I Do we need a reference frame in Quantum Hilbert space?

[Robert Shaw]

Entangled states are only separable relative to certain basis states. So does that mean that reference frames have importance beyond those in spacetime?

 

[facenian]

Are you asociating a basis of Hilber space to a specific inertial frame?

 

[atyy]

Whether entanglement is basis dependent or not is a matter of convention.

If one uses the entanglement entropy to describe entanglement, then entanglement is basis independent.
https://www-thphys.physics.ox.ac.uk/people/JohnCardy/seminars/born1.pdf

However, there are other, basis-dependent notions of entanglement.
https://arxiv.org/abs/quant-ph/0507189v4
https://arxiv.org/abs/1009.4147
https://arxiv.org/abs/1302.3509

 

[rubi]

Entanglement entropy is one way to recognize basis independence. But already the standard definition is basis independent: A vector $\psi\in\mathcal H\otimes\mathcal H$ is called separable if there are $\zeta,\xi\in\mathcal H$ such that $\psi = \zeta\otimes\xi$. Otherwise, it is called entangled. No reference to a basis is made in this definition.

 

[facenian]

Whether entanglement is basis dependent or not is a matter of convention.

Does this mean that entaglement as a physical phenomenon is conventional?

 

[Demystifier]

Once we split the system into subsystems, the entanglement does not depend on the choice of bases for each subsystem. However, there is no entanglement without the split into subsystems, and such a split may not be unique. If the choice of basis for the whole system is associated with such a split, it can be said that entanglement also depends on the choice of basis.

 

[Robert Shaw]

Does this mean that entaglement as a physical phenomenon is conventional?

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

 

[rubi]

Once we split the system into subsystems, the entanglement does not depend on the choice of bases for each subsystem. However, there is no entanglement without the split into subsystems, and such a split may not be unique. If the choice of basis for the whole system is associated with such a split, it can be said that entanglement also depends on the choice of basis.

Entanglement depends on the split of subsystems, but a split of subsystems does not fix a basis and a choice of basis does not induce a split of subsystems.

Edit:
To make this mathematicall precise: A split into subsystems of a Hilbert space $\mathcal H$ is a tuple $(\mathcal H_A, \mathcal H_B, U)$, where $\mathcal H_A, \mathcal H_B$ are Hilbert spaces and $U:\mathcal H\rightarrow\mathcal H_A\otimes\mathcal H_B$ is an isomorphism, i.e. a unitary operator. Such a split may be non-unique, but it is unrelated to a choice of basis.

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

No, because they still can't be written as $\zeta\otimes\xi$. The choice of basis doesn't affect this fact.

 

[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

Based on the other post's explanations, this is not the correct way to interpret entanglement.

 

[Robert Shaw]

Based on the other post's explanations, this is not the correct way to interpret entanglement.

Separability is a matter of choice as this paper makes clear https://arxiv.org/abs/1106.3047
("Entanglement or separability,the choice of how to factorise"...Bertlmann being John Bells friend and collaborator)

 

[Demystifier]

Entanglement depends on the split of subsystems, but a split of subsystems does not fix a basis and a choice of basis does not induce a split of subsystems.

If the system is split into subsystems $A$ and $B$, then it is natural (though not necessary) to choose a basis in which each member has a form $|\psi_A\rangle \otimes |\phi_B\rangle$. How would you take such a basis without the split?

 

[Robert Shaw]

Separability is a matter of choice as this paper makes clear https://arxiv.org/abs/1106.3047
("Entanglement or separability,the choice of how to factorise"...Bertlmann being John Bells friend and collaborator)

Please would you look at it this way.

Take two 2D vector spaces Zeta and Xi.

Consider the Cartesian product set (Zeta,Xi). One create maps from this set to the 4D vector space Psi.

In particular consider maps that are bilinear. There are many such maps.

An earthperson chooses one such map. They note that there are many vectors in the space Psi that are not separable.

A Martian chooses a different map. They choose to map to states that were not separable to the Earthperson.

So the Earthperson separable states are not separable to the Martian and vice versa.

The paper by Bertlmann and co give a fuller perspective.

 

[rubi]

If the system is split into subsystems $A$ and $B$, then it is natural (though not necessary) to choose a basis in which each member has a form $|\psi_A\rangle \otimes |\phi_B\rangle$. How would you take such a basis without the split?

Well, first of all, you still need to make infinitely many arbitrary choices here, because the vectors $\psi_A$ and $\phi_B$ are not fixed by the split, so the split does not induce a choice of basis. And the other way around, a choice of basis does not induce a split. It needn't even be compatible with any split. A split and a basis are very different concepts. What choice of basis is natural depends on the problem one wants to solve. It is most natural not to choose a basis at all and it is often natural to choose a basis, such that some important operator becomes diagonal.

 

[kith]

From the naive experimental point of view, talking about "chosing how to split a system" seems really strange.

