I 3-particle or more entanglement…

[bluecap]

Two particle entanglement such as bell state measurement for quantum teleportation is the example commonly used and in fact you can find everywhere. I understand it like how maximal entanglement can mean measuring one you can know the property of the other.. but it is not very useful for understanding billion-particle entanglement like what happens in macroscopic object like an apple entangling with the environment. So l want to be well versed in more than 2 particle entanglement as billion-particle entanglement is the more realistic everyday scenerio. Let me start with 2 particle entanglement and the math of it:

Entanglement is said to an extension of superposition to different systems. Here is the definition for two systems (this is the example Bhobba kept repeating over the years but he didn't mention the equations for more than 3 particle so I'll try to create the 3 particle equations after the following).

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 – it is said neither system 1 or system 2 are in a definite state - its in a peculiar non-classical state the combined systems are in.

Let me now generalize it to many systems:

Suppose three systems can be in state |a> and |b> and |c>.

*If system 1 is in state |a> and system 2 is in state |b> and system 3 is in state |c>,that is written as |a>|b>|c>.
*If system 1 is in state |a> and system 2 is in state |c> and system 3 is in state |b>,that is written as |a>|c>|b>
*If system 1 is in state |b> and system 2 is in state |a> and system 3 is in state |c>,that is written as |b>|a>|c>
*If system 1 is in state |b> and system 2 is in state |c> and system 3 is in state |a>,that is written as |b>|c>|a>
*If system 1 is in state |c> and system 2 is in state |a> and system 3 is in state |b>, that is written as |c>|a>|b>
*If system 1 is in state |c> and system 2 is in state |b> and system 3 is in state |a>, that is written as |c>|b>|a>

But we now apply the principle of superposition so that

c1*|a>|b>|c> + c2*|a>|c>|b>+ c3*|b>|a>|c> + c4*|b>|c>|a> + c5*|c>|a>|b> + c6*|c>|b>|a>
is a possible state.

The above convention or equation is correct right? (is the equation or the terms complete). I just read about the 2 particle convention and tried generalizing it to many particle convention above. So please confirm if it is indeed correct.

I’d like to know the following:

1. When you measure only system 2. Do you form new entanglement with system 2 (like you become system 4).. or do you break the entanglement of system 2 from system 1 and 3?
How can you tell if a measurement can break the existing entanglement or form new one?

2. If your measurement produce say c3*|b>|a>|c>, what happens to the other terms? Does the equation means only one term can happen and rest discarded?

3. Measuring c3*|b>|a>|c> means system 1 has state |b>, system 2 has state |a>, system 3 has state |c> right? So if you measure system 2, that means it is in |a>.

4. Let’s assume the state means up or down.. if you measure the system 2 and come up with c3*|b>|a>|c> and |a> is up. How do you know if |b> is up or down.. how do you know if |c> is up or down?

5. Now Many worlds or Branches.. for one particle superposition (of say spin up and down.. I understand the entanglement to observers or other systems is many worlds as taught by Peterdonis). But in the case of c1*|a>|b>|c> + c2*|a>|c>|b>+ c3*|b>|a>|c> + c4*|b>|c>|a> + c5*|c>|a>|b> + c6*|c>|b>|a>, is the many worlds each term?

6. Please answer using the above example and math.. 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.

7. Thanks very much for the help

 

[StevieTNZ]

Suppose three systems can be in state |a> and |b> and |c>.

*If system 1 is in state |a> and system 2 is in state |b> and system 3 is in state |c>,that is written as |a>|b>|c>.
*If system 1 is in state |a> and system 2 is in state |c> and system 3 is in state |b>,that is written as |a>|c>|b>
*If system 1 is in state |b> and system 2 is in state |a> and system 3 is in state |c>,that is written as |b>|a>|c>
*If system 1 is in state |b> and system 2 is in state |c> and system 3 is in state |a>,that is written as |b>|c>|a>
*If system 1 is in state |c> and system 2 is in state |a> and system 3 is in state |b>, that is written as |c>|a>|b>
*If system 1 is in state |c> and system 2 is in state |b> and system 3 is in state |a>, that is written as |c>|b>|a>

But we now apply the principle of superposition so that

c1*|a>|b>|c> + c2*|a>|c>|b>+ c3*|b>|a>|c> + c4*|b>|c>|a> + c5*|c>|a>|b> + c6*|c>|b>|a>
is a possible state.

