I 3-particle or more entanglement…

[bluecap]

In the simplest case, yes. The simplest case would be that each of the 8 terms is an eigenstate of the measurement being made, so each of the 8 terms would get another factor describing the state of the measuring device when it indicates the corresponding result. (In more complicated cases, the 8 terms you wrote down would not individually be eigenstates; but that just means we'd need to rewrite the state in a different basis. It doesn't change the principle of what's going on.)
Yes. More precisely, one of the 8 terms (with the measuring device factor multiplied in) would be randomly selected to be the one that the wave function collapsed into. The probabilities of the random selection would be the squares of the coefficients of the 8 terms (assuming the coefficients were properly normalized).
Yes, because you have a state with only one term, consisting of a single factor for each subsystem (each of the three particles and the measuring device), which is trivially separable.
No, it's because the collapse destroys all but one term, so it turns the wave function into one that is trivially separable. You can have classical objects in a no collapse interpretation like the MWI, and they can entangle with quantum systems; so "classicality" in itself doesn't prevent entanglement.
Wrong. If all but one branch is destroyed by wave function collapse, the resulting wave function is trivially separable, regardless of whether M is "classical" or not. See above.

What is mean by "trivially separable"? Why is it that if it's trivially separable, there would be no entanglement.. I think this is the key to why there is no entanglement in Copenhagen. Thanks

This is not correct. A collapse interpretation like Copenhagen is perfectly consistent with decoherence. It just adds wave function collapse as an extra ingredient (which is not experimentally testable).
No, then whatever was the end product of the explosion (gases, plasma, whatever) would be entangled with itself and the environment. Why would it be any different from the case of ashes?

 

[bluecap]

To the Bohmians, I have a question...

First, I think turning a wave function into one that "trivially separable" means there is no longer any wave to make all the entanglement so everything simply collapses (like there is no more waves and water in the ocean which makes the boats drop to the ocean floor)... right Peterdonis? :)
So if there is no entanglement in Copenhagen upon measurement, and there is in MWI..

how about in Bohmian Mechanics.. is there entanglement in BM upon measurements?? why and why not?

 

[Demystifier]

So if there is no entanglement in Copenhagen upon measurement, and there is in MWI..

how about in Bohmian Mechanics.. is there entanglement in BM upon measurements?? why and why not?

Bohmian mechanics is somewhere in between. There is entanglement in the wave function, but it does not influence the particles. Or in more precise language, there is entanglement in the universal wave function but not in the effective wave function.

 

[bluecap]

Bohmian mechanics is somewhere in between. There is entanglement in the wave function, but it does not influence the particles. Or in more precise language, there is entanglement in the universal wave function but not in the effective wave function.

As review. In BM, the wave function controls the quantum potential that pushes the particle into either the left or right slit of the experiment.
Now's let's use the example of 3 systems in 2 state of spin up and spin down. Peterdonis (i think) said it was true that in simplest case, entanglement in MWI means that when you measure this state state c1*|up>|up>|up>+c2*|up>|up>|down>+c3*|up>|down>|up>+c4*|up>|down>|down>+c5*|down>|up>|up>+c6*|down>|up>|down>+c7*|down>|down>|up>+c8*|down>|down>|down>
each of the 8 terms would be entangled with the measuring device (or system 4) wave function. How about in BM, does the wave function in BM also entangled with all the 8 terms at the same time? If so, then MWI and BM is the same with the difference that in BM, only one world exists as the quantum potential pushes it to select one of the 8 terms only?

Is all variant of BM the same where it's wave function can entangled with all 8 terms at the same time? or for some, does it only choose one term? Thanks.

 

[Demystifier]

How about in BM, does the wave function in BM also entangled with all the 8 terms at the same time?

Yes.

If so, then MWI and BM is the same with the difference that in BM, only one world exists as the quantum potential pushes it to select one of the 8 terms only?

Yes.

Is all variant of BM the same where it's wave function can entangled with all 8 terms at the same time?

Yes.

 

[bluecap]

Yes.Yes.Yes.

Thanks. It's so difficult to decide whether to pick up wave function with the power to collapse (Copenhagen), or all worlds co-existing (MWI) or a wave function that can push quantum potential (BM)... maybe if only Newton could just convince us all this are just smokes and mirrors.. I guess the Ensemble Intepretation is the closest to this.. I guess I'll listen more to Vanheez71 arguments that all are just statistics and there is no objective wave function that can collapse, or produce multiple worlds or push quantum potentials... because it is possible all this doesn't even exist..

