right> = \left<\phi,U^{-1} U O U^{-1} U \psi\right> = \left<\phi,O\psi\right>$
So all predictions of the theory are independent of any unitarily equivalent choices that can be made (such as choices of observers or choices of Hilbert spaces and so on). Since this is a general proof, it also applies to your specific situation.

 

[Robert Shaw]

Well, I have explained it already in my post #27. Vectors transform as $\psi^\prime = U\psi$ and the observables transform as $O^\prime = U O U^{-1}$. You can now calculate:
$\left<\phi^\prime,O^\prime \psi^\prime\right> = \left<U\phi,U O U^{-1} U \psi\right> = \left<\phi,U^\dagger U O U^{-1} U \psi\right> = \left<\phi,U^{-1} U O U^{-1} U \psi\right> = \left<\phi,O\psi\right>$
So all predictions of the theory are independent of any unitarily equivalent choices that can be made (such as choices of observers or choices of Hilbert spaces and so on). Since this is a general proof, it also applies to your specific situation.

You seem to be ignoring the fact that all the measurement operators in my example are shown as matrices using the same basis set.

I'll run you through.

Alice measures Alice(Up) or Alice(Down). Her operator is diag(1,1,-1,-1). Bob's operator is diag(1,-1,1,-1).

Ylc measures Ylc(Green) and Ylc(Blue) with operator antidiag(1,1,1,1) and Zog has antidiag(1,-1-1,1).

A state preparation procedure is found to prepare a state |1,0,0,0>.

Alice and Bob make observations. Every measurement is the same Alice(Up) and Bob(Up).

Ylg and Zog make observations. Ylg has a mix of 50/50 Green/Blue. So does Zog. However Zog always observes Blue whenever Ylg sees Green.

The important thing is that the matrices for the operators are defined using the same basis. All 4 operators are Hermitian and so are legitimate.

 

[rubi]

You seem to be ignoring the fact that all the measurement operators in my example are shown as matrices using the same basis set.

That's wrong. If you use transformed operators, you must also use transformed states. Otherwise, it is not a legitimate calculation.

The important thing is that the matrices for the operators are defined using the same basis.

You can write them down in any basis, since the result will be independent of the choice of basis. You just need to make sure that all quantities have been tranformed appropriately and you didn't do that in your example.

 

[Robert Shaw]

That's wrong. If you use transformed operators, you must also use transformed states. Otherwise, it is not a legitimate calculation.You can write them down in any basis, since the result will be independent of the choice of basis. You just need to make sure that all quantities have been tranformed appropriately and you didn't do that in your example.

For a qubit:

1) you can have different types of apparatus for measuring the state. They give different results.

2) for example we can have one that measures z-spin and another x-spin

3) The two different apparatus have two different operators, sigma-Z and sigma-X.

4) the operators are Hermitian and their eigenstates span Hilbert spaceFor my example:

1) there is an Earthperson apparatus and a Martian apparatus. They give different results for the same state. Nothing wrong with that.

2) the Earth apparatus measures Up/Down. The Martian apparatus measures Green/Blue (analogous to the Z and X)

3) The two types of apparatus have different operators (just as in the qubit example).

4) The operators are Hermitian and they span Hilbert space.

From these follow my calculations.

 

[Jilang]

Why would you expect all properities to be entangled?

 

[rubi]

2) the Earth apparatus measures Up/Down. The Martian apparatus measures Green/Blue (analogous to the Z and X)

If the Martians don't measure the same subsystems, i.e. the spins of Alice's and Bob's particles, then of course they needn't detect entanglement between their subsystems. That's not surprising at all and I had already said that in my post #8. But if the Martians measure the subsystems of Alice and Bob from their reference frame, they will of course still find that Alice's and Bob's subsystems are entangled. A change of reference frame plays no role as I have shown in #42.

 

[Zafa Pi]

That's a good point. Separability is a mathematical property.

The physical status of separability is not clear.

The math property: A pair of particles represented by the state {a,b,c,d] (in the 4D tensor product of two 2D Hilbert spaces) is separable if:
[a,b,c,d] = [u,v]⊗[x,y] =[ux,uy,vx,vy].
In this case the random variable given by an observable A applied to [u,v] is independent of of the random variable given by an observable B applied to [x,y].
When the proper correspondences are made to the lab/physical status (e.g. the particles are photons, the observables are polarization analyzers) then we still get independence.

If the state is not separable, i.e. entangled, then the respective random variables are not independent (except in rare cases).
And if I interpret @rubi correctly, any observer will notice that.

For Alice the partial measurement suboperator is diag(1,1,-1,-1) and Bob is diag(1,-1,1,-1).

I don't understand, the measurement operator/observable for Alice is is a 2D Hermitian matrix, like Z = diag(1,-1). In fact I don't understand your post #39 at all.

 

[Robert Shaw]

If the Martians don't measure the same subsystems, i.e. the spins of Alice's and Bob's particles, then of course they needn't detect entanglement between their subsystems. That's not surprising at all and I had already said that in my post #8. But if the Martians measure the subsystems of Alice and Bob from their reference frame, they will of course still find that Alice's and Bob's subsystems are entangled. A change of reference frame plays no role as I have shown in #42.

