# I Does quantum entanglement depend on the chosen basis?

#### microsansfil

Hi,

In this presentation about quantum optics it is mentioned that the same quantum state |Ψ> has different expressions in different mode bases : factorized state or entangled state.

This presentation is related to this video :

In some way entanglement isn't intrinsic. It depend on the way you look at the mode !

/Patrick

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#### mfb

Mentor
The factorized states consist of multiple particles. The property "the states of the two particles are entangled" is independent of the basis. You are just hiding it in the base chosen.

I discussed a similar example here where you can hide the time-evolution of a system by a clever choice of coordinates.

#### microsansfil

The factorized states consist of multiple particles.
Yes, Claude Fabre speak about Fock state : |p1 : f1, p2 : f2, ...> with pi number of photon.

However, in the example given :

|Ψ> = |1 : fx> ⊗ |1 : fy> which is not entangled state on the basis of polarization modes fx fy
The same state |Ψ> can also be written : |2 : g45>⊗ |0 : g-45> - |0 : g45>⊗ |1 : g-45> which is an entangled state on the basis of polarization modes g45 g-45

/Patrick

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#### stevendaryl

Staff Emeritus
I would say that two subsystems are entangled if there is no way to write the state as a product of states. So if there is some basis in which the systems are not entangled, then they are not entangled in any basis.

#### microsansfil

That's what I understood too. Now, Claude Fabre of the LKB is not a novice in this field, hence my astonishment.

In the context of quantum optics, the notion of entanglement would be more subtle ?

/Patrick

#### atyy

Entanglement can be basis dependent or basis independent, depending on the definition. I'll add references later.

#### DrChinese

Gold Member
So if there is some basis in which the systems are not entangled, then they are not entangled in any basis.
Note sure if this is what you are referring to or not. I believe there can be photons pairs that are momentum entangled, but not polarization entangled. Type I SPDC using only a single crystal would do that. See Figure 8, left diagram. (Not entirely clear that it is type I here.)

https://www.researchgate.net/publication/2186486_Photonic_Entanglement_for_Fundamental_Tests_and_Quantum_Communication

#### martinbn

Note sure if this is what you are referring to or not.
I don't think that's what he means. Take two quibits in the state $\psi=|-\rangle |-\rangle-|-\rangle |+\rangle+|+\rangle |-\rangle-|+\rangle |+\rangle$. It may look entangled but in the bases $\{|0\rangle,|1\rangle\}$, defined by $|0\rangle=|-\rangle+|+\rangle$ and $|1\rangle=|-\rangle-|+\rangle$, the state is $\psi=|0\rangle|1\rangle$. That's why the definition says that the state is not entangled if there is a basis in which the state is a pure tensor (product).

#### microsansfil

Take two quibits in the state $\psi=|-\rangle |-\rangle-|-\rangle |+\rangle+|+\rangle |-\rangle-|+\rangle |+\rangle$. It may look entangled
it seems to me that it isn't entangled because it could be factorized $\psi=(|-\rangle +|+\rangle)(|-\rangle -|+\rangle)$

/Patrick

#### martinbn

Yes, that's the point I am making. And that's what stevendaryl said.

#### vanhees71

Gold Member
I'd also say that it doesn't make the slightest sense to say something is "entangled". It's analogous to the sentence "Alice is bigger", which makes no sense of its own, because it doesn't say in which sense "Alice is bigger". You may immediately ask, "bigger compared to what?".

The same holds for "entanglement". You have to tell which observables of the quantum system under investigation are entangled to make sense of "entanglement". E.g., in the Stern-Gerlach apparatus the magnetic moment of an uncharged particle (or atom in the original setup of 1922) in quantization direction (i.e., in direction of the large homogeneous part of the magnetic field) gets entangled with the position of the particle.

Another often used example (because it's so easily experimentally realizable these days) are the polarizations of two photons in a single-photon state.

This shows that there's no more confusion about the choice of basis to decide whether a (pure) states describes "entanglement" or "no entanglement", because it's clear that you have to use the eigenbasis of the self-adjoint operators describing the observables to express the state in question to see whether it's describing entanglement between the observables in question or not, and then it's unique, whether these observables are entangled or not. By definition two observables are entangled in a given state if the state is not a pure product state in the uniquely defined orthonormal basis of the putatively entangled observables.