 

[PeterDonis]

What is mean by "trivially separable"?

Write down the math like I said. What does it look like for the case where the wave function collapses (i.e., where you eliminate all but one branch)?

 

[bluecap]

Write down the math like I said. What does it look like for the case where the wave function collapses (i.e., where you eliminate all but one branch)?

Here's the math for a quantum state $|\psi \rangle =\sum_{i} ci |wi \rangle$

with w1, w2, w3, etc. as the branches...

the wave function collapses from the full $|\psi>$ to just one of the basis eigenstates,$|\omega i>$

So you mean in collapse, the wave fragmented and the other parts disappear.. so you have incomplete wave pertaining to the collapsed value.. but why can't the measuring M or other system entangled with that fragment of the wave.. it's like fourier.. one can add 2 waves together.. or are you saying there is literally no more wave when it is collapse.. but note after measurement, the wave starts evolving again, it doesn't mean the wave just stays dead.

Also if entanglement is not possible in collapsed interpretation.. how come we can perform qubits entanglement (3 or more qubits) at the lab.. why isn't the above "trivially separable" invoked all the time so we shouldn't be able to make any entanglement experiment at all?

Thanks.

 

[PeterDonis]

with w1, w2, w3, etc. as the branches...

But we are talking about a total system that contains multiple subsystems; so what do each of the branch terms actually look like? Write them out in detail, including each subsystem.

the wave function collapses from the full $|\psi>$ to just one of the basis eigenstates, $|\omega i>$

And what does that basis eigenstate look like when you write it out in full including each subsystem?

 

[PeterDonis]

if entanglement is not possible in collapsed interpretation.. how come we can perform qubits entanglement (3 or more qubits) at the lab

Because in those experiments the wave function never collapses; everything is isolated so the qubits don't interact with anything except each other, and all the interactions are reversible.

 

[bluecap]

But we are talking about a total system that contains multiple subsystems; so what do each of the branch terms actually look like? Write them out in detail, including each subsystem.

And what does that basis eigenstate look like when you write it out in full including each subsystem?

ok.. for the 3 systems and 2 states composing of spin up and spin down

$\psi \rangle = \sum_i c_i | \phi_i \rangle$

where $\phi_i = |s_1\rangle |s_2\rangle |s_3\rangle$ or the spin states for example $|up\rangle |down\rangle |up\rangle$

so

$\psi \rangle = \sum_i c_i | \phi_i \rangle$ = $\sum_i c_i (|s_1\rangle |s_2\rangle |s_3\rangle)$= $c_1(|up\rangle|up\rangle|up\rangle)+c_2(|up\rangle|up\rangle|down\rangle)+c_3(|up\rangle|down\rangle|up\rangle)+c_4(|up\rangle|down\rangle|down\rangle)+c_5(|down\rangle|up\rangle|up\rangle)+c_6(|down\rangle|up\rangle|down\rangle)+c_7(|down\rangle|down\rangle|up\rangle)+c_8(|down\rangle|down\rangle|down\rangle)$

All wave functions are assumed to be normalized; the total probability of measuring all possible states is unity:
$\langle \psi|\psi \rangle = \sum_i |c_i|^2 = 1$

Upon measurement, the wave function collapses from the full $|\psi \rangle$ to just one of the basis eigenstates, $|\phi_i\rangle$ , that is:

$|\psi\rangle \rightarrow |\phi_i\rangle$

so in the case of the states given above.. only one basis eigenstate.. for example.. $c_3(|up\rangle|down\rangle|up\rangle)$ pops up and the rest vanishes (at least in Copenhagen)..

So since only one eigenstate comes out and the rest of the terms collapse.. are you saying that since the system is in only one eigenstate $|up\rangle|down\rangle|up\rangle$ (should this be written as simply "up,down,up" since up,down,up is no longer in superposition?), then the measuring device M or system 4 won't affect or change the system or like a non-demolition measurement.. hence they are not entangled?

 

[PeterDonis]

for the 3 systems and 2 states composing of spin up and spin down...

Yes, you have a linear combination of 8 eigenstates, where each eigenstate is a product of 3 kets, one for each of the 3 particles. So is this state $\vert \psi \rangle$ entangled? (Hint: the answer is yes. But can you see why?)

