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.