Multiple entanglements question

[brajesh]

TL;DR

What happens when multiple particles are entangled and then one is observed? What does it tell us about the rest?

I understand if two particles are entangled, and one is observed, then the other would have the opposite spin without being observed.
But what happens when multiple particles are entangled and then one particle is observed?
What does it tell us about the rest of them?

 

[Nugatory]

I understand if two particles are entangled, and one is observed, then the other would have the opposite spin without being observed.

That’s not quite right. We start with a single quantum system whose wave function is a superposition of “A is up, B is down” and “A is down, B is up”. This is an entangled state, but not because A and B have opposite spins, it’s an entangled state because both branches of the superposition involve both particles (the mathematical formalism of the theory states this idea more clearly and without ambiguity).

A measurement collapses the wave function of this single quantum system down to one branch; instead of a superposition the post-measurement wave function is either “A is up, B is down” or “ A is down, B is up”.

But what happens when multiple particles are entangled and then one particle is observed?

It’s the same thing. The entangled state might be a superposition of the states “A is red, B is green, C is yellow” and “A is blue, B is purple, C is orange”, and a measurement will collapse it to one of these.

 

[StevieTNZ]

It’s the same thing. The entangled state might be a superposition of the states “A is red, B is green, C is yellow” and “A is blue, B is purple, C is orange”, and a measurement will collapse it to one of these.

What about in the case of a GHZ state |H>|H>|H> + |V>|V>|V>, with one photon measured in the H/V axis and the other two in the 45/135 basis?

 

[brajesh]

That’s not quite right. We start with a single quantum system whose wave function is a superposition of “A is up, B is down” and “A is down, B is up”. This is an entangled state, but not because A and B have opposite spins, it’s an entangled state because both branches of the superposition involve both particles (the mathematical formalism of the theory states this idea more clearly and without ambiguity).

A measurement collapses the wave function of this single quantum system down to one branch; instead of a superposition the post-measurement wave function is either “A is up, B is down” or “ A is down, B is up”.
It’s the same thing. The entangled state might be a superposition of the states “A is red, B is green, C is yellow” and “A is blue, B is purple, C is orange”, and a measurement will collapse it to one of these.

I do understand what you wrote, but I'm a bit confused because I thought the observed states could only be up and down, so I'm not sure where the colors came from?

 

[PeroK]

I do understand what you wrote, but I'm a bit confused because I thought the observed states could only be up and down, so I'm not sure where the colors came from?

The colours come from quantum chromodynamics, where each quark has a colour charge of red, green or blue. @Nugatory got a bit carried away by adding yellow, purple and orange!

Up and down states apply to measurements of spin on a spin-1/2 particle, such as the electron. There is more to QM than that!

 

[Nugatory]

I do understand what you wrote, but I'm a bit confused because I thought the observed states could only be up and down, so I'm not sure where the colors came from?

Entanglement works with any observable property, not just spin (although it may be difficult or impossible to prepare a system in that is entangled on some observables). For example, when a particle decays into two other particles the momentum of the daughter particles is entangled: the state is a superposition (with an infinite number of terms because there is an infinity of possible directions for the vector $\vec{p_A}$ to point) of terms of the form “daughter particle A has momentum $\vec{p_A}$ and daughter particle B has momentum $\vec{p_B}=-\vec{p_A}$”. Measuring the momentum of one particle tells us the momentum of the other. The kinetic energy and angular momentum will be similarly entangled.

So I was just using colors as labels for some hypothetical particle property that can have multiple values. The point is that no matter what properties we’re talking about and how many particles are involved, if the initial state is a superposition of terms that involve multiple particles then a measurement will collapse the wave function to just one of those terms.
(There are some subtleties here involving what a “term” is and I’m deliberately glossing over these. But here’s an challenge: @StevieTNZ wrote down the state of an entangled three-photon system in which measurement of one polarization determines the polarization of the other two. Can you write down a state in which measurement of one photon’s polarization leaves the other two still entangled?)

 

[DrChinese]

What about in the case of a GHZ state |H>|H>|H> + |V>|V>|V>, with one photon measured in the H/V axis and the other two in the 45/135 basis?

But what happens when multiple particles are entangled and then one particle is observed?
What does it tell us about the rest of them?

With a 3 particle GHZ entangled state, a measurement on one particle places the other 2 into separable (unentangled) states.

With a 3 particle W entangled state, a measurement on one particle places the other 2 into a Bell (entangled) state.