Upon measurement, the wave function collapses from the full $|\psi \rangle$ to just one of the basis eigenstates

Yes, so now the state is just a single eigenstate, which is a product of 3 kets, one for each of the three particles. So is this state entangled, or separable? (Hint: the answer is, it's separable--but can you see why? And can you see how that's obvious, just from the description of the state that I just gave?)

 

[bluecap]

Yes, you have a linear combination of 8 eigenstates, where each eigenstate is a product of 3 kets, one for each of the 3 particles. So is this state $\vert \psi \rangle$ entangled? (Hint: the answer is yes. But can you see why?)

There is interference or phase coherence? Anyway. You mentioned about linear combination. Why, is there unlinear combination?
Also earlier DrChinese wrote: "
Generally, an entangled state of N particles will have some number of permutations, where there is some known or conserved observable.

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."

Does it mean the 8 terms would be less and won't this cause loss of coherence?

Yes, so now the state is just a single eigenstate, which is a product of 3 kets, one for each of the three particles. So is this state entangled, or separable? (Hint: the answer is, it's separable--but can you see why? And can you see how that's obvious, just from the description of the state that I just gave?)

Separable means because there is no interference or loss of phase coherence. But it's kinda weird. It's as if coherence can trap the particles and imprison them so they are not separable? Maybe entangled is like Siamese twin?

 

[DrChinese]

There is interference or phase coherence? ... Also earlier DrChinese wrote:...

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.

 

[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.

How do you tell which of the terms is eliminated?

$c_1(|up\rangle|up\rangle|up\rangle)+c_2(|up\rangle|up\rangle|down\rangle)+c_3(|up\rangle|down\rangle|up\rangle)+c_4(|up\rangle|down\rangle|down\rangle)+c_5(|down\rangle|up\rangle|up\rangle)+c_6(|down\rangle|up\rangle|down\rangle)+c_7(|down\rangle|down\rangle|up\rangle)+c_8(|down\rangle|down\rangle|down\rangle)$

Or let's make it in bit form:

000
001
010
011
100
101
110
111

You stated that if you measure +1, the remaining is +1,-1 and in superposition while if you measured -1, the remaining is +1,+1 and is not in superposition, how do you make the 8 terms reduce to only 4?

Thanks a lot.

 

[PeterDonis]

There is interference or phase coherence?

No. Go back to basics. What is the definition of an entangled state?

You mentioned about linear combination. Why, is there unlinear combination?

Not according to standard QM. Standard QM is linear: you can add any two quantum states to get another quantum state (with appropriate normalization). So in this context, "linear combination" is just a fancy word for "sum".

Separable means because there is no interference or loss of phase coherence.

No. Again, go back to basics. A separable state is a state that is not entangled. What is the definition of an entangled state?

 

[bluecap]

No. Go back to basics. What is the definition of an entangled state?
Not according to standard QM. Standard QM is linear: you can add any two quantum states to get another quantum state (with appropriate normalization). So in this context, "linear combination" is just a fancy word for "sum".
No. Again, go back to basics. A separable state is a state that is not entangled. What is the definition of an entangled state?

according to Wikipedia... "An entangled system is defined to be one whose quantum state cannot be factored as a product of states of its local constituents; that is to say, they are not individual particles but are an inseparable whole. In entanglement, one constituent cannot be fully described without considering the other(s). Note that the state of a composite system is always expressible as a sum, or superposition, of products of states of local constituents; it is entangled if this sum necessarily has more than one term."

So when the $c_3(|up\rangle|down\rangle|up\rangle)$ interacts with system 4, they are product states or separable. So entanglement is when there are at least two subsystem that can't be factored or be separated or in other words only the whole has superposition.. and this is the only way another system like the measuring device can be entangled with it. I think it's correct now. right? lol.. thanks so much Peterdonis.

 

[bluecap]

according to Wikipedia... "An entangled system is defined to be one whose quantum state cannot be factored as a product of states of its local constituents; that is to say, they are not individual particles but are an inseparable whole. In entanglement, one constituent cannot be fully described without considering the other(s). Note that the state of a composite system is always expressible as a sum, or superposition, of products of states of local constituents; it is entangled if this sum necessarily has more than one term."