There are 27 possible states for the last "equation".

Once those are written, the answers to your questions will become evident.

 

[DrChinese]

I am not going to try to explain it in your terms, because there is a lot more to specify. 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.

At any rate, there are experiments involving entanglement of large N, where N>>1000. Please keep in mind that there is also monogamy of entanglement. A particle A that is maximally entangled with B, cannot also be entangled with C on the same basis.

 

[bluecap]

There are 27 possible states for the last "equation".

Once those are written, the answers to your questions will become evident.

For the 2 particle case, Let’s give the observable “spin up or down” to the |a> or |b> state.

if system 1 is in state |up> and system 2 is in state |down> that is written as c1*|up>|down>
if system 1 is in state |down> and system 2 is in state |up> that is written as c2 |down>|up>
we now apply the principle of superposition so that c1*|up>|down> + c2*|down>\up> is a possible state,
the 2 doesn’t become 4. It is not like 2^2 = 4.. so why are you using 3^3?
Note
c1 * system 1 system 2 + c2 * system 1 system 2 = 80% * |up>|down> + 20% * |down>\up> (for example)..
there are only 2 terms out of the two state |a> and |b>
so how can 3 state becomes 27 terms?

 

[bluecap]

There are 27 possible states for the last "equation".

Once those are written, the answers to your questions will become evident.

Thinking it over. For the 2 particle case.. so 2 state for each system produce 4 overall terms. The following is correct?

If system 1 is in state |up> and system 2 is in state |down> that is written as |up>|down>.
If system 1 is in state |down> and system 2 is in state |up> that is written as |down>|up>.
If system 1 is in state |up> and system 2 is in state |up> that is written as |up>|up>.
If system 1 is in state |down> and system 2 is in state |down> that is written as |down>|down>.

But we now apply the principle of superposition so that c1*|up>|down> + c2*|down>|up> + c3|up>|up> + c4|down>|down> is a possible state.

there are 4 terms?
Now generalizing this to 3 particles.. since there are only 2 possibilities “spin up”, “spin down”, there it should be 2^3 or 8 only

|up>|up>|up>
|up>|up>|down>
|up>|down>|up>
|up>|down>|down>
|down>|up>|up>
|down>|up>|down>
|down>|down>|up>
|down>|down>|down>

So for 2 spin, the terms should be 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>

There are only 8 terms.. not 27.. right (others)?

 

[StevieTNZ]

For the 2 particle case, Let’s give the observable “spin up or down” to the |a> or |b> state.

if system 1 is in state |up> and system 2 is in state |down> that is written as c1*|up>|down>
if system 1 is in state |down> and system 2 is in state |up> that is written as c2 |down>|up>
we now apply the principle of superposition so that c1*|up>|down> + c2*|down>\up> is a possible state,
the 2 doesn’t become 4. It is not like 2^2 = 4.. so why are you using 3^3?

I am referring to the case where you have three quantum systems in play, with each able to take on state |a>, |b> or |c>.

 

[bluecap]

I am referring to the case where you have three quantum systems in play, with each able to take on state |a>, |b> or |c>.

Let's say I have 3 systems of particles that can be spin up and spin down.. so it's only 2 state or |a> and |b>?

But to write the 3 systems in superposition.. is it not you use the following equation?

c1*|a>|b>|c> + c2*|a>|c>|b>+ c3*|b>|a>|c> + c4*|b>|c>|a> + c5*|c>|a>|b> + c6*|c>|b>|a>

or 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> ??

if not.. any clue what is the right equation?

