I 3-particle or more entanglement…

[PeterDonis]

if the measuring device $|\psi \rangle$ interacts with $c_3(|up\rangle|down\rangle|up\rangle)$, will the result be product $(c_3(|up\rangle|down\rangle|up\rangle)) |\psi \rangle$ or won't it just turn into an entangled $c_1(|up\rangle|down\rangle|up\rangle+ c_2|\psi \rangle$

Neither. To know what state comes out of the interaction, you have to know what the interaction is--how it entangles the measuring device with the measured system. There is no one standard formula that expresses that. It depends on the interaction.

 

[bluecap]

Just to be clear: My example was describing when you start with 3 entangled particles, and then measure 1 of the 3. What are you left with? You have the original 8 cases (if that's what you start with) getting reduced to 4.

Note this is not a homework. But I’ll to start solving for it with actual example. Returning to your message #3

“For example, you might have total spin be +1.
If you have 3 particles combining to be +1, and measure one to be +1, you know the other two are a permutation where one is +1 and the other is -1. This is also a spin entangled entangled state. If you instead measured one to be -1, you know the other two are +1 and the other is +1. This is not a spin entangled entangled state because there is no superposition.”

Let’s apply it to the following where the combined spin must b +1

abc (total is +1 or 1…)
000
001
010
011 If measure a then no superposition. If measure b or c (with superposition) spin entangled entangled state
100
101 If measure b then no superposition. If measure a or c (with superposition) spin entangled entangled state
110 If measure c then no superposition. If measure a or b (with superposition) spin entangled entangled state
111

Please give me clue which of the above 4 is eliminated. But if we will retain those with two 1's or 11 then what is elimited is 000, 010, 001, 100 and even 111 (this is because when you measure either, the rest is not in superposition) but these are 5 and not the 4 terms you mentioned above.

Or in case DrChinese doesn’t know exactly how to do it (because he said 3 particle entanglement is notoriously difficult and it may not help with transmitting codes from Alice to Bob). Can others please share so we can wrap up the thread as this is the only question left unanswered.. lol.. many thanks!

 

[Zafa Pi]

Below are two things you want to know.

I still don’t understand the GHZ entanglement even after reading many website about it but understanding the math above would enable me to start attempting to understand it. Thanks

How can you tell if a measurement can break the existing entanglement or form new one?

I will use standard q-computing/information notation. If you can't find it (say Nielsen&Chuang) I'll explain.
I disagree with @DrChinese that GHZ

is NOTORIOUSLY difficult to understand,

A bit of familiarity with simple (2D algebraic) tensor products and how q-measurements are made is sufficient. No dynamics is necessary.

Physical set up for GHZ:
Alice, Bob, Carol, and Eve are all mutually one light minute apart. At noon Eve simultaneously sends a light signal to each of A, B, and C. When A receives her signal she flips a fair coin. If it comes up heads she selects (via some objective process) a value of either 1 or -1 and calls that Ah. If she flips tails she selects 1 or -1 and calls it At. This takes her less than ½ minute. Each of B and C do the same, calling their selections Bh, Bt, and Ch, Ct.

We assume that no influence or information can go faster than the speed of light (called locality) then none of the three know what the others flipped, nor can one's selection influence another's.

GHZ Theorem: Let's assume that if only one of A, B, or C flipped a head then the product of their selections equals -1. I.e., we assume -1 = Ah•Bt•Ct = At•Bh•Ct = At•Bt•Ch.
Then we may conclude if all three flipped heads their product would be -1.
I.e., -1 = Ah•Bh•Ch.

Proof: -1 = (Ah•Bt•Ct)•(At•Bh•Ct)•(At•Bt•Ch) = At²•Bt²•Ct²•Ah•Bh•Ch which implies that -1 = Ah•Bh•Ch. QED

If Eve sent each of A, B, C one photon from the entangled triple state |ψ⟩ = √½(|000⟩ + |111⟩) and if flipping a head selects the value obtained by measuring a photon with Pauli X and flipping a tail means measuring with Pauli Y, then X⊗Y⊗Y and Y⊗X⊗Y and Y⊗Y⊗X each operating on |ψ⟩ yield -1, so the hypothesis of the Theorem is satisfied. However, X⊗X⊗X operating on |ψ⟩ gives 1, contradicting the conclusion.

Lab tests show the QM predictions are correct. So what's wrong?

This takes care of the first of the questions.

