Is Multi-Particle Entanglement Possible in Quantum Systems?

[lucas_]

I read the following in the April copy of Sci-Am in the article on Firewall:

"It is a further consequence of quantum theory that a particle can be fully entangled only with one other: if particle B is entangled with particle A, then it cannot also be entangled with particle C. Entanglement is monogamous".

Is this not contradicted by multi particle entanglement? Can't you really entangle 3 photons or more than 2 particles? Why?

 

[craigi]

I read the following in the April copy of Sci-Am in the article on Firewall:

"It is a further consequence of quantum theory that a particle can be fully entangled only with one other: if particle B is entangled with particle A, then it cannot also be entangled with particle C. Entanglement is monogamous".

Is this not contradicted by multi particle entanglement? Can't you really entangle 3 photons or more than 2 particles? Why?

Yes, multi-particle entanglement is possible, but those particles aren't fully entangled, in the 2-particle entanglement sense. In fact, 3 particles can exhibit 3-particle entanglement without any pair of particles exhibitting 2-particle entanglement.

More generally, the same applies if we consider systems instead of particles.

 

[DrewD]

It should have read: "A qubit can only be entangled with one other qubit". This is an important fact, but it only applies (I shouldn't say only, there may be special cases) to a qubit which is a two state system.

 

[lucas_]

Yes, multi-particle entanglement is possible, but those particles aren't fully entangled, in the 2-particle entanglement sense. In fact, 3 particles can exhibit 3-particle entanglement without any pair of particles exhibitting 2-particle entanglement.

More generally, the same applies if we consider systems instead of particles.

Can you give an example of each case? 2 particle genuine entanglement and more than 2 particle entanglement?

 

[craigi]

Can you give an example of each case? 2 particle genuine entanglement and more than 2 particle entanglement?

Are you looking for an experimental arrangement or more detail from a theoretical perspective?

 

[bhobba]

"It is a further consequence of quantum theory that a particle can be fully entangled only with one other: if particle B is entangled with particle A, then it cannot also be entangled with particle C. Entanglement is monogamous".

You can entangle as many particles as you want, and obviously so from its definition.

Entanglement is an extension of superposition to different systems. Here is the definition for two systems. Suppose two systems can be in state |a> and |b>. If system 1 is in state |a> and system 2 is in state |b> that is written as |a>|b>. If system 1 is in state |b> and system 2 is in state |a> that is written as |b>|a>. But we now apply the principle of superposition so that c1*|a>|b> + c2*|b>|a> is a possible state, The systems are entangled - neither system 1 or system 2 are in a definite state - its in a peculiar non-classical state the combined systems are in.

Generalising it to many systems is obvious and immediate.

Thanks
Bill

 

[bhobba]

Yes, multi-particle entanglement is possible, but those particles aren't fully entangled

That's obviously incorrect, as its definition shows.

Thanks
Bill

 

[lucas_]

You can entangle as many particles as you want, and obviously so from its definition.

Entanglement is an extension of superposition to different systems. Here is the definition for two systems. Suppose two systems can be in state |a> and |b>. If system 1 is in state |a> and system 2 is in state |b> that is written as |a>|b>. If system 1 is in state |b> and system 2 is in state |a> that is written as |b>|a>. But we now apply the principle of superposition so that c1*|a>|b> + c2*|b>|a> is a possible state, The systems are entangled - neither system 1 or system 2 are in a definite state - its in a peculiar non-classical state the combined systems are in.

Generalising it to many systems is obvious and immediate.

Thanks
Bill

Are you saying the sci-am article is wrong about its statement that only 2 particles can be entangled? The following is the context of it in the April Sci-Am article
Burning Rings of Fire by Joseph Polchinski:

"It is a further consequence of quantum theory that a particle can be fully entangled only with one other: if particle B is entangled with particle A, then it cannot also be entangled with particle C. Entanglement is monogamous".

In the case of the black hole, think about a Hawking photon; call it "B," emitted after the black hole is at least halfway evaporated. The Hawking process implies that B is part of a pair; call its partner that falls into the black hole "A." A and B are entangled. Furthermore, the information that originally fell into the black hole has been encoded into all the Hawking radiation particles. Now, if information is not lost, and the outgoing Hawking photon B ends up in a definite quantum state, then B must be entangled with some combination, "C," of the other Hawking particles that already escaped (otherwise, the output would not preserve the information). But then we have a contradiction: polygamy!

The price of saving quantum mechanics, keeping the entanglement between B and C and not having anything else out of the ordinary on the outside of the black hole, is the loss of entanglement between A and B. The Hawking photons A and B began just inside and outside the horizon when they arose as an ephemeral particle-antiparticle pair. In quantum theory, the cost of breaking this entanglement, like the cost of breaking a chemical bond, is energy. Breaking the entanglement for all the Hawking pairs implies that the horizon is a wall of high-energy particles, which we termed a firewall. An infalling astronaut, rather than moving freely through the horizon, encounters something dramatic.