 

[vanhees71]

What is a "W entangled state"?

 

[DrChinese]

What is a "W entangled state"?

If the GHZ state is represented as:
|0>|0>|0> + |1>|1>|1>

Then the W state is represented as:
|1>|0>|0> + |0>|1>|0> + |0>|0>|1>

 

[StevieTNZ]

What about with a GHZ state, one photon measured in the 45/135 basis, and the other two in H/V basis? Surely the other two would be either |H>|H> or |V>|V>, not |H>|V> or |V>|H> as if the collapse of the photon measured in the 45/135 basis caused the other two photons to take on either |45> or |135>.

 

[brajesh]

Entanglement works with any observable property, not just spin (although it may be difficult or impossible to prepare a system in that is entangled on some observables). For example, when a particle decays into two other particles the momentum of the daughter particles is entangled: the state is a superposition (with an infinite number of terms because there is an infinity of possible directions for the vector $\vec{p_A}$ to point) of terms of the form “daughter particle A has momentum $\vec{p_A}$ and daughter particle B has momentum $\vec{p_B}=-\vec{p_A}$”. Measuring the momentum of one particle tells us the momentum of the other. The kinetic energy and angular momentum will be similarly entangled.

So I was just using colors as labels for some hypothetical particle property that can have multiple values. The point is that no matter what properties we’re talking about and how many particles are involved, if the initial state is a superposition of terms that involve multiple particles then a measurement will collapse the wave function to just one of those terms.
(There are some subtleties here involving what a “term” is and I’m deliberately glossing over these. But here’s an challenge: @StevieTNZ wrote down the state of an entangled three-photon system in which measurement of one polarization determines the polarization of the other two. Can you write down a state in which measurement of one photon’s polarization leaves the other two still entangled?)

Thanks I'm still not clear.

Let's take 3 particles that are entangled, where 1 represents spin up and 0 represents spin down.
So there would be superposition of these 8 states for these particles.

state 1 - 000
state 2 - 001
state 3 -010
state 4 -011
state 5 -100
state 6 -101
state 7 -110
state 8 -111

Now if I examine one of them and find it to be in a spin of 1, then does it mean the other two particles now "lock" into some spin? But I've no idea what those spins would be because I had 8 possible states in the beginning?

 

[PeroK]

Thanks I'm still not clear.

Let's take 3 particles that are entangled, where 1 represents spin up and 0 represents spin down.
So there would be superposition of these 8 states for these particles.

state 1 - 000
state 2 - 001
state 3 -010
state 4 -011
state 5 -100
state 6 -101
state 7 -110
state 8 -111

Now if I examine one of them and find it to be in a spin of 1, then does it mean the other two particles now "lock" into some spin? But I've no idea what those spins would be because I had 8 possible states in the beginning?

If we assume an equal superposition of all eight states, and we measure the first particle, then we have two possibilities:

First particle is measured to be $0$. The remaining particles are in an equal superposition of $00$, $01$, $10$ and $11$.

First particle is measured to be $1$. The remaining particles are in an equal superposition of $00$, $01$, $10$ and $11$.

So, in fact, whatever the measurement on the first particle, the second and third particles are in the same superposition of four possible states.

 

[PeroK]

PS that's not an entangled state, as it can be written as a product of individual states. Entanglement and superposition are not the same thing.

 

[Nugatory]

So there would be superposition of these 8 states for these particles.

Only if you've prepared the system to be in that particular state: an equal superposition of all eight of those basis states.

You might also want to think about the case in which the initial state is a superposition of states 2, 3, 6, and 7 and the case in which the initial state is a superposition of 0 and 5.

 

[brajesh]

If we assume an equal superposition of all eight states, and we measure the first particle, then we have two possibilities:

First particle is measured to be $0$. The remaining particles are in an equal superposition of $00$, $01$, $10$ and $11$.

First particle is measured to be $1$. The remaining particles are in an equal superposition of $00$, $01$, $10$ and $11$.

So, in fact, whatever the measurement on the first particle, the second and third particles are in the same superposition of four possible states.

got it, thank you :)

 

[DrChinese]

What about with a GHZ state, one photon measured in the 45/135 basis, and the other two in H/V basis? Surely the other two would be either |H>|H> or |V>|V>, not |H>|V> or |V>|H> as if the collapse of the photon measured in the 45/135 basis caused the other two photons to take on either |45> or |135>.