 

[bluecap]

Bill Hobba, (@Hobba)

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

For my slow mind, it's not obvious and immediately and StevenTNZ is confusing me. So for 3 systems that has spin up and spin down.. do you use only |a> and |b> or also |c> and |d> to signify the 4 possibilities of 1. up,up 2. up, down 3. up,up 4. down, down? Is the following correct?

c1*|a>|b>|c> + c2*|a>|c>|b>+ c3*|b>|a>|c> + c4*|b>|c>|a> + c5*|c>|a>|b> + c6*|c>|b>|a>?

But it's only 3 systems with state up and down.. it seems wrong.. what then is the correct equation for 3 systems that has the particle that can be spin up and spin down? I've been thinking of this for 2 days. Thanks!

 

[bhobba]

But it's only 3 systems with state up and down.. it seems wrong.. what then is the correct equation for 3 systems that has the particle that can be spin up and spin down? I've been thinking of this for 2 days. Thanks!

I have zero idea where you got my quote from, but its context looks like a general comment about entanglement and superposition.

However what you wrote is for 3 different states |a>, |b>, |c> - up and down is just 2.

I suggest you think a bit more.

Thanks
Bill

 

[bluecap]

I have zero idea where you got my quote from, but its context looks like a general comment about entanglement and superposition.

However what you wrote is for 3 different states |a>, |b>, |c> - up and down is just 2.

I suggest you think a bit more.

Thanks
Bill

So how do you write 3 system entanglement with two states up and down.. please just share the first 3 terms.. lol... thanks..

 

[bhobba]

So how do you write 3 system entanglement with two states up and down.. please just share the first 3 terms.. lol... thanks..

Mate I will leave it to you - you have three objects with two states - its just a matter of writing the permutations - tedious but trivial.

Thanks
Bill

 

[bluecap]

Mate I will leave it to you - you have three objects with two states - its just a matter of writing the permutations - tedious but trivial.

Thanks
Bill

I've been figuring it out for a couple of days.. if the equation c1*|a>|b> + c2*|b>|a> is for 2 state and 2 system.. I thought 3 system would make it |a>, |b>, |c>.. if it remains as |a> and |b>, how do you turn the c1*|a>|b> + c2*|b>|a> into a 3 system.. it doesn't make sense to make it c1*|a>|b> + c2*|b>|a> + c3*|a>|b>, any clue?

 

[Nugatory]

I've been figuring it out for a couple of days.. if the equation c1*|a>|b> + c2*|b>|a> is for 2 state and 2 system.. I thought 3 system would make it |a>, |b>, |c>.. if it remains as |a> and |b>, how do you turn the c1*|a>|b> + c2*|b>|a> into a 3 system.. it doesn't make sense to make it c1*|a>|b> + c2*|b>|a> + c3*|a>|b>, any clue?

You want to look at the tensor product of Hilbert spaces - that's the starting point for this stuff.

 

[bluecap]

You want to look at the tensor product of Hilbert spaces - that's the starting point for this stuff.

Oh. Actually I was also studying your convention yesterday in msg #6 in https://www.physicsforums.com/threads/entangled-particles.868970/#post-5455273 where you wrote:

"In this framework it's a lot easier to see how three-particle entanglement might work. If we have three observables commuting A, B, and C each taking on either of two possible values, then an eigenstate will be written as something like $|+--\rangle$ where the three signs represent the values of A, B, C in that order. A state like $\frac{\sqrt{2}}{2}(|+--\rangle+|-++\rangle$ or $\frac{\sqrt{2}}{2}(|+--\rangle+|-+-\rangle$
is entangled; a measurement of A will cause the wave function to collapse to either $|+--\rangle$ or $|-++\rangle$, states in which we know more about what will happen when and if we subsequently measure B or C. (Be warned that this is a highly contrived example; in any realistic situation the wave function would be such that even after collapse B and C would remain entanged)."