Exactly.

The system is separable from Alice Bob frame of reference.

It is entangled from Martian frame of reference; as you say the Martians see different subsystems.

Hence the title question for this post.

The answer is that separability and entanglement are dependent on the frame of reference.

There have been several references posted in this thread to academic papers that discuss this issue. They are well worth reading.

 

[rubi]

Exactly.

The system is separable from Alice Bob frame of reference.

It is entangled from Martian frame of reference because they see different subsystems.

No, that's false, read my post again, I said exactly the opposite. The entanglement is independent of the reference frame. Every observer in every reference frame will conclude that Alice and Bob's particles are entangled. See post #42 for a proof. If the Martians make a conclusion about Alice and Bob's subsystems, the don't see different subsystems. They see the same subsystems from from a different reference frame.

Hence the title of this post.

The answer to the question is that separability and entanglement are dependent on the frame of reference.

Entanglement is independent of the reference frame. A change of reference frame is induced by unitary transformation and as post #42 proves, unitary transformations won't change any physically detectable fact about the world. If you deny this, then please point out the mistake in the utterly trivial proof I have given.

 

[facenian]

Consider the 4 maximally entangled states, eg |ud>-|du>. If a Martian chose them for basis states then he would see them as separable

I see a tautology here, |ud>-|du> is en entangled state because it cannot be put in the form $| a>|b>$ where a and b are vectors in their respective spaces. I you redefine |ud>-|du> as a new vector in the product space you have simply mathematically erased entanglement by redefining the Hilbert space that now, in a sense, is no longer a product space because now |ud>-|du> is by definition a single vector.
This way of looking at entanglement have no physical implications.

 

[Robert Shaw]

No, that's false, read my post again, I said exactly the opposite. The entanglement is independent of the reference frame. Every observer in every reference frame will conclude that Alice and Bob's particles are entangled. See post #42 for a proof. If the Martians make a conclusion about Alice and Bob's subsystems, the don't see different subsystems. They see the same subsystems from from a different reference frame.Entanglement is independent of the reference frame. A change of reference frame is induced by unitary transformation and as post #42 proves, unitary transformations won't change any physically detectable fact about the world. If you deny this, then please point out the mistake in the utterly trivial proof I have given.

Please find some references below that appear to contradict your position:Observables can be tailored to change the entanglement of any pure state
N. L. Harshman and Kedar S. Ranade
Phys. Rev. A 84, 012303 – Published 5 July 2011

Observables and entanglement in the two-body system
N. L. Harshman
AIP Conference Proceedings 1508, 386 (2012)

Entanglement or separability: The choice of how to factorize the algebra of a density matrix
Walter Thirring, Reinhold A. Bertlmann, Philipp Köhler, Heide Narnhofer
https://arxiv.org/abs/1106.3047

Factorization and entanglement in quantum systems

Jon Eakins and George Jaroszkiewicz

Published 17 December 2002 • Journal of Physics A: Mathematical and General,Volume 36, Number 2

 

[rubi]

Please find some references below that appear to contradict your position:Observables can be tailored to change the entanglement of any pure state
N. L. Harshman and Kedar S. Ranade
Phys. Rev. A 84, 012303 – Published 5 July 2011

Observables and entanglement in the two-body system
N. L. Harshman
AIP Conference Proceedings 1508, 386 (2012)

Entanglement or separability: The choice of how to factorize the algebra of a density matrix
Walter Thirring, Reinhold A. Bertlmann, Philipp Köhler, Heide Narnhofer
https://arxiv.org/abs/1106.3047

Factorization and entanglement in quantum systems

Jon Eakins and George Jaroszkiewicz

Published 17 December 2002 • Journal of Physics A: Mathematical and General,Volume 36, Number 2

All these references support my position, because they are concerned with different choices of subsystems. A change of reference frame, however, does not change the subsystem and hence does not affect the entanglement of the choosen subsystems. There are only four equal signs in my proof in post #42. If you disagree, then please explain, which one of them is incorrect.

 

[Robert Shaw]

Please find some references below that appear to contradict your position:Observables can be tailored to change the entanglement of any pure state
N. L. Harshman and Kedar S. Ranade
Phys. Rev. A 84, 012303 – Published 5 July 2011

Observables and entanglement in the two-body system
N. L. Harshman
AIP Conference Proceedings 1508, 386 (2012)

Entanglement or separability: The choice of how to factorize the algebra of a density matrix
Walter Thirring, Reinhold A. Bertlmann, Philipp Köhler, Heide Narnhofer
https://arxiv.org/abs/1106.3047

Factorization and entanglement in quantum systems

Jon Eakins and George Jaroszkiewicz

Published 17 December 2002 • Journal of Physics A: Mathematical and General,Volume 36, Number 2

You seem to be linking "reference frame" to "unitary transformation" in a way that makes no sense in the context of my question.

Of course a unitary transformation will trivially produce the result you cite. That is not the point I am making.

I am talking about two Earth observers and two Martian observers. Their observable operators are different.