If Pauli Z is used to measure all three photons from |ψ⟩ then one gets either 1, 1, 1, or -1, -1, -1. Thus if you measure the left photon with Z and get 1 then you will also get 1 if you then measure the other two. So the other two cannot be entangled. Any measurement ruins entanglement.

 

[Simon Phoenix]

Any measurement ruins entanglement.

You have to be a bit careful here. If I have a GHZ state of 3 qubits $$ | \psi \rangle = \frac 1 {\sqrt{2}} \left( | 000 \rangle + | 111 \rangle \right) $$ where this is expressed in terms of "spin-z" eigenstates, then if I measure "spin-x" for qubit 1, I end up with qubits 2 and 3 being in a maximally entangled state.

This measurement certainly reduces the entanglement from 3 bits (initially) to 2 bits, but it does not completely remove the entanglement - for that you need to measure "spin-z".

 

[Simon Phoenix]

By transferring it to something else. For example, if A is maximally entangled with B, you can make A and C interact in such a way that C is now maximally entangled with B and A is not entangled with anything. (But this only works if C was not entangled with anything to start with.)

There's a nice example of this that has been demonstrated experimentally. You start with a very high-Q cavity with 'nothing' in it (so just the single mode vacuum). Send an excited 2-level atom, resonant with the cavity mode, through the cavity and tailor the interaction time such that there's a 1/2 probability of spontaneous emission into the cavity (in other words there's a 1/2 probability of finding a photon in the cavity should we decide to measure this). After this the field and atom are maximally entangled.

Now send a second 2-level atom through, also resonant with the cavity mode, and tailor the interaction time such that there's a unit probability of absorption. What we end up with is the 2 atoms being maximally entangled and the field in the cavity has "decoupled" and returned to the vacuum.

It's a kind of pre-cursor to entanglement swapping - the 2 atoms never directly interact with one another yet end up being entangled.

 

[vanhees71]

You have to be a bit careful here. If I have a GHZ state of 3 qubits $$ | \psi \rangle = \frac 1 {\sqrt{2}} \left( | 000 \rangle + | 111 \rangle \right) $$ where this is expressed in terms of "spin-z" eigenstates, then if I measure "spin-x" for qubit 1, I end up with qubits 2 and 3 being in a maximally entangled state.
This measurement certainly reduces the entanglement from 3 bits (initially) to 2 bits, but it does not completely remove the entanglement - for that you need to measure "spin-z".

You mean if you prepare the spin-$x$ component of qubit 1, and indeed that's what makes the GHZ states so interesting.

 

[Simon Phoenix]

You mean if you prepare the spin-$x$ component of qubit 1, and indeed that's what makes the GHZ states so interesting.

Lol - yes, I can prepare it by making a measurement

 

[vanhees71]

Nowadays it's very important to be very clear concerning this point. What you mean are von-Neumann filter measurements, which are indeed a preparation procedure, but what "measuring" does to the state of a system depends on the details of the measurement device used, and it does not imply the preparation of the system in the corresponding eigenstate!

 

[Simon Phoenix]

What you mean are von-Neumann filter measurements

Well any measurement of the "spin-x" of particle 1 should do the trick and leave you with some entanglement of qubits 2 and 3 (unless it's a useless measurement). We can handle the case of an imperfect measurement using the POVM formalism and we can also work out what the corresponding density operator will be for qubits 2 and 3, given a particular result. You should end up with a non-separable density operator, given a particular result, for qubits 2 and 3. Good question though - I'll have to work this one out.

 

[Zafa Pi]

You have to be a bit careful here. If I have a GHZ state of 3 qubits $$ | \psi \rangle = \frac 1 {\sqrt{2}} \left( | 000 \rangle + | 111 \rangle \right) $$ where this is expressed in terms of "spin-z" eigenstates, then if I measure "spin-x" for qubit 1, I end up with qubits 2 and 3 being in a maximally entangled state.

This measurement certainly reduces the entanglement from 3 bits (initially) to 2 bits, but it does not completely remove the entanglement - for that you need to measure "spin-z".

Right you are., good looking out. I missed that, and should have said, "A measurement may ruin entanglement."

Indeed, if the left qubit/photon is measured with X and you get 1 the state of the other two is √½(|++⟩ + |--⟩) and if you got -1 then the other two have state √½(|+-⟩ + |-+⟩). Where +,- are eigenstates in the X basis.