Nugatory, For 2 state, your used $\frac{\sqrt{2}}{2}(|+--\rangle+|-++\rangle$ so I thought it's equal to |a>, |b>, |c> in Bhobba convention...

anyway, can you please share the first 2 terms to know how entanglement of 3 system with 2 state (spin up and down) look like.. I don't know the meaning of permutations nor tensor so no idea how to start.. thank you.

 

[StevieTNZ]

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> ??

if not.. any clue what is the right equation?

Yes, that is the state I would write. That's only if it is allowable by QM (e.g. in this case, the spin states).

 

[bluecap]

Yes, that is the state I would write. That's only if it is allowable by QM (e.g. in this case, the spin states).

So I was right all along.. and writing it in up, down is not the same as |a>, |b>,|c.. I have some questions:

1. What do you mean if it's allowable by QM.. of course the spin states up and down are allowable...
2. So in measurement, only one of the terms occurs and the rest collapse?
3. So the 8 terms make up 8 many worlds or branches?

 

[bluecap]

I am not going to try to explain it in your terms, because there is a lot more to specify. 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.

At any rate, there are experiments involving entanglement of large N, where N>>1000. Please keep in mind that there is also monogamy of entanglement. A particle A that is maximally entangled with B, cannot also be entangled with C on the same basis.

Is this permutations thing related to GHZ? Or also there in billion-particle entanglement.. are you saying the permutations can make the terms a little less.. but how would this decrease the terms in 3 systems and 2 states (up and down)...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> ?

I read in Amir Aczel book Entanglement in page 222 "The most amazing thing about three-particle entanglement, and the main reason for the interest taken in the GHZ proposal, is that it can be used to prove Bell's theorem without the cumbersome use of inequalities", what can you say about this?

and in page 231, "Furthermore, actual experiments have shown that the quantum theory is correct, and therefore Einstein's local realism is not. The GHZ theorem proves the contradiction in a much more direct, easier to understand, and non-statistical way, as compared with Bell's original theorem."

Do you agree?

 

[DrChinese]

So I was right all along...

Thanks for making my day with this one.

 

[DrChinese]

I read in Amir Aczel book Entanglement in page 222 "The most amazing thing about three-particle entanglement, and the main reason for the interest taken in the GHZ proposal, is that it can be used to prove Bell's theorem without the cumbersome use of inequalities", what can you say about this?

and in page 231, "Furthermore, actual experiments have shown that the quantum theory is correct, and therefore Einstein's local realism is not. The GHZ theorem proves the contradiction in a much more direct, easier to understand, and non-statistical way, as compared with Bell's original theorem."

Do you agree?

GHZ is an amazing theorem. It is NOTORIOUSLY difficult to understand, nothing like the simplicity of Bell. Honestly GHZ would be a subject for another thread. But yes, there is no need for inequalities going down this path. In principle, the outcome of every trial is itself a strike against local realism. Local realism predicts +1, where QM predicts -1 in every case. So no statistcs, averages, etc. are involved. QM says one thing, LR says the diametric opposite.

 

[DrChinese]

... billion-particle entanglement is the more realistic everyday scenerio.

There is no reasonable way to describe the entanglement of everyday objects as being entangled as part of large N groups. Decoherence implies that entanglement of open systems dissipates fairly rapidly over time. Interactions - such as heat transfer - normally sever any entanglement between objects. And they must first be entangled before they can decohere.

 

[StevieTNZ]

So I was right all along.. and writing it in up, down is not the same as |a>, |b>,|c.. I have some questions:

1. What do you mean if it's allowable by QM.. of course the spin states up and down are allowable...

Yes, spin states of up and down are allowable, but I am not sure when it comes to three quantum systems.

2. So in measurement, only one of the terms occurs and the rest collapse?

If you measure system A and it is spin down, systems B and C may be found as |down>|down>, |up>|up>, |down>|up> or |up>|down>. Whether systems B and C collapse into one of those states at the time of measurement of system A I don't think anyone can really answer.