 

[atyy]

I read the following in the April copy of Sci-Am in the article on Firewall:

"It is a further consequence of quantum theory that a particle can be fully entangled only with one other: if particle B is entangled with particle A, then it cannot also be entangled with particle C. Entanglement is monogamous".

Is this not contradicted by multi particle entanglement? Can't you really entangle 3 photons or more than 2 particles? Why?

It is not contradicted by multi-particle entanglement because of the qualification "fully entangled". It is true that maximal entanglement is monogamous.

 

[DrewD]

Again, it should have said qubit, not particle. Two state systems can only be entangled between two systems*. Since photons are polarized, they can be considered a qubit and therefore the monogamy theorem applies.

*I think that maybe I should be saying observable with a two dimensional Hilbert space here instead?

 

[StevieTNZ]

Are you saying the sci-am article is wrong about its statement that only 2 particles can be entangled?.

Three photons can be entangled with each other (GHZ state -- http://en.wikipedia.org/wiki/Greenberger–Horne–Zeilinger_state).

Google searching should bring up the experiments performed to date.

EDIT: see http://www.wired.com/2012/02/new-quantum-record-physicists-entangle-8-photons/ and http://physicsworld.com/cws/article/news/2012/oct/25/physicists-entangle-100000-photons

 

[atyy]

cragi's answer in post #2 is correct.

 

[bhobba]

Are you saying the sci-am article is wrong about its statement that only 2 particles can be entangled?

That's exactly what I am saying. And it's obviously the case from the definition of entanglement.

Note however as with any article context is everything. You might like to reread the article and see if within its context its true.

Thanks
Bill

 

[DrewD]

Again, it should have said qubit, not particle. Two state systems can only be entangled between two systems*. Since photons are polarized, they can be considered a qubit and therefore the monogamy theorem applies.

*I think that maybe I should be saying observable with a two dimensional Hilbert space here instead?

I went back and reread some on monogamy. Not only does it have to be a qubit, but specifically the sum of the measures of entanglement of the two subsystems (ab and bc say) have to be less than the entanglement of the first system with the other combined system. So there is a degree to which the three particles can be entangled, but I think the measure of entanglement cannot be equal... I'm still refreshing on this. It is not something that I learned well the first time around.

 

[atyy]

Monogamy of entanglement is explained in these slides by Patrick Hayden http://simons.berkeley.edu/sites/default/files/docs/1301/slideshayden.pdf. Take a look at slide #14.

 

[f95toli]

I don't think it is helpful to talk about "particles" in this context. There are plenty of experiments which use large ensembles as single qubits; and those ensembles can then be entangled with another qubit (which could be another ensemble). It is much better to talk about "systems" than particles. See e.g. the Tavis-Cummings Hamiltonian.

Another (trivial) example of this would be any experiment showing entanglement between macroscopic qubits (say entanglement between two superconducting qubits), each qubit is obviously made up a large (~ 10^23) number of particles. .

In every case the "trick" is of course to arrange it so that multiple particles behaves like single system.

 

[Strilanc]

If two spins are maximally entangled with each other, measuring a third spin won't decrease that amount of entanglement. For example, in the state $(\left|00\right\rangle + \left|11\right\rangle)(\left|0\right\rangle + \left|1\right\rangle)$ the first qubit is more entangled with the second qubit than in the GHZ state $\left|000\right\rangle + \left|111\right\rangle$.

A pragmatic reason to want GHZ states to have a lower measure of entanglement than bell pairs is that it's harder to use GHZ states to violate classical limits on correlation. GHZ states have just as much "agreeing-when-measured" between the first two qubits, but they will act "more classical" than bell pairs because the third qubit acts like a measurement on them (preventing interference effects). For example, you can't do superdense coding with GHZ states (without first cancelling out the third wheel).

 

[lucas_]

In Decoherence, is it maximal entanglement between 2 particles in the environment and system or multi-particle entanglement between them?

For example. when the dust atom interact with a photon from the environment.. can other photons in the environment interact with the same atom?

 

[StevieTNZ]

If I have two entangled photons, and I send one to an apparatus to be measured, the entanglement is now between the apparatus and the other photon.

 

[atyy]

In Decoherence, is it maximal entanglement between 2 particles in the environment and system or multi-particle entanglement between them?

For example. when the dust atom interact with a photon from the environment.. can other photons in the environment interact with the same atom?

If you treat the environment as a single system, and the quantum system as another system, then the case of "ideal decoherence" means that the environment and the quantum system do become maximally entangled. (There is usually not ideal decoherence, but ideal decoherence is the heuristic which real decoherence is supposed to be a very good approximation to.)

However, if you treat the environment as consisting of many subsystems, then the quantum system does not become maximally entangled with each subsystem of the environment.