Let's say we want to study the behaviour of two particles in the lab. This physical situation by itself already fixes the "split" of the combined system of both particles into the two subsystems of the individual particles. Whether there's entanglement or not, is nothing which depends on our (or someone else's) description of the situation. We simply perform the necessary measurements on the subsystems and we either get non-classical correlations or we don't.

If, on the other hand, we want to study the behavior of the combined system as a whole, introducing a split at all isn't physically meaningful. As long as we are not performing experiments on subsystems, it is a purely formal act to write the Hilbert space as a tensor product of two other spaces and it is thus a purely formal act to discuss the separability of a state with respect to any such factorization.

In the example above, the experimenter has two subsystems (the two particles). He doesn't have a big system which he can arbitrarily split into subsystems. So two descriptions which describe the state as either entangled or separable are not on equal footing. If we go to the lab and do the measurements, we only see one kind of behaviour.

So are there non-trivial examples where the experimenter can chose different splits of a system into subsystems (corresponding to different factorizations of the Hilbert space) such that he can observe both kinds of behaviour?

 

[kith]

Disregarding experiments, it may be interesting to discuss different factorizations anyway, especially when it comes to the ontology of QM. You may be interested in the following (arxiv-only) article which has been discussed here before:

https://arxiv.org/abs/1210.8447
"Nothing happens in the Universe of the Everett Interpretation" by J. Schwindt

 

[Zafa Pi]

Take two 2D vector spaces Zeta and Xi.

Consider the Cartesian product set (Zeta,Xi). One create maps from this set to the 4D vector space Psi.

A vector ψ∈H⊗Hψ∈H⊗H\psi\in\mathcal H\otimes\mathcal H is called separable if there are ζ,ξ∈Hζ,ξ∈H\zeta,\xi\in\mathcal H such that ψ=ζ⊗ξψ=ζ⊗ξ\psi = \zeta\otimes\xi. Otherwise, it is called entangled. No reference to a basis is made in this definition.

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.

 

[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.

I was merely following the definition of a tensor product: In mathematics, the tensor product V ⊗ W of two vector spaces V and W (over the same field) is itself a vector space, together with an operation of bilinear composition, denoted by ⊗, from ordered pairs in the Cartesian product V × W into V ⊗ W, in a way that generalizes the outer product.

Happy to furnish an example

 

[Robert Shaw]

I was merely following the definition of a tensor product: In mathematics, the tensor product V ⊗ W of two vector spaces V and W (over the same field) is itself a vector space, together with an operation of bilinear composition, denoted by ⊗, from ordered pairs in the Cartesian product V × W into V ⊗ W, in a way that generalizes the outer product.

Happy to furnish an example

Also a Cartesian product is just an ordered pair not a vector. So [1,0]X[0,1] is not a 4 vector, it only relates to 4vectors as a consequence of a bilinear map being defined.

 

[Robert Shaw]

Also a Cartesian product is just an ordered pair not a vector. So [1,0]X[0,1] is not a 4 vector, it only relates to 4vectors as a consequence of a bilinear map being defined.

A Cartesian product is something in set theory, it has no intrinsic vector properties until they are created by the bilinear map

 

[Robert Shaw]

A Cartesian product is something in set theory, it has no intrinsic vector properties until they are created by the bilinear map

EXAMPLE

The map is fully specified in 2x2 case when 4 relationships are defined. Bilinearity then fills the space.

Let's choose to define maps using vectors A and B from first space and K and L from the second. For simplicity let's assume A/B and K/L are orthogonal. There are 4 ordered pairs, AK, AL, BK and BL.

The earthperson chooses vectors R,S,T,U from the 4 space. They are orthogonal. He maps AK to R and so on. Bilinearity then fills the rest.

The Martian chooses vectors V,W,X,Y that are orthogonal. They are Bell vectors, eg on Earth basis (0,1,1,0) etc. He maps AK to V etc. This fulfils the requirements of a bilinear map and the 4space can be spanned in this way too.

Both bilinear maps are legitimate, because there is no unique mapping, there are infinitely many possible bilinear maps.

What is interesting is that cartesian pairs do not cover all the 4space. The Cartesian pairs only represent vectors that are separable (to use quantum terminology). The rest of the 4space is non-separable vectors, or in quantum terms entangled.

 

[Robert Shaw]

Disregarding experiments, it may be interesting to discuss different factorizations anyway, especially when it comes to the ontology of QM. You may be interested in the following (arxiv-only) article which has been discussed here before:

https://arxiv.org/abs/1210.8447
"Nothing happens in the Universe of the Everett Interpretation" by J. Schwindt

Excellent paper, thanks for the reference

 

[Robert Shaw]

A Cartesian product is something in set theory, it has no intrinsic vector properties until they are created by the bilinear map

As Schwindt puts it (see reference above)

"A state vector gets the property of “representing a structure” only with respect toan external observer who measures the state according to a specific factorization
and basis."