I haven't fully understood the above example of polarization states, but I guess it's the following: In the first example the observables under consideration are either polarization in 0 and 90-degree direction (relative to an arbitrary direction in the plane perpendicular to the photons' momenta) vs. polarization in $\pm 45^{\circ}$. In the latter case the state is obviously entangled, while it's not in the former. Of course, there's no contradiction in this, because the measurement of polarization in 0 and $90^{\circ}$ direction is incompatible with the measurement of polarization in $\pm 45^{\circ}$ direction. The most simple experimental setup is to use a polaroid foil directed in $0$ direction or $45^{\circ}$ direction. You cannot have both direction in the same setup. In the first setup, if the measured photon gets through its polarization is in $0^{\circ}$ direction with certainty, but it's polarized with 50% in $45^{\circ}$ and with 50% in $-45^{\circ}$ direction an vice versa. That's why there's no contradiction in the observation that the above discussed two-photon state is not entangled with respect to the observable "single-photon polarizations in $0$ or $90^{\circ}$ direction" but that it is entangled with respect to the observable "single-photon polarizations in $+45^{\circ}$ or $-45^{\circ}$ direction.

Of course, things become more involved if the eigenvalues of the involved observable operators are degenerate. Then I'd define that the observables are entangled if and only if there's no complete orthonormal basis wrt. which the state is a product state.

#### stevendaryl

Staff Emeritus
I'd also say that it doesn't make the slightest sense to say something is "entangled". It's analogous to the sentence "Alice is bigger", which makes no sense of its own, because it doesn't say in which sense "Alice is bigger". You may immediately ask, "bigger compared to what?".

The same holds for "entanglement". You have to tell which observables of the quantum system under investigation are entangled to make sense of "entanglement". E.g., in the Stern-Gerlach apparatus the magnetic moment of an uncharged particle (or atom in the original setup of 1922) in quantization direction (i.e., in direction of the large homogeneous part of the magnetic field) gets entangled with the position of the particle.
Well, the definition that two systems are entangled if their state cannot be written as a product makes some sense as a definition. However, spatially separated identical particles are always entangled by this definition. Two electrons will always have a composite state that looks like $\frac{1}{\sqrt{2}} ( |\psi\rangle |\phi\rangle - |\phi\rangle |\psi\rangle)$.

#### microsansfil

Well, the definition that two systems are entangled if their state cannot be written as a product
I learned the same mathematical definition.

vanhees71 said:
You have to tell which observables of the quantum system under investigation are entangled to make sense of "entanglement".
What can be the definition of entanglement concerning observables? It is about Operators' tensor product ? Two observables are entangled if they cannot be written as a product ?

/Patrick

#### vanhees71

Gold Member
Well, the definition that two systems are entangled if their state cannot be written as a product makes some sense as a definition. However, spatially separated identical particles are always entangled by this definition. Two electrons will always have a composite state that looks like $\frac{1}{\sqrt{2}} ( |\psi\rangle |\phi\rangle - |\phi\rangle |\psi\rangle)$.
As I tried to explain above your first sentence is at best incomplete. You have to state with respect to which basis, i.e., with respect to which observables you are considering the question of entanglement, the pure-state-representing vector is a product state. Of course, indistinguishable paricle's state vectors are (almost) always entangled states with respect to any one-particle basis due to being fermions or bosons, since the Hilbert space is spanned by antisymmetrized or symmetrized products of single-particle basis vectors. Only bosons can be in unentangled states, namely the product state made of only one single-particle state.

#### stevendaryl

Staff Emeritus
As I tried to explain above your first sentence is at best incomplete. You have to state with respect to which basis,
Not if your definition is "Two subsystems are entangled if there does not exist a basis in which the state is a product state." That's not incomplete.

#### vanhees71

Gold Member
That IS incomplete. You have to state with respect to which observables a given state of a system describes entanglement. The example, this entire thread is about is a simple example for that:

With respect to the single-photon polarization states $|H \rangle$, $|V \rangle$, which belong to the observable, say $|H \rangle \langle H|$, which tests, whether a photon is horizontally polarized (it describes an ideal polarization filter, letting a photon with certainty through if it's horizontally polarized). Then the OP considered the two-photon-polarization state
$$|\Psi \rangle=|H \rangle \otimes |V \rangle,$$
which clearly shows that the two photons's $H$-$V$ polarization observables are not entangled, because it's a product state in the corresponding eigenbasis.

Now what about the "vertically" polarization observable? Its basis is
$$| \pm \rangle=\frac{1}{\sqrt{2}} (|H \rangle \pm |V \rangle).$$
Obviously we have
$$|\Psi \rangle=\frac{1}{2} (|+\rangle +|-\rangle) \otimes (|+\rangle - |- \rangle),$$
which obviously is not a product state in the $|\pm \rangle$ basis, and thus wrt. the "vertical polarization" observable the same state is entangled.