So when the $c_3(|up\rangle|down\rangle|up\rangle)$ interacts with system 4, they are product states or separable. So entanglement is when there are at least two subsystem that can't be factored or be separated or in other words only the whole has superposition.. and this is the only way another system like the measuring device can be entangled with it. I think it's correct now. right? lol.. thanks so much Peterdonis.

but 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 $c_1(|up\rangle|down\rangle|up\rangle+$ $c_2|\psi \rangle$, how do you control whether its the former or latter that would form?

 

[PeterDonis]

An entangled system is defined to be one whose quantum state cannot be factored as a product of states of its local constituents

Yes. Here the "local constituents" are the 3 particles. So you have a state $\vert \psi \rangle$ which is a sum of 8 terms, each of the form $\vert s_1 \rangle \vert s_2 \rangle \vert s_3 \rangle$, where $s_1$, $s_2$, and $s_3$ tell whether particles 1, 2, and 3, respectively, are "up" or "down", and where each term has an arbitrary complex coefficient $c$ in front of it (with the constraints that the sum of the squares of all the coefficients must be 1). Is there any way to factor that state into a product of states of the 3 particles? In other words, is there any way to express those 8 terms as a product of three factors, such that factor 1 only contains kets of the form $\vert s_1 \rangle$ (i.e., only tells you about the spin of particle 1), factor 2 only contains kets of the form $\vert s_2 \rangle$, and factor 3 only contains kets of the form $\vert s_3 \rangle$? (The answer, as we've already seen, is "no", meaning that this state is entangled. But again, can you see why?)

Now answer the same question for the state after wave function collapse (using a collapse interpretation), i.e., for a single term of the form $\vert s_1 \rangle \vert s_2 \rangle \vert s_3 \rangle$. Can that state be factored as a product of kets of the form $\vert s_1 \rangle$, $\vert s_2 \rangle$, and $\vert s_3 \rangle$? (The answer, as we've already seen, is "yes", meaning that this state is separable. But now it should be obvious why that answer is obvious--i.e., why I said such a state was "trivially" separable, because it's trivial to see that it is.)

So when the $c_3(|up\rangle|down\rangle|up\rangle)$ interacts...

You're going too fast. First you need to get clear about why the state $\vert \psi \rangle$ with the 8 terms is entangled, while one individual term picked out from it is not. See above.

 

[bluecap]

Yes. Here the "local constituents" are the 3 particles. So you have a state $\vert \psi \rangle$ which is a sum of 8 terms, each of the form $\vert s_1 \rangle \vert s_2 \rangle \vert s_3 \rangle$, where $s_1$, $s_2$, and $s_3$ tell whether particles 1, 2, and 3, respectively, are "up" or "down", and where each term has an arbitrary complex coefficient $c$ in front of it (with the constraints that the sum of the squares of all the coefficients must be 1). Is there any way to factor that state into a product of states of the 3 particles? In other words, is there any way to express those 8 terms as a product of three factors, such that factor 1 only contains kets of the form $\vert s_1 \rangle$ (i.e., only tells you about the spin of particle 1), factor 2 only contains kets of the form $\vert s_2 \rangle$, and factor 3 only contains kets of the form $\vert s_3 \rangle$? (The answer, as we've already seen, is "no", meaning that this state is entangled. But again, can you see why?)

Now answer the same question for the state after wave function collapse (using a collapse interpretation), i.e., for a single term of the form $\vert s_1 \rangle \vert s_2 \rangle \vert s_3 \rangle$. Can that state be factored as a product of kets of the form $\vert s_1 \rangle$, $\vert s_2 \rangle$, and $\vert s_3 \rangle$? (The answer, as we've already seen, is "yes", meaning that this state is separable. But now it should be obvious why that answer is obvious--i.e., why I said such a state was "trivially" separable, because it's trivial to see that it is.)
You're going too fast. First you need to get clear about why the state $\vert \psi \rangle$ with the 8 terms is entangled, while one individual term picked out from it is not. See above.

I understood the above..

but if the measuring device $|\psi \rangle$ interacts with the only collapsed eigenstate $c_3(|up\rangle|down\rangle|up\rangle)$, will the result be product $(c_3(|up\rangle|down\rangle|up\rangle))$ $|\psi \rangle$ or won't it just turn into an entangled $c_1(|up\rangle|down\rangle|up\rangle+$ $c_2|\psi \rangle$, how do you control whether its the former or latter that would form?

 

[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.