Certainly, the order of measurement is not relevant - that is true with any entanglement scenario. Let's assume you get H when measured on the H/V basis. The other 2 are therefore H as well. So if you measure on the 45/135 basis, the chance of the others being any particular combination of the following are equal and random. On other bases, the odds would vary.

HH
HV
VH
VV

 

[Heidi]

We saw different exemples of multiplease entangled systems.
The W system 100 + 010 + 001 and
the GHZ system 000 + 111
i read in the monogamy article
https://en.wikipedia.org/wiki/Monogamy_of_entanglement
that if a A particles ismaximally entangled a second B particle then it cannot be entangled with a third on C.
How can we see in the given examples if A and B are maximally entangles? and in the general case?
does it mean that when a qbit is entangled with a macroscopic system, the information is not on one particle of the big system but ON it?
another question how is monogamy related to the no cloning principle? could we clone if there was no monogamy?

 

[DrChinese]

We saw different exemples of multiplease entangled systems.
The W system 100 + 010 + 001 and
the GHZ system 000 + 111
i read in the monogamy article
https://en.wikipedia.org/wiki/Monogamy_of_entanglement
that if a A particles ismaximally entangled a second B particle then it cannot be entangled with a third on C.
How can we see in the given examples if A and B are maximally entangles? and in the general case?

No 2 of the 3 particles in those states (GHZ or W) are maximally entangled - otherwise they would be an EPR pair (which are maximally entangled). That is true even if the 3 particle state itself is considered maximally entangled as a whole.

Suppose you have a GHZ state consisting of particles A, B and C. If A and B were themselves truly maximally entangled, then a measurement on C should have no impact on A - but that is not the case if C is entangled with A & B. Put another way, you are asking if within a GHZ state: A & B can be maximally entangled while A & C are also maximally entangled (or B & C). Clearly, the answer to that must be NO. So "maximally entangled" for N particles (such as N=3) is not the same as maximally entangled for 2 particles.

As to cloning: I should point out that although you cannot clone an unknown quantum state, you can teleport that state elsewhere. If it were really "cloned", you would have 2 identical unknown states when you are finished. That doesn't happen - only 1 at most.

 

[Heidi]

How to translate this in formulas?

 

[StevieTNZ]

Certainly, the order of measurement is not relevant - that is true with any entanglement scenario. Let's assume you get H when measured on the H/V basis. The other 2 are therefore H as well. So if you measure on the 45/135 basis, the chance of the others being any particular combination of the following are equal and random. On other bases, the odds would vary.

HH
HV
VH
VV

I would have thought the only combinations available for the three photons would be:
Either |45>|H>|H>
Or |135>|H>|H>
Or |45>|V>|V>
Or |135>|V>|V>.

 

[DrChinese]

I would have thought the only combinations available for the three photons would be:
Either |45>|H>|H>
Or |135>|H>|H>
Or |45>|V>|V>
Or |135>|V>|V>.

Well, once you measure one of the 3 in the GHZ state (on H/V i.e. 0/90 basis) and get (say) H>, then the other 2 are now in a separable Product state - and they are also H>. If you measure any H> photon on the 45/135 basis, it has a 50-50 chance of coming out as H> or V>. Measure 2 of those photons, and you have 4 (2x2) distinct permutations.

 

[StevieTNZ]

Well, once you measure one of the 3 in the GHZ state (on H/V i.e. 0/90 basis) and get (say) H>, then the other 2 are now in a separable Product state - and they are also H>. If you measure any H> photon on the 45/135 basis, it has a 50-50 chance of coming out as H> or V>. Measure 2 of those photons, and you have 4 (2x2) distinct permutations.

I'm thinking more along the lines of measuring one in the 45/135 basis, and the other two in the 0/90 basis.

 

[DrChinese]

I'm thinking more along the lines of measuring one in the 45/135 basis, and the other two in the 0/90 basis.

a. If you measure one of 3 GHZ photons in the 45/135 basis and get either H: the 4 outcome permutations for the other 2 (both on the 0/90 basis) are (equally):

HH
HV
VH
VV

b. If you measure one of 3 GHZ photons in the 0/90 basis and get H: the 2 outcome permutations for the other 2 (first listed below on the 0/90 basis, the second on the 45/135 basis) are (equally):

HH
HV

c. Another way to think of the above: there are only 4 total outcome permutations with the angle settings you specified:

H: HH
H: HV
V:VH
V:VV