3. So the 8 terms make up 8 many worlds or branches?

Yes.

 

[PeterDonis]

Interactions - such as heat transfer - normally sever any entanglement between objects.

I don't think this is correct as a statement about QM itself. It's a correct statement of some interpretations--the ones that have wave function collapse--but it is not a correct statement of other interpretations. In the MWI, for example, interactions entangle the interacting systems.

In the particular case where we have two systems, A and B, that are maximally entangled, then if, say, A interacts with another system C, then at least a portion of its entanglement with B will be severed. But that's because at least a portion of the entanglement gets transferred to C.

 

[PeterDonis]

spin states of up and down are allowable, but I am not sure when it comes to three quantum systems

If all three quantum systems are spin-1/2 fermions, then each one of them can be up or down (more precisely, each one has a Hilbert space with those states as a basis). It doesn't matter how many there are.

You might be confusing this with the case of a quantum system that has more than two allowable spin states to begin with, such as a massive spin-1 boson (which has three). But that's true even if there's only one of them

 

[bluecap]

I don't think this is correct as a statement about QM itself. It's a correct statement of some interpretations--the ones that have wave function collapse--but it is not a correct statement of other interpretations. In the MWI, for example, interactions entangle the interacting systems.

In the particular case where we have two systems, A and B, that are maximally entangled, then if, say, A interacts with another system C, then at least a portion of its entanglement with B will be severed. But that's because at least a portion of the entanglement gets transferred to C.

Is it a rule that all kinds of interactions and measurement can only form entanglement between the existing system and the new system.. is there absolute no exception.. how do you completely break off entanglement then instead of forming new one (even if a tiny portion of C in your example above is entangled to A and B.. it is still called A, B, C entanglement)? All I can think of is a gamma photon knocking off a particle in the atom that would cut the entanglement of the particle in the atom.. is this the only way to completely cut any existing entanglement, by mechanically separating them?

 

[PeterDonis]

how do you completely break off entanglement then

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.)

But cases like this, practically speaking, only occur in carefully controlled laboratory experiments (for example, the kinds of things that are done in quantum computing). In all other cases, it's basically impossible to remove all entanglement from some particular quantum system.

(Note that the above does not make use of any particular interpretation of QM. In collapse interpretations, wave function collapse does remove entanglement, because it takes away all but one of the branches of the wave function. But there is no way to test this experimentally, because collapse interpretations and no collapse interpretations make the same predictions for all experiments.)

 

[bluecap]

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.)

But cases like this, practically speaking, only occur in carefully controlled laboratory experiments (for example, the kinds of things that are done in quantum computing). In all other cases, it's basically impossible to remove all entanglement from some particular quantum system.

(Note that the above does not make use of any particular interpretation of QM. In collapse interpretations, wave function collapse does remove entanglement, because it takes away all but one of the branches of the wave function. But there is no way to test this experimentally, because collapse interpretations and no collapse interpretations make the same predictions for all experiments.)

In the case of A and B, and C entangling with A and B.. how do you make the concept of branches enter into it.. I mean, in collapse interpretation, how does wave function collapse stop the entangling of C to A and B in contrast to no collapse interpretation.. there are no branches in the A, B, and C interaction.. is there?

But separately from the interpretations.. if you shut down the power in the lab of say the aspect experiment.. then you will cut the entanglement.. is it not? And for the apple entangling with itself and the environment.. if you burn the apple.. it will also cut the entanglement in the atoms inside and to the environment.. is it not.. or do you mean the ashes are also in entanglement?

 

[PeterDonis]

In the case of A and B, and C entangling with A and B.. how do you make the concept of branches enter into it.. I mean, in collapse interpretation, how does wave function collapse stop the entangling of C to A and B in contrast to no collapse interpretation.. there are no branches in the A, B, and C interaction.. is there?

Rather than my answering this question for you, I'm going to make a suggestion: write down the math. This is an "I" level thread, so you should have enough background to do that. First write down a state in which A and B are entangled, and C is not. Then write down a state in which B and C are entangled but A is not. Then see if you can imagine a unitary interaction that will transform one into the other.