When the Martians make observations on the same state as the Earthpersons they get different results and draw different conclusions about separability.

You seem to be making an assumption that the reduced states for Earth and Mars are the same but that is not correct in the case of the model we are discussing.

 

[Robert Shaw]

You seem to be linking "reference frame" to "unitary transformation" in a way that makes no sense in the context of my question.

Of course a unitary transformation will trivially produce the result you cite. That is not the point I am making.

I am talking about two Earth observers and two Martian observers. Their observable operators are different.

When the Martians make observations on the same state as the Earthpersons they get different results and draw different conclusions about separability.

You seem to be making an assumption that the reduced states for Earth and Mars are the same but that is not correct in the case of the model we are discussing.

In your terminology, you are assuming that subsystems for Earth and Mars are the same. That assumption is neither necessary nor correct.

 

[rubi]

You seem to be linking "reference frame" to "unitary transformation" in a way that makes no sense in the context of my question.

A change of reference frame is always induced by a unitary transformation. It absolutely is relevant to your question and you should make an effort to understand its relevance. It seems to me that you haven't understood how reference frames are dealt with in quantum theory.

I am talking about two Earth observers and two Martian observers. Their observable operators are different.

If they are not related by a unitary transformation, then they concern different subsystems and then trivially, the notion of entanglement changes as I have already said in my post #8. However, it has absolutely nothing to do with reference frames. A choice of reference frame still does not influence entanglement in the slightest.

You seem to be making an assumption that the reduced states for Earth and Mars are the same but that is not correct in the case of the model we are discussing.

If the same subsystems are considered, then the Martians will obtain the same reduced states (up to unitary equivalence). It's trivial that different subsystems needn't be entangled. Just because Alice is entangled to Bob, it doesn't mean Eve is entangled to Charlie. Not everything is entangled with everything. So what? However, Eve and Charlie will still consider Alice and Bob to be entangled.

In your terminology, you are assuming that subsystems for Earth and Mars are the same. That assumption is neither necessary nor correct.

It has nothing to do with Earth and Mars. Alice is one subsystem and Bob is another subsystem. It doesn't matter whether you observe those subsystems from Earth or Mars.

 

[facenian]

You seem to be linking "reference frame" to "unitary transformation" in a way that makes no sense in the context of my question.

A change of reference frame is always induced by a unitary transformation. It absolutely is relevant to your question and you should make an effort to understand its relevance. It seems to me that you haven't understood how reference frames are dealt with in quantum theory

I believe rubi is right, you can pick an arbitrary observer but this does not mean you can select an arbitrary sate, different observers are constrained by unitary transformations by Wigner's theorem.

 

[kith]

I think you are talking past each because Robert partly uses the terminology in a nonstandard way. Wenn Robert says "reference frame", he refers to what is more commonly called "tensor product factorization of the Hilbert space". (I think everybody here agrees that for a given factorization, the amount of entanglement of a state doesn't depend on the basis)

The second reference in my post #34 gives an example where different factorizations seem to be at least conceptually meaningful: we can look at the hydrogen atom as a two-particle system which consists of an electron and a proton or we can look at it as a center-of-mass degree of freedom and an "effective" electron in a potential. In the first factorization, a typical state is entangled, in the second, it is not (the COM movement is independent of what happens internally in the absence of external fields).

But in Robert's example, what does it mean to use the wildly different factorization of the Martians? It would yield strange non-local subsystems (two degrees of freedem each of which somehow includes half of Alice's particle and half of Bob's particle). Robert sidesteppes this important physical point by saying that the Earthians and the Martians "observe the same state" (of the combined system). But neither the Earthians nor the Martians observe states, they observe the behavior of subsystems.

So what's missing is how could measurements on the subsystems which are consistent with the Martian factorization be implemented? As far as I can see, the cited references don't claim that this is possible but instead note that what they do is mostly of theoretical interest and may lead to better calculation techniques.

 

[kith]

Also to be clear: if spatially localized observers like Alice and Bob don't have access to such factorizations in principle, this would imply that the Martians themselves have to be non-local entities. Which would kind of put this example into the realm of esoterism. ;-)

 

[Robert Shaw]

Also to be clear: if spatially localized observers like Alice and Bob don't have access to such factorizations in principle, this would imply that the Martians themselves have to be non-local entities. Which would kind of put this example into the realm of esoterism. ;-)

Localisation does not exist in a qubit model. Alice and Bob are not spatial. To make them so would require a much more complex model, with position variables as well as Up/Down. Same for two qubit models.

Quantum information theory studies such simple models because they lead to significant discoveries. I would not dismiss them as esoteric.

They are certainly idealisations. But everything in physics is an idealisation. Engineers try to model reality, and do a pretty good job.That's why I set up the simple toy model. It's what we physicists use, we seldom study highly complex multiparameter models.

Separability is not problem-free in the way you suggest. Even in classical physics 6N dynamic variables could either be N objects with 6 parameters or 6 objects with N parameters. "Particle" has no universally accepted definition in physics.

There is much to discover concerning the problem of separability. It certainly isn't fully explored.