#### stevendaryl

Staff Emeritus
That IS incomplete. You have to state with respect to which observables a given state of a system describes entanglement.
No, you don't. You can say that if there exists a basis under which they can be written as a product state, then they are not entangled.

#### vanhees71

Gold Member
Hm, that's not very concrete and adds to the confusion rather than helping with the somewhat unintuitive physics of entanglement, and then people again come and say QT is weird :-(.

#### stevendaryl

Staff Emeritus
Hm, that's not very concrete and adds to the confusion rather than helping with the somewhat unintuitive physics of entanglement, and then people again come and say QT is weird :-(.
I would say on the contrary that if something is only entangled in a particular basis, then none of the interesting properties of entanglement come into play.

The interesting thing about the state $\frac{1}{\sqrt{2}}(|u\rangle |d\rangle - |d\rangle |u\rangle)$ is that it doesn't matter what basis you choose; the state cannot be written as a product state. In contrast, the state $\frac{1}{2}(|u\rangle |u\rangle - |d\rangle |u\rangle - |u\rangle |d\rangle + |d\rangle |d\rangle)$ may look entangled, but there are no interesting consequences of that entanglement.

#### vanhees71

Gold Member
This is special, because it's a state of total angular momentum (or spin in this case) 0. Of course that's a rotaionally invariant state, and that's why it's entangled wrt. any angular-moment direction. I'd call this a "Bell state" or "maximally entangled" state because of this property. It's of course the most beautiful entangled state because of this high symmetry.

#### stevendaryl

Staff Emeritus
This is special, because it's a state of total angular momentum (or spin in this case) 0. Of course that's a rotaionally invariant state, and that's why it's entangled wrt. any angular-moment direction. I'd call this a "Bell state" or "maximally entangled" state because of this property. It's of course the most beautiful entangled state because of this high symmetry.
But there is nothing particularly interesting (as far as I know) about states that are only entangled with respect to a single basis. What's interesting about entanglement is that there is a state for the composite system, but the component systems cannot be said to have a state (unless you go to density matrices).

#### PeterDonis

Mentor
which obviously is not a product state in the $|\pm \rangle$ basis, and thus wrt. the "vertical polarization" observable the same state is entangled.
No, it isn't, because measuring, say, $|+\rangle$ on one subsystem tells you nothing about the state of the other subsystem: it's equally likely to be $|+\rangle$ or $|-\rangle$. If the two were entangled, measuring one would give you at least some information about the other. This is obvious from the fact that the state factorizes into a tensor product of states on each subsystem. The fact that the two factor states are not single kets does not make the state as a whole entangled.

#### vanhees71

Gold Member
But there is nothing particularly interesting (as far as I know) about states that are only entangled with respect to a single basis. What's interesting about entanglement is that there is a state for the composite system, but the component systems cannot be said to have a state (unless you go to density matrices).
Well, take the SG experiment, where the entanglement of position and the spin component in direction of the field is the very point of the entire experiment, exemplifying the nature of measurement as an entanglement between pointer observables (position of the particle) and the measured quantity (spin/magnetic-moment component (which, BTW, is necessarily only FAPP as ideal as qualitatively (!) argued in textbooks!). I think entanglement (be it complete as in the Bell tests of the last decades or partial as in the SG experiment) is at the heart of understanding quantum theory!

Of course, a quantum state is always represented uniquely by a statistical operator (which in the position representation is the density matrix). Pure states are special cases, representing completely determined systems (i.e., systems prepared in a common eigenstate of a complete set of compatible observables).

#### vanhees71

Gold Member
No, it isn't, because measuring, say, $|+\rangle$ on one subsystem tells you nothing about the state of the other subsystem: it's equally likely to be $|+\rangle$ or $|-\rangle$. If the two were entangled, measuring one would give you at least some information about the other. This is obvious from the fact that the state factorizes into a tensor product of states on each subsystem. The fact that the two factor states are not single kets does not make the state as a whole entangled.
Then I'd need a clear definition, what you mean by "entanglement". I've explained what I understand under "entanglement" above, but of course I can be wrong. It's interesting that there seems not to be a clear definition in the literature, such that we have to argue about its definition to begin with.

#### PeterDonis

Mentor
Then I'd need a clear definition, what you mean by "entanglement".
An entangled state is a state of a quantum system that cannot be factorized into a product of states of its subsystems. Writing a state in a different basis doesn't change whether this is true or not.

It's interesting that there seems not to be a clear definition in the literature
The definition I've just given is the one I've seen in every piece of fairly recent literature I've read.

"Does quantum entanglement depend on the chosen basis?"

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