With regard to branches, consider a simpler case: we have a system A and a measuring device M. By using M to make a measurement of A, we entangle A and M. What does the resulting state look like? This will be the state after measurement according to a no collapse interpretation like the MWI. Then suppose the wave function collapses. What does that do to the state? Is the resulting state entangled?

if you shut down the power in the lab of say the aspect experiment.. then you will cut the entanglement.. is it not?

Cut the entanglement of what? If the power is shut down, the lab equipment is not producing any photons to begin with, so it's meaningless to ask whether they are entangled or not.

for the apple entangling with itself and the environment.. if you burn the apple.. it will also cut the entanglement in the atoms inside and to the environment.. is it not..

No.

or do you mean the ashes are also in entanglement?

What do you think?

 

[StevieTNZ]

In the case of A and B, and C entangling with A and B.. how do you make the concept of branches enter into it.. I mean, in collapse interpretation, how does wave function collapse stop the entangling of C to A and B in contrast to no collapse interpretation.. there are no branches in the A, B, and C interaction.. is there?

Experiments have been done where three photons were entangled with each other. I believe the max to date has been 10 photons entangled with each other (but it may actually be more than 10).

 

[bluecap]

Rather than my answering this question for you, I'm going to make a suggestion: write down the math. This is an "I" level thread, so you should have enough background to do that. First write down a state in which A and B are entangled, and C is not. Then write down a state in which B and C are entangled but A is not. Then see if you can imagine a unitary interaction that will transform one into the other.

the above is no problem.. I can write it but don't know how to put into latex.. I was not questioning how C can cut off the entanglement of A and B.. I understood the concept.. what I was questioning was about the involvement of the branches (see below)

With regard to branches, consider a simpler case: we have a system A and a measuring device M. By using M to make a measurement of A, we entangle A and M. What does the resulting state look like? This will be the state after measurement according to a no collapse interpretation like the MWI. Then suppose the wave function collapses. What does that do to the state? Is the resulting state entangled?

For this equation of 3 systems and 2 state (up and down). I use 3 system because I already have the equation typed:

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> ?

A measurement in MWI would mean the new system or 4th system would be entangled with each of the 8 terms making up 8 branches (according to StevenTNZ this is true, right?)

But in collapse interpretation, the measuring device or 4th system is classical, so it would collapse the wave function of the 8 terms creating only one branch.. and there is no entanglement between the one branch and the measuring device.. therefore the reason there is no entanglement in collapse interpretation is because the device is classical and doesn't entangle with the system.. and not because one branch is chosen.. because even if one branch is chosen.. there is still entanglement if the M device is not classical (remember Copenhagen has the quantum-classical divide.. so Copenhagen in the first place doesn't allow decoherence to exist at all). is this what you mean?

Cut the entanglement of what? If the power is shut down, the lab equipment is not producing any photons to begin with, so it's meaningless to ask whether they are entangled or not.
No.
What do you think?

You may be right the ashes as substance is still entangled with itself and the environment, the only way to avoid is to get vaporized at the heart of a nuclear explosion. (?)

 

[PeterDonis]

A measurement in MWI would mean the new system or 4th system would be entangled with each of the 8 terms making up 8 branches

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.)

But in collapse interpretation, the measuring device or 4th system is classical, so it would collapse the wave function of the 8 terms creating only one branch

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).

and there is no entanglement between the one branch and the measuring device

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.

the reason there is no entanglement in collapse interpretation is because the device is classical and doesn't entangle with the system

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.

even if one branch is chosen.. there is still entanglement if the M device is not classical

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.

Copenhagen has the quantum-classical divide.. so Copenhagen in the first place doesn't allow decoherence to exist at all

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).

You may be right the ashes as substance is still entangled with itself and the environment, the only way to avoid is to get vaporized at the heart of a nuclear explosion.

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]

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?