 

[rubi]

It's not really the factorization that is needed. The factorization is arbitrary and has no physical relevance, since all unitarily equivalent quantum theories describe the same physics. A quantum theory always comes with an algebra of observables and a subsystem of a quantum system is characterized by a subalgebra of this algebra (i.e. take the subalgebra of observables of particle 1 only). If you have a pure quantum state on the full algebra, it may become mixed when restricted to a subalgebra. The Schwindt paper doesn't take the observables into account and is thus pretty useless.

 

[Robert Shaw]

It's not really the factorization that is needed. The factorization is arbitrary and has no physical relevance, since all unitarily equivalent quantum theories describe the same physics. A quantum theory always comes with an algebra of observables and a subsystem of a quantum system is characterized by a subalgebra of this algebra (i.e. take the subalgebra of observables of particle 1 only). If you have a pure quantum state on the full algebra, it may become mixed when restricted to a subalgebra. The Schwindt paper doesn't take the observables into account and is thus pretty useless.

That's a very good point. The algebra of observables is critical.

However, how many textbooks on QM make that point?

The usual approach in most textbooks is to focus on states and discuss separability as though it's an absolute, rather than relative to the algebra of observables.

Do you have any recommendations on textbooks that make clear the importance of the algebra of observables in understanding QM?

 

[Robert Shaw]

That's a very good point. The algebra of observables is critical.

However, how many textbooks on QM make that point?

The usual approach in most textbooks is to focus on states and discuss separability as though it's an absolute, rather than relative to the algebra of observables.

Do you have any recommendations on textbooks that make clear the importance of the algebra of observables in understanding QM?

e.g. in Ballentine's book, separability is NOT introduced in the context of observables but rather as if it is a property of the state space itself.

 

[Robert Shaw]

It's not really the factorization that is needed. The factorization is arbitrary and has no physical relevance, since all unitarily equivalent quantum theories describe the same physics. A quantum theory always comes with an algebra of observables and a subsystem of a quantum system is characterized by a subalgebra of this algebra (i.e. take the subalgebra of observables of particle 1 only). If you have a pure quantum state on the full algebra, it may become mixed when restricted to a subalgebra. The Schwindt paper doesn't take the observables into account and is thus pretty useless.

On reflection, the algebra of observables does not resolve the puzzle.

Earthperson would have a particle 1 and 2 suboperators E1 and E2. The state (1,0,0,0) on Earth basis would be measured as up for particle 1 and 2.

Martian could have suboperator M1 and M2. It would measure up for particle 1 and 2 for state (1,0,0,1) on Earth basis. Earthperson would see the same state as entangled.

Maybe the resolution is Quantum Censorship - to forbid alternative measurement systems (forbidding martians to speak)

 

[rubi]

Do you have any recommendations on textbooks that make clear the importance of the algebra of observables in understanding QM?

A good textbook that highlights the algebraic foundations of quantum mechanics is "An Introduction to the Mathematical Structure of Quantum Mechanics" by Strocchi.

Martian could have suboperator M1 and M2. It would measure up for particle 1 and 2 for state (1,0,0,1) on Earth basis. Earthperson would see the same state as entangled.

No, because the Martian would have to use the transformed operator if he uses an unitarily equivalent Hilbert space. Both states ($\psi \rightarrow U \psi$) and operators ($O \rightarrow U O U^{-1}$) need to be transformed and physics is invariant under such unitary equivalences. The structure of the Hilbert space can't be detected by any experiment.

 

[vanhees71]

This would be strange already from a mathematical point of view since there's essentially only one separable Hilbert space, which is the arena of (non-relativistic) QM.

 

[facenian]

This would be strange already from a mathematical point of view since there's essentially only one separable Hilbert space, which is the arena of (non-relativistic) QM.

Does this mean that the arena for QFT is different?

 

[vanhees71]

I'm not so sure about this. For a properly regularized Fock space, I'd say so, but maybe not in the infinite-volume limit. I guess the more mathematical forum users are more qualified to answer this.

 

[A. Neumaier]

there's essentially only one separable Hilbert space, which is the arena of (non-relativistic) QM.

... 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.

 

[Robert Shaw]

A good textbook that highlights the algebraic foundations of quantum mechanics is "An Introduction to the Mathematical Structure of Quantum Mechanics" by Strocchi.No, because the Martian would have to use the transformed operator if he uses an unitarily equivalent Hilbert space. Both states ($\psi \rightarrow U \psi$) and operators ($O \rightarrow U O U^{-1}$) need to be transformed and physics is invariant under such unitary equivalences. The structure of the Hilbert space can't be detected by any experiment.

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.

 

[Robert Shaw]

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.

Many thanks for the book recommendation, will get a copy

 

[kith]

Let me point out two more references regarding the factorization of the Hilbert space.

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)

 

[Robert Shaw]

Let me point out two more references regarding the factorization of the Hilbert space.

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)

Thanks, I've seen them already and they are excellent

 

[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.